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Find each product.
1. 5w( −3w2 + 2w − 4)
SOLUTION:
2. 6g2(3g
3 + 4g2 + 10g − 1)
SOLUTION:
3. 4km2(8km
2 + 2k2m + 5k)
SOLUTION:
4. −3p4r3(2p
2r4 − 6p 6
r3 − 5)
SOLUTION:
5. 2ab(7a4b
2 + a5b − 2a)
SOLUTION:
6. c2d
3(5cd
7 − 3c3d
2 − 4d3)
SOLUTION:
Simplify each expression.
7. t(4t2 + 15t + 4) − 4(3t − 1)
SOLUTION:
8. x(3x2 + 4) + 2(7x − 3)
SOLUTION:
9. −2d(d3c
2 − 4dc2 + 2d
2c) + c
2(dc
2 − 3d4)
SOLUTION:
10. −5w2(8w
2x − 11wx
2) + 6x(9wx
4 − 4w − 3x2)
SOLUTION:
11. GRIDDED RESPONSE Marlene is buying a new plasma television. The height of the screen of the television is one half the width plus 5 inches. The width is 30 inches. Find the height of the screen in inches.
SOLUTION: Let h be the height and let w be the width.
Let w = 30.
The height is 20 inches.
Solve each equation.
12. −6(11 − 2c) = 7(−2 − 2c)
SOLUTION:
13. t(2t + 3) + 20 = 2t(t − 3)
SOLUTION:
14. −2(w + 1) + w = 7 − 4w
SOLUTION:
15. 3(y − 2) + 2y = 4y + 14
SOLUTION:
16. a(a + 3) + a(a − 6) + 35 = a(a − 5) + a(a + 7)
SOLUTION:
17. n(n − 4) + n(n + 8) = n(n − 13) + n(n + 1) + 16
SOLUTION:
Find each product.
18. b(b2 − 12b + 1)
SOLUTION:
19. f (f2 + 2f + 25)
SOLUTION:
20. −3m3(2m
3 − 12m2 + 2m + 25)
SOLUTION:
21. 2j 2(5j
3 − 15j2 + 2j + 2)
SOLUTION:
22. 2pr2(2pr + 5p
2r − 15p )
SOLUTION:
23. 4t3u(2t
2u
2 − 10tu4 + 2)
SOLUTION:
Simplify each expression.
24. −3(5x2 + 2x + 9) + x(2x − 3)
SOLUTION:
25. a(−8a2 + 2a + 4) + 3(6a
2 − 4)
SOLUTION:
26. −4d(5d2 − 12) + 7(d + 5)
SOLUTION:
27. −9g(−2g + g2) + 3(g
2 + 4)
SOLUTION:
28. 2j (7j2k
2 + j k2 + 5k) − 9k(−2j
2k
2 + 2k2 + 3j )
SOLUTION:
29. 4n(2n3p
2 − 3np2 + 5n) + 4p (6n
2p − 2np
2 + 3p )
SOLUTION:
30. DAMS A new dam being built has the shape of a trapezoid. The base at the bottom of the dam is 2 times the
height. The base at the top of the dam is times the height minus 30 feet.
a. Write an expression to find the area of the trapezoidal cross section of the dam. b. If the height of the dam is 180 feet, find the area of this cross section.
SOLUTION:
a. The equation for the area of a trapezoid is .
b1
= 2h
b2 =
b. Let h = 180.
So, the area is 32,940 square feet.
Solve each equation.
31. 7(t2 + 5t − 9) + t = t(7t − 2) + 13
SOLUTION:
32. w(4w + 6) + 2w = 2(2w2 + 7w − 3)
SOLUTION:
33. 5(4z + 6) − 2(z − 4) = 7z(z + 4) − z(7z − 2) − 48
SOLUTION:
34. 9c(c − 11) + 10(5c − 3) = 3c(c + 5) + c(6c − 3) − 30
SOLUTION:
35. 2f (5f − 2) − 10(f2 − 3f + 6) = −8f (f + 4) + 4(2f
2 − 7f )
SOLUTION:
36. 2k(−3k + 4) + 6(k2 + 10) = k(4k + 8) − 2k(2k + 5)
SOLUTION:
Simplify each expression.
37. np2(30p
2 + 9n
2p − 12)
SOLUTION:
38. r2t(10r
3 + 5rt
3 + 15t
2)
SOLUTION:
39. −5q2w
3(4q + 7w) + 4qw
2(7q
2w + 2q) − 3qw(3q
2w
2 + 9)
SOLUTION:
40. −x2z(2z
2 + 4xz3) + xz
2(xz + 5x
3z) + x2
z3(3x
2z + 4xz)
SOLUTION:
41. PARKING A parking garage charges $30 per month plus $0.50 per daytime hour and $0.25 per hour during nights and weekends. Suppose Trent parks in the garage for 47 hours in January and h of those are night and weekend hours. a. Find an expression for Trent’s January bill. b. Find the cost if Trent had 12 hours of night and weekend hours.
SOLUTION: a. First, multiply the price per hour times the number of hours. Let h represent the number of weekend and night hours. For the weekend and night hours, he would spend 0.25h. The number of daytime hours is 47 – h, so he spent 0.50(47 - h) = 23.50 – 0.50h. To find the total cost, add these two values to the cost per month. 30 + (23.50 − 0.50h) + 0.25h = 53.50 − 0.25h. b. If Trent parked 12 night and weekend hours, h = 12
So, the cost for January is $50.50.
42. CCSS MODELING Che is building a dog house for his new puppy. The upper face of the dog house is a trapezoid. If the height of the trapezoid is 12 inches, find the area of the face of this piece of the dog house.
SOLUTION:
The formula for the area of a trapezoid is .
Now substitute h = 12 inches into the equation.
Therefore, the area of the trapezoid is 318 square inches.
43. TENNIS The tennis club is building a new tennis court with a path around it.
a. Write an expression for the area of the tennis court. b. Write an expression for the area of the path. c. What is the perimeter of the outside of the path?
SOLUTION: a. The area of the tennis court can be found by the equation . Substitute the values of the length and width and simplify.
b. The area of the path is the area of the large rectangle minus the area of the smaller rectangle, which is the area ofthe tennis court.
Area of the large rectangle = 2.5x(x + 6) or 2.5x2 + 15x.
Now subtract the area of the tennis court to find the area of just the path.
c. The perimeter of the outside of the path is the perimeter of the larger rectangle.
If x = 36 feet, then the perimeter of the outside of the path is 7(36) + 12 or 264 ft.
44. MULTIPLE REPRESENTATIONS In this problem, you will investigate the degree of the product of a monomialand a polynomial. a. TABULAR Write three monomials of different degrees and three polynomials of different degrees. Determine the degree of each monomial and polynomial. Multiply the monomials by the polynomials. Determine the degree of each product. Record your results in a table like the one below.
b. VERBAL Make a conjecture about the degree of the product of a monomial and a polynomial. What is the degree of the product of a monomial of degree a and a polynomial of degree b?
SOLUTION: a. There possible monomials are 2x, 3x
2, 4x
3. Three polynomials are x2 − 1, x5 + 1, x
6 + 1.
The degree for the monomial is the sum of the exponents. For the example given, case they are 1,2, and 3. The degree for the polynomial is the largest degree of the monomials. For the example given, they are 2, 5, and 6. Use the distributive property to find the product of the monomial and the polynomial. For the example give, they
are2x3 − 2x, 3x
7 + 3x
2, and 4x
9 + 4x. For the example given, the degree of the new polynomials are 3,7 and 9.
b. The degree of the product is the sum of the degree of the monomial and the degree of the polynomial, or a + b.
45. ERROR ANALYSIS Pearl and Ted both worked on this problem. Is either of them correct? Explain your reasoning.
SOLUTION:
Ted’s solution is correct. Pearl made a mistake when distributing on the middle term. She did not add the exponents correctly.
46. CCSS PERSEVERANCE Find p such that 3xp(4x
2p+3 + 2x3p−2
) = 12x12 + 6x
10.
SOLUTION: First simplify the left side of the equation.
Because the coefficients of the terms are the same, the corresponding terms’ exponents can be set equal to each other.
47. CHALLENGE Find 4x−3
y2(2x
5y−4 + 6x
−7y
6 − 4x0y
-2).
SOLUTION:
48. REASONING Is there a value for x that makes the statement (x + 2)2 = x2 + 22
true? If so, find a value for x. Explain your reasoning.
SOLUTION: Solve for x.
When 0 is substituted in for x in the equation, both sides are 22 or 4, which makes the equation true.
49. OPEN ENDED Write a monomial and a polynomial using n as the variable. Find their product.
SOLUTION: Choose a monomial with variable n and a polynomial with variable n. For example, let the monomial be 3n and the polynomial be 4n + 1. Find the product.
50. WRITING IN MATH Describe the steps to multiply a polynomial by a monomial.
SOLUTION: To multiply a polynomial by a monomial, use the Distributive Property. Multiply each term of the polynomial by the monomial. Then simplify by multiplying the coefficients together and using the Product of Powers Property for the variables. Consider the following example:
51. Every week a store sells j jeans and t T-shirts. The store makes $8 for each T-shirt and $12 for each pair of jeans. Which of the following expressions represents the total amount of money, in dollars, the store makes every week? A 8j + 12t B 12j + 8t C 20(j + t) D 96jt
SOLUTION: 12j is the amount made from jeans. 8t is the amount made from T-shirts. The sum, or 12j + 8t, represents the total amount the store makes every week. So, the correct choice is B.
52. If a = 5x + 7y and b = 2y − 3x, what is a + b?F 2x − 9y G 3y + 4x H 2x + 9y J 2x − 5y
SOLUTION:
So, the correct choice is H.
53. GEOMETRY A triangle has sides of length 5 inches and 8.5 inches. Which of the following cannot be the length of the third side? A 3.5 inches B 4 inches C 5.5 inches D 12 inches
SOLUTION: The sum of any two sides of a triangle must be greater than the length of the third side. Choice A: 3.5 + 5 = 8.5 8.5 Choice B: 3.5 + 5 = 8.5 > 8.5 Choice C: 5.5 + 5 = 10.5 > 8.5 Choice D: 12 + 5 = 17 > 8.5 Let x be the third side of the triangle.
3.5 cannot be the third side of the triangle. So, the correct choice is A.
54. SHORT RESPONSE Write an equation in which x varies directly as the cube of y and inversely as the square of z .
SOLUTION: If x varies directly as the cube of y , then both x and y must be in the numerators on either side of the equal sign.
When x varies inversely with the square of z. then the z must be in the denominator. Then . The
combination of the two is .
Find each sum or difference.
55. (2x2 − 7) + (8 − 5x
2)
SOLUTION:
56. (3z2 + 2z − 1) + (z
2 − 6)
SOLUTION:
57. (2a − 4a2 + 1) − (5a
2 − 2a − 6)
SOLUTION:
58. (a3 − 3a
2 + 4) − (4a
2 + 7)
SOLUTION:
59. (2ab − 3a + 4b) + (5a + 4ab)
SOLUTION:
60. (8c3 − 3c
2 + c − 2) − (3c
3 + 9)
SOLUTION:
Write a recursive formula for each sequence.61. 16, 2, –12, –26, ...
SOLUTION: Determine if there is a common difference. 2 – 16 = –14 –12 – 2 = –14 –26 – (–12) = –14 The common difference is –14. The first term is 16. To find the next term, subtract 14 from the previous term.
62. –5, 3, 11, 19, ...
SOLUTION: Determine if there is a common difference. 3 – (–5) = 8 11 – 3 = 8 19 – 11 = 8 The common difference is 8. The first term is 3. To find the next term, add 8 to the previous term.
63. 27, 43, 59, 75, ...
SOLUTION: Determine if there is a common difference. 43 – 27 = 16 59 – 43 = 16 75 – 59 = 16 The common difference is 16. The first term is 43. To find the next term, add 16 to the previous term.
64. 80, –200, 500, –1250, ...
SOLUTION: Check for a common difference. –200 – 80 = –280 500 – (–200) = 700 There is no common difference. Check for a common ratio. –200 ÷ 80 = –2.5 500 ÷ (–200) = –2.5 –1250 ÷ 500 = –2.5 The common ratio is –2.5. The first term is 80. To find the next term, multiply the previous term by –2.5.
65. 100, 60, 36, 21.6, ...
SOLUTION: Check for a common difference. 60 – 100 = –40 36 – 60 = –24 There is no common difference. Check for a common ratio. 60 ÷ 100 = 0.6 36 ÷ 60 = 0.6 21.6 ÷ 36 = 0.6 The common ratio is 0.6. The first term is 100. To find the next term, multiply the previous term by 0.6.
66.
SOLUTION: Check for a common difference.
There is no common difference. Check for a common ratio.
4 ÷ 1 = 4 The common ratio is 4.
The first term is . To find the next term, multiply the previous term by 4.
67. TRAVEL In 2003, about 9.5 million people took cruises. Between 2003 and 2008, the number increased by about 740,000 each year. Write the point-slope form of an equation to find the total number of people y taking a cruise for any year x. Estimate the number of people who took a cruise in 2010.
SOLUTION: The statement “increased by 740,000 each year” is a constant rate, or the slope. Now substitute the slope and the point (2003, 9,500,000) into the point slope equation.
To find the number of people who took a cruise in 2010, substitute x = 2010.
So, approximately 14,680,000 people took a cruise in 2010.
Write an equation in function notation for each relation.
68.
SOLUTION: The y-intercept is 0. Find the slope by choosing two points. Choose (0, 0) and (1, 4).
So the slope intercept form is f (x) = 4x + 0 or f (x) = 4x.
69.
SOLUTION:
The y-intercept is 0. Find the slope by choosing two points. Choose (0, 0) and (2, −1).
So in slope intercept form is f (x) = −0.5x + 0 or f (x) = −0.5x.
Simplify.
70. b(b2)(b
3)
SOLUTION:
71. 2y(3y2)
SOLUTION:
72. −y4(−2y
3)
SOLUTION:
73. −3z3(−5z
4 + 2z)
SOLUTION:
74. 2m(−4m4) − 3(−5m
3)
SOLUTION:
75. 4p 2(−2p
3) + 2p
4(5p
6)
SOLUTION:
eSolutions Manual - Powered by Cognero Page 1
8-2 Multiplying a Polynomial by a Monomial
Find each product.
1. 5w( −3w2 + 2w − 4)
SOLUTION:
2. 6g2(3g
3 + 4g2 + 10g − 1)
SOLUTION:
3. 4km2(8km
2 + 2k2m + 5k)
SOLUTION:
4. −3p4r3(2p
2r4 − 6p 6
r3 − 5)
SOLUTION:
5. 2ab(7a4b
2 + a5b − 2a)
SOLUTION:
6. c2d
3(5cd
7 − 3c3d
2 − 4d3)
SOLUTION:
Simplify each expression.
7. t(4t2 + 15t + 4) − 4(3t − 1)
SOLUTION:
8. x(3x2 + 4) + 2(7x − 3)
SOLUTION:
9. −2d(d3c
2 − 4dc2 + 2d
2c) + c
2(dc
2 − 3d4)
SOLUTION:
10. −5w2(8w
2x − 11wx
2) + 6x(9wx
4 − 4w − 3x2)
SOLUTION:
11. GRIDDED RESPONSE Marlene is buying a new plasma television. The height of the screen of the television is one half the width plus 5 inches. The width is 30 inches. Find the height of the screen in inches.
SOLUTION: Let h be the height and let w be the width.
Let w = 30.
The height is 20 inches.
Solve each equation.
12. −6(11 − 2c) = 7(−2 − 2c)
SOLUTION:
13. t(2t + 3) + 20 = 2t(t − 3)
SOLUTION:
14. −2(w + 1) + w = 7 − 4w
SOLUTION:
15. 3(y − 2) + 2y = 4y + 14
SOLUTION:
16. a(a + 3) + a(a − 6) + 35 = a(a − 5) + a(a + 7)
SOLUTION:
17. n(n − 4) + n(n + 8) = n(n − 13) + n(n + 1) + 16
SOLUTION:
Find each product.
18. b(b2 − 12b + 1)
SOLUTION:
19. f (f2 + 2f + 25)
SOLUTION:
20. −3m3(2m
3 − 12m2 + 2m + 25)
SOLUTION:
21. 2j 2(5j
3 − 15j2 + 2j + 2)
SOLUTION:
22. 2pr2(2pr + 5p
2r − 15p )
SOLUTION:
23. 4t3u(2t
2u
2 − 10tu4 + 2)
SOLUTION:
Simplify each expression.
24. −3(5x2 + 2x + 9) + x(2x − 3)
SOLUTION:
25. a(−8a2 + 2a + 4) + 3(6a
2 − 4)
SOLUTION:
26. −4d(5d2 − 12) + 7(d + 5)
SOLUTION:
27. −9g(−2g + g2) + 3(g
2 + 4)
SOLUTION:
28. 2j (7j2k
2 + j k2 + 5k) − 9k(−2j
2k
2 + 2k2 + 3j )
SOLUTION:
29. 4n(2n3p
2 − 3np2 + 5n) + 4p (6n
2p − 2np
2 + 3p )
SOLUTION:
30. DAMS A new dam being built has the shape of a trapezoid. The base at the bottom of the dam is 2 times the
height. The base at the top of the dam is times the height minus 30 feet.
a. Write an expression to find the area of the trapezoidal cross section of the dam. b. If the height of the dam is 180 feet, find the area of this cross section.
SOLUTION:
a. The equation for the area of a trapezoid is .
b1
= 2h
b2 =
b. Let h = 180.
So, the area is 32,940 square feet.
Solve each equation.
31. 7(t2 + 5t − 9) + t = t(7t − 2) + 13
SOLUTION:
32. w(4w + 6) + 2w = 2(2w2 + 7w − 3)
SOLUTION:
33. 5(4z + 6) − 2(z − 4) = 7z(z + 4) − z(7z − 2) − 48
SOLUTION:
34. 9c(c − 11) + 10(5c − 3) = 3c(c + 5) + c(6c − 3) − 30
SOLUTION:
35. 2f (5f − 2) − 10(f2 − 3f + 6) = −8f (f + 4) + 4(2f
2 − 7f )
SOLUTION:
36. 2k(−3k + 4) + 6(k2 + 10) = k(4k + 8) − 2k(2k + 5)
SOLUTION:
Simplify each expression.
37. np2(30p
2 + 9n
2p − 12)
SOLUTION:
38. r2t(10r
3 + 5rt
3 + 15t
2)
SOLUTION:
39. −5q2w
3(4q + 7w) + 4qw
2(7q
2w + 2q) − 3qw(3q
2w
2 + 9)
SOLUTION:
40. −x2z(2z
2 + 4xz3) + xz
2(xz + 5x
3z) + x2
z3(3x
2z + 4xz)
SOLUTION:
41. PARKING A parking garage charges $30 per month plus $0.50 per daytime hour and $0.25 per hour during nights and weekends. Suppose Trent parks in the garage for 47 hours in January and h of those are night and weekend hours. a. Find an expression for Trent’s January bill. b. Find the cost if Trent had 12 hours of night and weekend hours.
SOLUTION: a. First, multiply the price per hour times the number of hours. Let h represent the number of weekend and night hours. For the weekend and night hours, he would spend 0.25h. The number of daytime hours is 47 – h, so he spent 0.50(47 - h) = 23.50 – 0.50h. To find the total cost, add these two values to the cost per month. 30 + (23.50 − 0.50h) + 0.25h = 53.50 − 0.25h. b. If Trent parked 12 night and weekend hours, h = 12
So, the cost for January is $50.50.
42. CCSS MODELING Che is building a dog house for his new puppy. The upper face of the dog house is a trapezoid. If the height of the trapezoid is 12 inches, find the area of the face of this piece of the dog house.
SOLUTION:
The formula for the area of a trapezoid is .
Now substitute h = 12 inches into the equation.
Therefore, the area of the trapezoid is 318 square inches.
43. TENNIS The tennis club is building a new tennis court with a path around it.
a. Write an expression for the area of the tennis court. b. Write an expression for the area of the path. c. What is the perimeter of the outside of the path?
SOLUTION: a. The area of the tennis court can be found by the equation . Substitute the values of the length and width and simplify.
b. The area of the path is the area of the large rectangle minus the area of the smaller rectangle, which is the area ofthe tennis court.
Area of the large rectangle = 2.5x(x + 6) or 2.5x2 + 15x.
Now subtract the area of the tennis court to find the area of just the path.
c. The perimeter of the outside of the path is the perimeter of the larger rectangle.
If x = 36 feet, then the perimeter of the outside of the path is 7(36) + 12 or 264 ft.
44. MULTIPLE REPRESENTATIONS In this problem, you will investigate the degree of the product of a monomialand a polynomial. a. TABULAR Write three monomials of different degrees and three polynomials of different degrees. Determine the degree of each monomial and polynomial. Multiply the monomials by the polynomials. Determine the degree of each product. Record your results in a table like the one below.
b. VERBAL Make a conjecture about the degree of the product of a monomial and a polynomial. What is the degree of the product of a monomial of degree a and a polynomial of degree b?
SOLUTION: a. There possible monomials are 2x, 3x
2, 4x
3. Three polynomials are x2 − 1, x5 + 1, x
6 + 1.
The degree for the monomial is the sum of the exponents. For the example given, case they are 1,2, and 3. The degree for the polynomial is the largest degree of the monomials. For the example given, they are 2, 5, and 6. Use the distributive property to find the product of the monomial and the polynomial. For the example give, they
are2x3 − 2x, 3x
7 + 3x
2, and 4x
9 + 4x. For the example given, the degree of the new polynomials are 3,7 and 9.
b. The degree of the product is the sum of the degree of the monomial and the degree of the polynomial, or a + b.
45. ERROR ANALYSIS Pearl and Ted both worked on this problem. Is either of them correct? Explain your reasoning.
SOLUTION:
Ted’s solution is correct. Pearl made a mistake when distributing on the middle term. She did not add the exponents correctly.
46. CCSS PERSEVERANCE Find p such that 3xp(4x
2p+3 + 2x3p−2
) = 12x12 + 6x
10.
SOLUTION: First simplify the left side of the equation.
Because the coefficients of the terms are the same, the corresponding terms’ exponents can be set equal to each other.
47. CHALLENGE Find 4x−3
y2(2x
5y−4 + 6x
−7y
6 − 4x0y
-2).
SOLUTION:
48. REASONING Is there a value for x that makes the statement (x + 2)2 = x2 + 22
true? If so, find a value for x. Explain your reasoning.
SOLUTION: Solve for x.
When 0 is substituted in for x in the equation, both sides are 22 or 4, which makes the equation true.
49. OPEN ENDED Write a monomial and a polynomial using n as the variable. Find their product.
SOLUTION: Choose a monomial with variable n and a polynomial with variable n. For example, let the monomial be 3n and the polynomial be 4n + 1. Find the product.
50. WRITING IN MATH Describe the steps to multiply a polynomial by a monomial.
SOLUTION: To multiply a polynomial by a monomial, use the Distributive Property. Multiply each term of the polynomial by the monomial. Then simplify by multiplying the coefficients together and using the Product of Powers Property for the variables. Consider the following example:
51. Every week a store sells j jeans and t T-shirts. The store makes $8 for each T-shirt and $12 for each pair of jeans. Which of the following expressions represents the total amount of money, in dollars, the store makes every week? A 8j + 12t B 12j + 8t C 20(j + t) D 96jt
SOLUTION: 12j is the amount made from jeans. 8t is the amount made from T-shirts. The sum, or 12j + 8t, represents the total amount the store makes every week. So, the correct choice is B.
52. If a = 5x + 7y and b = 2y − 3x, what is a + b?F 2x − 9y G 3y + 4x H 2x + 9y J 2x − 5y
SOLUTION:
So, the correct choice is H.
53. GEOMETRY A triangle has sides of length 5 inches and 8.5 inches. Which of the following cannot be the length of the third side? A 3.5 inches B 4 inches C 5.5 inches D 12 inches
SOLUTION: The sum of any two sides of a triangle must be greater than the length of the third side. Choice A: 3.5 + 5 = 8.5 8.5 Choice B: 3.5 + 5 = 8.5 > 8.5 Choice C: 5.5 + 5 = 10.5 > 8.5 Choice D: 12 + 5 = 17 > 8.5 Let x be the third side of the triangle.
3.5 cannot be the third side of the triangle. So, the correct choice is A.
54. SHORT RESPONSE Write an equation in which x varies directly as the cube of y and inversely as the square of z .
SOLUTION: If x varies directly as the cube of y , then both x and y must be in the numerators on either side of the equal sign.
When x varies inversely with the square of z. then the z must be in the denominator. Then . The
combination of the two is .
Find each sum or difference.
55. (2x2 − 7) + (8 − 5x
2)
SOLUTION:
56. (3z2 + 2z − 1) + (z
2 − 6)
SOLUTION:
57. (2a − 4a2 + 1) − (5a
2 − 2a − 6)
SOLUTION:
58. (a3 − 3a
2 + 4) − (4a
2 + 7)
SOLUTION:
59. (2ab − 3a + 4b) + (5a + 4ab)
SOLUTION:
60. (8c3 − 3c
2 + c − 2) − (3c
3 + 9)
SOLUTION:
Write a recursive formula for each sequence.61. 16, 2, –12, –26, ...
SOLUTION: Determine if there is a common difference. 2 – 16 = –14 –12 – 2 = –14 –26 – (–12) = –14 The common difference is –14. The first term is 16. To find the next term, subtract 14 from the previous term.
62. –5, 3, 11, 19, ...
SOLUTION: Determine if there is a common difference. 3 – (–5) = 8 11 – 3 = 8 19 – 11 = 8 The common difference is 8. The first term is 3. To find the next term, add 8 to the previous term.
63. 27, 43, 59, 75, ...
SOLUTION: Determine if there is a common difference. 43 – 27 = 16 59 – 43 = 16 75 – 59 = 16 The common difference is 16. The first term is 43. To find the next term, add 16 to the previous term.
64. 80, –200, 500, –1250, ...
SOLUTION: Check for a common difference. –200 – 80 = –280 500 – (–200) = 700 There is no common difference. Check for a common ratio. –200 ÷ 80 = –2.5 500 ÷ (–200) = –2.5 –1250 ÷ 500 = –2.5 The common ratio is –2.5. The first term is 80. To find the next term, multiply the previous term by –2.5.
65. 100, 60, 36, 21.6, ...
SOLUTION: Check for a common difference. 60 – 100 = –40 36 – 60 = –24 There is no common difference. Check for a common ratio. 60 ÷ 100 = 0.6 36 ÷ 60 = 0.6 21.6 ÷ 36 = 0.6 The common ratio is 0.6. The first term is 100. To find the next term, multiply the previous term by 0.6.
66.
SOLUTION: Check for a common difference.
There is no common difference. Check for a common ratio.
4 ÷ 1 = 4 The common ratio is 4.
The first term is . To find the next term, multiply the previous term by 4.
67. TRAVEL In 2003, about 9.5 million people took cruises. Between 2003 and 2008, the number increased by about 740,000 each year. Write the point-slope form of an equation to find the total number of people y taking a cruise for any year x. Estimate the number of people who took a cruise in 2010.
SOLUTION: The statement “increased by 740,000 each year” is a constant rate, or the slope. Now substitute the slope and the point (2003, 9,500,000) into the point slope equation.
To find the number of people who took a cruise in 2010, substitute x = 2010.
So, approximately 14,680,000 people took a cruise in 2010.
Write an equation in function notation for each relation.
68.
SOLUTION: The y-intercept is 0. Find the slope by choosing two points. Choose (0, 0) and (1, 4).
So the slope intercept form is f (x) = 4x + 0 or f (x) = 4x.
69.
SOLUTION:
The y-intercept is 0. Find the slope by choosing two points. Choose (0, 0) and (2, −1).
So in slope intercept form is f (x) = −0.5x + 0 or f (x) = −0.5x.
Simplify.
70. b(b2)(b
3)
SOLUTION:
71. 2y(3y2)
SOLUTION:
72. −y4(−2y
3)
SOLUTION:
73. −3z3(−5z
4 + 2z)
SOLUTION:
74. 2m(−4m4) − 3(−5m
3)
SOLUTION:
75. 4p 2(−2p
3) + 2p
4(5p
6)
SOLUTION:
eSolutions Manual - Powered by Cognero Page 2
8-2 Multiplying a Polynomial by a Monomial
Find each product.
1. 5w( −3w2 + 2w − 4)
SOLUTION:
2. 6g2(3g
3 + 4g2 + 10g − 1)
SOLUTION:
3. 4km2(8km
2 + 2k2m + 5k)
SOLUTION:
4. −3p4r3(2p
2r4 − 6p 6
r3 − 5)
SOLUTION:
5. 2ab(7a4b
2 + a5b − 2a)
SOLUTION:
6. c2d
3(5cd
7 − 3c3d
2 − 4d3)
SOLUTION:
Simplify each expression.
7. t(4t2 + 15t + 4) − 4(3t − 1)
SOLUTION:
8. x(3x2 + 4) + 2(7x − 3)
SOLUTION:
9. −2d(d3c
2 − 4dc2 + 2d
2c) + c
2(dc
2 − 3d4)
SOLUTION:
10. −5w2(8w
2x − 11wx
2) + 6x(9wx
4 − 4w − 3x2)
SOLUTION:
11. GRIDDED RESPONSE Marlene is buying a new plasma television. The height of the screen of the television is one half the width plus 5 inches. The width is 30 inches. Find the height of the screen in inches.
SOLUTION: Let h be the height and let w be the width.
Let w = 30.
The height is 20 inches.
Solve each equation.
12. −6(11 − 2c) = 7(−2 − 2c)
SOLUTION:
13. t(2t + 3) + 20 = 2t(t − 3)
SOLUTION:
14. −2(w + 1) + w = 7 − 4w
SOLUTION:
15. 3(y − 2) + 2y = 4y + 14
SOLUTION:
16. a(a + 3) + a(a − 6) + 35 = a(a − 5) + a(a + 7)
SOLUTION:
17. n(n − 4) + n(n + 8) = n(n − 13) + n(n + 1) + 16
SOLUTION:
Find each product.
18. b(b2 − 12b + 1)
SOLUTION:
19. f (f2 + 2f + 25)
SOLUTION:
20. −3m3(2m
3 − 12m2 + 2m + 25)
SOLUTION:
21. 2j 2(5j
3 − 15j2 + 2j + 2)
SOLUTION:
22. 2pr2(2pr + 5p
2r − 15p )
SOLUTION:
23. 4t3u(2t
2u
2 − 10tu4 + 2)
SOLUTION:
Simplify each expression.
24. −3(5x2 + 2x + 9) + x(2x − 3)
SOLUTION:
25. a(−8a2 + 2a + 4) + 3(6a
2 − 4)
SOLUTION:
26. −4d(5d2 − 12) + 7(d + 5)
SOLUTION:
27. −9g(−2g + g2) + 3(g
2 + 4)
SOLUTION:
28. 2j (7j2k
2 + j k2 + 5k) − 9k(−2j
2k
2 + 2k2 + 3j )
SOLUTION:
29. 4n(2n3p
2 − 3np2 + 5n) + 4p (6n
2p − 2np
2 + 3p )
SOLUTION:
30. DAMS A new dam being built has the shape of a trapezoid. The base at the bottom of the dam is 2 times the
height. The base at the top of the dam is times the height minus 30 feet.
a. Write an expression to find the area of the trapezoidal cross section of the dam. b. If the height of the dam is 180 feet, find the area of this cross section.
SOLUTION:
a. The equation for the area of a trapezoid is .
b1
= 2h
b2 =
b. Let h = 180.
So, the area is 32,940 square feet.
Solve each equation.
31. 7(t2 + 5t − 9) + t = t(7t − 2) + 13
SOLUTION:
32. w(4w + 6) + 2w = 2(2w2 + 7w − 3)
SOLUTION:
33. 5(4z + 6) − 2(z − 4) = 7z(z + 4) − z(7z − 2) − 48
SOLUTION:
34. 9c(c − 11) + 10(5c − 3) = 3c(c + 5) + c(6c − 3) − 30
SOLUTION:
35. 2f (5f − 2) − 10(f2 − 3f + 6) = −8f (f + 4) + 4(2f
2 − 7f )
SOLUTION:
36. 2k(−3k + 4) + 6(k2 + 10) = k(4k + 8) − 2k(2k + 5)
SOLUTION:
Simplify each expression.
37. np2(30p
2 + 9n
2p − 12)
SOLUTION:
38. r2t(10r
3 + 5rt
3 + 15t
2)
SOLUTION:
39. −5q2w
3(4q + 7w) + 4qw
2(7q
2w + 2q) − 3qw(3q
2w
2 + 9)
SOLUTION:
40. −x2z(2z
2 + 4xz3) + xz
2(xz + 5x
3z) + x2
z3(3x
2z + 4xz)
SOLUTION:
41. PARKING A parking garage charges $30 per month plus $0.50 per daytime hour and $0.25 per hour during nights and weekends. Suppose Trent parks in the garage for 47 hours in January and h of those are night and weekend hours. a. Find an expression for Trent’s January bill. b. Find the cost if Trent had 12 hours of night and weekend hours.
SOLUTION: a. First, multiply the price per hour times the number of hours. Let h represent the number of weekend and night hours. For the weekend and night hours, he would spend 0.25h. The number of daytime hours is 47 – h, so he spent 0.50(47 - h) = 23.50 – 0.50h. To find the total cost, add these two values to the cost per month. 30 + (23.50 − 0.50h) + 0.25h = 53.50 − 0.25h. b. If Trent parked 12 night and weekend hours, h = 12
So, the cost for January is $50.50.
42. CCSS MODELING Che is building a dog house for his new puppy. The upper face of the dog house is a trapezoid. If the height of the trapezoid is 12 inches, find the area of the face of this piece of the dog house.
SOLUTION:
The formula for the area of a trapezoid is .
Now substitute h = 12 inches into the equation.
Therefore, the area of the trapezoid is 318 square inches.
43. TENNIS The tennis club is building a new tennis court with a path around it.
a. Write an expression for the area of the tennis court. b. Write an expression for the area of the path. c. What is the perimeter of the outside of the path?
SOLUTION: a. The area of the tennis court can be found by the equation . Substitute the values of the length and width and simplify.
b. The area of the path is the area of the large rectangle minus the area of the smaller rectangle, which is the area ofthe tennis court.
Area of the large rectangle = 2.5x(x + 6) or 2.5x2 + 15x.
Now subtract the area of the tennis court to find the area of just the path.
c. The perimeter of the outside of the path is the perimeter of the larger rectangle.
If x = 36 feet, then the perimeter of the outside of the path is 7(36) + 12 or 264 ft.
44. MULTIPLE REPRESENTATIONS In this problem, you will investigate the degree of the product of a monomialand a polynomial. a. TABULAR Write three monomials of different degrees and three polynomials of different degrees. Determine the degree of each monomial and polynomial. Multiply the monomials by the polynomials. Determine the degree of each product. Record your results in a table like the one below.
b. VERBAL Make a conjecture about the degree of the product of a monomial and a polynomial. What is the degree of the product of a monomial of degree a and a polynomial of degree b?
SOLUTION: a. There possible monomials are 2x, 3x
2, 4x
3. Three polynomials are x2 − 1, x5 + 1, x
6 + 1.
The degree for the monomial is the sum of the exponents. For the example given, case they are 1,2, and 3. The degree for the polynomial is the largest degree of the monomials. For the example given, they are 2, 5, and 6. Use the distributive property to find the product of the monomial and the polynomial. For the example give, they
are2x3 − 2x, 3x
7 + 3x
2, and 4x
9 + 4x. For the example given, the degree of the new polynomials are 3,7 and 9.
b. The degree of the product is the sum of the degree of the monomial and the degree of the polynomial, or a + b.
45. ERROR ANALYSIS Pearl and Ted both worked on this problem. Is either of them correct? Explain your reasoning.
SOLUTION:
Ted’s solution is correct. Pearl made a mistake when distributing on the middle term. She did not add the exponents correctly.
46. CCSS PERSEVERANCE Find p such that 3xp(4x
2p+3 + 2x3p−2
) = 12x12 + 6x
10.
SOLUTION: First simplify the left side of the equation.
Because the coefficients of the terms are the same, the corresponding terms’ exponents can be set equal to each other.
47. CHALLENGE Find 4x−3
y2(2x
5y−4 + 6x
−7y
6 − 4x0y
-2).
SOLUTION:
48. REASONING Is there a value for x that makes the statement (x + 2)2 = x2 + 22
true? If so, find a value for x. Explain your reasoning.
SOLUTION: Solve for x.
When 0 is substituted in for x in the equation, both sides are 22 or 4, which makes the equation true.
49. OPEN ENDED Write a monomial and a polynomial using n as the variable. Find their product.
SOLUTION: Choose a monomial with variable n and a polynomial with variable n. For example, let the monomial be 3n and the polynomial be 4n + 1. Find the product.
50. WRITING IN MATH Describe the steps to multiply a polynomial by a monomial.
SOLUTION: To multiply a polynomial by a monomial, use the Distributive Property. Multiply each term of the polynomial by the monomial. Then simplify by multiplying the coefficients together and using the Product of Powers Property for the variables. Consider the following example:
51. Every week a store sells j jeans and t T-shirts. The store makes $8 for each T-shirt and $12 for each pair of jeans. Which of the following expressions represents the total amount of money, in dollars, the store makes every week? A 8j + 12t B 12j + 8t C 20(j + t) D 96jt
SOLUTION: 12j is the amount made from jeans. 8t is the amount made from T-shirts. The sum, or 12j + 8t, represents the total amount the store makes every week. So, the correct choice is B.
52. If a = 5x + 7y and b = 2y − 3x, what is a + b?F 2x − 9y G 3y + 4x H 2x + 9y J 2x − 5y
SOLUTION:
So, the correct choice is H.
53. GEOMETRY A triangle has sides of length 5 inches and 8.5 inches. Which of the following cannot be the length of the third side? A 3.5 inches B 4 inches C 5.5 inches D 12 inches
SOLUTION: The sum of any two sides of a triangle must be greater than the length of the third side. Choice A: 3.5 + 5 = 8.5 8.5 Choice B: 3.5 + 5 = 8.5 > 8.5 Choice C: 5.5 + 5 = 10.5 > 8.5 Choice D: 12 + 5 = 17 > 8.5 Let x be the third side of the triangle.
3.5 cannot be the third side of the triangle. So, the correct choice is A.
54. SHORT RESPONSE Write an equation in which x varies directly as the cube of y and inversely as the square of z .
SOLUTION: If x varies directly as the cube of y , then both x and y must be in the numerators on either side of the equal sign.
When x varies inversely with the square of z. then the z must be in the denominator. Then . The
combination of the two is .
Find each sum or difference.
55. (2x2 − 7) + (8 − 5x
2)
SOLUTION:
56. (3z2 + 2z − 1) + (z
2 − 6)
SOLUTION:
57. (2a − 4a2 + 1) − (5a
2 − 2a − 6)
SOLUTION:
58. (a3 − 3a
2 + 4) − (4a
2 + 7)
SOLUTION:
59. (2ab − 3a + 4b) + (5a + 4ab)
SOLUTION:
60. (8c3 − 3c
2 + c − 2) − (3c
3 + 9)
SOLUTION:
Write a recursive formula for each sequence.61. 16, 2, –12, –26, ...
SOLUTION: Determine if there is a common difference. 2 – 16 = –14 –12 – 2 = –14 –26 – (–12) = –14 The common difference is –14. The first term is 16. To find the next term, subtract 14 from the previous term.
62. –5, 3, 11, 19, ...
SOLUTION: Determine if there is a common difference. 3 – (–5) = 8 11 – 3 = 8 19 – 11 = 8 The common difference is 8. The first term is 3. To find the next term, add 8 to the previous term.
63. 27, 43, 59, 75, ...
SOLUTION: Determine if there is a common difference. 43 – 27 = 16 59 – 43 = 16 75 – 59 = 16 The common difference is 16. The first term is 43. To find the next term, add 16 to the previous term.
64. 80, –200, 500, –1250, ...
SOLUTION: Check for a common difference. –200 – 80 = –280 500 – (–200) = 700 There is no common difference. Check for a common ratio. –200 ÷ 80 = –2.5 500 ÷ (–200) = –2.5 –1250 ÷ 500 = –2.5 The common ratio is –2.5. The first term is 80. To find the next term, multiply the previous term by –2.5.
65. 100, 60, 36, 21.6, ...
SOLUTION: Check for a common difference. 60 – 100 = –40 36 – 60 = –24 There is no common difference. Check for a common ratio. 60 ÷ 100 = 0.6 36 ÷ 60 = 0.6 21.6 ÷ 36 = 0.6 The common ratio is 0.6. The first term is 100. To find the next term, multiply the previous term by 0.6.
66.
SOLUTION: Check for a common difference.
There is no common difference. Check for a common ratio.
4 ÷ 1 = 4 The common ratio is 4.
The first term is . To find the next term, multiply the previous term by 4.
67. TRAVEL In 2003, about 9.5 million people took cruises. Between 2003 and 2008, the number increased by about 740,000 each year. Write the point-slope form of an equation to find the total number of people y taking a cruise for any year x. Estimate the number of people who took a cruise in 2010.
SOLUTION: The statement “increased by 740,000 each year” is a constant rate, or the slope. Now substitute the slope and the point (2003, 9,500,000) into the point slope equation.
To find the number of people who took a cruise in 2010, substitute x = 2010.
So, approximately 14,680,000 people took a cruise in 2010.
Write an equation in function notation for each relation.
68.
SOLUTION: The y-intercept is 0. Find the slope by choosing two points. Choose (0, 0) and (1, 4).
So the slope intercept form is f (x) = 4x + 0 or f (x) = 4x.
69.
SOLUTION:
The y-intercept is 0. Find the slope by choosing two points. Choose (0, 0) and (2, −1).
So in slope intercept form is f (x) = −0.5x + 0 or f (x) = −0.5x.
Simplify.
70. b(b2)(b
3)
SOLUTION:
71. 2y(3y2)
SOLUTION:
72. −y4(−2y
3)
SOLUTION:
73. −3z3(−5z
4 + 2z)
SOLUTION:
74. 2m(−4m4) − 3(−5m
3)
SOLUTION:
75. 4p 2(−2p
3) + 2p
4(5p
6)
SOLUTION:
eSolutions Manual - Powered by Cognero Page 3
8-2 Multiplying a Polynomial by a Monomial
Find each product.
1. 5w( −3w2 + 2w − 4)
SOLUTION:
2. 6g2(3g
3 + 4g2 + 10g − 1)
SOLUTION:
3. 4km2(8km
2 + 2k2m + 5k)
SOLUTION:
4. −3p4r3(2p
2r4 − 6p 6
r3 − 5)
SOLUTION:
5. 2ab(7a4b
2 + a5b − 2a)
SOLUTION:
6. c2d
3(5cd
7 − 3c3d
2 − 4d3)
SOLUTION:
Simplify each expression.
7. t(4t2 + 15t + 4) − 4(3t − 1)
SOLUTION:
8. x(3x2 + 4) + 2(7x − 3)
SOLUTION:
9. −2d(d3c
2 − 4dc2 + 2d
2c) + c
2(dc
2 − 3d4)
SOLUTION:
10. −5w2(8w
2x − 11wx
2) + 6x(9wx
4 − 4w − 3x2)
SOLUTION:
11. GRIDDED RESPONSE Marlene is buying a new plasma television. The height of the screen of the television is one half the width plus 5 inches. The width is 30 inches. Find the height of the screen in inches.
SOLUTION: Let h be the height and let w be the width.
Let w = 30.
The height is 20 inches.
Solve each equation.
12. −6(11 − 2c) = 7(−2 − 2c)
SOLUTION:
13. t(2t + 3) + 20 = 2t(t − 3)
SOLUTION:
14. −2(w + 1) + w = 7 − 4w
SOLUTION:
15. 3(y − 2) + 2y = 4y + 14
SOLUTION:
16. a(a + 3) + a(a − 6) + 35 = a(a − 5) + a(a + 7)
SOLUTION:
17. n(n − 4) + n(n + 8) = n(n − 13) + n(n + 1) + 16
SOLUTION:
Find each product.
18. b(b2 − 12b + 1)
SOLUTION:
19. f (f2 + 2f + 25)
SOLUTION:
20. −3m3(2m
3 − 12m2 + 2m + 25)
SOLUTION:
21. 2j 2(5j
3 − 15j2 + 2j + 2)
SOLUTION:
22. 2pr2(2pr + 5p
2r − 15p )
SOLUTION:
23. 4t3u(2t
2u
2 − 10tu4 + 2)
SOLUTION:
Simplify each expression.
24. −3(5x2 + 2x + 9) + x(2x − 3)
SOLUTION:
25. a(−8a2 + 2a + 4) + 3(6a
2 − 4)
SOLUTION:
26. −4d(5d2 − 12) + 7(d + 5)
SOLUTION:
27. −9g(−2g + g2) + 3(g
2 + 4)
SOLUTION:
28. 2j (7j2k
2 + j k2 + 5k) − 9k(−2j
2k
2 + 2k2 + 3j )
SOLUTION:
29. 4n(2n3p
2 − 3np2 + 5n) + 4p (6n
2p − 2np
2 + 3p )
SOLUTION:
30. DAMS A new dam being built has the shape of a trapezoid. The base at the bottom of the dam is 2 times the
height. The base at the top of the dam is times the height minus 30 feet.
a. Write an expression to find the area of the trapezoidal cross section of the dam. b. If the height of the dam is 180 feet, find the area of this cross section.
SOLUTION:
a. The equation for the area of a trapezoid is .
b1
= 2h
b2 =
b. Let h = 180.
So, the area is 32,940 square feet.
Solve each equation.
31. 7(t2 + 5t − 9) + t = t(7t − 2) + 13
SOLUTION:
32. w(4w + 6) + 2w = 2(2w2 + 7w − 3)
SOLUTION:
33. 5(4z + 6) − 2(z − 4) = 7z(z + 4) − z(7z − 2) − 48
SOLUTION:
34. 9c(c − 11) + 10(5c − 3) = 3c(c + 5) + c(6c − 3) − 30
SOLUTION:
35. 2f (5f − 2) − 10(f2 − 3f + 6) = −8f (f + 4) + 4(2f
2 − 7f )
SOLUTION:
36. 2k(−3k + 4) + 6(k2 + 10) = k(4k + 8) − 2k(2k + 5)
SOLUTION:
Simplify each expression.
37. np2(30p
2 + 9n
2p − 12)
SOLUTION:
38. r2t(10r
3 + 5rt
3 + 15t
2)
SOLUTION:
39. −5q2w
3(4q + 7w) + 4qw
2(7q
2w + 2q) − 3qw(3q
2w
2 + 9)
SOLUTION:
40. −x2z(2z
2 + 4xz3) + xz
2(xz + 5x
3z) + x2
z3(3x
2z + 4xz)
SOLUTION:
41. PARKING A parking garage charges $30 per month plus $0.50 per daytime hour and $0.25 per hour during nights and weekends. Suppose Trent parks in the garage for 47 hours in January and h of those are night and weekend hours. a. Find an expression for Trent’s January bill. b. Find the cost if Trent had 12 hours of night and weekend hours.
SOLUTION: a. First, multiply the price per hour times the number of hours. Let h represent the number of weekend and night hours. For the weekend and night hours, he would spend 0.25h. The number of daytime hours is 47 – h, so he spent 0.50(47 - h) = 23.50 – 0.50h. To find the total cost, add these two values to the cost per month. 30 + (23.50 − 0.50h) + 0.25h = 53.50 − 0.25h. b. If Trent parked 12 night and weekend hours, h = 12
So, the cost for January is $50.50.
42. CCSS MODELING Che is building a dog house for his new puppy. The upper face of the dog house is a trapezoid. If the height of the trapezoid is 12 inches, find the area of the face of this piece of the dog house.
SOLUTION:
The formula for the area of a trapezoid is .
Now substitute h = 12 inches into the equation.
Therefore, the area of the trapezoid is 318 square inches.
43. TENNIS The tennis club is building a new tennis court with a path around it.
a. Write an expression for the area of the tennis court. b. Write an expression for the area of the path. c. What is the perimeter of the outside of the path?
SOLUTION: a. The area of the tennis court can be found by the equation . Substitute the values of the length and width and simplify.
b. The area of the path is the area of the large rectangle minus the area of the smaller rectangle, which is the area ofthe tennis court.
Area of the large rectangle = 2.5x(x + 6) or 2.5x2 + 15x.
Now subtract the area of the tennis court to find the area of just the path.
c. The perimeter of the outside of the path is the perimeter of the larger rectangle.
If x = 36 feet, then the perimeter of the outside of the path is 7(36) + 12 or 264 ft.
44. MULTIPLE REPRESENTATIONS In this problem, you will investigate the degree of the product of a monomialand a polynomial. a. TABULAR Write three monomials of different degrees and three polynomials of different degrees. Determine the degree of each monomial and polynomial. Multiply the monomials by the polynomials. Determine the degree of each product. Record your results in a table like the one below.
b. VERBAL Make a conjecture about the degree of the product of a monomial and a polynomial. What is the degree of the product of a monomial of degree a and a polynomial of degree b?
SOLUTION: a. There possible monomials are 2x, 3x
2, 4x
3. Three polynomials are x2 − 1, x5 + 1, x
6 + 1.
The degree for the monomial is the sum of the exponents. For the example given, case they are 1,2, and 3. The degree for the polynomial is the largest degree of the monomials. For the example given, they are 2, 5, and 6. Use the distributive property to find the product of the monomial and the polynomial. For the example give, they
are2x3 − 2x, 3x
7 + 3x
2, and 4x
9 + 4x. For the example given, the degree of the new polynomials are 3,7 and 9.
b. The degree of the product is the sum of the degree of the monomial and the degree of the polynomial, or a + b.
45. ERROR ANALYSIS Pearl and Ted both worked on this problem. Is either of them correct? Explain your reasoning.
SOLUTION:
Ted’s solution is correct. Pearl made a mistake when distributing on the middle term. She did not add the exponents correctly.
46. CCSS PERSEVERANCE Find p such that 3xp(4x
2p+3 + 2x3p−2
) = 12x12 + 6x
10.
SOLUTION: First simplify the left side of the equation.
Because the coefficients of the terms are the same, the corresponding terms’ exponents can be set equal to each other.
47. CHALLENGE Find 4x−3
y2(2x
5y−4 + 6x
−7y
6 − 4x0y
-2).
SOLUTION:
48. REASONING Is there a value for x that makes the statement (x + 2)2 = x2 + 22
true? If so, find a value for x. Explain your reasoning.
SOLUTION: Solve for x.
When 0 is substituted in for x in the equation, both sides are 22 or 4, which makes the equation true.
49. OPEN ENDED Write a monomial and a polynomial using n as the variable. Find their product.
SOLUTION: Choose a monomial with variable n and a polynomial with variable n. For example, let the monomial be 3n and the polynomial be 4n + 1. Find the product.
50. WRITING IN MATH Describe the steps to multiply a polynomial by a monomial.
SOLUTION: To multiply a polynomial by a monomial, use the Distributive Property. Multiply each term of the polynomial by the monomial. Then simplify by multiplying the coefficients together and using the Product of Powers Property for the variables. Consider the following example:
51. Every week a store sells j jeans and t T-shirts. The store makes $8 for each T-shirt and $12 for each pair of jeans. Which of the following expressions represents the total amount of money, in dollars, the store makes every week? A 8j + 12t B 12j + 8t C 20(j + t) D 96jt
SOLUTION: 12j is the amount made from jeans. 8t is the amount made from T-shirts. The sum, or 12j + 8t, represents the total amount the store makes every week. So, the correct choice is B.
52. If a = 5x + 7y and b = 2y − 3x, what is a + b?F 2x − 9y G 3y + 4x H 2x + 9y J 2x − 5y
SOLUTION:
So, the correct choice is H.
53. GEOMETRY A triangle has sides of length 5 inches and 8.5 inches. Which of the following cannot be the length of the third side? A 3.5 inches B 4 inches C 5.5 inches D 12 inches
SOLUTION: The sum of any two sides of a triangle must be greater than the length of the third side. Choice A: 3.5 + 5 = 8.5 8.5 Choice B: 3.5 + 5 = 8.5 > 8.5 Choice C: 5.5 + 5 = 10.5 > 8.5 Choice D: 12 + 5 = 17 > 8.5 Let x be the third side of the triangle.
3.5 cannot be the third side of the triangle. So, the correct choice is A.
54. SHORT RESPONSE Write an equation in which x varies directly as the cube of y and inversely as the square of z .
SOLUTION: If x varies directly as the cube of y , then both x and y must be in the numerators on either side of the equal sign.
When x varies inversely with the square of z. then the z must be in the denominator. Then . The
combination of the two is .
Find each sum or difference.
55. (2x2 − 7) + (8 − 5x
2)
SOLUTION:
56. (3z2 + 2z − 1) + (z
2 − 6)
SOLUTION:
57. (2a − 4a2 + 1) − (5a
2 − 2a − 6)
SOLUTION:
58. (a3 − 3a
2 + 4) − (4a
2 + 7)
SOLUTION:
59. (2ab − 3a + 4b) + (5a + 4ab)
SOLUTION:
60. (8c3 − 3c
2 + c − 2) − (3c
3 + 9)
SOLUTION:
Write a recursive formula for each sequence.61. 16, 2, –12, –26, ...
SOLUTION: Determine if there is a common difference. 2 – 16 = –14 –12 – 2 = –14 –26 – (–12) = –14 The common difference is –14. The first term is 16. To find the next term, subtract 14 from the previous term.
62. –5, 3, 11, 19, ...
SOLUTION: Determine if there is a common difference. 3 – (–5) = 8 11 – 3 = 8 19 – 11 = 8 The common difference is 8. The first term is 3. To find the next term, add 8 to the previous term.
63. 27, 43, 59, 75, ...
SOLUTION: Determine if there is a common difference. 43 – 27 = 16 59 – 43 = 16 75 – 59 = 16 The common difference is 16. The first term is 43. To find the next term, add 16 to the previous term.
64. 80, –200, 500, –1250, ...
SOLUTION: Check for a common difference. –200 – 80 = –280 500 – (–200) = 700 There is no common difference. Check for a common ratio. –200 ÷ 80 = –2.5 500 ÷ (–200) = –2.5 –1250 ÷ 500 = –2.5 The common ratio is –2.5. The first term is 80. To find the next term, multiply the previous term by –2.5.
65. 100, 60, 36, 21.6, ...
SOLUTION: Check for a common difference. 60 – 100 = –40 36 – 60 = –24 There is no common difference. Check for a common ratio. 60 ÷ 100 = 0.6 36 ÷ 60 = 0.6 21.6 ÷ 36 = 0.6 The common ratio is 0.6. The first term is 100. To find the next term, multiply the previous term by 0.6.
66.
SOLUTION: Check for a common difference.
There is no common difference. Check for a common ratio.
4 ÷ 1 = 4 The common ratio is 4.
The first term is . To find the next term, multiply the previous term by 4.
67. TRAVEL In 2003, about 9.5 million people took cruises. Between 2003 and 2008, the number increased by about 740,000 each year. Write the point-slope form of an equation to find the total number of people y taking a cruise for any year x. Estimate the number of people who took a cruise in 2010.
SOLUTION: The statement “increased by 740,000 each year” is a constant rate, or the slope. Now substitute the slope and the point (2003, 9,500,000) into the point slope equation.
To find the number of people who took a cruise in 2010, substitute x = 2010.
So, approximately 14,680,000 people took a cruise in 2010.
Write an equation in function notation for each relation.
68.
SOLUTION: The y-intercept is 0. Find the slope by choosing two points. Choose (0, 0) and (1, 4).
So the slope intercept form is f (x) = 4x + 0 or f (x) = 4x.
69.
SOLUTION:
The y-intercept is 0. Find the slope by choosing two points. Choose (0, 0) and (2, −1).
So in slope intercept form is f (x) = −0.5x + 0 or f (x) = −0.5x.
Simplify.
70. b(b2)(b
3)
SOLUTION:
71. 2y(3y2)
SOLUTION:
72. −y4(−2y
3)
SOLUTION:
73. −3z3(−5z
4 + 2z)
SOLUTION:
74. 2m(−4m4) − 3(−5m
3)
SOLUTION:
75. 4p 2(−2p
3) + 2p
4(5p
6)
SOLUTION:
eSolutions Manual - Powered by Cognero Page 4
8-2 Multiplying a Polynomial by a Monomial
Find each product.
1. 5w( −3w2 + 2w − 4)
SOLUTION:
2. 6g2(3g
3 + 4g2 + 10g − 1)
SOLUTION:
3. 4km2(8km
2 + 2k2m + 5k)
SOLUTION:
4. −3p4r3(2p
2r4 − 6p 6
r3 − 5)
SOLUTION:
5. 2ab(7a4b
2 + a5b − 2a)
SOLUTION:
6. c2d
3(5cd
7 − 3c3d
2 − 4d3)
SOLUTION:
Simplify each expression.
7. t(4t2 + 15t + 4) − 4(3t − 1)
SOLUTION:
8. x(3x2 + 4) + 2(7x − 3)
SOLUTION:
9. −2d(d3c
2 − 4dc2 + 2d
2c) + c
2(dc
2 − 3d4)
SOLUTION:
10. −5w2(8w
2x − 11wx
2) + 6x(9wx
4 − 4w − 3x2)
SOLUTION:
11. GRIDDED RESPONSE Marlene is buying a new plasma television. The height of the screen of the television is one half the width plus 5 inches. The width is 30 inches. Find the height of the screen in inches.
SOLUTION: Let h be the height and let w be the width.
Let w = 30.
The height is 20 inches.
Solve each equation.
12. −6(11 − 2c) = 7(−2 − 2c)
SOLUTION:
13. t(2t + 3) + 20 = 2t(t − 3)
SOLUTION:
14. −2(w + 1) + w = 7 − 4w
SOLUTION:
15. 3(y − 2) + 2y = 4y + 14
SOLUTION:
16. a(a + 3) + a(a − 6) + 35 = a(a − 5) + a(a + 7)
SOLUTION:
17. n(n − 4) + n(n + 8) = n(n − 13) + n(n + 1) + 16
SOLUTION:
Find each product.
18. b(b2 − 12b + 1)
SOLUTION:
19. f (f2 + 2f + 25)
SOLUTION:
20. −3m3(2m
3 − 12m2 + 2m + 25)
SOLUTION:
21. 2j 2(5j
3 − 15j2 + 2j + 2)
SOLUTION:
22. 2pr2(2pr + 5p
2r − 15p )
SOLUTION:
23. 4t3u(2t
2u
2 − 10tu4 + 2)
SOLUTION:
Simplify each expression.
24. −3(5x2 + 2x + 9) + x(2x − 3)
SOLUTION:
25. a(−8a2 + 2a + 4) + 3(6a
2 − 4)
SOLUTION:
26. −4d(5d2 − 12) + 7(d + 5)
SOLUTION:
27. −9g(−2g + g2) + 3(g
2 + 4)
SOLUTION:
28. 2j (7j2k
2 + j k2 + 5k) − 9k(−2j
2k
2 + 2k2 + 3j )
SOLUTION:
29. 4n(2n3p
2 − 3np2 + 5n) + 4p (6n
2p − 2np
2 + 3p )
SOLUTION:
30. DAMS A new dam being built has the shape of a trapezoid. The base at the bottom of the dam is 2 times the
height. The base at the top of the dam is times the height minus 30 feet.
a. Write an expression to find the area of the trapezoidal cross section of the dam. b. If the height of the dam is 180 feet, find the area of this cross section.
SOLUTION:
a. The equation for the area of a trapezoid is .
b1
= 2h
b2 =
b. Let h = 180.
So, the area is 32,940 square feet.
Solve each equation.
31. 7(t2 + 5t − 9) + t = t(7t − 2) + 13
SOLUTION:
32. w(4w + 6) + 2w = 2(2w2 + 7w − 3)
SOLUTION:
33. 5(4z + 6) − 2(z − 4) = 7z(z + 4) − z(7z − 2) − 48
SOLUTION:
34. 9c(c − 11) + 10(5c − 3) = 3c(c + 5) + c(6c − 3) − 30
SOLUTION:
35. 2f (5f − 2) − 10(f2 − 3f + 6) = −8f (f + 4) + 4(2f
2 − 7f )
SOLUTION:
36. 2k(−3k + 4) + 6(k2 + 10) = k(4k + 8) − 2k(2k + 5)
SOLUTION:
Simplify each expression.
37. np2(30p
2 + 9n
2p − 12)
SOLUTION:
38. r2t(10r
3 + 5rt
3 + 15t
2)
SOLUTION:
39. −5q2w
3(4q + 7w) + 4qw
2(7q
2w + 2q) − 3qw(3q
2w
2 + 9)
SOLUTION:
40. −x2z(2z
2 + 4xz3) + xz
2(xz + 5x
3z) + x2
z3(3x
2z + 4xz)
SOLUTION:
41. PARKING A parking garage charges $30 per month plus $0.50 per daytime hour and $0.25 per hour during nights and weekends. Suppose Trent parks in the garage for 47 hours in January and h of those are night and weekend hours. a. Find an expression for Trent’s January bill. b. Find the cost if Trent had 12 hours of night and weekend hours.
SOLUTION: a. First, multiply the price per hour times the number of hours. Let h represent the number of weekend and night hours. For the weekend and night hours, he would spend 0.25h. The number of daytime hours is 47 – h, so he spent 0.50(47 - h) = 23.50 – 0.50h. To find the total cost, add these two values to the cost per month. 30 + (23.50 − 0.50h) + 0.25h = 53.50 − 0.25h. b. If Trent parked 12 night and weekend hours, h = 12
So, the cost for January is $50.50.
42. CCSS MODELING Che is building a dog house for his new puppy. The upper face of the dog house is a trapezoid. If the height of the trapezoid is 12 inches, find the area of the face of this piece of the dog house.
SOLUTION:
The formula for the area of a trapezoid is .
Now substitute h = 12 inches into the equation.
Therefore, the area of the trapezoid is 318 square inches.
43. TENNIS The tennis club is building a new tennis court with a path around it.
a. Write an expression for the area of the tennis court. b. Write an expression for the area of the path. c. What is the perimeter of the outside of the path?
SOLUTION: a. The area of the tennis court can be found by the equation . Substitute the values of the length and width and simplify.
b. The area of the path is the area of the large rectangle minus the area of the smaller rectangle, which is the area ofthe tennis court.
Area of the large rectangle = 2.5x(x + 6) or 2.5x2 + 15x.
Now subtract the area of the tennis court to find the area of just the path.
c. The perimeter of the outside of the path is the perimeter of the larger rectangle.
If x = 36 feet, then the perimeter of the outside of the path is 7(36) + 12 or 264 ft.
44. MULTIPLE REPRESENTATIONS In this problem, you will investigate the degree of the product of a monomialand a polynomial. a. TABULAR Write three monomials of different degrees and three polynomials of different degrees. Determine the degree of each monomial and polynomial. Multiply the monomials by the polynomials. Determine the degree of each product. Record your results in a table like the one below.
b. VERBAL Make a conjecture about the degree of the product of a monomial and a polynomial. What is the degree of the product of a monomial of degree a and a polynomial of degree b?
SOLUTION: a. There possible monomials are 2x, 3x
2, 4x
3. Three polynomials are x2 − 1, x5 + 1, x
6 + 1.
The degree for the monomial is the sum of the exponents. For the example given, case they are 1,2, and 3. The degree for the polynomial is the largest degree of the monomials. For the example given, they are 2, 5, and 6. Use the distributive property to find the product of the monomial and the polynomial. For the example give, they
are2x3 − 2x, 3x
7 + 3x
2, and 4x
9 + 4x. For the example given, the degree of the new polynomials are 3,7 and 9.
b. The degree of the product is the sum of the degree of the monomial and the degree of the polynomial, or a + b.
45. ERROR ANALYSIS Pearl and Ted both worked on this problem. Is either of them correct? Explain your reasoning.
SOLUTION:
Ted’s solution is correct. Pearl made a mistake when distributing on the middle term. She did not add the exponents correctly.
46. CCSS PERSEVERANCE Find p such that 3xp(4x
2p+3 + 2x3p−2
) = 12x12 + 6x
10.
SOLUTION: First simplify the left side of the equation.
Because the coefficients of the terms are the same, the corresponding terms’ exponents can be set equal to each other.
47. CHALLENGE Find 4x−3
y2(2x
5y−4 + 6x
−7y
6 − 4x0y
-2).
SOLUTION:
48. REASONING Is there a value for x that makes the statement (x + 2)2 = x2 + 22
true? If so, find a value for x. Explain your reasoning.
SOLUTION: Solve for x.
When 0 is substituted in for x in the equation, both sides are 22 or 4, which makes the equation true.
49. OPEN ENDED Write a monomial and a polynomial using n as the variable. Find their product.
SOLUTION: Choose a monomial with variable n and a polynomial with variable n. For example, let the monomial be 3n and the polynomial be 4n + 1. Find the product.
50. WRITING IN MATH Describe the steps to multiply a polynomial by a monomial.
SOLUTION: To multiply a polynomial by a monomial, use the Distributive Property. Multiply each term of the polynomial by the monomial. Then simplify by multiplying the coefficients together and using the Product of Powers Property for the variables. Consider the following example:
51. Every week a store sells j jeans and t T-shirts. The store makes $8 for each T-shirt and $12 for each pair of jeans. Which of the following expressions represents the total amount of money, in dollars, the store makes every week? A 8j + 12t B 12j + 8t C 20(j + t) D 96jt
SOLUTION: 12j is the amount made from jeans. 8t is the amount made from T-shirts. The sum, or 12j + 8t, represents the total amount the store makes every week. So, the correct choice is B.
52. If a = 5x + 7y and b = 2y − 3x, what is a + b?F 2x − 9y G 3y + 4x H 2x + 9y J 2x − 5y
SOLUTION:
So, the correct choice is H.
53. GEOMETRY A triangle has sides of length 5 inches and 8.5 inches. Which of the following cannot be the length of the third side? A 3.5 inches B 4 inches C 5.5 inches D 12 inches
SOLUTION: The sum of any two sides of a triangle must be greater than the length of the third side. Choice A: 3.5 + 5 = 8.5 8.5 Choice B: 3.5 + 5 = 8.5 > 8.5 Choice C: 5.5 + 5 = 10.5 > 8.5 Choice D: 12 + 5 = 17 > 8.5 Let x be the third side of the triangle.
3.5 cannot be the third side of the triangle. So, the correct choice is A.
54. SHORT RESPONSE Write an equation in which x varies directly as the cube of y and inversely as the square of z .
SOLUTION: If x varies directly as the cube of y , then both x and y must be in the numerators on either side of the equal sign.
When x varies inversely with the square of z. then the z must be in the denominator. Then . The
combination of the two is .
Find each sum or difference.
55. (2x2 − 7) + (8 − 5x
2)
SOLUTION:
56. (3z2 + 2z − 1) + (z
2 − 6)
SOLUTION:
57. (2a − 4a2 + 1) − (5a
2 − 2a − 6)
SOLUTION:
58. (a3 − 3a
2 + 4) − (4a
2 + 7)
SOLUTION:
59. (2ab − 3a + 4b) + (5a + 4ab)
SOLUTION:
60. (8c3 − 3c
2 + c − 2) − (3c
3 + 9)
SOLUTION:
Write a recursive formula for each sequence.61. 16, 2, –12, –26, ...
SOLUTION: Determine if there is a common difference. 2 – 16 = –14 –12 – 2 = –14 –26 – (–12) = –14 The common difference is –14. The first term is 16. To find the next term, subtract 14 from the previous term.
62. –5, 3, 11, 19, ...
SOLUTION: Determine if there is a common difference. 3 – (–5) = 8 11 – 3 = 8 19 – 11 = 8 The common difference is 8. The first term is 3. To find the next term, add 8 to the previous term.
63. 27, 43, 59, 75, ...
SOLUTION: Determine if there is a common difference. 43 – 27 = 16 59 – 43 = 16 75 – 59 = 16 The common difference is 16. The first term is 43. To find the next term, add 16 to the previous term.
64. 80, –200, 500, –1250, ...
SOLUTION: Check for a common difference. –200 – 80 = –280 500 – (–200) = 700 There is no common difference. Check for a common ratio. –200 ÷ 80 = –2.5 500 ÷ (–200) = –2.5 –1250 ÷ 500 = –2.5 The common ratio is –2.5. The first term is 80. To find the next term, multiply the previous term by –2.5.
65. 100, 60, 36, 21.6, ...
SOLUTION: Check for a common difference. 60 – 100 = –40 36 – 60 = –24 There is no common difference. Check for a common ratio. 60 ÷ 100 = 0.6 36 ÷ 60 = 0.6 21.6 ÷ 36 = 0.6 The common ratio is 0.6. The first term is 100. To find the next term, multiply the previous term by 0.6.
66.
SOLUTION: Check for a common difference.
There is no common difference. Check for a common ratio.
4 ÷ 1 = 4 The common ratio is 4.
The first term is . To find the next term, multiply the previous term by 4.
67. TRAVEL In 2003, about 9.5 million people took cruises. Between 2003 and 2008, the number increased by about 740,000 each year. Write the point-slope form of an equation to find the total number of people y taking a cruise for any year x. Estimate the number of people who took a cruise in 2010.
SOLUTION: The statement “increased by 740,000 each year” is a constant rate, or the slope. Now substitute the slope and the point (2003, 9,500,000) into the point slope equation.
To find the number of people who took a cruise in 2010, substitute x = 2010.
So, approximately 14,680,000 people took a cruise in 2010.
Write an equation in function notation for each relation.
68.
SOLUTION: The y-intercept is 0. Find the slope by choosing two points. Choose (0, 0) and (1, 4).
So the slope intercept form is f (x) = 4x + 0 or f (x) = 4x.
69.
SOLUTION:
The y-intercept is 0. Find the slope by choosing two points. Choose (0, 0) and (2, −1).
So in slope intercept form is f (x) = −0.5x + 0 or f (x) = −0.5x.
Simplify.
70. b(b2)(b
3)
SOLUTION:
71. 2y(3y2)
SOLUTION:
72. −y4(−2y
3)
SOLUTION:
73. −3z3(−5z
4 + 2z)
SOLUTION:
74. 2m(−4m4) − 3(−5m
3)
SOLUTION:
75. 4p 2(−2p
3) + 2p
4(5p
6)
SOLUTION:
eSolutions Manual - Powered by Cognero Page 5
8-2 Multiplying a Polynomial by a Monomial
Find each product.
1. 5w( −3w2 + 2w − 4)
SOLUTION:
2. 6g2(3g
3 + 4g2 + 10g − 1)
SOLUTION:
3. 4km2(8km
2 + 2k2m + 5k)
SOLUTION:
4. −3p4r3(2p
2r4 − 6p 6
r3 − 5)
SOLUTION:
5. 2ab(7a4b
2 + a5b − 2a)
SOLUTION:
6. c2d
3(5cd
7 − 3c3d
2 − 4d3)
SOLUTION:
Simplify each expression.
7. t(4t2 + 15t + 4) − 4(3t − 1)
SOLUTION:
8. x(3x2 + 4) + 2(7x − 3)
SOLUTION:
9. −2d(d3c
2 − 4dc2 + 2d
2c) + c
2(dc
2 − 3d4)
SOLUTION:
10. −5w2(8w
2x − 11wx
2) + 6x(9wx
4 − 4w − 3x2)
SOLUTION:
11. GRIDDED RESPONSE Marlene is buying a new plasma television. The height of the screen of the television is one half the width plus 5 inches. The width is 30 inches. Find the height of the screen in inches.
SOLUTION: Let h be the height and let w be the width.
Let w = 30.
The height is 20 inches.
Solve each equation.
12. −6(11 − 2c) = 7(−2 − 2c)
SOLUTION:
13. t(2t + 3) + 20 = 2t(t − 3)
SOLUTION:
14. −2(w + 1) + w = 7 − 4w
SOLUTION:
15. 3(y − 2) + 2y = 4y + 14
SOLUTION:
16. a(a + 3) + a(a − 6) + 35 = a(a − 5) + a(a + 7)
SOLUTION:
17. n(n − 4) + n(n + 8) = n(n − 13) + n(n + 1) + 16
SOLUTION:
Find each product.
18. b(b2 − 12b + 1)
SOLUTION:
19. f (f2 + 2f + 25)
SOLUTION:
20. −3m3(2m
3 − 12m2 + 2m + 25)
SOLUTION:
21. 2j 2(5j
3 − 15j2 + 2j + 2)
SOLUTION:
22. 2pr2(2pr + 5p
2r − 15p )
SOLUTION:
23. 4t3u(2t
2u
2 − 10tu4 + 2)
SOLUTION:
Simplify each expression.
24. −3(5x2 + 2x + 9) + x(2x − 3)
SOLUTION:
25. a(−8a2 + 2a + 4) + 3(6a
2 − 4)
SOLUTION:
26. −4d(5d2 − 12) + 7(d + 5)
SOLUTION:
27. −9g(−2g + g2) + 3(g
2 + 4)
SOLUTION:
28. 2j (7j2k
2 + j k2 + 5k) − 9k(−2j
2k
2 + 2k2 + 3j )
SOLUTION:
29. 4n(2n3p
2 − 3np2 + 5n) + 4p (6n
2p − 2np
2 + 3p )
SOLUTION:
30. DAMS A new dam being built has the shape of a trapezoid. The base at the bottom of the dam is 2 times the
height. The base at the top of the dam is times the height minus 30 feet.
a. Write an expression to find the area of the trapezoidal cross section of the dam. b. If the height of the dam is 180 feet, find the area of this cross section.
SOLUTION:
a. The equation for the area of a trapezoid is .
b1
= 2h
b2 =
b. Let h = 180.
So, the area is 32,940 square feet.
Solve each equation.
31. 7(t2 + 5t − 9) + t = t(7t − 2) + 13
SOLUTION:
32. w(4w + 6) + 2w = 2(2w2 + 7w − 3)
SOLUTION:
33. 5(4z + 6) − 2(z − 4) = 7z(z + 4) − z(7z − 2) − 48
SOLUTION:
34. 9c(c − 11) + 10(5c − 3) = 3c(c + 5) + c(6c − 3) − 30
SOLUTION:
35. 2f (5f − 2) − 10(f2 − 3f + 6) = −8f (f + 4) + 4(2f
2 − 7f )
SOLUTION:
36. 2k(−3k + 4) + 6(k2 + 10) = k(4k + 8) − 2k(2k + 5)
SOLUTION:
Simplify each expression.
37. np2(30p
2 + 9n
2p − 12)
SOLUTION:
38. r2t(10r
3 + 5rt
3 + 15t
2)
SOLUTION:
39. −5q2w
3(4q + 7w) + 4qw
2(7q
2w + 2q) − 3qw(3q
2w
2 + 9)
SOLUTION:
40. −x2z(2z
2 + 4xz3) + xz
2(xz + 5x
3z) + x2
z3(3x
2z + 4xz)
SOLUTION:
41. PARKING A parking garage charges $30 per month plus $0.50 per daytime hour and $0.25 per hour during nights and weekends. Suppose Trent parks in the garage for 47 hours in January and h of those are night and weekend hours. a. Find an expression for Trent’s January bill. b. Find the cost if Trent had 12 hours of night and weekend hours.
SOLUTION: a. First, multiply the price per hour times the number of hours. Let h represent the number of weekend and night hours. For the weekend and night hours, he would spend 0.25h. The number of daytime hours is 47 – h, so he spent 0.50(47 - h) = 23.50 – 0.50h. To find the total cost, add these two values to the cost per month. 30 + (23.50 − 0.50h) + 0.25h = 53.50 − 0.25h. b. If Trent parked 12 night and weekend hours, h = 12
So, the cost for January is $50.50.
42. CCSS MODELING Che is building a dog house for his new puppy. The upper face of the dog house is a trapezoid. If the height of the trapezoid is 12 inches, find the area of the face of this piece of the dog house.
SOLUTION:
The formula for the area of a trapezoid is .
Now substitute h = 12 inches into the equation.
Therefore, the area of the trapezoid is 318 square inches.
43. TENNIS The tennis club is building a new tennis court with a path around it.
a. Write an expression for the area of the tennis court. b. Write an expression for the area of the path. c. What is the perimeter of the outside of the path?
SOLUTION: a. The area of the tennis court can be found by the equation . Substitute the values of the length and width and simplify.
b. The area of the path is the area of the large rectangle minus the area of the smaller rectangle, which is the area ofthe tennis court.
Area of the large rectangle = 2.5x(x + 6) or 2.5x2 + 15x.
Now subtract the area of the tennis court to find the area of just the path.
c. The perimeter of the outside of the path is the perimeter of the larger rectangle.
If x = 36 feet, then the perimeter of the outside of the path is 7(36) + 12 or 264 ft.
44. MULTIPLE REPRESENTATIONS In this problem, you will investigate the degree of the product of a monomialand a polynomial. a. TABULAR Write three monomials of different degrees and three polynomials of different degrees. Determine the degree of each monomial and polynomial. Multiply the monomials by the polynomials. Determine the degree of each product. Record your results in a table like the one below.
b. VERBAL Make a conjecture about the degree of the product of a monomial and a polynomial. What is the degree of the product of a monomial of degree a and a polynomial of degree b?
SOLUTION: a. There possible monomials are 2x, 3x
2, 4x
3. Three polynomials are x2 − 1, x5 + 1, x
6 + 1.
The degree for the monomial is the sum of the exponents. For the example given, case they are 1,2, and 3. The degree for the polynomial is the largest degree of the monomials. For the example given, they are 2, 5, and 6. Use the distributive property to find the product of the monomial and the polynomial. For the example give, they
are2x3 − 2x, 3x
7 + 3x
2, and 4x
9 + 4x. For the example given, the degree of the new polynomials are 3,7 and 9.
b. The degree of the product is the sum of the degree of the monomial and the degree of the polynomial, or a + b.
45. ERROR ANALYSIS Pearl and Ted both worked on this problem. Is either of them correct? Explain your reasoning.
SOLUTION:
Ted’s solution is correct. Pearl made a mistake when distributing on the middle term. She did not add the exponents correctly.
46. CCSS PERSEVERANCE Find p such that 3xp(4x
2p+3 + 2x3p−2
) = 12x12 + 6x
10.
SOLUTION: First simplify the left side of the equation.
Because the coefficients of the terms are the same, the corresponding terms’ exponents can be set equal to each other.
47. CHALLENGE Find 4x−3
y2(2x
5y−4 + 6x
−7y
6 − 4x0y
-2).
SOLUTION:
48. REASONING Is there a value for x that makes the statement (x + 2)2 = x2 + 22
true? If so, find a value for x. Explain your reasoning.
SOLUTION: Solve for x.
When 0 is substituted in for x in the equation, both sides are 22 or 4, which makes the equation true.
49. OPEN ENDED Write a monomial and a polynomial using n as the variable. Find their product.
SOLUTION: Choose a monomial with variable n and a polynomial with variable n. For example, let the monomial be 3n and the polynomial be 4n + 1. Find the product.
50. WRITING IN MATH Describe the steps to multiply a polynomial by a monomial.
SOLUTION: To multiply a polynomial by a monomial, use the Distributive Property. Multiply each term of the polynomial by the monomial. Then simplify by multiplying the coefficients together and using the Product of Powers Property for the variables. Consider the following example:
51. Every week a store sells j jeans and t T-shirts. The store makes $8 for each T-shirt and $12 for each pair of jeans. Which of the following expressions represents the total amount of money, in dollars, the store makes every week? A 8j + 12t B 12j + 8t C 20(j + t) D 96jt
SOLUTION: 12j is the amount made from jeans. 8t is the amount made from T-shirts. The sum, or 12j + 8t, represents the total amount the store makes every week. So, the correct choice is B.
52. If a = 5x + 7y and b = 2y − 3x, what is a + b?F 2x − 9y G 3y + 4x H 2x + 9y J 2x − 5y
SOLUTION:
So, the correct choice is H.
53. GEOMETRY A triangle has sides of length 5 inches and 8.5 inches. Which of the following cannot be the length of the third side? A 3.5 inches B 4 inches C 5.5 inches D 12 inches
SOLUTION: The sum of any two sides of a triangle must be greater than the length of the third side. Choice A: 3.5 + 5 = 8.5 8.5 Choice B: 3.5 + 5 = 8.5 > 8.5 Choice C: 5.5 + 5 = 10.5 > 8.5 Choice D: 12 + 5 = 17 > 8.5 Let x be the third side of the triangle.
3.5 cannot be the third side of the triangle. So, the correct choice is A.
54. SHORT RESPONSE Write an equation in which x varies directly as the cube of y and inversely as the square of z .
SOLUTION: If x varies directly as the cube of y , then both x and y must be in the numerators on either side of the equal sign.
When x varies inversely with the square of z. then the z must be in the denominator. Then . The
combination of the two is .
Find each sum or difference.
55. (2x2 − 7) + (8 − 5x
2)
SOLUTION:
56. (3z2 + 2z − 1) + (z
2 − 6)
SOLUTION:
57. (2a − 4a2 + 1) − (5a
2 − 2a − 6)
SOLUTION:
58. (a3 − 3a
2 + 4) − (4a
2 + 7)
SOLUTION:
59. (2ab − 3a + 4b) + (5a + 4ab)
SOLUTION:
60. (8c3 − 3c
2 + c − 2) − (3c
3 + 9)
SOLUTION:
Write a recursive formula for each sequence.61. 16, 2, –12, –26, ...
SOLUTION: Determine if there is a common difference. 2 – 16 = –14 –12 – 2 = –14 –26 – (–12) = –14 The common difference is –14. The first term is 16. To find the next term, subtract 14 from the previous term.
62. –5, 3, 11, 19, ...
SOLUTION: Determine if there is a common difference. 3 – (–5) = 8 11 – 3 = 8 19 – 11 = 8 The common difference is 8. The first term is 3. To find the next term, add 8 to the previous term.
63. 27, 43, 59, 75, ...
SOLUTION: Determine if there is a common difference. 43 – 27 = 16 59 – 43 = 16 75 – 59 = 16 The common difference is 16. The first term is 43. To find the next term, add 16 to the previous term.
64. 80, –200, 500, –1250, ...
SOLUTION: Check for a common difference. –200 – 80 = –280 500 – (–200) = 700 There is no common difference. Check for a common ratio. –200 ÷ 80 = –2.5 500 ÷ (–200) = –2.5 –1250 ÷ 500 = –2.5 The common ratio is –2.5. The first term is 80. To find the next term, multiply the previous term by –2.5.
65. 100, 60, 36, 21.6, ...
SOLUTION: Check for a common difference. 60 – 100 = –40 36 – 60 = –24 There is no common difference. Check for a common ratio. 60 ÷ 100 = 0.6 36 ÷ 60 = 0.6 21.6 ÷ 36 = 0.6 The common ratio is 0.6. The first term is 100. To find the next term, multiply the previous term by 0.6.
66.
SOLUTION: Check for a common difference.
There is no common difference. Check for a common ratio.
4 ÷ 1 = 4 The common ratio is 4.
The first term is . To find the next term, multiply the previous term by 4.
67. TRAVEL In 2003, about 9.5 million people took cruises. Between 2003 and 2008, the number increased by about 740,000 each year. Write the point-slope form of an equation to find the total number of people y taking a cruise for any year x. Estimate the number of people who took a cruise in 2010.
SOLUTION: The statement “increased by 740,000 each year” is a constant rate, or the slope. Now substitute the slope and the point (2003, 9,500,000) into the point slope equation.
To find the number of people who took a cruise in 2010, substitute x = 2010.
So, approximately 14,680,000 people took a cruise in 2010.
Write an equation in function notation for each relation.
68.
SOLUTION: The y-intercept is 0. Find the slope by choosing two points. Choose (0, 0) and (1, 4).
So the slope intercept form is f (x) = 4x + 0 or f (x) = 4x.
69.
SOLUTION:
The y-intercept is 0. Find the slope by choosing two points. Choose (0, 0) and (2, −1).
So in slope intercept form is f (x) = −0.5x + 0 or f (x) = −0.5x.
Simplify.
70. b(b2)(b
3)
SOLUTION:
71. 2y(3y2)
SOLUTION:
72. −y4(−2y
3)
SOLUTION:
73. −3z3(−5z
4 + 2z)
SOLUTION:
74. 2m(−4m4) − 3(−5m
3)
SOLUTION:
75. 4p 2(−2p
3) + 2p
4(5p
6)
SOLUTION:
eSolutions Manual - Powered by Cognero Page 6
8-2 Multiplying a Polynomial by a Monomial
Find each product.
1. 5w( −3w2 + 2w − 4)
SOLUTION:
2. 6g2(3g
3 + 4g2 + 10g − 1)
SOLUTION:
3. 4km2(8km
2 + 2k2m + 5k)
SOLUTION:
4. −3p4r3(2p
2r4 − 6p 6
r3 − 5)
SOLUTION:
5. 2ab(7a4b
2 + a5b − 2a)
SOLUTION:
6. c2d
3(5cd
7 − 3c3d
2 − 4d3)
SOLUTION:
Simplify each expression.
7. t(4t2 + 15t + 4) − 4(3t − 1)
SOLUTION:
8. x(3x2 + 4) + 2(7x − 3)
SOLUTION:
9. −2d(d3c
2 − 4dc2 + 2d
2c) + c
2(dc
2 − 3d4)
SOLUTION:
10. −5w2(8w
2x − 11wx
2) + 6x(9wx
4 − 4w − 3x2)
SOLUTION:
11. GRIDDED RESPONSE Marlene is buying a new plasma television. The height of the screen of the television is one half the width plus 5 inches. The width is 30 inches. Find the height of the screen in inches.
SOLUTION: Let h be the height and let w be the width.
Let w = 30.
The height is 20 inches.
Solve each equation.
12. −6(11 − 2c) = 7(−2 − 2c)
SOLUTION:
13. t(2t + 3) + 20 = 2t(t − 3)
SOLUTION:
14. −2(w + 1) + w = 7 − 4w
SOLUTION:
15. 3(y − 2) + 2y = 4y + 14
SOLUTION:
16. a(a + 3) + a(a − 6) + 35 = a(a − 5) + a(a + 7)
SOLUTION:
17. n(n − 4) + n(n + 8) = n(n − 13) + n(n + 1) + 16
SOLUTION:
Find each product.
18. b(b2 − 12b + 1)
SOLUTION:
19. f (f2 + 2f + 25)
SOLUTION:
20. −3m3(2m
3 − 12m2 + 2m + 25)
SOLUTION:
21. 2j 2(5j
3 − 15j2 + 2j + 2)
SOLUTION:
22. 2pr2(2pr + 5p
2r − 15p )
SOLUTION:
23. 4t3u(2t
2u
2 − 10tu4 + 2)
SOLUTION:
Simplify each expression.
24. −3(5x2 + 2x + 9) + x(2x − 3)
SOLUTION:
25. a(−8a2 + 2a + 4) + 3(6a
2 − 4)
SOLUTION:
26. −4d(5d2 − 12) + 7(d + 5)
SOLUTION:
27. −9g(−2g + g2) + 3(g
2 + 4)
SOLUTION:
28. 2j (7j2k
2 + j k2 + 5k) − 9k(−2j
2k
2 + 2k2 + 3j )
SOLUTION:
29. 4n(2n3p
2 − 3np2 + 5n) + 4p (6n
2p − 2np
2 + 3p )
SOLUTION:
30. DAMS A new dam being built has the shape of a trapezoid. The base at the bottom of the dam is 2 times the
height. The base at the top of the dam is times the height minus 30 feet.
a. Write an expression to find the area of the trapezoidal cross section of the dam. b. If the height of the dam is 180 feet, find the area of this cross section.
SOLUTION:
a. The equation for the area of a trapezoid is .
b1
= 2h
b2 =
b. Let h = 180.
So, the area is 32,940 square feet.
Solve each equation.
31. 7(t2 + 5t − 9) + t = t(7t − 2) + 13
SOLUTION:
32. w(4w + 6) + 2w = 2(2w2 + 7w − 3)
SOLUTION:
33. 5(4z + 6) − 2(z − 4) = 7z(z + 4) − z(7z − 2) − 48
SOLUTION:
34. 9c(c − 11) + 10(5c − 3) = 3c(c + 5) + c(6c − 3) − 30
SOLUTION:
35. 2f (5f − 2) − 10(f2 − 3f + 6) = −8f (f + 4) + 4(2f
2 − 7f )
SOLUTION:
36. 2k(−3k + 4) + 6(k2 + 10) = k(4k + 8) − 2k(2k + 5)
SOLUTION:
Simplify each expression.
37. np2(30p
2 + 9n
2p − 12)
SOLUTION:
38. r2t(10r
3 + 5rt
3 + 15t
2)
SOLUTION:
39. −5q2w
3(4q + 7w) + 4qw
2(7q
2w + 2q) − 3qw(3q
2w
2 + 9)
SOLUTION:
40. −x2z(2z
2 + 4xz3) + xz
2(xz + 5x
3z) + x2
z3(3x
2z + 4xz)
SOLUTION:
41. PARKING A parking garage charges $30 per month plus $0.50 per daytime hour and $0.25 per hour during nights and weekends. Suppose Trent parks in the garage for 47 hours in January and h of those are night and weekend hours. a. Find an expression for Trent’s January bill. b. Find the cost if Trent had 12 hours of night and weekend hours.
SOLUTION: a. First, multiply the price per hour times the number of hours. Let h represent the number of weekend and night hours. For the weekend and night hours, he would spend 0.25h. The number of daytime hours is 47 – h, so he spent 0.50(47 - h) = 23.50 – 0.50h. To find the total cost, add these two values to the cost per month. 30 + (23.50 − 0.50h) + 0.25h = 53.50 − 0.25h. b. If Trent parked 12 night and weekend hours, h = 12
So, the cost for January is $50.50.
42. CCSS MODELING Che is building a dog house for his new puppy. The upper face of the dog house is a trapezoid. If the height of the trapezoid is 12 inches, find the area of the face of this piece of the dog house.
SOLUTION:
The formula for the area of a trapezoid is .
Now substitute h = 12 inches into the equation.
Therefore, the area of the trapezoid is 318 square inches.
43. TENNIS The tennis club is building a new tennis court with a path around it.
a. Write an expression for the area of the tennis court. b. Write an expression for the area of the path. c. What is the perimeter of the outside of the path?
SOLUTION: a. The area of the tennis court can be found by the equation . Substitute the values of the length and width and simplify.
b. The area of the path is the area of the large rectangle minus the area of the smaller rectangle, which is the area ofthe tennis court.
Area of the large rectangle = 2.5x(x + 6) or 2.5x2 + 15x.
Now subtract the area of the tennis court to find the area of just the path.
c. The perimeter of the outside of the path is the perimeter of the larger rectangle.
If x = 36 feet, then the perimeter of the outside of the path is 7(36) + 12 or 264 ft.
44. MULTIPLE REPRESENTATIONS In this problem, you will investigate the degree of the product of a monomialand a polynomial. a. TABULAR Write three monomials of different degrees and three polynomials of different degrees. Determine the degree of each monomial and polynomial. Multiply the monomials by the polynomials. Determine the degree of each product. Record your results in a table like the one below.
b. VERBAL Make a conjecture about the degree of the product of a monomial and a polynomial. What is the degree of the product of a monomial of degree a and a polynomial of degree b?
SOLUTION: a. There possible monomials are 2x, 3x
2, 4x
3. Three polynomials are x2 − 1, x5 + 1, x
6 + 1.
The degree for the monomial is the sum of the exponents. For the example given, case they are 1,2, and 3. The degree for the polynomial is the largest degree of the monomials. For the example given, they are 2, 5, and 6. Use the distributive property to find the product of the monomial and the polynomial. For the example give, they
are2x3 − 2x, 3x
7 + 3x
2, and 4x
9 + 4x. For the example given, the degree of the new polynomials are 3,7 and 9.
b. The degree of the product is the sum of the degree of the monomial and the degree of the polynomial, or a + b.
45. ERROR ANALYSIS Pearl and Ted both worked on this problem. Is either of them correct? Explain your reasoning.
SOLUTION:
Ted’s solution is correct. Pearl made a mistake when distributing on the middle term. She did not add the exponents correctly.
46. CCSS PERSEVERANCE Find p such that 3xp(4x
2p+3 + 2x3p−2
) = 12x12 + 6x
10.
SOLUTION: First simplify the left side of the equation.
Because the coefficients of the terms are the same, the corresponding terms’ exponents can be set equal to each other.
47. CHALLENGE Find 4x−3
y2(2x
5y−4 + 6x
−7y
6 − 4x0y
-2).
SOLUTION:
48. REASONING Is there a value for x that makes the statement (x + 2)2 = x2 + 22
true? If so, find a value for x. Explain your reasoning.
SOLUTION: Solve for x.
When 0 is substituted in for x in the equation, both sides are 22 or 4, which makes the equation true.
49. OPEN ENDED Write a monomial and a polynomial using n as the variable. Find their product.
SOLUTION: Choose a monomial with variable n and a polynomial with variable n. For example, let the monomial be 3n and the polynomial be 4n + 1. Find the product.
50. WRITING IN MATH Describe the steps to multiply a polynomial by a monomial.
SOLUTION: To multiply a polynomial by a monomial, use the Distributive Property. Multiply each term of the polynomial by the monomial. Then simplify by multiplying the coefficients together and using the Product of Powers Property for the variables. Consider the following example:
51. Every week a store sells j jeans and t T-shirts. The store makes $8 for each T-shirt and $12 for each pair of jeans. Which of the following expressions represents the total amount of money, in dollars, the store makes every week? A 8j + 12t B 12j + 8t C 20(j + t) D 96jt
SOLUTION: 12j is the amount made from jeans. 8t is the amount made from T-shirts. The sum, or 12j + 8t, represents the total amount the store makes every week. So, the correct choice is B.
52. If a = 5x + 7y and b = 2y − 3x, what is a + b?F 2x − 9y G 3y + 4x H 2x + 9y J 2x − 5y
SOLUTION:
So, the correct choice is H.
53. GEOMETRY A triangle has sides of length 5 inches and 8.5 inches. Which of the following cannot be the length of the third side? A 3.5 inches B 4 inches C 5.5 inches D 12 inches
SOLUTION: The sum of any two sides of a triangle must be greater than the length of the third side. Choice A: 3.5 + 5 = 8.5 8.5 Choice B: 3.5 + 5 = 8.5 > 8.5 Choice C: 5.5 + 5 = 10.5 > 8.5 Choice D: 12 + 5 = 17 > 8.5 Let x be the third side of the triangle.
3.5 cannot be the third side of the triangle. So, the correct choice is A.
54. SHORT RESPONSE Write an equation in which x varies directly as the cube of y and inversely as the square of z .
SOLUTION: If x varies directly as the cube of y , then both x and y must be in the numerators on either side of the equal sign.
When x varies inversely with the square of z. then the z must be in the denominator. Then . The
combination of the two is .
Find each sum or difference.
55. (2x2 − 7) + (8 − 5x
2)
SOLUTION:
56. (3z2 + 2z − 1) + (z
2 − 6)
SOLUTION:
57. (2a − 4a2 + 1) − (5a
2 − 2a − 6)
SOLUTION:
58. (a3 − 3a
2 + 4) − (4a
2 + 7)
SOLUTION:
59. (2ab − 3a + 4b) + (5a + 4ab)
SOLUTION:
60. (8c3 − 3c
2 + c − 2) − (3c
3 + 9)
SOLUTION:
Write a recursive formula for each sequence.61. 16, 2, –12, –26, ...
SOLUTION: Determine if there is a common difference. 2 – 16 = –14 –12 – 2 = –14 –26 – (–12) = –14 The common difference is –14. The first term is 16. To find the next term, subtract 14 from the previous term.
62. –5, 3, 11, 19, ...
SOLUTION: Determine if there is a common difference. 3 – (–5) = 8 11 – 3 = 8 19 – 11 = 8 The common difference is 8. The first term is 3. To find the next term, add 8 to the previous term.
63. 27, 43, 59, 75, ...
SOLUTION: Determine if there is a common difference. 43 – 27 = 16 59 – 43 = 16 75 – 59 = 16 The common difference is 16. The first term is 43. To find the next term, add 16 to the previous term.
64. 80, –200, 500, –1250, ...
SOLUTION: Check for a common difference. –200 – 80 = –280 500 – (–200) = 700 There is no common difference. Check for a common ratio. –200 ÷ 80 = –2.5 500 ÷ (–200) = –2.5 –1250 ÷ 500 = –2.5 The common ratio is –2.5. The first term is 80. To find the next term, multiply the previous term by –2.5.
65. 100, 60, 36, 21.6, ...
SOLUTION: Check for a common difference. 60 – 100 = –40 36 – 60 = –24 There is no common difference. Check for a common ratio. 60 ÷ 100 = 0.6 36 ÷ 60 = 0.6 21.6 ÷ 36 = 0.6 The common ratio is 0.6. The first term is 100. To find the next term, multiply the previous term by 0.6.
66.
SOLUTION: Check for a common difference.
There is no common difference. Check for a common ratio.
4 ÷ 1 = 4 The common ratio is 4.
The first term is . To find the next term, multiply the previous term by 4.
67. TRAVEL In 2003, about 9.5 million people took cruises. Between 2003 and 2008, the number increased by about 740,000 each year. Write the point-slope form of an equation to find the total number of people y taking a cruise for any year x. Estimate the number of people who took a cruise in 2010.
SOLUTION: The statement “increased by 740,000 each year” is a constant rate, or the slope. Now substitute the slope and the point (2003, 9,500,000) into the point slope equation.
To find the number of people who took a cruise in 2010, substitute x = 2010.
So, approximately 14,680,000 people took a cruise in 2010.
Write an equation in function notation for each relation.
68.
SOLUTION: The y-intercept is 0. Find the slope by choosing two points. Choose (0, 0) and (1, 4).
So the slope intercept form is f (x) = 4x + 0 or f (x) = 4x.
69.
SOLUTION:
The y-intercept is 0. Find the slope by choosing two points. Choose (0, 0) and (2, −1).
So in slope intercept form is f (x) = −0.5x + 0 or f (x) = −0.5x.
Simplify.
70. b(b2)(b
3)
SOLUTION:
71. 2y(3y2)
SOLUTION:
72. −y4(−2y
3)
SOLUTION:
73. −3z3(−5z
4 + 2z)
SOLUTION:
74. 2m(−4m4) − 3(−5m
3)
SOLUTION:
75. 4p 2(−2p
3) + 2p
4(5p
6)
SOLUTION:
eSolutions Manual - Powered by Cognero Page 7
8-2 Multiplying a Polynomial by a Monomial
Find each product.
1. 5w( −3w2 + 2w − 4)
SOLUTION:
2. 6g2(3g
3 + 4g2 + 10g − 1)
SOLUTION:
3. 4km2(8km
2 + 2k2m + 5k)
SOLUTION:
4. −3p4r3(2p
2r4 − 6p 6
r3 − 5)
SOLUTION:
5. 2ab(7a4b
2 + a5b − 2a)
SOLUTION:
6. c2d
3(5cd
7 − 3c3d
2 − 4d3)
SOLUTION:
Simplify each expression.
7. t(4t2 + 15t + 4) − 4(3t − 1)
SOLUTION:
8. x(3x2 + 4) + 2(7x − 3)
SOLUTION:
9. −2d(d3c
2 − 4dc2 + 2d
2c) + c
2(dc
2 − 3d4)
SOLUTION:
10. −5w2(8w
2x − 11wx
2) + 6x(9wx
4 − 4w − 3x2)
SOLUTION:
11. GRIDDED RESPONSE Marlene is buying a new plasma television. The height of the screen of the television is one half the width plus 5 inches. The width is 30 inches. Find the height of the screen in inches.
SOLUTION: Let h be the height and let w be the width.
Let w = 30.
The height is 20 inches.
Solve each equation.
12. −6(11 − 2c) = 7(−2 − 2c)
SOLUTION:
13. t(2t + 3) + 20 = 2t(t − 3)
SOLUTION:
14. −2(w + 1) + w = 7 − 4w
SOLUTION:
15. 3(y − 2) + 2y = 4y + 14
SOLUTION:
16. a(a + 3) + a(a − 6) + 35 = a(a − 5) + a(a + 7)
SOLUTION:
17. n(n − 4) + n(n + 8) = n(n − 13) + n(n + 1) + 16
SOLUTION:
Find each product.
18. b(b2 − 12b + 1)
SOLUTION:
19. f (f2 + 2f + 25)
SOLUTION:
20. −3m3(2m
3 − 12m2 + 2m + 25)
SOLUTION:
21. 2j 2(5j
3 − 15j2 + 2j + 2)
SOLUTION:
22. 2pr2(2pr + 5p
2r − 15p )
SOLUTION:
23. 4t3u(2t
2u
2 − 10tu4 + 2)
SOLUTION:
Simplify each expression.
24. −3(5x2 + 2x + 9) + x(2x − 3)
SOLUTION:
25. a(−8a2 + 2a + 4) + 3(6a
2 − 4)
SOLUTION:
26. −4d(5d2 − 12) + 7(d + 5)
SOLUTION:
27. −9g(−2g + g2) + 3(g
2 + 4)
SOLUTION:
28. 2j (7j2k
2 + j k2 + 5k) − 9k(−2j
2k
2 + 2k2 + 3j )
SOLUTION:
29. 4n(2n3p
2 − 3np2 + 5n) + 4p (6n
2p − 2np
2 + 3p )
SOLUTION:
30. DAMS A new dam being built has the shape of a trapezoid. The base at the bottom of the dam is 2 times the
height. The base at the top of the dam is times the height minus 30 feet.
a. Write an expression to find the area of the trapezoidal cross section of the dam. b. If the height of the dam is 180 feet, find the area of this cross section.
SOLUTION:
a. The equation for the area of a trapezoid is .
b1
= 2h
b2 =
b. Let h = 180.
So, the area is 32,940 square feet.
Solve each equation.
31. 7(t2 + 5t − 9) + t = t(7t − 2) + 13
SOLUTION:
32. w(4w + 6) + 2w = 2(2w2 + 7w − 3)
SOLUTION:
33. 5(4z + 6) − 2(z − 4) = 7z(z + 4) − z(7z − 2) − 48
SOLUTION:
34. 9c(c − 11) + 10(5c − 3) = 3c(c + 5) + c(6c − 3) − 30
SOLUTION:
35. 2f (5f − 2) − 10(f2 − 3f + 6) = −8f (f + 4) + 4(2f
2 − 7f )
SOLUTION:
36. 2k(−3k + 4) + 6(k2 + 10) = k(4k + 8) − 2k(2k + 5)
SOLUTION:
Simplify each expression.
37. np2(30p
2 + 9n
2p − 12)
SOLUTION:
38. r2t(10r
3 + 5rt
3 + 15t
2)
SOLUTION:
39. −5q2w
3(4q + 7w) + 4qw
2(7q
2w + 2q) − 3qw(3q
2w
2 + 9)
SOLUTION:
40. −x2z(2z
2 + 4xz3) + xz
2(xz + 5x
3z) + x2
z3(3x
2z + 4xz)
SOLUTION:
41. PARKING A parking garage charges $30 per month plus $0.50 per daytime hour and $0.25 per hour during nights and weekends. Suppose Trent parks in the garage for 47 hours in January and h of those are night and weekend hours. a. Find an expression for Trent’s January bill. b. Find the cost if Trent had 12 hours of night and weekend hours.
SOLUTION: a. First, multiply the price per hour times the number of hours. Let h represent the number of weekend and night hours. For the weekend and night hours, he would spend 0.25h. The number of daytime hours is 47 – h, so he spent 0.50(47 - h) = 23.50 – 0.50h. To find the total cost, add these two values to the cost per month. 30 + (23.50 − 0.50h) + 0.25h = 53.50 − 0.25h. b. If Trent parked 12 night and weekend hours, h = 12
So, the cost for January is $50.50.
42. CCSS MODELING Che is building a dog house for his new puppy. The upper face of the dog house is a trapezoid. If the height of the trapezoid is 12 inches, find the area of the face of this piece of the dog house.
SOLUTION:
The formula for the area of a trapezoid is .
Now substitute h = 12 inches into the equation.
Therefore, the area of the trapezoid is 318 square inches.
43. TENNIS The tennis club is building a new tennis court with a path around it.
a. Write an expression for the area of the tennis court. b. Write an expression for the area of the path. c. What is the perimeter of the outside of the path?
SOLUTION: a. The area of the tennis court can be found by the equation . Substitute the values of the length and width and simplify.
b. The area of the path is the area of the large rectangle minus the area of the smaller rectangle, which is the area ofthe tennis court.
Area of the large rectangle = 2.5x(x + 6) or 2.5x2 + 15x.
Now subtract the area of the tennis court to find the area of just the path.
c. The perimeter of the outside of the path is the perimeter of the larger rectangle.
If x = 36 feet, then the perimeter of the outside of the path is 7(36) + 12 or 264 ft.
44. MULTIPLE REPRESENTATIONS In this problem, you will investigate the degree of the product of a monomialand a polynomial. a. TABULAR Write three monomials of different degrees and three polynomials of different degrees. Determine the degree of each monomial and polynomial. Multiply the monomials by the polynomials. Determine the degree of each product. Record your results in a table like the one below.
b. VERBAL Make a conjecture about the degree of the product of a monomial and a polynomial. What is the degree of the product of a monomial of degree a and a polynomial of degree b?
SOLUTION: a. There possible monomials are 2x, 3x
2, 4x
3. Three polynomials are x2 − 1, x5 + 1, x
6 + 1.
The degree for the monomial is the sum of the exponents. For the example given, case they are 1,2, and 3. The degree for the polynomial is the largest degree of the monomials. For the example given, they are 2, 5, and 6. Use the distributive property to find the product of the monomial and the polynomial. For the example give, they
are2x3 − 2x, 3x
7 + 3x
2, and 4x
9 + 4x. For the example given, the degree of the new polynomials are 3,7 and 9.
b. The degree of the product is the sum of the degree of the monomial and the degree of the polynomial, or a + b.
45. ERROR ANALYSIS Pearl and Ted both worked on this problem. Is either of them correct? Explain your reasoning.
SOLUTION:
Ted’s solution is correct. Pearl made a mistake when distributing on the middle term. She did not add the exponents correctly.
46. CCSS PERSEVERANCE Find p such that 3xp(4x
2p+3 + 2x3p−2
) = 12x12 + 6x
10.
SOLUTION: First simplify the left side of the equation.
Because the coefficients of the terms are the same, the corresponding terms’ exponents can be set equal to each other.
47. CHALLENGE Find 4x−3
y2(2x
5y−4 + 6x
−7y
6 − 4x0y
-2).
SOLUTION:
48. REASONING Is there a value for x that makes the statement (x + 2)2 = x2 + 22
true? If so, find a value for x. Explain your reasoning.
SOLUTION: Solve for x.
When 0 is substituted in for x in the equation, both sides are 22 or 4, which makes the equation true.
49. OPEN ENDED Write a monomial and a polynomial using n as the variable. Find their product.
SOLUTION: Choose a monomial with variable n and a polynomial with variable n. For example, let the monomial be 3n and the polynomial be 4n + 1. Find the product.
50. WRITING IN MATH Describe the steps to multiply a polynomial by a monomial.
SOLUTION: To multiply a polynomial by a monomial, use the Distributive Property. Multiply each term of the polynomial by the monomial. Then simplify by multiplying the coefficients together and using the Product of Powers Property for the variables. Consider the following example:
51. Every week a store sells j jeans and t T-shirts. The store makes $8 for each T-shirt and $12 for each pair of jeans. Which of the following expressions represents the total amount of money, in dollars, the store makes every week? A 8j + 12t B 12j + 8t C 20(j + t) D 96jt
SOLUTION: 12j is the amount made from jeans. 8t is the amount made from T-shirts. The sum, or 12j + 8t, represents the total amount the store makes every week. So, the correct choice is B.
52. If a = 5x + 7y and b = 2y − 3x, what is a + b?F 2x − 9y G 3y + 4x H 2x + 9y J 2x − 5y
SOLUTION:
So, the correct choice is H.
53. GEOMETRY A triangle has sides of length 5 inches and 8.5 inches. Which of the following cannot be the length of the third side? A 3.5 inches B 4 inches C 5.5 inches D 12 inches
SOLUTION: The sum of any two sides of a triangle must be greater than the length of the third side. Choice A: 3.5 + 5 = 8.5 8.5 Choice B: 3.5 + 5 = 8.5 > 8.5 Choice C: 5.5 + 5 = 10.5 > 8.5 Choice D: 12 + 5 = 17 > 8.5 Let x be the third side of the triangle.
3.5 cannot be the third side of the triangle. So, the correct choice is A.
54. SHORT RESPONSE Write an equation in which x varies directly as the cube of y and inversely as the square of z .
SOLUTION: If x varies directly as the cube of y , then both x and y must be in the numerators on either side of the equal sign.
When x varies inversely with the square of z. then the z must be in the denominator. Then . The
combination of the two is .
Find each sum or difference.
55. (2x2 − 7) + (8 − 5x
2)
SOLUTION:
56. (3z2 + 2z − 1) + (z
2 − 6)
SOLUTION:
57. (2a − 4a2 + 1) − (5a
2 − 2a − 6)
SOLUTION:
58. (a3 − 3a
2 + 4) − (4a
2 + 7)
SOLUTION:
59. (2ab − 3a + 4b) + (5a + 4ab)
SOLUTION:
60. (8c3 − 3c
2 + c − 2) − (3c
3 + 9)
SOLUTION:
Write a recursive formula for each sequence.61. 16, 2, –12, –26, ...
SOLUTION: Determine if there is a common difference. 2 – 16 = –14 –12 – 2 = –14 –26 – (–12) = –14 The common difference is –14. The first term is 16. To find the next term, subtract 14 from the previous term.
62. –5, 3, 11, 19, ...
SOLUTION: Determine if there is a common difference. 3 – (–5) = 8 11 – 3 = 8 19 – 11 = 8 The common difference is 8. The first term is 3. To find the next term, add 8 to the previous term.
63. 27, 43, 59, 75, ...
SOLUTION: Determine if there is a common difference. 43 – 27 = 16 59 – 43 = 16 75 – 59 = 16 The common difference is 16. The first term is 43. To find the next term, add 16 to the previous term.
64. 80, –200, 500, –1250, ...
SOLUTION: Check for a common difference. –200 – 80 = –280 500 – (–200) = 700 There is no common difference. Check for a common ratio. –200 ÷ 80 = –2.5 500 ÷ (–200) = –2.5 –1250 ÷ 500 = –2.5 The common ratio is –2.5. The first term is 80. To find the next term, multiply the previous term by –2.5.
65. 100, 60, 36, 21.6, ...
SOLUTION: Check for a common difference. 60 – 100 = –40 36 – 60 = –24 There is no common difference. Check for a common ratio. 60 ÷ 100 = 0.6 36 ÷ 60 = 0.6 21.6 ÷ 36 = 0.6 The common ratio is 0.6. The first term is 100. To find the next term, multiply the previous term by 0.6.
66.
SOLUTION: Check for a common difference.
There is no common difference. Check for a common ratio.
4 ÷ 1 = 4 The common ratio is 4.
The first term is . To find the next term, multiply the previous term by 4.
67. TRAVEL In 2003, about 9.5 million people took cruises. Between 2003 and 2008, the number increased by about 740,000 each year. Write the point-slope form of an equation to find the total number of people y taking a cruise for any year x. Estimate the number of people who took a cruise in 2010.
SOLUTION: The statement “increased by 740,000 each year” is a constant rate, or the slope. Now substitute the slope and the point (2003, 9,500,000) into the point slope equation.
To find the number of people who took a cruise in 2010, substitute x = 2010.
So, approximately 14,680,000 people took a cruise in 2010.
Write an equation in function notation for each relation.
68.
SOLUTION: The y-intercept is 0. Find the slope by choosing two points. Choose (0, 0) and (1, 4).
So the slope intercept form is f (x) = 4x + 0 or f (x) = 4x.
69.
SOLUTION:
The y-intercept is 0. Find the slope by choosing two points. Choose (0, 0) and (2, −1).
So in slope intercept form is f (x) = −0.5x + 0 or f (x) = −0.5x.
Simplify.
70. b(b2)(b
3)
SOLUTION:
71. 2y(3y2)
SOLUTION:
72. −y4(−2y
3)
SOLUTION:
73. −3z3(−5z
4 + 2z)
SOLUTION:
74. 2m(−4m4) − 3(−5m
3)
SOLUTION:
75. 4p 2(−2p
3) + 2p
4(5p
6)
SOLUTION:
eSolutions Manual - Powered by Cognero Page 8
8-2 Multiplying a Polynomial by a Monomial
Find each product.
1. 5w( −3w2 + 2w − 4)
SOLUTION:
2. 6g2(3g
3 + 4g2 + 10g − 1)
SOLUTION:
3. 4km2(8km
2 + 2k2m + 5k)
SOLUTION:
4. −3p4r3(2p
2r4 − 6p 6
r3 − 5)
SOLUTION:
5. 2ab(7a4b
2 + a5b − 2a)
SOLUTION:
6. c2d
3(5cd
7 − 3c3d
2 − 4d3)
SOLUTION:
Simplify each expression.
7. t(4t2 + 15t + 4) − 4(3t − 1)
SOLUTION:
8. x(3x2 + 4) + 2(7x − 3)
SOLUTION:
9. −2d(d3c
2 − 4dc2 + 2d
2c) + c
2(dc
2 − 3d4)
SOLUTION:
10. −5w2(8w
2x − 11wx
2) + 6x(9wx
4 − 4w − 3x2)
SOLUTION:
11. GRIDDED RESPONSE Marlene is buying a new plasma television. The height of the screen of the television is one half the width plus 5 inches. The width is 30 inches. Find the height of the screen in inches.
SOLUTION: Let h be the height and let w be the width.
Let w = 30.
The height is 20 inches.
Solve each equation.
12. −6(11 − 2c) = 7(−2 − 2c)
SOLUTION:
13. t(2t + 3) + 20 = 2t(t − 3)
SOLUTION:
14. −2(w + 1) + w = 7 − 4w
SOLUTION:
15. 3(y − 2) + 2y = 4y + 14
SOLUTION:
16. a(a + 3) + a(a − 6) + 35 = a(a − 5) + a(a + 7)
SOLUTION:
17. n(n − 4) + n(n + 8) = n(n − 13) + n(n + 1) + 16
SOLUTION:
Find each product.
18. b(b2 − 12b + 1)
SOLUTION:
19. f (f2 + 2f + 25)
SOLUTION:
20. −3m3(2m
3 − 12m2 + 2m + 25)
SOLUTION:
21. 2j 2(5j
3 − 15j2 + 2j + 2)
SOLUTION:
22. 2pr2(2pr + 5p
2r − 15p )
SOLUTION:
23. 4t3u(2t
2u
2 − 10tu4 + 2)
SOLUTION:
Simplify each expression.
24. −3(5x2 + 2x + 9) + x(2x − 3)
SOLUTION:
25. a(−8a2 + 2a + 4) + 3(6a
2 − 4)
SOLUTION:
26. −4d(5d2 − 12) + 7(d + 5)
SOLUTION:
27. −9g(−2g + g2) + 3(g
2 + 4)
SOLUTION:
28. 2j (7j2k
2 + j k2 + 5k) − 9k(−2j
2k
2 + 2k2 + 3j )
SOLUTION:
29. 4n(2n3p
2 − 3np2 + 5n) + 4p (6n
2p − 2np
2 + 3p )
SOLUTION:
30. DAMS A new dam being built has the shape of a trapezoid. The base at the bottom of the dam is 2 times the
height. The base at the top of the dam is times the height minus 30 feet.
a. Write an expression to find the area of the trapezoidal cross section of the dam. b. If the height of the dam is 180 feet, find the area of this cross section.
SOLUTION:
a. The equation for the area of a trapezoid is .
b1
= 2h
b2 =
b. Let h = 180.
So, the area is 32,940 square feet.
Solve each equation.
31. 7(t2 + 5t − 9) + t = t(7t − 2) + 13
SOLUTION:
32. w(4w + 6) + 2w = 2(2w2 + 7w − 3)
SOLUTION:
33. 5(4z + 6) − 2(z − 4) = 7z(z + 4) − z(7z − 2) − 48
SOLUTION:
34. 9c(c − 11) + 10(5c − 3) = 3c(c + 5) + c(6c − 3) − 30
SOLUTION:
35. 2f (5f − 2) − 10(f2 − 3f + 6) = −8f (f + 4) + 4(2f
2 − 7f )
SOLUTION:
36. 2k(−3k + 4) + 6(k2 + 10) = k(4k + 8) − 2k(2k + 5)
SOLUTION:
Simplify each expression.
37. np2(30p
2 + 9n
2p − 12)
SOLUTION:
38. r2t(10r
3 + 5rt
3 + 15t
2)
SOLUTION:
39. −5q2w
3(4q + 7w) + 4qw
2(7q
2w + 2q) − 3qw(3q
2w
2 + 9)
SOLUTION:
40. −x2z(2z
2 + 4xz3) + xz
2(xz + 5x
3z) + x2
z3(3x
2z + 4xz)
SOLUTION:
41. PARKING A parking garage charges $30 per month plus $0.50 per daytime hour and $0.25 per hour during nights and weekends. Suppose Trent parks in the garage for 47 hours in January and h of those are night and weekend hours. a. Find an expression for Trent’s January bill. b. Find the cost if Trent had 12 hours of night and weekend hours.
SOLUTION: a. First, multiply the price per hour times the number of hours. Let h represent the number of weekend and night hours. For the weekend and night hours, he would spend 0.25h. The number of daytime hours is 47 – h, so he spent 0.50(47 - h) = 23.50 – 0.50h. To find the total cost, add these two values to the cost per month. 30 + (23.50 − 0.50h) + 0.25h = 53.50 − 0.25h. b. If Trent parked 12 night and weekend hours, h = 12
So, the cost for January is $50.50.
42. CCSS MODELING Che is building a dog house for his new puppy. The upper face of the dog house is a trapezoid. If the height of the trapezoid is 12 inches, find the area of the face of this piece of the dog house.
SOLUTION:
The formula for the area of a trapezoid is .
Now substitute h = 12 inches into the equation.
Therefore, the area of the trapezoid is 318 square inches.
43. TENNIS The tennis club is building a new tennis court with a path around it.
a. Write an expression for the area of the tennis court. b. Write an expression for the area of the path. c. What is the perimeter of the outside of the path?
SOLUTION: a. The area of the tennis court can be found by the equation . Substitute the values of the length and width and simplify.
b. The area of the path is the area of the large rectangle minus the area of the smaller rectangle, which is the area ofthe tennis court.
Area of the large rectangle = 2.5x(x + 6) or 2.5x2 + 15x.
Now subtract the area of the tennis court to find the area of just the path.
c. The perimeter of the outside of the path is the perimeter of the larger rectangle.
If x = 36 feet, then the perimeter of the outside of the path is 7(36) + 12 or 264 ft.
44. MULTIPLE REPRESENTATIONS In this problem, you will investigate the degree of the product of a monomialand a polynomial. a. TABULAR Write three monomials of different degrees and three polynomials of different degrees. Determine the degree of each monomial and polynomial. Multiply the monomials by the polynomials. Determine the degree of each product. Record your results in a table like the one below.
b. VERBAL Make a conjecture about the degree of the product of a monomial and a polynomial. What is the degree of the product of a monomial of degree a and a polynomial of degree b?
SOLUTION: a. There possible monomials are 2x, 3x
2, 4x
3. Three polynomials are x2 − 1, x5 + 1, x
6 + 1.
The degree for the monomial is the sum of the exponents. For the example given, case they are 1,2, and 3. The degree for the polynomial is the largest degree of the monomials. For the example given, they are 2, 5, and 6. Use the distributive property to find the product of the monomial and the polynomial. For the example give, they
are2x3 − 2x, 3x
7 + 3x
2, and 4x
9 + 4x. For the example given, the degree of the new polynomials are 3,7 and 9.
b. The degree of the product is the sum of the degree of the monomial and the degree of the polynomial, or a + b.
45. ERROR ANALYSIS Pearl and Ted both worked on this problem. Is either of them correct? Explain your reasoning.
SOLUTION:
Ted’s solution is correct. Pearl made a mistake when distributing on the middle term. She did not add the exponents correctly.
46. CCSS PERSEVERANCE Find p such that 3xp(4x
2p+3 + 2x3p−2
) = 12x12 + 6x
10.
SOLUTION: First simplify the left side of the equation.
Because the coefficients of the terms are the same, the corresponding terms’ exponents can be set equal to each other.
47. CHALLENGE Find 4x−3
y2(2x
5y−4 + 6x
−7y
6 − 4x0y
-2).
SOLUTION:
48. REASONING Is there a value for x that makes the statement (x + 2)2 = x2 + 22
true? If so, find a value for x. Explain your reasoning.
SOLUTION: Solve for x.
When 0 is substituted in for x in the equation, both sides are 22 or 4, which makes the equation true.
49. OPEN ENDED Write a monomial and a polynomial using n as the variable. Find their product.
SOLUTION: Choose a monomial with variable n and a polynomial with variable n. For example, let the monomial be 3n and the polynomial be 4n + 1. Find the product.
50. WRITING IN MATH Describe the steps to multiply a polynomial by a monomial.
SOLUTION: To multiply a polynomial by a monomial, use the Distributive Property. Multiply each term of the polynomial by the monomial. Then simplify by multiplying the coefficients together and using the Product of Powers Property for the variables. Consider the following example:
51. Every week a store sells j jeans and t T-shirts. The store makes $8 for each T-shirt and $12 for each pair of jeans. Which of the following expressions represents the total amount of money, in dollars, the store makes every week? A 8j + 12t B 12j + 8t C 20(j + t) D 96jt
SOLUTION: 12j is the amount made from jeans. 8t is the amount made from T-shirts. The sum, or 12j + 8t, represents the total amount the store makes every week. So, the correct choice is B.
52. If a = 5x + 7y and b = 2y − 3x, what is a + b?F 2x − 9y G 3y + 4x H 2x + 9y J 2x − 5y
SOLUTION:
So, the correct choice is H.
53. GEOMETRY A triangle has sides of length 5 inches and 8.5 inches. Which of the following cannot be the length of the third side? A 3.5 inches B 4 inches C 5.5 inches D 12 inches
SOLUTION: The sum of any two sides of a triangle must be greater than the length of the third side. Choice A: 3.5 + 5 = 8.5 8.5 Choice B: 3.5 + 5 = 8.5 > 8.5 Choice C: 5.5 + 5 = 10.5 > 8.5 Choice D: 12 + 5 = 17 > 8.5 Let x be the third side of the triangle.
3.5 cannot be the third side of the triangle. So, the correct choice is A.
54. SHORT RESPONSE Write an equation in which x varies directly as the cube of y and inversely as the square of z .
SOLUTION: If x varies directly as the cube of y , then both x and y must be in the numerators on either side of the equal sign.
When x varies inversely with the square of z. then the z must be in the denominator. Then . The
combination of the two is .
Find each sum or difference.
55. (2x2 − 7) + (8 − 5x
2)
SOLUTION:
56. (3z2 + 2z − 1) + (z
2 − 6)
SOLUTION:
57. (2a − 4a2 + 1) − (5a
2 − 2a − 6)
SOLUTION:
58. (a3 − 3a
2 + 4) − (4a
2 + 7)
SOLUTION:
59. (2ab − 3a + 4b) + (5a + 4ab)
SOLUTION:
60. (8c3 − 3c
2 + c − 2) − (3c
3 + 9)
SOLUTION:
Write a recursive formula for each sequence.61. 16, 2, –12, –26, ...
SOLUTION: Determine if there is a common difference. 2 – 16 = –14 –12 – 2 = –14 –26 – (–12) = –14 The common difference is –14. The first term is 16. To find the next term, subtract 14 from the previous term.
62. –5, 3, 11, 19, ...
SOLUTION: Determine if there is a common difference. 3 – (–5) = 8 11 – 3 = 8 19 – 11 = 8 The common difference is 8. The first term is 3. To find the next term, add 8 to the previous term.
63. 27, 43, 59, 75, ...
SOLUTION: Determine if there is a common difference. 43 – 27 = 16 59 – 43 = 16 75 – 59 = 16 The common difference is 16. The first term is 43. To find the next term, add 16 to the previous term.
64. 80, –200, 500, –1250, ...
SOLUTION: Check for a common difference. –200 – 80 = –280 500 – (–200) = 700 There is no common difference. Check for a common ratio. –200 ÷ 80 = –2.5 500 ÷ (–200) = –2.5 –1250 ÷ 500 = –2.5 The common ratio is –2.5. The first term is 80. To find the next term, multiply the previous term by –2.5.
65. 100, 60, 36, 21.6, ...
SOLUTION: Check for a common difference. 60 – 100 = –40 36 – 60 = –24 There is no common difference. Check for a common ratio. 60 ÷ 100 = 0.6 36 ÷ 60 = 0.6 21.6 ÷ 36 = 0.6 The common ratio is 0.6. The first term is 100. To find the next term, multiply the previous term by 0.6.
66.
SOLUTION: Check for a common difference.
There is no common difference. Check for a common ratio.
4 ÷ 1 = 4 The common ratio is 4.
The first term is . To find the next term, multiply the previous term by 4.
67. TRAVEL In 2003, about 9.5 million people took cruises. Between 2003 and 2008, the number increased by about 740,000 each year. Write the point-slope form of an equation to find the total number of people y taking a cruise for any year x. Estimate the number of people who took a cruise in 2010.
SOLUTION: The statement “increased by 740,000 each year” is a constant rate, or the slope. Now substitute the slope and the point (2003, 9,500,000) into the point slope equation.
To find the number of people who took a cruise in 2010, substitute x = 2010.
So, approximately 14,680,000 people took a cruise in 2010.
Write an equation in function notation for each relation.
68.
SOLUTION: The y-intercept is 0. Find the slope by choosing two points. Choose (0, 0) and (1, 4).
So the slope intercept form is f (x) = 4x + 0 or f (x) = 4x.
69.
SOLUTION:
The y-intercept is 0. Find the slope by choosing two points. Choose (0, 0) and (2, −1).
So in slope intercept form is f (x) = −0.5x + 0 or f (x) = −0.5x.
Simplify.
70. b(b2)(b
3)
SOLUTION:
71. 2y(3y2)
SOLUTION:
72. −y4(−2y
3)
SOLUTION:
73. −3z3(−5z
4 + 2z)
SOLUTION:
74. 2m(−4m4) − 3(−5m
3)
SOLUTION:
75. 4p 2(−2p
3) + 2p
4(5p
6)
SOLUTION:
eSolutions Manual - Powered by Cognero Page 9
8-2 Multiplying a Polynomial by a Monomial
Find each product.
1. 5w( −3w2 + 2w − 4)
SOLUTION:
2. 6g2(3g
3 + 4g2 + 10g − 1)
SOLUTION:
3. 4km2(8km
2 + 2k2m + 5k)
SOLUTION:
4. −3p4r3(2p
2r4 − 6p 6
r3 − 5)
SOLUTION:
5. 2ab(7a4b
2 + a5b − 2a)
SOLUTION:
6. c2d
3(5cd
7 − 3c3d
2 − 4d3)
SOLUTION:
Simplify each expression.
7. t(4t2 + 15t + 4) − 4(3t − 1)
SOLUTION:
8. x(3x2 + 4) + 2(7x − 3)
SOLUTION:
9. −2d(d3c
2 − 4dc2 + 2d
2c) + c
2(dc
2 − 3d4)
SOLUTION:
10. −5w2(8w
2x − 11wx
2) + 6x(9wx
4 − 4w − 3x2)
SOLUTION:
11. GRIDDED RESPONSE Marlene is buying a new plasma television. The height of the screen of the television is one half the width plus 5 inches. The width is 30 inches. Find the height of the screen in inches.
SOLUTION: Let h be the height and let w be the width.
Let w = 30.
The height is 20 inches.
Solve each equation.
12. −6(11 − 2c) = 7(−2 − 2c)
SOLUTION:
13. t(2t + 3) + 20 = 2t(t − 3)
SOLUTION:
14. −2(w + 1) + w = 7 − 4w
SOLUTION:
15. 3(y − 2) + 2y = 4y + 14
SOLUTION:
16. a(a + 3) + a(a − 6) + 35 = a(a − 5) + a(a + 7)
SOLUTION:
17. n(n − 4) + n(n + 8) = n(n − 13) + n(n + 1) + 16
SOLUTION:
Find each product.
18. b(b2 − 12b + 1)
SOLUTION:
19. f (f2 + 2f + 25)
SOLUTION:
20. −3m3(2m
3 − 12m2 + 2m + 25)
SOLUTION:
21. 2j 2(5j
3 − 15j2 + 2j + 2)
SOLUTION:
22. 2pr2(2pr + 5p
2r − 15p )
SOLUTION:
23. 4t3u(2t
2u
2 − 10tu4 + 2)
SOLUTION:
Simplify each expression.
24. −3(5x2 + 2x + 9) + x(2x − 3)
SOLUTION:
25. a(−8a2 + 2a + 4) + 3(6a
2 − 4)
SOLUTION:
26. −4d(5d2 − 12) + 7(d + 5)
SOLUTION:
27. −9g(−2g + g2) + 3(g
2 + 4)
SOLUTION:
28. 2j (7j2k
2 + j k2 + 5k) − 9k(−2j
2k
2 + 2k2 + 3j )
SOLUTION:
29. 4n(2n3p
2 − 3np2 + 5n) + 4p (6n
2p − 2np
2 + 3p )
SOLUTION:
30. DAMS A new dam being built has the shape of a trapezoid. The base at the bottom of the dam is 2 times the
height. The base at the top of the dam is times the height minus 30 feet.
a. Write an expression to find the area of the trapezoidal cross section of the dam. b. If the height of the dam is 180 feet, find the area of this cross section.
SOLUTION:
a. The equation for the area of a trapezoid is .
b1
= 2h
b2 =
b. Let h = 180.
So, the area is 32,940 square feet.
Solve each equation.
31. 7(t2 + 5t − 9) + t = t(7t − 2) + 13
SOLUTION:
32. w(4w + 6) + 2w = 2(2w2 + 7w − 3)
SOLUTION:
33. 5(4z + 6) − 2(z − 4) = 7z(z + 4) − z(7z − 2) − 48
SOLUTION:
34. 9c(c − 11) + 10(5c − 3) = 3c(c + 5) + c(6c − 3) − 30
SOLUTION:
35. 2f (5f − 2) − 10(f2 − 3f + 6) = −8f (f + 4) + 4(2f
2 − 7f )
SOLUTION:
36. 2k(−3k + 4) + 6(k2 + 10) = k(4k + 8) − 2k(2k + 5)
SOLUTION:
Simplify each expression.
37. np2(30p
2 + 9n
2p − 12)
SOLUTION:
38. r2t(10r
3 + 5rt
3 + 15t
2)
SOLUTION:
39. −5q2w
3(4q + 7w) + 4qw
2(7q
2w + 2q) − 3qw(3q
2w
2 + 9)
SOLUTION:
40. −x2z(2z
2 + 4xz3) + xz
2(xz + 5x
3z) + x2
z3(3x
2z + 4xz)
SOLUTION:
41. PARKING A parking garage charges $30 per month plus $0.50 per daytime hour and $0.25 per hour during nights and weekends. Suppose Trent parks in the garage for 47 hours in January and h of those are night and weekend hours. a. Find an expression for Trent’s January bill. b. Find the cost if Trent had 12 hours of night and weekend hours.
SOLUTION: a. First, multiply the price per hour times the number of hours. Let h represent the number of weekend and night hours. For the weekend and night hours, he would spend 0.25h. The number of daytime hours is 47 – h, so he spent 0.50(47 - h) = 23.50 – 0.50h. To find the total cost, add these two values to the cost per month. 30 + (23.50 − 0.50h) + 0.25h = 53.50 − 0.25h. b. If Trent parked 12 night and weekend hours, h = 12
So, the cost for January is $50.50.
42. CCSS MODELING Che is building a dog house for his new puppy. The upper face of the dog house is a trapezoid. If the height of the trapezoid is 12 inches, find the area of the face of this piece of the dog house.
SOLUTION:
The formula for the area of a trapezoid is .
Now substitute h = 12 inches into the equation.
Therefore, the area of the trapezoid is 318 square inches.
43. TENNIS The tennis club is building a new tennis court with a path around it.
a. Write an expression for the area of the tennis court. b. Write an expression for the area of the path. c. What is the perimeter of the outside of the path?
SOLUTION: a. The area of the tennis court can be found by the equation . Substitute the values of the length and width and simplify.
b. The area of the path is the area of the large rectangle minus the area of the smaller rectangle, which is the area ofthe tennis court.
Area of the large rectangle = 2.5x(x + 6) or 2.5x2 + 15x.
Now subtract the area of the tennis court to find the area of just the path.
c. The perimeter of the outside of the path is the perimeter of the larger rectangle.
If x = 36 feet, then the perimeter of the outside of the path is 7(36) + 12 or 264 ft.
44. MULTIPLE REPRESENTATIONS In this problem, you will investigate the degree of the product of a monomialand a polynomial. a. TABULAR Write three monomials of different degrees and three polynomials of different degrees. Determine the degree of each monomial and polynomial. Multiply the monomials by the polynomials. Determine the degree of each product. Record your results in a table like the one below.
b. VERBAL Make a conjecture about the degree of the product of a monomial and a polynomial. What is the degree of the product of a monomial of degree a and a polynomial of degree b?
SOLUTION: a. There possible monomials are 2x, 3x
2, 4x
3. Three polynomials are x2 − 1, x5 + 1, x
6 + 1.
The degree for the monomial is the sum of the exponents. For the example given, case they are 1,2, and 3. The degree for the polynomial is the largest degree of the monomials. For the example given, they are 2, 5, and 6. Use the distributive property to find the product of the monomial and the polynomial. For the example give, they
are2x3 − 2x, 3x
7 + 3x
2, and 4x
9 + 4x. For the example given, the degree of the new polynomials are 3,7 and 9.
b. The degree of the product is the sum of the degree of the monomial and the degree of the polynomial, or a + b.
45. ERROR ANALYSIS Pearl and Ted both worked on this problem. Is either of them correct? Explain your reasoning.
SOLUTION:
Ted’s solution is correct. Pearl made a mistake when distributing on the middle term. She did not add the exponents correctly.
46. CCSS PERSEVERANCE Find p such that 3xp(4x
2p+3 + 2x3p−2
) = 12x12 + 6x
10.
SOLUTION: First simplify the left side of the equation.
Because the coefficients of the terms are the same, the corresponding terms’ exponents can be set equal to each other.
47. CHALLENGE Find 4x−3
y2(2x
5y−4 + 6x
−7y
6 − 4x0y
-2).
SOLUTION:
48. REASONING Is there a value for x that makes the statement (x + 2)2 = x2 + 22
true? If so, find a value for x. Explain your reasoning.
SOLUTION: Solve for x.
When 0 is substituted in for x in the equation, both sides are 22 or 4, which makes the equation true.
49. OPEN ENDED Write a monomial and a polynomial using n as the variable. Find their product.
SOLUTION: Choose a monomial with variable n and a polynomial with variable n. For example, let the monomial be 3n and the polynomial be 4n + 1. Find the product.
50. WRITING IN MATH Describe the steps to multiply a polynomial by a monomial.
SOLUTION: To multiply a polynomial by a monomial, use the Distributive Property. Multiply each term of the polynomial by the monomial. Then simplify by multiplying the coefficients together and using the Product of Powers Property for the variables. Consider the following example:
51. Every week a store sells j jeans and t T-shirts. The store makes $8 for each T-shirt and $12 for each pair of jeans. Which of the following expressions represents the total amount of money, in dollars, the store makes every week? A 8j + 12t B 12j + 8t C 20(j + t) D 96jt
SOLUTION: 12j is the amount made from jeans. 8t is the amount made from T-shirts. The sum, or 12j + 8t, represents the total amount the store makes every week. So, the correct choice is B.
52. If a = 5x + 7y and b = 2y − 3x, what is a + b?F 2x − 9y G 3y + 4x H 2x + 9y J 2x − 5y
SOLUTION:
So, the correct choice is H.
53. GEOMETRY A triangle has sides of length 5 inches and 8.5 inches. Which of the following cannot be the length of the third side? A 3.5 inches B 4 inches C 5.5 inches D 12 inches
SOLUTION: The sum of any two sides of a triangle must be greater than the length of the third side. Choice A: 3.5 + 5 = 8.5 8.5 Choice B: 3.5 + 5 = 8.5 > 8.5 Choice C: 5.5 + 5 = 10.5 > 8.5 Choice D: 12 + 5 = 17 > 8.5 Let x be the third side of the triangle.
3.5 cannot be the third side of the triangle. So, the correct choice is A.
54. SHORT RESPONSE Write an equation in which x varies directly as the cube of y and inversely as the square of z .
SOLUTION: If x varies directly as the cube of y , then both x and y must be in the numerators on either side of the equal sign.
When x varies inversely with the square of z. then the z must be in the denominator. Then . The
combination of the two is .
Find each sum or difference.
55. (2x2 − 7) + (8 − 5x
2)
SOLUTION:
56. (3z2 + 2z − 1) + (z
2 − 6)
SOLUTION:
57. (2a − 4a2 + 1) − (5a
2 − 2a − 6)
SOLUTION:
58. (a3 − 3a
2 + 4) − (4a
2 + 7)
SOLUTION:
59. (2ab − 3a + 4b) + (5a + 4ab)
SOLUTION:
60. (8c3 − 3c
2 + c − 2) − (3c
3 + 9)
SOLUTION:
Write a recursive formula for each sequence.61. 16, 2, –12, –26, ...
SOLUTION: Determine if there is a common difference. 2 – 16 = –14 –12 – 2 = –14 –26 – (–12) = –14 The common difference is –14. The first term is 16. To find the next term, subtract 14 from the previous term.
62. –5, 3, 11, 19, ...
SOLUTION: Determine if there is a common difference. 3 – (–5) = 8 11 – 3 = 8 19 – 11 = 8 The common difference is 8. The first term is 3. To find the next term, add 8 to the previous term.
63. 27, 43, 59, 75, ...
SOLUTION: Determine if there is a common difference. 43 – 27 = 16 59 – 43 = 16 75 – 59 = 16 The common difference is 16. The first term is 43. To find the next term, add 16 to the previous term.
64. 80, –200, 500, –1250, ...
SOLUTION: Check for a common difference. –200 – 80 = –280 500 – (–200) = 700 There is no common difference. Check for a common ratio. –200 ÷ 80 = –2.5 500 ÷ (–200) = –2.5 –1250 ÷ 500 = –2.5 The common ratio is –2.5. The first term is 80. To find the next term, multiply the previous term by –2.5.
65. 100, 60, 36, 21.6, ...
SOLUTION: Check for a common difference. 60 – 100 = –40 36 – 60 = –24 There is no common difference. Check for a common ratio. 60 ÷ 100 = 0.6 36 ÷ 60 = 0.6 21.6 ÷ 36 = 0.6 The common ratio is 0.6. The first term is 100. To find the next term, multiply the previous term by 0.6.
66.
SOLUTION: Check for a common difference.
There is no common difference. Check for a common ratio.
4 ÷ 1 = 4 The common ratio is 4.
The first term is . To find the next term, multiply the previous term by 4.
67. TRAVEL In 2003, about 9.5 million people took cruises. Between 2003 and 2008, the number increased by about 740,000 each year. Write the point-slope form of an equation to find the total number of people y taking a cruise for any year x. Estimate the number of people who took a cruise in 2010.
SOLUTION: The statement “increased by 740,000 each year” is a constant rate, or the slope. Now substitute the slope and the point (2003, 9,500,000) into the point slope equation.
To find the number of people who took a cruise in 2010, substitute x = 2010.
So, approximately 14,680,000 people took a cruise in 2010.
Write an equation in function notation for each relation.
68.
SOLUTION: The y-intercept is 0. Find the slope by choosing two points. Choose (0, 0) and (1, 4).
So the slope intercept form is f (x) = 4x + 0 or f (x) = 4x.
69.
SOLUTION:
The y-intercept is 0. Find the slope by choosing two points. Choose (0, 0) and (2, −1).
So in slope intercept form is f (x) = −0.5x + 0 or f (x) = −0.5x.
Simplify.
70. b(b2)(b
3)
SOLUTION:
71. 2y(3y2)
SOLUTION:
72. −y4(−2y
3)
SOLUTION:
73. −3z3(−5z
4 + 2z)
SOLUTION:
74. 2m(−4m4) − 3(−5m
3)
SOLUTION:
75. 4p 2(−2p
3) + 2p
4(5p
6)
SOLUTION:
eSolutions Manual - Powered by Cognero Page 10
8-2 Multiplying a Polynomial by a Monomial
Find each product.
1. 5w( −3w2 + 2w − 4)
SOLUTION:
2. 6g2(3g
3 + 4g2 + 10g − 1)
SOLUTION:
3. 4km2(8km
2 + 2k2m + 5k)
SOLUTION:
4. −3p4r3(2p
2r4 − 6p 6
r3 − 5)
SOLUTION:
5. 2ab(7a4b
2 + a5b − 2a)
SOLUTION:
6. c2d
3(5cd
7 − 3c3d
2 − 4d3)
SOLUTION:
Simplify each expression.
7. t(4t2 + 15t + 4) − 4(3t − 1)
SOLUTION:
8. x(3x2 + 4) + 2(7x − 3)
SOLUTION:
9. −2d(d3c
2 − 4dc2 + 2d
2c) + c
2(dc
2 − 3d4)
SOLUTION:
10. −5w2(8w
2x − 11wx
2) + 6x(9wx
4 − 4w − 3x2)
SOLUTION:
11. GRIDDED RESPONSE Marlene is buying a new plasma television. The height of the screen of the television is one half the width plus 5 inches. The width is 30 inches. Find the height of the screen in inches.
SOLUTION: Let h be the height and let w be the width.
Let w = 30.
The height is 20 inches.
Solve each equation.
12. −6(11 − 2c) = 7(−2 − 2c)
SOLUTION:
13. t(2t + 3) + 20 = 2t(t − 3)
SOLUTION:
14. −2(w + 1) + w = 7 − 4w
SOLUTION:
15. 3(y − 2) + 2y = 4y + 14
SOLUTION:
16. a(a + 3) + a(a − 6) + 35 = a(a − 5) + a(a + 7)
SOLUTION:
17. n(n − 4) + n(n + 8) = n(n − 13) + n(n + 1) + 16
SOLUTION:
Find each product.
18. b(b2 − 12b + 1)
SOLUTION:
19. f (f2 + 2f + 25)
SOLUTION:
20. −3m3(2m
3 − 12m2 + 2m + 25)
SOLUTION:
21. 2j 2(5j
3 − 15j2 + 2j + 2)
SOLUTION:
22. 2pr2(2pr + 5p
2r − 15p )
SOLUTION:
23. 4t3u(2t
2u
2 − 10tu4 + 2)
SOLUTION:
Simplify each expression.
24. −3(5x2 + 2x + 9) + x(2x − 3)
SOLUTION:
25. a(−8a2 + 2a + 4) + 3(6a
2 − 4)
SOLUTION:
26. −4d(5d2 − 12) + 7(d + 5)
SOLUTION:
27. −9g(−2g + g2) + 3(g
2 + 4)
SOLUTION:
28. 2j (7j2k
2 + j k2 + 5k) − 9k(−2j
2k
2 + 2k2 + 3j )
SOLUTION:
29. 4n(2n3p
2 − 3np2 + 5n) + 4p (6n
2p − 2np
2 + 3p )
SOLUTION:
30. DAMS A new dam being built has the shape of a trapezoid. The base at the bottom of the dam is 2 times the
height. The base at the top of the dam is times the height minus 30 feet.
a. Write an expression to find the area of the trapezoidal cross section of the dam. b. If the height of the dam is 180 feet, find the area of this cross section.
SOLUTION:
a. The equation for the area of a trapezoid is .
b1
= 2h
b2 =
b. Let h = 180.
So, the area is 32,940 square feet.
Solve each equation.
31. 7(t2 + 5t − 9) + t = t(7t − 2) + 13
SOLUTION:
32. w(4w + 6) + 2w = 2(2w2 + 7w − 3)
SOLUTION:
33. 5(4z + 6) − 2(z − 4) = 7z(z + 4) − z(7z − 2) − 48
SOLUTION:
34. 9c(c − 11) + 10(5c − 3) = 3c(c + 5) + c(6c − 3) − 30
SOLUTION:
35. 2f (5f − 2) − 10(f2 − 3f + 6) = −8f (f + 4) + 4(2f
2 − 7f )
SOLUTION:
36. 2k(−3k + 4) + 6(k2 + 10) = k(4k + 8) − 2k(2k + 5)
SOLUTION:
Simplify each expression.
37. np2(30p
2 + 9n
2p − 12)
SOLUTION:
38. r2t(10r
3 + 5rt
3 + 15t
2)
SOLUTION:
39. −5q2w
3(4q + 7w) + 4qw
2(7q
2w + 2q) − 3qw(3q
2w
2 + 9)
SOLUTION:
40. −x2z(2z
2 + 4xz3) + xz
2(xz + 5x
3z) + x2
z3(3x
2z + 4xz)
SOLUTION:
41. PARKING A parking garage charges $30 per month plus $0.50 per daytime hour and $0.25 per hour during nights and weekends. Suppose Trent parks in the garage for 47 hours in January and h of those are night and weekend hours. a. Find an expression for Trent’s January bill. b. Find the cost if Trent had 12 hours of night and weekend hours.
SOLUTION: a. First, multiply the price per hour times the number of hours. Let h represent the number of weekend and night hours. For the weekend and night hours, he would spend 0.25h. The number of daytime hours is 47 – h, so he spent 0.50(47 - h) = 23.50 – 0.50h. To find the total cost, add these two values to the cost per month. 30 + (23.50 − 0.50h) + 0.25h = 53.50 − 0.25h. b. If Trent parked 12 night and weekend hours, h = 12
So, the cost for January is $50.50.
42. CCSS MODELING Che is building a dog house for his new puppy. The upper face of the dog house is a trapezoid. If the height of the trapezoid is 12 inches, find the area of the face of this piece of the dog house.
SOLUTION:
The formula for the area of a trapezoid is .
Now substitute h = 12 inches into the equation.
Therefore, the area of the trapezoid is 318 square inches.
43. TENNIS The tennis club is building a new tennis court with a path around it.
a. Write an expression for the area of the tennis court. b. Write an expression for the area of the path. c. What is the perimeter of the outside of the path?
SOLUTION: a. The area of the tennis court can be found by the equation . Substitute the values of the length and width and simplify.
b. The area of the path is the area of the large rectangle minus the area of the smaller rectangle, which is the area ofthe tennis court.
Area of the large rectangle = 2.5x(x + 6) or 2.5x2 + 15x.
Now subtract the area of the tennis court to find the area of just the path.
c. The perimeter of the outside of the path is the perimeter of the larger rectangle.
If x = 36 feet, then the perimeter of the outside of the path is 7(36) + 12 or 264 ft.
44. MULTIPLE REPRESENTATIONS In this problem, you will investigate the degree of the product of a monomialand a polynomial. a. TABULAR Write three monomials of different degrees and three polynomials of different degrees. Determine the degree of each monomial and polynomial. Multiply the monomials by the polynomials. Determine the degree of each product. Record your results in a table like the one below.
b. VERBAL Make a conjecture about the degree of the product of a monomial and a polynomial. What is the degree of the product of a monomial of degree a and a polynomial of degree b?
SOLUTION: a. There possible monomials are 2x, 3x
2, 4x
3. Three polynomials are x2 − 1, x5 + 1, x
6 + 1.
The degree for the monomial is the sum of the exponents. For the example given, case they are 1,2, and 3. The degree for the polynomial is the largest degree of the monomials. For the example given, they are 2, 5, and 6. Use the distributive property to find the product of the monomial and the polynomial. For the example give, they
are2x3 − 2x, 3x
7 + 3x
2, and 4x
9 + 4x. For the example given, the degree of the new polynomials are 3,7 and 9.
b. The degree of the product is the sum of the degree of the monomial and the degree of the polynomial, or a + b.
45. ERROR ANALYSIS Pearl and Ted both worked on this problem. Is either of them correct? Explain your reasoning.
SOLUTION:
Ted’s solution is correct. Pearl made a mistake when distributing on the middle term. She did not add the exponents correctly.
46. CCSS PERSEVERANCE Find p such that 3xp(4x
2p+3 + 2x3p−2
) = 12x12 + 6x
10.
SOLUTION: First simplify the left side of the equation.
Because the coefficients of the terms are the same, the corresponding terms’ exponents can be set equal to each other.
47. CHALLENGE Find 4x−3
y2(2x
5y−4 + 6x
−7y
6 − 4x0y
-2).
SOLUTION:
48. REASONING Is there a value for x that makes the statement (x + 2)2 = x2 + 22
true? If so, find a value for x. Explain your reasoning.
SOLUTION: Solve for x.
When 0 is substituted in for x in the equation, both sides are 22 or 4, which makes the equation true.
49. OPEN ENDED Write a monomial and a polynomial using n as the variable. Find their product.
SOLUTION: Choose a monomial with variable n and a polynomial with variable n. For example, let the monomial be 3n and the polynomial be 4n + 1. Find the product.
50. WRITING IN MATH Describe the steps to multiply a polynomial by a monomial.
SOLUTION: To multiply a polynomial by a monomial, use the Distributive Property. Multiply each term of the polynomial by the monomial. Then simplify by multiplying the coefficients together and using the Product of Powers Property for the variables. Consider the following example:
51. Every week a store sells j jeans and t T-shirts. The store makes $8 for each T-shirt and $12 for each pair of jeans. Which of the following expressions represents the total amount of money, in dollars, the store makes every week? A 8j + 12t B 12j + 8t C 20(j + t) D 96jt
SOLUTION: 12j is the amount made from jeans. 8t is the amount made from T-shirts. The sum, or 12j + 8t, represents the total amount the store makes every week. So, the correct choice is B.
52. If a = 5x + 7y and b = 2y − 3x, what is a + b?F 2x − 9y G 3y + 4x H 2x + 9y J 2x − 5y
SOLUTION:
So, the correct choice is H.
53. GEOMETRY A triangle has sides of length 5 inches and 8.5 inches. Which of the following cannot be the length of the third side? A 3.5 inches B 4 inches C 5.5 inches D 12 inches
SOLUTION: The sum of any two sides of a triangle must be greater than the length of the third side. Choice A: 3.5 + 5 = 8.5 8.5 Choice B: 3.5 + 5 = 8.5 > 8.5 Choice C: 5.5 + 5 = 10.5 > 8.5 Choice D: 12 + 5 = 17 > 8.5 Let x be the third side of the triangle.
3.5 cannot be the third side of the triangle. So, the correct choice is A.
54. SHORT RESPONSE Write an equation in which x varies directly as the cube of y and inversely as the square of z .
SOLUTION: If x varies directly as the cube of y , then both x and y must be in the numerators on either side of the equal sign.
When x varies inversely with the square of z. then the z must be in the denominator. Then . The
combination of the two is .
Find each sum or difference.
55. (2x2 − 7) + (8 − 5x
2)
SOLUTION:
56. (3z2 + 2z − 1) + (z
2 − 6)
SOLUTION:
57. (2a − 4a2 + 1) − (5a
2 − 2a − 6)
SOLUTION:
58. (a3 − 3a
2 + 4) − (4a
2 + 7)
SOLUTION:
59. (2ab − 3a + 4b) + (5a + 4ab)
SOLUTION:
60. (8c3 − 3c
2 + c − 2) − (3c
3 + 9)
SOLUTION:
Write a recursive formula for each sequence.61. 16, 2, –12, –26, ...
SOLUTION: Determine if there is a common difference. 2 – 16 = –14 –12 – 2 = –14 –26 – (–12) = –14 The common difference is –14. The first term is 16. To find the next term, subtract 14 from the previous term.
62. –5, 3, 11, 19, ...
SOLUTION: Determine if there is a common difference. 3 – (–5) = 8 11 – 3 = 8 19 – 11 = 8 The common difference is 8. The first term is 3. To find the next term, add 8 to the previous term.
63. 27, 43, 59, 75, ...
SOLUTION: Determine if there is a common difference. 43 – 27 = 16 59 – 43 = 16 75 – 59 = 16 The common difference is 16. The first term is 43. To find the next term, add 16 to the previous term.
64. 80, –200, 500, –1250, ...
SOLUTION: Check for a common difference. –200 – 80 = –280 500 – (–200) = 700 There is no common difference. Check for a common ratio. –200 ÷ 80 = –2.5 500 ÷ (–200) = –2.5 –1250 ÷ 500 = –2.5 The common ratio is –2.5. The first term is 80. To find the next term, multiply the previous term by –2.5.
65. 100, 60, 36, 21.6, ...
SOLUTION: Check for a common difference. 60 – 100 = –40 36 – 60 = –24 There is no common difference. Check for a common ratio. 60 ÷ 100 = 0.6 36 ÷ 60 = 0.6 21.6 ÷ 36 = 0.6 The common ratio is 0.6. The first term is 100. To find the next term, multiply the previous term by 0.6.
66.
SOLUTION: Check for a common difference.
There is no common difference. Check for a common ratio.
4 ÷ 1 = 4 The common ratio is 4.
The first term is . To find the next term, multiply the previous term by 4.
67. TRAVEL In 2003, about 9.5 million people took cruises. Between 2003 and 2008, the number increased by about 740,000 each year. Write the point-slope form of an equation to find the total number of people y taking a cruise for any year x. Estimate the number of people who took a cruise in 2010.
SOLUTION: The statement “increased by 740,000 each year” is a constant rate, or the slope. Now substitute the slope and the point (2003, 9,500,000) into the point slope equation.
To find the number of people who took a cruise in 2010, substitute x = 2010.
So, approximately 14,680,000 people took a cruise in 2010.
Write an equation in function notation for each relation.
68.
SOLUTION: The y-intercept is 0. Find the slope by choosing two points. Choose (0, 0) and (1, 4).
So the slope intercept form is f (x) = 4x + 0 or f (x) = 4x.
69.
SOLUTION:
The y-intercept is 0. Find the slope by choosing two points. Choose (0, 0) and (2, −1).
So in slope intercept form is f (x) = −0.5x + 0 or f (x) = −0.5x.
Simplify.
70. b(b2)(b
3)
SOLUTION:
71. 2y(3y2)
SOLUTION:
72. −y4(−2y
3)
SOLUTION:
73. −3z3(−5z
4 + 2z)
SOLUTION:
74. 2m(−4m4) − 3(−5m
3)
SOLUTION:
75. 4p 2(−2p
3) + 2p
4(5p
6)
SOLUTION:
eSolutions Manual - Powered by Cognero Page 11
8-2 Multiplying a Polynomial by a Monomial
Find each product.
1. 5w( −3w2 + 2w − 4)
SOLUTION:
2. 6g2(3g
3 + 4g2 + 10g − 1)
SOLUTION:
3. 4km2(8km
2 + 2k2m + 5k)
SOLUTION:
4. −3p4r3(2p
2r4 − 6p 6
r3 − 5)
SOLUTION:
5. 2ab(7a4b
2 + a5b − 2a)
SOLUTION:
6. c2d
3(5cd
7 − 3c3d
2 − 4d3)
SOLUTION:
Simplify each expression.
7. t(4t2 + 15t + 4) − 4(3t − 1)
SOLUTION:
8. x(3x2 + 4) + 2(7x − 3)
SOLUTION:
9. −2d(d3c
2 − 4dc2 + 2d
2c) + c
2(dc
2 − 3d4)
SOLUTION:
10. −5w2(8w
2x − 11wx
2) + 6x(9wx
4 − 4w − 3x2)
SOLUTION:
11. GRIDDED RESPONSE Marlene is buying a new plasma television. The height of the screen of the television is one half the width plus 5 inches. The width is 30 inches. Find the height of the screen in inches.
SOLUTION: Let h be the height and let w be the width.
Let w = 30.
The height is 20 inches.
Solve each equation.
12. −6(11 − 2c) = 7(−2 − 2c)
SOLUTION:
13. t(2t + 3) + 20 = 2t(t − 3)
SOLUTION:
14. −2(w + 1) + w = 7 − 4w
SOLUTION:
15. 3(y − 2) + 2y = 4y + 14
SOLUTION:
16. a(a + 3) + a(a − 6) + 35 = a(a − 5) + a(a + 7)
SOLUTION:
17. n(n − 4) + n(n + 8) = n(n − 13) + n(n + 1) + 16
SOLUTION:
Find each product.
18. b(b2 − 12b + 1)
SOLUTION:
19. f (f2 + 2f + 25)
SOLUTION:
20. −3m3(2m
3 − 12m2 + 2m + 25)
SOLUTION:
21. 2j 2(5j
3 − 15j2 + 2j + 2)
SOLUTION:
22. 2pr2(2pr + 5p
2r − 15p )
SOLUTION:
23. 4t3u(2t
2u
2 − 10tu4 + 2)
SOLUTION:
Simplify each expression.
24. −3(5x2 + 2x + 9) + x(2x − 3)
SOLUTION:
25. a(−8a2 + 2a + 4) + 3(6a
2 − 4)
SOLUTION:
26. −4d(5d2 − 12) + 7(d + 5)
SOLUTION:
27. −9g(−2g + g2) + 3(g
2 + 4)
SOLUTION:
28. 2j (7j2k
2 + j k2 + 5k) − 9k(−2j
2k
2 + 2k2 + 3j )
SOLUTION:
29. 4n(2n3p
2 − 3np2 + 5n) + 4p (6n
2p − 2np
2 + 3p )
SOLUTION:
30. DAMS A new dam being built has the shape of a trapezoid. The base at the bottom of the dam is 2 times the
height. The base at the top of the dam is times the height minus 30 feet.
a. Write an expression to find the area of the trapezoidal cross section of the dam. b. If the height of the dam is 180 feet, find the area of this cross section.
SOLUTION:
a. The equation for the area of a trapezoid is .
b1
= 2h
b2 =
b. Let h = 180.
So, the area is 32,940 square feet.
Solve each equation.
31. 7(t2 + 5t − 9) + t = t(7t − 2) + 13
SOLUTION:
32. w(4w + 6) + 2w = 2(2w2 + 7w − 3)
SOLUTION:
33. 5(4z + 6) − 2(z − 4) = 7z(z + 4) − z(7z − 2) − 48
SOLUTION:
34. 9c(c − 11) + 10(5c − 3) = 3c(c + 5) + c(6c − 3) − 30
SOLUTION:
35. 2f (5f − 2) − 10(f2 − 3f + 6) = −8f (f + 4) + 4(2f
2 − 7f )
SOLUTION:
36. 2k(−3k + 4) + 6(k2 + 10) = k(4k + 8) − 2k(2k + 5)
SOLUTION:
Simplify each expression.
37. np2(30p
2 + 9n
2p − 12)
SOLUTION:
38. r2t(10r
3 + 5rt
3 + 15t
2)
SOLUTION:
39. −5q2w
3(4q + 7w) + 4qw
2(7q
2w + 2q) − 3qw(3q
2w
2 + 9)
SOLUTION:
40. −x2z(2z
2 + 4xz3) + xz
2(xz + 5x
3z) + x2
z3(3x
2z + 4xz)
SOLUTION:
41. PARKING A parking garage charges $30 per month plus $0.50 per daytime hour and $0.25 per hour during nights and weekends. Suppose Trent parks in the garage for 47 hours in January and h of those are night and weekend hours. a. Find an expression for Trent’s January bill. b. Find the cost if Trent had 12 hours of night and weekend hours.
SOLUTION: a. First, multiply the price per hour times the number of hours. Let h represent the number of weekend and night hours. For the weekend and night hours, he would spend 0.25h. The number of daytime hours is 47 – h, so he spent 0.50(47 - h) = 23.50 – 0.50h. To find the total cost, add these two values to the cost per month. 30 + (23.50 − 0.50h) + 0.25h = 53.50 − 0.25h. b. If Trent parked 12 night and weekend hours, h = 12
So, the cost for January is $50.50.
42. CCSS MODELING Che is building a dog house for his new puppy. The upper face of the dog house is a trapezoid. If the height of the trapezoid is 12 inches, find the area of the face of this piece of the dog house.
SOLUTION:
The formula for the area of a trapezoid is .
Now substitute h = 12 inches into the equation.
Therefore, the area of the trapezoid is 318 square inches.
43. TENNIS The tennis club is building a new tennis court with a path around it.
a. Write an expression for the area of the tennis court. b. Write an expression for the area of the path. c. What is the perimeter of the outside of the path?
SOLUTION: a. The area of the tennis court can be found by the equation . Substitute the values of the length and width and simplify.
b. The area of the path is the area of the large rectangle minus the area of the smaller rectangle, which is the area ofthe tennis court.
Area of the large rectangle = 2.5x(x + 6) or 2.5x2 + 15x.
Now subtract the area of the tennis court to find the area of just the path.
c. The perimeter of the outside of the path is the perimeter of the larger rectangle.
If x = 36 feet, then the perimeter of the outside of the path is 7(36) + 12 or 264 ft.
44. MULTIPLE REPRESENTATIONS In this problem, you will investigate the degree of the product of a monomialand a polynomial. a. TABULAR Write three monomials of different degrees and three polynomials of different degrees. Determine the degree of each monomial and polynomial. Multiply the monomials by the polynomials. Determine the degree of each product. Record your results in a table like the one below.
b. VERBAL Make a conjecture about the degree of the product of a monomial and a polynomial. What is the degree of the product of a monomial of degree a and a polynomial of degree b?
SOLUTION: a. There possible monomials are 2x, 3x
2, 4x
3. Three polynomials are x2 − 1, x5 + 1, x
6 + 1.
The degree for the monomial is the sum of the exponents. For the example given, case they are 1,2, and 3. The degree for the polynomial is the largest degree of the monomials. For the example given, they are 2, 5, and 6. Use the distributive property to find the product of the monomial and the polynomial. For the example give, they
are2x3 − 2x, 3x
7 + 3x
2, and 4x
9 + 4x. For the example given, the degree of the new polynomials are 3,7 and 9.
b. The degree of the product is the sum of the degree of the monomial and the degree of the polynomial, or a + b.
45. ERROR ANALYSIS Pearl and Ted both worked on this problem. Is either of them correct? Explain your reasoning.
SOLUTION:
Ted’s solution is correct. Pearl made a mistake when distributing on the middle term. She did not add the exponents correctly.
46. CCSS PERSEVERANCE Find p such that 3xp(4x
2p+3 + 2x3p−2
) = 12x12 + 6x
10.
SOLUTION: First simplify the left side of the equation.
Because the coefficients of the terms are the same, the corresponding terms’ exponents can be set equal to each other.
47. CHALLENGE Find 4x−3
y2(2x
5y−4 + 6x
−7y
6 − 4x0y
-2).
SOLUTION:
48. REASONING Is there a value for x that makes the statement (x + 2)2 = x2 + 22
true? If so, find a value for x. Explain your reasoning.
SOLUTION: Solve for x.
When 0 is substituted in for x in the equation, both sides are 22 or 4, which makes the equation true.
49. OPEN ENDED Write a monomial and a polynomial using n as the variable. Find their product.
SOLUTION: Choose a monomial with variable n and a polynomial with variable n. For example, let the monomial be 3n and the polynomial be 4n + 1. Find the product.
50. WRITING IN MATH Describe the steps to multiply a polynomial by a monomial.
SOLUTION: To multiply a polynomial by a monomial, use the Distributive Property. Multiply each term of the polynomial by the monomial. Then simplify by multiplying the coefficients together and using the Product of Powers Property for the variables. Consider the following example:
51. Every week a store sells j jeans and t T-shirts. The store makes $8 for each T-shirt and $12 for each pair of jeans. Which of the following expressions represents the total amount of money, in dollars, the store makes every week? A 8j + 12t B 12j + 8t C 20(j + t) D 96jt
SOLUTION: 12j is the amount made from jeans. 8t is the amount made from T-shirts. The sum, or 12j + 8t, represents the total amount the store makes every week. So, the correct choice is B.
52. If a = 5x + 7y and b = 2y − 3x, what is a + b?F 2x − 9y G 3y + 4x H 2x + 9y J 2x − 5y
SOLUTION:
So, the correct choice is H.
53. GEOMETRY A triangle has sides of length 5 inches and 8.5 inches. Which of the following cannot be the length of the third side? A 3.5 inches B 4 inches C 5.5 inches D 12 inches
SOLUTION: The sum of any two sides of a triangle must be greater than the length of the third side. Choice A: 3.5 + 5 = 8.5 8.5 Choice B: 3.5 + 5 = 8.5 > 8.5 Choice C: 5.5 + 5 = 10.5 > 8.5 Choice D: 12 + 5 = 17 > 8.5 Let x be the third side of the triangle.
3.5 cannot be the third side of the triangle. So, the correct choice is A.
54. SHORT RESPONSE Write an equation in which x varies directly as the cube of y and inversely as the square of z .
SOLUTION: If x varies directly as the cube of y , then both x and y must be in the numerators on either side of the equal sign.
When x varies inversely with the square of z. then the z must be in the denominator. Then . The
combination of the two is .
Find each sum or difference.
55. (2x2 − 7) + (8 − 5x
2)
SOLUTION:
56. (3z2 + 2z − 1) + (z
2 − 6)
SOLUTION:
57. (2a − 4a2 + 1) − (5a
2 − 2a − 6)
SOLUTION:
58. (a3 − 3a
2 + 4) − (4a
2 + 7)
SOLUTION:
59. (2ab − 3a + 4b) + (5a + 4ab)
SOLUTION:
60. (8c3 − 3c
2 + c − 2) − (3c
3 + 9)
SOLUTION:
Write a recursive formula for each sequence.61. 16, 2, –12, –26, ...
SOLUTION: Determine if there is a common difference. 2 – 16 = –14 –12 – 2 = –14 –26 – (–12) = –14 The common difference is –14. The first term is 16. To find the next term, subtract 14 from the previous term.
62. –5, 3, 11, 19, ...
SOLUTION: Determine if there is a common difference. 3 – (–5) = 8 11 – 3 = 8 19 – 11 = 8 The common difference is 8. The first term is 3. To find the next term, add 8 to the previous term.
63. 27, 43, 59, 75, ...
SOLUTION: Determine if there is a common difference. 43 – 27 = 16 59 – 43 = 16 75 – 59 = 16 The common difference is 16. The first term is 43. To find the next term, add 16 to the previous term.
64. 80, –200, 500, –1250, ...
SOLUTION: Check for a common difference. –200 – 80 = –280 500 – (–200) = 700 There is no common difference. Check for a common ratio. –200 ÷ 80 = –2.5 500 ÷ (–200) = –2.5 –1250 ÷ 500 = –2.5 The common ratio is –2.5. The first term is 80. To find the next term, multiply the previous term by –2.5.
65. 100, 60, 36, 21.6, ...
SOLUTION: Check for a common difference. 60 – 100 = –40 36 – 60 = –24 There is no common difference. Check for a common ratio. 60 ÷ 100 = 0.6 36 ÷ 60 = 0.6 21.6 ÷ 36 = 0.6 The common ratio is 0.6. The first term is 100. To find the next term, multiply the previous term by 0.6.
66.
SOLUTION: Check for a common difference.
There is no common difference. Check for a common ratio.
4 ÷ 1 = 4 The common ratio is 4.
The first term is . To find the next term, multiply the previous term by 4.
67. TRAVEL In 2003, about 9.5 million people took cruises. Between 2003 and 2008, the number increased by about 740,000 each year. Write the point-slope form of an equation to find the total number of people y taking a cruise for any year x. Estimate the number of people who took a cruise in 2010.
SOLUTION: The statement “increased by 740,000 each year” is a constant rate, or the slope. Now substitute the slope and the point (2003, 9,500,000) into the point slope equation.
To find the number of people who took a cruise in 2010, substitute x = 2010.
So, approximately 14,680,000 people took a cruise in 2010.
Write an equation in function notation for each relation.
68.
SOLUTION: The y-intercept is 0. Find the slope by choosing two points. Choose (0, 0) and (1, 4).
So the slope intercept form is f (x) = 4x + 0 or f (x) = 4x.
69.
SOLUTION:
The y-intercept is 0. Find the slope by choosing two points. Choose (0, 0) and (2, −1).
So in slope intercept form is f (x) = −0.5x + 0 or f (x) = −0.5x.
Simplify.
70. b(b2)(b
3)
SOLUTION:
71. 2y(3y2)
SOLUTION:
72. −y4(−2y
3)
SOLUTION:
73. −3z3(−5z
4 + 2z)
SOLUTION:
74. 2m(−4m4) − 3(−5m
3)
SOLUTION:
75. 4p 2(−2p
3) + 2p
4(5p
6)
SOLUTION:
eSolutions Manual - Powered by Cognero Page 12
8-2 Multiplying a Polynomial by a Monomial
Find each product.
1. 5w( −3w2 + 2w − 4)
SOLUTION:
2. 6g2(3g
3 + 4g2 + 10g − 1)
SOLUTION:
3. 4km2(8km
2 + 2k2m + 5k)
SOLUTION:
4. −3p4r3(2p
2r4 − 6p 6
r3 − 5)
SOLUTION:
5. 2ab(7a4b
2 + a5b − 2a)
SOLUTION:
6. c2d
3(5cd
7 − 3c3d
2 − 4d3)
SOLUTION:
Simplify each expression.
7. t(4t2 + 15t + 4) − 4(3t − 1)
SOLUTION:
8. x(3x2 + 4) + 2(7x − 3)
SOLUTION:
9. −2d(d3c
2 − 4dc2 + 2d
2c) + c
2(dc
2 − 3d4)
SOLUTION:
10. −5w2(8w
2x − 11wx
2) + 6x(9wx
4 − 4w − 3x2)
SOLUTION:
11. GRIDDED RESPONSE Marlene is buying a new plasma television. The height of the screen of the television is one half the width plus 5 inches. The width is 30 inches. Find the height of the screen in inches.
SOLUTION: Let h be the height and let w be the width.
Let w = 30.
The height is 20 inches.
Solve each equation.
12. −6(11 − 2c) = 7(−2 − 2c)
SOLUTION:
13. t(2t + 3) + 20 = 2t(t − 3)
SOLUTION:
14. −2(w + 1) + w = 7 − 4w
SOLUTION:
15. 3(y − 2) + 2y = 4y + 14
SOLUTION:
16. a(a + 3) + a(a − 6) + 35 = a(a − 5) + a(a + 7)
SOLUTION:
17. n(n − 4) + n(n + 8) = n(n − 13) + n(n + 1) + 16
SOLUTION:
Find each product.
18. b(b2 − 12b + 1)
SOLUTION:
19. f (f2 + 2f + 25)
SOLUTION:
20. −3m3(2m
3 − 12m2 + 2m + 25)
SOLUTION:
21. 2j 2(5j
3 − 15j2 + 2j + 2)
SOLUTION:
22. 2pr2(2pr + 5p
2r − 15p )
SOLUTION:
23. 4t3u(2t
2u
2 − 10tu4 + 2)
SOLUTION:
Simplify each expression.
24. −3(5x2 + 2x + 9) + x(2x − 3)
SOLUTION:
25. a(−8a2 + 2a + 4) + 3(6a
2 − 4)
SOLUTION:
26. −4d(5d2 − 12) + 7(d + 5)
SOLUTION:
27. −9g(−2g + g2) + 3(g
2 + 4)
SOLUTION:
28. 2j (7j2k
2 + j k2 + 5k) − 9k(−2j
2k
2 + 2k2 + 3j )
SOLUTION:
29. 4n(2n3p
2 − 3np2 + 5n) + 4p (6n
2p − 2np
2 + 3p )
SOLUTION:
30. DAMS A new dam being built has the shape of a trapezoid. The base at the bottom of the dam is 2 times the
height. The base at the top of the dam is times the height minus 30 feet.
a. Write an expression to find the area of the trapezoidal cross section of the dam. b. If the height of the dam is 180 feet, find the area of this cross section.
SOLUTION:
a. The equation for the area of a trapezoid is .
b1
= 2h
b2 =
b. Let h = 180.
So, the area is 32,940 square feet.
Solve each equation.
31. 7(t2 + 5t − 9) + t = t(7t − 2) + 13
SOLUTION:
32. w(4w + 6) + 2w = 2(2w2 + 7w − 3)
SOLUTION:
33. 5(4z + 6) − 2(z − 4) = 7z(z + 4) − z(7z − 2) − 48
SOLUTION:
34. 9c(c − 11) + 10(5c − 3) = 3c(c + 5) + c(6c − 3) − 30
SOLUTION:
35. 2f (5f − 2) − 10(f2 − 3f + 6) = −8f (f + 4) + 4(2f
2 − 7f )
SOLUTION:
36. 2k(−3k + 4) + 6(k2 + 10) = k(4k + 8) − 2k(2k + 5)
SOLUTION:
Simplify each expression.
37. np2(30p
2 + 9n
2p − 12)
SOLUTION:
38. r2t(10r
3 + 5rt
3 + 15t
2)
SOLUTION:
39. −5q2w
3(4q + 7w) + 4qw
2(7q
2w + 2q) − 3qw(3q
2w
2 + 9)
SOLUTION:
40. −x2z(2z
2 + 4xz3) + xz
2(xz + 5x
3z) + x2
z3(3x
2z + 4xz)
SOLUTION:
41. PARKING A parking garage charges $30 per month plus $0.50 per daytime hour and $0.25 per hour during nights and weekends. Suppose Trent parks in the garage for 47 hours in January and h of those are night and weekend hours. a. Find an expression for Trent’s January bill. b. Find the cost if Trent had 12 hours of night and weekend hours.
SOLUTION: a. First, multiply the price per hour times the number of hours. Let h represent the number of weekend and night hours. For the weekend and night hours, he would spend 0.25h. The number of daytime hours is 47 – h, so he spent 0.50(47 - h) = 23.50 – 0.50h. To find the total cost, add these two values to the cost per month. 30 + (23.50 − 0.50h) + 0.25h = 53.50 − 0.25h. b. If Trent parked 12 night and weekend hours, h = 12
So, the cost for January is $50.50.
42. CCSS MODELING Che is building a dog house for his new puppy. The upper face of the dog house is a trapezoid. If the height of the trapezoid is 12 inches, find the area of the face of this piece of the dog house.
SOLUTION:
The formula for the area of a trapezoid is .
Now substitute h = 12 inches into the equation.
Therefore, the area of the trapezoid is 318 square inches.
43. TENNIS The tennis club is building a new tennis court with a path around it.
a. Write an expression for the area of the tennis court. b. Write an expression for the area of the path. c. What is the perimeter of the outside of the path?
SOLUTION: a. The area of the tennis court can be found by the equation . Substitute the values of the length and width and simplify.
b. The area of the path is the area of the large rectangle minus the area of the smaller rectangle, which is the area ofthe tennis court.
Area of the large rectangle = 2.5x(x + 6) or 2.5x2 + 15x.
Now subtract the area of the tennis court to find the area of just the path.
c. The perimeter of the outside of the path is the perimeter of the larger rectangle.
If x = 36 feet, then the perimeter of the outside of the path is 7(36) + 12 or 264 ft.
44. MULTIPLE REPRESENTATIONS In this problem, you will investigate the degree of the product of a monomialand a polynomial. a. TABULAR Write three monomials of different degrees and three polynomials of different degrees. Determine the degree of each monomial and polynomial. Multiply the monomials by the polynomials. Determine the degree of each product. Record your results in a table like the one below.
b. VERBAL Make a conjecture about the degree of the product of a monomial and a polynomial. What is the degree of the product of a monomial of degree a and a polynomial of degree b?
SOLUTION: a. There possible monomials are 2x, 3x
2, 4x
3. Three polynomials are x2 − 1, x5 + 1, x
6 + 1.
The degree for the monomial is the sum of the exponents. For the example given, case they are 1,2, and 3. The degree for the polynomial is the largest degree of the monomials. For the example given, they are 2, 5, and 6. Use the distributive property to find the product of the monomial and the polynomial. For the example give, they
are2x3 − 2x, 3x
7 + 3x
2, and 4x
9 + 4x. For the example given, the degree of the new polynomials are 3,7 and 9.
b. The degree of the product is the sum of the degree of the monomial and the degree of the polynomial, or a + b.
45. ERROR ANALYSIS Pearl and Ted both worked on this problem. Is either of them correct? Explain your reasoning.
SOLUTION:
Ted’s solution is correct. Pearl made a mistake when distributing on the middle term. She did not add the exponents correctly.
46. CCSS PERSEVERANCE Find p such that 3xp(4x
2p+3 + 2x3p−2
) = 12x12 + 6x
10.
SOLUTION: First simplify the left side of the equation.
Because the coefficients of the terms are the same, the corresponding terms’ exponents can be set equal to each other.
47. CHALLENGE Find 4x−3
y2(2x
5y−4 + 6x
−7y
6 − 4x0y
-2).
SOLUTION:
48. REASONING Is there a value for x that makes the statement (x + 2)2 = x2 + 22
true? If so, find a value for x. Explain your reasoning.
SOLUTION: Solve for x.
When 0 is substituted in for x in the equation, both sides are 22 or 4, which makes the equation true.
49. OPEN ENDED Write a monomial and a polynomial using n as the variable. Find their product.
SOLUTION: Choose a monomial with variable n and a polynomial with variable n. For example, let the monomial be 3n and the polynomial be 4n + 1. Find the product.
50. WRITING IN MATH Describe the steps to multiply a polynomial by a monomial.
SOLUTION: To multiply a polynomial by a monomial, use the Distributive Property. Multiply each term of the polynomial by the monomial. Then simplify by multiplying the coefficients together and using the Product of Powers Property for the variables. Consider the following example:
51. Every week a store sells j jeans and t T-shirts. The store makes $8 for each T-shirt and $12 for each pair of jeans. Which of the following expressions represents the total amount of money, in dollars, the store makes every week? A 8j + 12t B 12j + 8t C 20(j + t) D 96jt
SOLUTION: 12j is the amount made from jeans. 8t is the amount made from T-shirts. The sum, or 12j + 8t, represents the total amount the store makes every week. So, the correct choice is B.
52. If a = 5x + 7y and b = 2y − 3x, what is a + b?F 2x − 9y G 3y + 4x H 2x + 9y J 2x − 5y
SOLUTION:
So, the correct choice is H.
53. GEOMETRY A triangle has sides of length 5 inches and 8.5 inches. Which of the following cannot be the length of the third side? A 3.5 inches B 4 inches C 5.5 inches D 12 inches
SOLUTION: The sum of any two sides of a triangle must be greater than the length of the third side. Choice A: 3.5 + 5 = 8.5 8.5 Choice B: 3.5 + 5 = 8.5 > 8.5 Choice C: 5.5 + 5 = 10.5 > 8.5 Choice D: 12 + 5 = 17 > 8.5 Let x be the third side of the triangle.
3.5 cannot be the third side of the triangle. So, the correct choice is A.
54. SHORT RESPONSE Write an equation in which x varies directly as the cube of y and inversely as the square of z .
SOLUTION: If x varies directly as the cube of y , then both x and y must be in the numerators on either side of the equal sign.
When x varies inversely with the square of z. then the z must be in the denominator. Then . The
combination of the two is .
Find each sum or difference.
55. (2x2 − 7) + (8 − 5x
2)
SOLUTION:
56. (3z2 + 2z − 1) + (z
2 − 6)
SOLUTION:
57. (2a − 4a2 + 1) − (5a
2 − 2a − 6)
SOLUTION:
58. (a3 − 3a
2 + 4) − (4a
2 + 7)
SOLUTION:
59. (2ab − 3a + 4b) + (5a + 4ab)
SOLUTION:
60. (8c3 − 3c
2 + c − 2) − (3c
3 + 9)
SOLUTION:
Write a recursive formula for each sequence.61. 16, 2, –12, –26, ...
SOLUTION: Determine if there is a common difference. 2 – 16 = –14 –12 – 2 = –14 –26 – (–12) = –14 The common difference is –14. The first term is 16. To find the next term, subtract 14 from the previous term.
62. –5, 3, 11, 19, ...
SOLUTION: Determine if there is a common difference. 3 – (–5) = 8 11 – 3 = 8 19 – 11 = 8 The common difference is 8. The first term is 3. To find the next term, add 8 to the previous term.
63. 27, 43, 59, 75, ...
SOLUTION: Determine if there is a common difference. 43 – 27 = 16 59 – 43 = 16 75 – 59 = 16 The common difference is 16. The first term is 43. To find the next term, add 16 to the previous term.
64. 80, –200, 500, –1250, ...
SOLUTION: Check for a common difference. –200 – 80 = –280 500 – (–200) = 700 There is no common difference. Check for a common ratio. –200 ÷ 80 = –2.5 500 ÷ (–200) = –2.5 –1250 ÷ 500 = –2.5 The common ratio is –2.5. The first term is 80. To find the next term, multiply the previous term by –2.5.
65. 100, 60, 36, 21.6, ...
SOLUTION: Check for a common difference. 60 – 100 = –40 36 – 60 = –24 There is no common difference. Check for a common ratio. 60 ÷ 100 = 0.6 36 ÷ 60 = 0.6 21.6 ÷ 36 = 0.6 The common ratio is 0.6. The first term is 100. To find the next term, multiply the previous term by 0.6.
66.
SOLUTION: Check for a common difference.
There is no common difference. Check for a common ratio.
4 ÷ 1 = 4 The common ratio is 4.
The first term is . To find the next term, multiply the previous term by 4.
67. TRAVEL In 2003, about 9.5 million people took cruises. Between 2003 and 2008, the number increased by about 740,000 each year. Write the point-slope form of an equation to find the total number of people y taking a cruise for any year x. Estimate the number of people who took a cruise in 2010.
SOLUTION: The statement “increased by 740,000 each year” is a constant rate, or the slope. Now substitute the slope and the point (2003, 9,500,000) into the point slope equation.
To find the number of people who took a cruise in 2010, substitute x = 2010.
So, approximately 14,680,000 people took a cruise in 2010.
Write an equation in function notation for each relation.
68.
SOLUTION: The y-intercept is 0. Find the slope by choosing two points. Choose (0, 0) and (1, 4).
So the slope intercept form is f (x) = 4x + 0 or f (x) = 4x.
69.
SOLUTION:
The y-intercept is 0. Find the slope by choosing two points. Choose (0, 0) and (2, −1).
So in slope intercept form is f (x) = −0.5x + 0 or f (x) = −0.5x.
Simplify.
70. b(b2)(b
3)
SOLUTION:
71. 2y(3y2)
SOLUTION:
72. −y4(−2y
3)
SOLUTION:
73. −3z3(−5z
4 + 2z)
SOLUTION:
74. 2m(−4m4) − 3(−5m
3)
SOLUTION:
75. 4p 2(−2p
3) + 2p
4(5p
6)
SOLUTION:
eSolutions Manual - Powered by Cognero Page 13
8-2 Multiplying a Polynomial by a Monomial
Find each product.
1. 5w( −3w2 + 2w − 4)
SOLUTION:
2. 6g2(3g
3 + 4g2 + 10g − 1)
SOLUTION:
3. 4km2(8km
2 + 2k2m + 5k)
SOLUTION:
4. −3p4r3(2p
2r4 − 6p 6
r3 − 5)
SOLUTION:
5. 2ab(7a4b
2 + a5b − 2a)
SOLUTION:
6. c2d
3(5cd
7 − 3c3d
2 − 4d3)
SOLUTION:
Simplify each expression.
7. t(4t2 + 15t + 4) − 4(3t − 1)
SOLUTION:
8. x(3x2 + 4) + 2(7x − 3)
SOLUTION:
9. −2d(d3c
2 − 4dc2 + 2d
2c) + c
2(dc
2 − 3d4)
SOLUTION:
10. −5w2(8w
2x − 11wx
2) + 6x(9wx
4 − 4w − 3x2)
SOLUTION:
11. GRIDDED RESPONSE Marlene is buying a new plasma television. The height of the screen of the television is one half the width plus 5 inches. The width is 30 inches. Find the height of the screen in inches.
SOLUTION: Let h be the height and let w be the width.
Let w = 30.
The height is 20 inches.
Solve each equation.
12. −6(11 − 2c) = 7(−2 − 2c)
SOLUTION:
13. t(2t + 3) + 20 = 2t(t − 3)
SOLUTION:
14. −2(w + 1) + w = 7 − 4w
SOLUTION:
15. 3(y − 2) + 2y = 4y + 14
SOLUTION:
16. a(a + 3) + a(a − 6) + 35 = a(a − 5) + a(a + 7)
SOLUTION:
17. n(n − 4) + n(n + 8) = n(n − 13) + n(n + 1) + 16
SOLUTION:
Find each product.
18. b(b2 − 12b + 1)
SOLUTION:
19. f (f2 + 2f + 25)
SOLUTION:
20. −3m3(2m
3 − 12m2 + 2m + 25)
SOLUTION:
21. 2j 2(5j
3 − 15j2 + 2j + 2)
SOLUTION:
22. 2pr2(2pr + 5p
2r − 15p )
SOLUTION:
23. 4t3u(2t
2u
2 − 10tu4 + 2)
SOLUTION:
Simplify each expression.
24. −3(5x2 + 2x + 9) + x(2x − 3)
SOLUTION:
25. a(−8a2 + 2a + 4) + 3(6a
2 − 4)
SOLUTION:
26. −4d(5d2 − 12) + 7(d + 5)
SOLUTION:
27. −9g(−2g + g2) + 3(g
2 + 4)
SOLUTION:
28. 2j (7j2k
2 + j k2 + 5k) − 9k(−2j
2k
2 + 2k2 + 3j )
SOLUTION:
29. 4n(2n3p
2 − 3np2 + 5n) + 4p (6n
2p − 2np
2 + 3p )
SOLUTION:
30. DAMS A new dam being built has the shape of a trapezoid. The base at the bottom of the dam is 2 times the
height. The base at the top of the dam is times the height minus 30 feet.
a. Write an expression to find the area of the trapezoidal cross section of the dam. b. If the height of the dam is 180 feet, find the area of this cross section.
SOLUTION:
a. The equation for the area of a trapezoid is .
b1
= 2h
b2 =
b. Let h = 180.
So, the area is 32,940 square feet.
Solve each equation.
31. 7(t2 + 5t − 9) + t = t(7t − 2) + 13
SOLUTION:
32. w(4w + 6) + 2w = 2(2w2 + 7w − 3)
SOLUTION:
33. 5(4z + 6) − 2(z − 4) = 7z(z + 4) − z(7z − 2) − 48
SOLUTION:
34. 9c(c − 11) + 10(5c − 3) = 3c(c + 5) + c(6c − 3) − 30
SOLUTION:
35. 2f (5f − 2) − 10(f2 − 3f + 6) = −8f (f + 4) + 4(2f
2 − 7f )
SOLUTION:
36. 2k(−3k + 4) + 6(k2 + 10) = k(4k + 8) − 2k(2k + 5)
SOLUTION:
Simplify each expression.
37. np2(30p
2 + 9n
2p − 12)
SOLUTION:
38. r2t(10r
3 + 5rt
3 + 15t
2)
SOLUTION:
39. −5q2w
3(4q + 7w) + 4qw
2(7q
2w + 2q) − 3qw(3q
2w
2 + 9)
SOLUTION:
40. −x2z(2z
2 + 4xz3) + xz
2(xz + 5x
3z) + x2
z3(3x
2z + 4xz)
SOLUTION:
41. PARKING A parking garage charges $30 per month plus $0.50 per daytime hour and $0.25 per hour during nights and weekends. Suppose Trent parks in the garage for 47 hours in January and h of those are night and weekend hours. a. Find an expression for Trent’s January bill. b. Find the cost if Trent had 12 hours of night and weekend hours.
SOLUTION: a. First, multiply the price per hour times the number of hours. Let h represent the number of weekend and night hours. For the weekend and night hours, he would spend 0.25h. The number of daytime hours is 47 – h, so he spent 0.50(47 - h) = 23.50 – 0.50h. To find the total cost, add these two values to the cost per month. 30 + (23.50 − 0.50h) + 0.25h = 53.50 − 0.25h. b. If Trent parked 12 night and weekend hours, h = 12
So, the cost for January is $50.50.
42. CCSS MODELING Che is building a dog house for his new puppy. The upper face of the dog house is a trapezoid. If the height of the trapezoid is 12 inches, find the area of the face of this piece of the dog house.
SOLUTION:
The formula for the area of a trapezoid is .
Now substitute h = 12 inches into the equation.
Therefore, the area of the trapezoid is 318 square inches.
43. TENNIS The tennis club is building a new tennis court with a path around it.
a. Write an expression for the area of the tennis court. b. Write an expression for the area of the path. c. What is the perimeter of the outside of the path?
SOLUTION: a. The area of the tennis court can be found by the equation . Substitute the values of the length and width and simplify.
b. The area of the path is the area of the large rectangle minus the area of the smaller rectangle, which is the area ofthe tennis court.
Area of the large rectangle = 2.5x(x + 6) or 2.5x2 + 15x.
Now subtract the area of the tennis court to find the area of just the path.
c. The perimeter of the outside of the path is the perimeter of the larger rectangle.
If x = 36 feet, then the perimeter of the outside of the path is 7(36) + 12 or 264 ft.
44. MULTIPLE REPRESENTATIONS In this problem, you will investigate the degree of the product of a monomialand a polynomial. a. TABULAR Write three monomials of different degrees and three polynomials of different degrees. Determine the degree of each monomial and polynomial. Multiply the monomials by the polynomials. Determine the degree of each product. Record your results in a table like the one below.
b. VERBAL Make a conjecture about the degree of the product of a monomial and a polynomial. What is the degree of the product of a monomial of degree a and a polynomial of degree b?
SOLUTION: a. There possible monomials are 2x, 3x
2, 4x
3. Three polynomials are x2 − 1, x5 + 1, x
6 + 1.
The degree for the monomial is the sum of the exponents. For the example given, case they are 1,2, and 3. The degree for the polynomial is the largest degree of the monomials. For the example given, they are 2, 5, and 6. Use the distributive property to find the product of the monomial and the polynomial. For the example give, they
are2x3 − 2x, 3x
7 + 3x
2, and 4x
9 + 4x. For the example given, the degree of the new polynomials are 3,7 and 9.
b. The degree of the product is the sum of the degree of the monomial and the degree of the polynomial, or a + b.
45. ERROR ANALYSIS Pearl and Ted both worked on this problem. Is either of them correct? Explain your reasoning.
SOLUTION:
Ted’s solution is correct. Pearl made a mistake when distributing on the middle term. She did not add the exponents correctly.
46. CCSS PERSEVERANCE Find p such that 3xp(4x
2p+3 + 2x3p−2
) = 12x12 + 6x
10.
SOLUTION: First simplify the left side of the equation.
Because the coefficients of the terms are the same, the corresponding terms’ exponents can be set equal to each other.
47. CHALLENGE Find 4x−3
y2(2x
5y−4 + 6x
−7y
6 − 4x0y
-2).
SOLUTION:
48. REASONING Is there a value for x that makes the statement (x + 2)2 = x2 + 22
true? If so, find a value for x. Explain your reasoning.
SOLUTION: Solve for x.
When 0 is substituted in for x in the equation, both sides are 22 or 4, which makes the equation true.
49. OPEN ENDED Write a monomial and a polynomial using n as the variable. Find their product.
SOLUTION: Choose a monomial with variable n and a polynomial with variable n. For example, let the monomial be 3n and the polynomial be 4n + 1. Find the product.
50. WRITING IN MATH Describe the steps to multiply a polynomial by a monomial.
SOLUTION: To multiply a polynomial by a monomial, use the Distributive Property. Multiply each term of the polynomial by the monomial. Then simplify by multiplying the coefficients together and using the Product of Powers Property for the variables. Consider the following example:
51. Every week a store sells j jeans and t T-shirts. The store makes $8 for each T-shirt and $12 for each pair of jeans. Which of the following expressions represents the total amount of money, in dollars, the store makes every week? A 8j + 12t B 12j + 8t C 20(j + t) D 96jt
SOLUTION: 12j is the amount made from jeans. 8t is the amount made from T-shirts. The sum, or 12j + 8t, represents the total amount the store makes every week. So, the correct choice is B.
52. If a = 5x + 7y and b = 2y − 3x, what is a + b?F 2x − 9y G 3y + 4x H 2x + 9y J 2x − 5y
SOLUTION:
So, the correct choice is H.
53. GEOMETRY A triangle has sides of length 5 inches and 8.5 inches. Which of the following cannot be the length of the third side? A 3.5 inches B 4 inches C 5.5 inches D 12 inches
SOLUTION: The sum of any two sides of a triangle must be greater than the length of the third side. Choice A: 3.5 + 5 = 8.5 8.5 Choice B: 3.5 + 5 = 8.5 > 8.5 Choice C: 5.5 + 5 = 10.5 > 8.5 Choice D: 12 + 5 = 17 > 8.5 Let x be the third side of the triangle.
3.5 cannot be the third side of the triangle. So, the correct choice is A.
54. SHORT RESPONSE Write an equation in which x varies directly as the cube of y and inversely as the square of z .
SOLUTION: If x varies directly as the cube of y , then both x and y must be in the numerators on either side of the equal sign.
When x varies inversely with the square of z. then the z must be in the denominator. Then . The
combination of the two is .
Find each sum or difference.
55. (2x2 − 7) + (8 − 5x
2)
SOLUTION:
56. (3z2 + 2z − 1) + (z
2 − 6)
SOLUTION:
57. (2a − 4a2 + 1) − (5a
2 − 2a − 6)
SOLUTION:
58. (a3 − 3a
2 + 4) − (4a
2 + 7)
SOLUTION:
59. (2ab − 3a + 4b) + (5a + 4ab)
SOLUTION:
60. (8c3 − 3c
2 + c − 2) − (3c
3 + 9)
SOLUTION:
Write a recursive formula for each sequence.61. 16, 2, –12, –26, ...
SOLUTION: Determine if there is a common difference. 2 – 16 = –14 –12 – 2 = –14 –26 – (–12) = –14 The common difference is –14. The first term is 16. To find the next term, subtract 14 from the previous term.
62. –5, 3, 11, 19, ...
SOLUTION: Determine if there is a common difference. 3 – (–5) = 8 11 – 3 = 8 19 – 11 = 8 The common difference is 8. The first term is 3. To find the next term, add 8 to the previous term.
63. 27, 43, 59, 75, ...
SOLUTION: Determine if there is a common difference. 43 – 27 = 16 59 – 43 = 16 75 – 59 = 16 The common difference is 16. The first term is 43. To find the next term, add 16 to the previous term.
64. 80, –200, 500, –1250, ...
SOLUTION: Check for a common difference. –200 – 80 = –280 500 – (–200) = 700 There is no common difference. Check for a common ratio. –200 ÷ 80 = –2.5 500 ÷ (–200) = –2.5 –1250 ÷ 500 = –2.5 The common ratio is –2.5. The first term is 80. To find the next term, multiply the previous term by –2.5.
65. 100, 60, 36, 21.6, ...
SOLUTION: Check for a common difference. 60 – 100 = –40 36 – 60 = –24 There is no common difference. Check for a common ratio. 60 ÷ 100 = 0.6 36 ÷ 60 = 0.6 21.6 ÷ 36 = 0.6 The common ratio is 0.6. The first term is 100. To find the next term, multiply the previous term by 0.6.
66.
SOLUTION: Check for a common difference.
There is no common difference. Check for a common ratio.
4 ÷ 1 = 4 The common ratio is 4.
The first term is . To find the next term, multiply the previous term by 4.
67. TRAVEL In 2003, about 9.5 million people took cruises. Between 2003 and 2008, the number increased by about 740,000 each year. Write the point-slope form of an equation to find the total number of people y taking a cruise for any year x. Estimate the number of people who took a cruise in 2010.
SOLUTION: The statement “increased by 740,000 each year” is a constant rate, or the slope. Now substitute the slope and the point (2003, 9,500,000) into the point slope equation.
To find the number of people who took a cruise in 2010, substitute x = 2010.
So, approximately 14,680,000 people took a cruise in 2010.
Write an equation in function notation for each relation.
68.
SOLUTION: The y-intercept is 0. Find the slope by choosing two points. Choose (0, 0) and (1, 4).
So the slope intercept form is f (x) = 4x + 0 or f (x) = 4x.
69.
SOLUTION:
The y-intercept is 0. Find the slope by choosing two points. Choose (0, 0) and (2, −1).
So in slope intercept form is f (x) = −0.5x + 0 or f (x) = −0.5x.
Simplify.
70. b(b2)(b
3)
SOLUTION:
71. 2y(3y2)
SOLUTION:
72. −y4(−2y
3)
SOLUTION:
73. −3z3(−5z
4 + 2z)
SOLUTION:
74. 2m(−4m4) − 3(−5m
3)
SOLUTION:
75. 4p 2(−2p
3) + 2p
4(5p
6)
SOLUTION:
eSolutions Manual - Powered by Cognero Page 14
8-2 Multiplying a Polynomial by a Monomial
Find each product.
1. 5w( −3w2 + 2w − 4)
SOLUTION:
2. 6g2(3g
3 + 4g2 + 10g − 1)
SOLUTION:
3. 4km2(8km
2 + 2k2m + 5k)
SOLUTION:
4. −3p4r3(2p
2r4 − 6p 6
r3 − 5)
SOLUTION:
5. 2ab(7a4b
2 + a5b − 2a)
SOLUTION:
6. c2d
3(5cd
7 − 3c3d
2 − 4d3)
SOLUTION:
Simplify each expression.
7. t(4t2 + 15t + 4) − 4(3t − 1)
SOLUTION:
8. x(3x2 + 4) + 2(7x − 3)
SOLUTION:
9. −2d(d3c
2 − 4dc2 + 2d
2c) + c
2(dc
2 − 3d4)
SOLUTION:
10. −5w2(8w
2x − 11wx
2) + 6x(9wx
4 − 4w − 3x2)
SOLUTION:
11. GRIDDED RESPONSE Marlene is buying a new plasma television. The height of the screen of the television is one half the width plus 5 inches. The width is 30 inches. Find the height of the screen in inches.
SOLUTION: Let h be the height and let w be the width.
Let w = 30.
The height is 20 inches.
Solve each equation.
12. −6(11 − 2c) = 7(−2 − 2c)
SOLUTION:
13. t(2t + 3) + 20 = 2t(t − 3)
SOLUTION:
14. −2(w + 1) + w = 7 − 4w
SOLUTION:
15. 3(y − 2) + 2y = 4y + 14
SOLUTION:
16. a(a + 3) + a(a − 6) + 35 = a(a − 5) + a(a + 7)
SOLUTION:
17. n(n − 4) + n(n + 8) = n(n − 13) + n(n + 1) + 16
SOLUTION:
Find each product.
18. b(b2 − 12b + 1)
SOLUTION:
19. f (f2 + 2f + 25)
SOLUTION:
20. −3m3(2m
3 − 12m2 + 2m + 25)
SOLUTION:
21. 2j 2(5j
3 − 15j2 + 2j + 2)
SOLUTION:
22. 2pr2(2pr + 5p
2r − 15p )
SOLUTION:
23. 4t3u(2t
2u
2 − 10tu4 + 2)
SOLUTION:
Simplify each expression.
24. −3(5x2 + 2x + 9) + x(2x − 3)
SOLUTION:
25. a(−8a2 + 2a + 4) + 3(6a
2 − 4)
SOLUTION:
26. −4d(5d2 − 12) + 7(d + 5)
SOLUTION:
27. −9g(−2g + g2) + 3(g
2 + 4)
SOLUTION:
28. 2j (7j2k
2 + j k2 + 5k) − 9k(−2j
2k
2 + 2k2 + 3j )
SOLUTION:
29. 4n(2n3p
2 − 3np2 + 5n) + 4p (6n
2p − 2np
2 + 3p )
SOLUTION:
30. DAMS A new dam being built has the shape of a trapezoid. The base at the bottom of the dam is 2 times the
height. The base at the top of the dam is times the height minus 30 feet.
a. Write an expression to find the area of the trapezoidal cross section of the dam. b. If the height of the dam is 180 feet, find the area of this cross section.
SOLUTION:
a. The equation for the area of a trapezoid is .
b1
= 2h
b2 =
b. Let h = 180.
So, the area is 32,940 square feet.
Solve each equation.
31. 7(t2 + 5t − 9) + t = t(7t − 2) + 13
SOLUTION:
32. w(4w + 6) + 2w = 2(2w2 + 7w − 3)
SOLUTION:
33. 5(4z + 6) − 2(z − 4) = 7z(z + 4) − z(7z − 2) − 48
SOLUTION:
34. 9c(c − 11) + 10(5c − 3) = 3c(c + 5) + c(6c − 3) − 30
SOLUTION:
35. 2f (5f − 2) − 10(f2 − 3f + 6) = −8f (f + 4) + 4(2f
2 − 7f )
SOLUTION:
36. 2k(−3k + 4) + 6(k2 + 10) = k(4k + 8) − 2k(2k + 5)
SOLUTION:
Simplify each expression.
37. np2(30p
2 + 9n
2p − 12)
SOLUTION:
38. r2t(10r
3 + 5rt
3 + 15t
2)
SOLUTION:
39. −5q2w
3(4q + 7w) + 4qw
2(7q
2w + 2q) − 3qw(3q
2w
2 + 9)
SOLUTION:
40. −x2z(2z
2 + 4xz3) + xz
2(xz + 5x
3z) + x2
z3(3x
2z + 4xz)
SOLUTION:
41. PARKING A parking garage charges $30 per month plus $0.50 per daytime hour and $0.25 per hour during nights and weekends. Suppose Trent parks in the garage for 47 hours in January and h of those are night and weekend hours. a. Find an expression for Trent’s January bill. b. Find the cost if Trent had 12 hours of night and weekend hours.
SOLUTION: a. First, multiply the price per hour times the number of hours. Let h represent the number of weekend and night hours. For the weekend and night hours, he would spend 0.25h. The number of daytime hours is 47 – h, so he spent 0.50(47 - h) = 23.50 – 0.50h. To find the total cost, add these two values to the cost per month. 30 + (23.50 − 0.50h) + 0.25h = 53.50 − 0.25h. b. If Trent parked 12 night and weekend hours, h = 12
So, the cost for January is $50.50.
42. CCSS MODELING Che is building a dog house for his new puppy. The upper face of the dog house is a trapezoid. If the height of the trapezoid is 12 inches, find the area of the face of this piece of the dog house.
SOLUTION:
The formula for the area of a trapezoid is .
Now substitute h = 12 inches into the equation.
Therefore, the area of the trapezoid is 318 square inches.
43. TENNIS The tennis club is building a new tennis court with a path around it.
a. Write an expression for the area of the tennis court. b. Write an expression for the area of the path. c. What is the perimeter of the outside of the path?
SOLUTION: a. The area of the tennis court can be found by the equation . Substitute the values of the length and width and simplify.
b. The area of the path is the area of the large rectangle minus the area of the smaller rectangle, which is the area ofthe tennis court.
Area of the large rectangle = 2.5x(x + 6) or 2.5x2 + 15x.
Now subtract the area of the tennis court to find the area of just the path.
c. The perimeter of the outside of the path is the perimeter of the larger rectangle.
If x = 36 feet, then the perimeter of the outside of the path is 7(36) + 12 or 264 ft.
44. MULTIPLE REPRESENTATIONS In this problem, you will investigate the degree of the product of a monomialand a polynomial. a. TABULAR Write three monomials of different degrees and three polynomials of different degrees. Determine the degree of each monomial and polynomial. Multiply the monomials by the polynomials. Determine the degree of each product. Record your results in a table like the one below.
b. VERBAL Make a conjecture about the degree of the product of a monomial and a polynomial. What is the degree of the product of a monomial of degree a and a polynomial of degree b?
SOLUTION: a. There possible monomials are 2x, 3x
2, 4x
3. Three polynomials are x2 − 1, x5 + 1, x
6 + 1.
The degree for the monomial is the sum of the exponents. For the example given, case they are 1,2, and 3. The degree for the polynomial is the largest degree of the monomials. For the example given, they are 2, 5, and 6. Use the distributive property to find the product of the monomial and the polynomial. For the example give, they
are2x3 − 2x, 3x
7 + 3x
2, and 4x
9 + 4x. For the example given, the degree of the new polynomials are 3,7 and 9.
b. The degree of the product is the sum of the degree of the monomial and the degree of the polynomial, or a + b.
45. ERROR ANALYSIS Pearl and Ted both worked on this problem. Is either of them correct? Explain your reasoning.
SOLUTION:
Ted’s solution is correct. Pearl made a mistake when distributing on the middle term. She did not add the exponents correctly.
46. CCSS PERSEVERANCE Find p such that 3xp(4x
2p+3 + 2x3p−2
) = 12x12 + 6x
10.
SOLUTION: First simplify the left side of the equation.
Because the coefficients of the terms are the same, the corresponding terms’ exponents can be set equal to each other.
47. CHALLENGE Find 4x−3
y2(2x
5y−4 + 6x
−7y
6 − 4x0y
-2).
SOLUTION:
48. REASONING Is there a value for x that makes the statement (x + 2)2 = x2 + 22
true? If so, find a value for x. Explain your reasoning.
SOLUTION: Solve for x.
When 0 is substituted in for x in the equation, both sides are 22 or 4, which makes the equation true.
49. OPEN ENDED Write a monomial and a polynomial using n as the variable. Find their product.
SOLUTION: Choose a monomial with variable n and a polynomial with variable n. For example, let the monomial be 3n and the polynomial be 4n + 1. Find the product.
50. WRITING IN MATH Describe the steps to multiply a polynomial by a monomial.
SOLUTION: To multiply a polynomial by a monomial, use the Distributive Property. Multiply each term of the polynomial by the monomial. Then simplify by multiplying the coefficients together and using the Product of Powers Property for the variables. Consider the following example:
51. Every week a store sells j jeans and t T-shirts. The store makes $8 for each T-shirt and $12 for each pair of jeans. Which of the following expressions represents the total amount of money, in dollars, the store makes every week? A 8j + 12t B 12j + 8t C 20(j + t) D 96jt
SOLUTION: 12j is the amount made from jeans. 8t is the amount made from T-shirts. The sum, or 12j + 8t, represents the total amount the store makes every week. So, the correct choice is B.
52. If a = 5x + 7y and b = 2y − 3x, what is a + b?F 2x − 9y G 3y + 4x H 2x + 9y J 2x − 5y
SOLUTION:
So, the correct choice is H.
53. GEOMETRY A triangle has sides of length 5 inches and 8.5 inches. Which of the following cannot be the length of the third side? A 3.5 inches B 4 inches C 5.5 inches D 12 inches
SOLUTION: The sum of any two sides of a triangle must be greater than the length of the third side. Choice A: 3.5 + 5 = 8.5 8.5 Choice B: 3.5 + 5 = 8.5 > 8.5 Choice C: 5.5 + 5 = 10.5 > 8.5 Choice D: 12 + 5 = 17 > 8.5 Let x be the third side of the triangle.
3.5 cannot be the third side of the triangle. So, the correct choice is A.
54. SHORT RESPONSE Write an equation in which x varies directly as the cube of y and inversely as the square of z .
SOLUTION: If x varies directly as the cube of y , then both x and y must be in the numerators on either side of the equal sign.
When x varies inversely with the square of z. then the z must be in the denominator. Then . The
combination of the two is .
Find each sum or difference.
55. (2x2 − 7) + (8 − 5x
2)
SOLUTION:
56. (3z2 + 2z − 1) + (z
2 − 6)
SOLUTION:
57. (2a − 4a2 + 1) − (5a
2 − 2a − 6)
SOLUTION:
58. (a3 − 3a
2 + 4) − (4a
2 + 7)
SOLUTION:
59. (2ab − 3a + 4b) + (5a + 4ab)
SOLUTION:
60. (8c3 − 3c
2 + c − 2) − (3c
3 + 9)
SOLUTION:
Write a recursive formula for each sequence.61. 16, 2, –12, –26, ...
SOLUTION: Determine if there is a common difference. 2 – 16 = –14 –12 – 2 = –14 –26 – (–12) = –14 The common difference is –14. The first term is 16. To find the next term, subtract 14 from the previous term.
62. –5, 3, 11, 19, ...
SOLUTION: Determine if there is a common difference. 3 – (–5) = 8 11 – 3 = 8 19 – 11 = 8 The common difference is 8. The first term is 3. To find the next term, add 8 to the previous term.
63. 27, 43, 59, 75, ...
SOLUTION: Determine if there is a common difference. 43 – 27 = 16 59 – 43 = 16 75 – 59 = 16 The common difference is 16. The first term is 43. To find the next term, add 16 to the previous term.
64. 80, –200, 500, –1250, ...
SOLUTION: Check for a common difference. –200 – 80 = –280 500 – (–200) = 700 There is no common difference. Check for a common ratio. –200 ÷ 80 = –2.5 500 ÷ (–200) = –2.5 –1250 ÷ 500 = –2.5 The common ratio is –2.5. The first term is 80. To find the next term, multiply the previous term by –2.5.
65. 100, 60, 36, 21.6, ...
SOLUTION: Check for a common difference. 60 – 100 = –40 36 – 60 = –24 There is no common difference. Check for a common ratio. 60 ÷ 100 = 0.6 36 ÷ 60 = 0.6 21.6 ÷ 36 = 0.6 The common ratio is 0.6. The first term is 100. To find the next term, multiply the previous term by 0.6.
66.
SOLUTION: Check for a common difference.
There is no common difference. Check for a common ratio.
4 ÷ 1 = 4 The common ratio is 4.
The first term is . To find the next term, multiply the previous term by 4.
67. TRAVEL In 2003, about 9.5 million people took cruises. Between 2003 and 2008, the number increased by about 740,000 each year. Write the point-slope form of an equation to find the total number of people y taking a cruise for any year x. Estimate the number of people who took a cruise in 2010.
SOLUTION: The statement “increased by 740,000 each year” is a constant rate, or the slope. Now substitute the slope and the point (2003, 9,500,000) into the point slope equation.
To find the number of people who took a cruise in 2010, substitute x = 2010.
So, approximately 14,680,000 people took a cruise in 2010.
Write an equation in function notation for each relation.
68.
SOLUTION: The y-intercept is 0. Find the slope by choosing two points. Choose (0, 0) and (1, 4).
So the slope intercept form is f (x) = 4x + 0 or f (x) = 4x.
69.
SOLUTION:
The y-intercept is 0. Find the slope by choosing two points. Choose (0, 0) and (2, −1).
So in slope intercept form is f (x) = −0.5x + 0 or f (x) = −0.5x.
Simplify.
70. b(b2)(b
3)
SOLUTION:
71. 2y(3y2)
SOLUTION:
72. −y4(−2y
3)
SOLUTION:
73. −3z3(−5z
4 + 2z)
SOLUTION:
74. 2m(−4m4) − 3(−5m
3)
SOLUTION:
75. 4p 2(−2p
3) + 2p
4(5p
6)
SOLUTION:
eSolutions Manual - Powered by Cognero Page 15
8-2 Multiplying a Polynomial by a Monomial
Find each product.
1. 5w( −3w2 + 2w − 4)
SOLUTION:
2. 6g2(3g
3 + 4g2 + 10g − 1)
SOLUTION:
3. 4km2(8km
2 + 2k2m + 5k)
SOLUTION:
4. −3p4r3(2p
2r4 − 6p 6
r3 − 5)
SOLUTION:
5. 2ab(7a4b
2 + a5b − 2a)
SOLUTION:
6. c2d
3(5cd
7 − 3c3d
2 − 4d3)
SOLUTION:
Simplify each expression.
7. t(4t2 + 15t + 4) − 4(3t − 1)
SOLUTION:
8. x(3x2 + 4) + 2(7x − 3)
SOLUTION:
9. −2d(d3c
2 − 4dc2 + 2d
2c) + c
2(dc
2 − 3d4)
SOLUTION:
10. −5w2(8w
2x − 11wx
2) + 6x(9wx
4 − 4w − 3x2)
SOLUTION:
11. GRIDDED RESPONSE Marlene is buying a new plasma television. The height of the screen of the television is one half the width plus 5 inches. The width is 30 inches. Find the height of the screen in inches.
SOLUTION: Let h be the height and let w be the width.
Let w = 30.
The height is 20 inches.
Solve each equation.
12. −6(11 − 2c) = 7(−2 − 2c)
SOLUTION:
13. t(2t + 3) + 20 = 2t(t − 3)
SOLUTION:
14. −2(w + 1) + w = 7 − 4w
SOLUTION:
15. 3(y − 2) + 2y = 4y + 14
SOLUTION:
16. a(a + 3) + a(a − 6) + 35 = a(a − 5) + a(a + 7)
SOLUTION:
17. n(n − 4) + n(n + 8) = n(n − 13) + n(n + 1) + 16
SOLUTION:
Find each product.
18. b(b2 − 12b + 1)
SOLUTION:
19. f (f2 + 2f + 25)
SOLUTION:
20. −3m3(2m
3 − 12m2 + 2m + 25)
SOLUTION:
21. 2j 2(5j
3 − 15j2 + 2j + 2)
SOLUTION:
22. 2pr2(2pr + 5p
2r − 15p )
SOLUTION:
23. 4t3u(2t
2u
2 − 10tu4 + 2)
SOLUTION:
Simplify each expression.
24. −3(5x2 + 2x + 9) + x(2x − 3)
SOLUTION:
25. a(−8a2 + 2a + 4) + 3(6a
2 − 4)
SOLUTION:
26. −4d(5d2 − 12) + 7(d + 5)
SOLUTION:
27. −9g(−2g + g2) + 3(g
2 + 4)
SOLUTION:
28. 2j (7j2k
2 + j k2 + 5k) − 9k(−2j
2k
2 + 2k2 + 3j )
SOLUTION:
29. 4n(2n3p
2 − 3np2 + 5n) + 4p (6n
2p − 2np
2 + 3p )
SOLUTION:
30. DAMS A new dam being built has the shape of a trapezoid. The base at the bottom of the dam is 2 times the
height. The base at the top of the dam is times the height minus 30 feet.
a. Write an expression to find the area of the trapezoidal cross section of the dam. b. If the height of the dam is 180 feet, find the area of this cross section.
SOLUTION:
a. The equation for the area of a trapezoid is .
b1
= 2h
b2 =
b. Let h = 180.
So, the area is 32,940 square feet.
Solve each equation.
31. 7(t2 + 5t − 9) + t = t(7t − 2) + 13
SOLUTION:
32. w(4w + 6) + 2w = 2(2w2 + 7w − 3)
SOLUTION:
33. 5(4z + 6) − 2(z − 4) = 7z(z + 4) − z(7z − 2) − 48
SOLUTION:
34. 9c(c − 11) + 10(5c − 3) = 3c(c + 5) + c(6c − 3) − 30
SOLUTION:
35. 2f (5f − 2) − 10(f2 − 3f + 6) = −8f (f + 4) + 4(2f
2 − 7f )
SOLUTION:
36. 2k(−3k + 4) + 6(k2 + 10) = k(4k + 8) − 2k(2k + 5)
SOLUTION:
Simplify each expression.
37. np2(30p
2 + 9n
2p − 12)
SOLUTION:
38. r2t(10r
3 + 5rt
3 + 15t
2)
SOLUTION:
39. −5q2w
3(4q + 7w) + 4qw
2(7q
2w + 2q) − 3qw(3q
2w
2 + 9)
SOLUTION:
40. −x2z(2z
2 + 4xz3) + xz
2(xz + 5x
3z) + x2
z3(3x
2z + 4xz)
SOLUTION:
41. PARKING A parking garage charges $30 per month plus $0.50 per daytime hour and $0.25 per hour during nights and weekends. Suppose Trent parks in the garage for 47 hours in January and h of those are night and weekend hours. a. Find an expression for Trent’s January bill. b. Find the cost if Trent had 12 hours of night and weekend hours.
SOLUTION: a. First, multiply the price per hour times the number of hours. Let h represent the number of weekend and night hours. For the weekend and night hours, he would spend 0.25h. The number of daytime hours is 47 – h, so he spent 0.50(47 - h) = 23.50 – 0.50h. To find the total cost, add these two values to the cost per month. 30 + (23.50 − 0.50h) + 0.25h = 53.50 − 0.25h. b. If Trent parked 12 night and weekend hours, h = 12
So, the cost for January is $50.50.
42. CCSS MODELING Che is building a dog house for his new puppy. The upper face of the dog house is a trapezoid. If the height of the trapezoid is 12 inches, find the area of the face of this piece of the dog house.
SOLUTION:
The formula for the area of a trapezoid is .
Now substitute h = 12 inches into the equation.
Therefore, the area of the trapezoid is 318 square inches.
43. TENNIS The tennis club is building a new tennis court with a path around it.
a. Write an expression for the area of the tennis court. b. Write an expression for the area of the path. c. What is the perimeter of the outside of the path?
SOLUTION: a. The area of the tennis court can be found by the equation . Substitute the values of the length and width and simplify.
b. The area of the path is the area of the large rectangle minus the area of the smaller rectangle, which is the area ofthe tennis court.
Area of the large rectangle = 2.5x(x + 6) or 2.5x2 + 15x.
Now subtract the area of the tennis court to find the area of just the path.
c. The perimeter of the outside of the path is the perimeter of the larger rectangle.
If x = 36 feet, then the perimeter of the outside of the path is 7(36) + 12 or 264 ft.
44. MULTIPLE REPRESENTATIONS In this problem, you will investigate the degree of the product of a monomialand a polynomial. a. TABULAR Write three monomials of different degrees and three polynomials of different degrees. Determine the degree of each monomial and polynomial. Multiply the monomials by the polynomials. Determine the degree of each product. Record your results in a table like the one below.
b. VERBAL Make a conjecture about the degree of the product of a monomial and a polynomial. What is the degree of the product of a monomial of degree a and a polynomial of degree b?
SOLUTION: a. There possible monomials are 2x, 3x
2, 4x
3. Three polynomials are x2 − 1, x5 + 1, x
6 + 1.
The degree for the monomial is the sum of the exponents. For the example given, case they are 1,2, and 3. The degree for the polynomial is the largest degree of the monomials. For the example given, they are 2, 5, and 6. Use the distributive property to find the product of the monomial and the polynomial. For the example give, they
are2x3 − 2x, 3x
7 + 3x
2, and 4x
9 + 4x. For the example given, the degree of the new polynomials are 3,7 and 9.
b. The degree of the product is the sum of the degree of the monomial and the degree of the polynomial, or a + b.
45. ERROR ANALYSIS Pearl and Ted both worked on this problem. Is either of them correct? Explain your reasoning.
SOLUTION:
Ted’s solution is correct. Pearl made a mistake when distributing on the middle term. She did not add the exponents correctly.
46. CCSS PERSEVERANCE Find p such that 3xp(4x
2p+3 + 2x3p−2
) = 12x12 + 6x
10.
SOLUTION: First simplify the left side of the equation.
Because the coefficients of the terms are the same, the corresponding terms’ exponents can be set equal to each other.
47. CHALLENGE Find 4x−3
y2(2x
5y−4 + 6x
−7y
6 − 4x0y
-2).
SOLUTION:
48. REASONING Is there a value for x that makes the statement (x + 2)2 = x2 + 22
true? If so, find a value for x. Explain your reasoning.
SOLUTION: Solve for x.
When 0 is substituted in for x in the equation, both sides are 22 or 4, which makes the equation true.
49. OPEN ENDED Write a monomial and a polynomial using n as the variable. Find their product.
SOLUTION: Choose a monomial with variable n and a polynomial with variable n. For example, let the monomial be 3n and the polynomial be 4n + 1. Find the product.
50. WRITING IN MATH Describe the steps to multiply a polynomial by a monomial.
SOLUTION: To multiply a polynomial by a monomial, use the Distributive Property. Multiply each term of the polynomial by the monomial. Then simplify by multiplying the coefficients together and using the Product of Powers Property for the variables. Consider the following example:
51. Every week a store sells j jeans and t T-shirts. The store makes $8 for each T-shirt and $12 for each pair of jeans. Which of the following expressions represents the total amount of money, in dollars, the store makes every week? A 8j + 12t B 12j + 8t C 20(j + t) D 96jt
SOLUTION: 12j is the amount made from jeans. 8t is the amount made from T-shirts. The sum, or 12j + 8t, represents the total amount the store makes every week. So, the correct choice is B.
52. If a = 5x + 7y and b = 2y − 3x, what is a + b?F 2x − 9y G 3y + 4x H 2x + 9y J 2x − 5y
SOLUTION:
So, the correct choice is H.
53. GEOMETRY A triangle has sides of length 5 inches and 8.5 inches. Which of the following cannot be the length of the third side? A 3.5 inches B 4 inches C 5.5 inches D 12 inches
SOLUTION: The sum of any two sides of a triangle must be greater than the length of the third side. Choice A: 3.5 + 5 = 8.5 8.5 Choice B: 3.5 + 5 = 8.5 > 8.5 Choice C: 5.5 + 5 = 10.5 > 8.5 Choice D: 12 + 5 = 17 > 8.5 Let x be the third side of the triangle.
3.5 cannot be the third side of the triangle. So, the correct choice is A.
54. SHORT RESPONSE Write an equation in which x varies directly as the cube of y and inversely as the square of z .
SOLUTION: If x varies directly as the cube of y , then both x and y must be in the numerators on either side of the equal sign.
When x varies inversely with the square of z. then the z must be in the denominator. Then . The
combination of the two is .
Find each sum or difference.
55. (2x2 − 7) + (8 − 5x
2)
SOLUTION:
56. (3z2 + 2z − 1) + (z
2 − 6)
SOLUTION:
57. (2a − 4a2 + 1) − (5a
2 − 2a − 6)
SOLUTION:
58. (a3 − 3a
2 + 4) − (4a
2 + 7)
SOLUTION:
59. (2ab − 3a + 4b) + (5a + 4ab)
SOLUTION:
60. (8c3 − 3c
2 + c − 2) − (3c
3 + 9)
SOLUTION:
Write a recursive formula for each sequence.61. 16, 2, –12, –26, ...
SOLUTION: Determine if there is a common difference. 2 – 16 = –14 –12 – 2 = –14 –26 – (–12) = –14 The common difference is –14. The first term is 16. To find the next term, subtract 14 from the previous term.
62. –5, 3, 11, 19, ...
SOLUTION: Determine if there is a common difference. 3 – (–5) = 8 11 – 3 = 8 19 – 11 = 8 The common difference is 8. The first term is 3. To find the next term, add 8 to the previous term.
63. 27, 43, 59, 75, ...
SOLUTION: Determine if there is a common difference. 43 – 27 = 16 59 – 43 = 16 75 – 59 = 16 The common difference is 16. The first term is 43. To find the next term, add 16 to the previous term.
64. 80, –200, 500, –1250, ...
SOLUTION: Check for a common difference. –200 – 80 = –280 500 – (–200) = 700 There is no common difference. Check for a common ratio. –200 ÷ 80 = –2.5 500 ÷ (–200) = –2.5 –1250 ÷ 500 = –2.5 The common ratio is –2.5. The first term is 80. To find the next term, multiply the previous term by –2.5.
65. 100, 60, 36, 21.6, ...
SOLUTION: Check for a common difference. 60 – 100 = –40 36 – 60 = –24 There is no common difference. Check for a common ratio. 60 ÷ 100 = 0.6 36 ÷ 60 = 0.6 21.6 ÷ 36 = 0.6 The common ratio is 0.6. The first term is 100. To find the next term, multiply the previous term by 0.6.
66.
SOLUTION: Check for a common difference.
There is no common difference. Check for a common ratio.
4 ÷ 1 = 4 The common ratio is 4.
The first term is . To find the next term, multiply the previous term by 4.
67. TRAVEL In 2003, about 9.5 million people took cruises. Between 2003 and 2008, the number increased by about 740,000 each year. Write the point-slope form of an equation to find the total number of people y taking a cruise for any year x. Estimate the number of people who took a cruise in 2010.
SOLUTION: The statement “increased by 740,000 each year” is a constant rate, or the slope. Now substitute the slope and the point (2003, 9,500,000) into the point slope equation.
To find the number of people who took a cruise in 2010, substitute x = 2010.
So, approximately 14,680,000 people took a cruise in 2010.
Write an equation in function notation for each relation.
68.
SOLUTION: The y-intercept is 0. Find the slope by choosing two points. Choose (0, 0) and (1, 4).
So the slope intercept form is f (x) = 4x + 0 or f (x) = 4x.
69.
SOLUTION:
The y-intercept is 0. Find the slope by choosing two points. Choose (0, 0) and (2, −1).
So in slope intercept form is f (x) = −0.5x + 0 or f (x) = −0.5x.
Simplify.
70. b(b2)(b
3)
SOLUTION:
71. 2y(3y2)
SOLUTION:
72. −y4(−2y
3)
SOLUTION:
73. −3z3(−5z
4 + 2z)
SOLUTION:
74. 2m(−4m4) − 3(−5m
3)
SOLUTION:
75. 4p 2(−2p
3) + 2p
4(5p
6)
SOLUTION:
eSolutions Manual - Powered by Cognero Page 16
8-2 Multiplying a Polynomial by a Monomial
Find each product.
1. 5w( −3w2 + 2w − 4)
SOLUTION:
2. 6g2(3g
3 + 4g2 + 10g − 1)
SOLUTION:
3. 4km2(8km
2 + 2k2m + 5k)
SOLUTION:
4. −3p4r3(2p
2r4 − 6p 6
r3 − 5)
SOLUTION:
5. 2ab(7a4b
2 + a5b − 2a)
SOLUTION:
6. c2d
3(5cd
7 − 3c3d
2 − 4d3)
SOLUTION:
Simplify each expression.
7. t(4t2 + 15t + 4) − 4(3t − 1)
SOLUTION:
8. x(3x2 + 4) + 2(7x − 3)
SOLUTION:
9. −2d(d3c
2 − 4dc2 + 2d
2c) + c
2(dc
2 − 3d4)
SOLUTION:
10. −5w2(8w
2x − 11wx
2) + 6x(9wx
4 − 4w − 3x2)
SOLUTION:
11. GRIDDED RESPONSE Marlene is buying a new plasma television. The height of the screen of the television is one half the width plus 5 inches. The width is 30 inches. Find the height of the screen in inches.
SOLUTION: Let h be the height and let w be the width.
Let w = 30.
The height is 20 inches.
Solve each equation.
12. −6(11 − 2c) = 7(−2 − 2c)
SOLUTION:
13. t(2t + 3) + 20 = 2t(t − 3)
SOLUTION:
14. −2(w + 1) + w = 7 − 4w
SOLUTION:
15. 3(y − 2) + 2y = 4y + 14
SOLUTION:
16. a(a + 3) + a(a − 6) + 35 = a(a − 5) + a(a + 7)
SOLUTION:
17. n(n − 4) + n(n + 8) = n(n − 13) + n(n + 1) + 16
SOLUTION:
Find each product.
18. b(b2 − 12b + 1)
SOLUTION:
19. f (f2 + 2f + 25)
SOLUTION:
20. −3m3(2m
3 − 12m2 + 2m + 25)
SOLUTION:
21. 2j 2(5j
3 − 15j2 + 2j + 2)
SOLUTION:
22. 2pr2(2pr + 5p
2r − 15p )
SOLUTION:
23. 4t3u(2t
2u
2 − 10tu4 + 2)
SOLUTION:
Simplify each expression.
24. −3(5x2 + 2x + 9) + x(2x − 3)
SOLUTION:
25. a(−8a2 + 2a + 4) + 3(6a
2 − 4)
SOLUTION:
26. −4d(5d2 − 12) + 7(d + 5)
SOLUTION:
27. −9g(−2g + g2) + 3(g
2 + 4)
SOLUTION:
28. 2j (7j2k
2 + j k2 + 5k) − 9k(−2j
2k
2 + 2k2 + 3j )
SOLUTION:
29. 4n(2n3p
2 − 3np2 + 5n) + 4p (6n
2p − 2np
2 + 3p )
SOLUTION:
30. DAMS A new dam being built has the shape of a trapezoid. The base at the bottom of the dam is 2 times the
height. The base at the top of the dam is times the height minus 30 feet.
a. Write an expression to find the area of the trapezoidal cross section of the dam. b. If the height of the dam is 180 feet, find the area of this cross section.
SOLUTION:
a. The equation for the area of a trapezoid is .
b1
= 2h
b2 =
b. Let h = 180.
So, the area is 32,940 square feet.
Solve each equation.
31. 7(t2 + 5t − 9) + t = t(7t − 2) + 13
SOLUTION:
32. w(4w + 6) + 2w = 2(2w2 + 7w − 3)
SOLUTION:
33. 5(4z + 6) − 2(z − 4) = 7z(z + 4) − z(7z − 2) − 48
SOLUTION:
34. 9c(c − 11) + 10(5c − 3) = 3c(c + 5) + c(6c − 3) − 30
SOLUTION:
35. 2f (5f − 2) − 10(f2 − 3f + 6) = −8f (f + 4) + 4(2f
2 − 7f )
SOLUTION:
36. 2k(−3k + 4) + 6(k2 + 10) = k(4k + 8) − 2k(2k + 5)
SOLUTION:
Simplify each expression.
37. np2(30p
2 + 9n
2p − 12)
SOLUTION:
38. r2t(10r
3 + 5rt
3 + 15t
2)
SOLUTION:
39. −5q2w
3(4q + 7w) + 4qw
2(7q
2w + 2q) − 3qw(3q
2w
2 + 9)
SOLUTION:
40. −x2z(2z
2 + 4xz3) + xz
2(xz + 5x
3z) + x2
z3(3x
2z + 4xz)
SOLUTION:
41. PARKING A parking garage charges $30 per month plus $0.50 per daytime hour and $0.25 per hour during nights and weekends. Suppose Trent parks in the garage for 47 hours in January and h of those are night and weekend hours. a. Find an expression for Trent’s January bill. b. Find the cost if Trent had 12 hours of night and weekend hours.
SOLUTION: a. First, multiply the price per hour times the number of hours. Let h represent the number of weekend and night hours. For the weekend and night hours, he would spend 0.25h. The number of daytime hours is 47 – h, so he spent 0.50(47 - h) = 23.50 – 0.50h. To find the total cost, add these two values to the cost per month. 30 + (23.50 − 0.50h) + 0.25h = 53.50 − 0.25h. b. If Trent parked 12 night and weekend hours, h = 12
So, the cost for January is $50.50.
42. CCSS MODELING Che is building a dog house for his new puppy. The upper face of the dog house is a trapezoid. If the height of the trapezoid is 12 inches, find the area of the face of this piece of the dog house.
SOLUTION:
The formula for the area of a trapezoid is .
Now substitute h = 12 inches into the equation.
Therefore, the area of the trapezoid is 318 square inches.
43. TENNIS The tennis club is building a new tennis court with a path around it.
a. Write an expression for the area of the tennis court. b. Write an expression for the area of the path. c. What is the perimeter of the outside of the path?
SOLUTION: a. The area of the tennis court can be found by the equation . Substitute the values of the length and width and simplify.
b. The area of the path is the area of the large rectangle minus the area of the smaller rectangle, which is the area ofthe tennis court.
Area of the large rectangle = 2.5x(x + 6) or 2.5x2 + 15x.
Now subtract the area of the tennis court to find the area of just the path.
c. The perimeter of the outside of the path is the perimeter of the larger rectangle.
If x = 36 feet, then the perimeter of the outside of the path is 7(36) + 12 or 264 ft.
44. MULTIPLE REPRESENTATIONS In this problem, you will investigate the degree of the product of a monomialand a polynomial. a. TABULAR Write three monomials of different degrees and three polynomials of different degrees. Determine the degree of each monomial and polynomial. Multiply the monomials by the polynomials. Determine the degree of each product. Record your results in a table like the one below.
b. VERBAL Make a conjecture about the degree of the product of a monomial and a polynomial. What is the degree of the product of a monomial of degree a and a polynomial of degree b?
SOLUTION: a. There possible monomials are 2x, 3x
2, 4x
3. Three polynomials are x2 − 1, x5 + 1, x
6 + 1.
The degree for the monomial is the sum of the exponents. For the example given, case they are 1,2, and 3. The degree for the polynomial is the largest degree of the monomials. For the example given, they are 2, 5, and 6. Use the distributive property to find the product of the monomial and the polynomial. For the example give, they
are2x3 − 2x, 3x
7 + 3x
2, and 4x
9 + 4x. For the example given, the degree of the new polynomials are 3,7 and 9.
b. The degree of the product is the sum of the degree of the monomial and the degree of the polynomial, or a + b.
45. ERROR ANALYSIS Pearl and Ted both worked on this problem. Is either of them correct? Explain your reasoning.
SOLUTION:
Ted’s solution is correct. Pearl made a mistake when distributing on the middle term. She did not add the exponents correctly.
46. CCSS PERSEVERANCE Find p such that 3xp(4x
2p+3 + 2x3p−2
) = 12x12 + 6x
10.
SOLUTION: First simplify the left side of the equation.
Because the coefficients of the terms are the same, the corresponding terms’ exponents can be set equal to each other.
47. CHALLENGE Find 4x−3
y2(2x
5y−4 + 6x
−7y
6 − 4x0y
-2).
SOLUTION:
48. REASONING Is there a value for x that makes the statement (x + 2)2 = x2 + 22
true? If so, find a value for x. Explain your reasoning.
SOLUTION: Solve for x.
When 0 is substituted in for x in the equation, both sides are 22 or 4, which makes the equation true.
49. OPEN ENDED Write a monomial and a polynomial using n as the variable. Find their product.
SOLUTION: Choose a monomial with variable n and a polynomial with variable n. For example, let the monomial be 3n and the polynomial be 4n + 1. Find the product.
50. WRITING IN MATH Describe the steps to multiply a polynomial by a monomial.
SOLUTION: To multiply a polynomial by a monomial, use the Distributive Property. Multiply each term of the polynomial by the monomial. Then simplify by multiplying the coefficients together and using the Product of Powers Property for the variables. Consider the following example:
51. Every week a store sells j jeans and t T-shirts. The store makes $8 for each T-shirt and $12 for each pair of jeans. Which of the following expressions represents the total amount of money, in dollars, the store makes every week? A 8j + 12t B 12j + 8t C 20(j + t) D 96jt
SOLUTION: 12j is the amount made from jeans. 8t is the amount made from T-shirts. The sum, or 12j + 8t, represents the total amount the store makes every week. So, the correct choice is B.
52. If a = 5x + 7y and b = 2y − 3x, what is a + b?F 2x − 9y G 3y + 4x H 2x + 9y J 2x − 5y
SOLUTION:
So, the correct choice is H.
53. GEOMETRY A triangle has sides of length 5 inches and 8.5 inches. Which of the following cannot be the length of the third side? A 3.5 inches B 4 inches C 5.5 inches D 12 inches
SOLUTION: The sum of any two sides of a triangle must be greater than the length of the third side. Choice A: 3.5 + 5 = 8.5 8.5 Choice B: 3.5 + 5 = 8.5 > 8.5 Choice C: 5.5 + 5 = 10.5 > 8.5 Choice D: 12 + 5 = 17 > 8.5 Let x be the third side of the triangle.
3.5 cannot be the third side of the triangle. So, the correct choice is A.
54. SHORT RESPONSE Write an equation in which x varies directly as the cube of y and inversely as the square of z .
SOLUTION: If x varies directly as the cube of y , then both x and y must be in the numerators on either side of the equal sign.
When x varies inversely with the square of z. then the z must be in the denominator. Then . The
combination of the two is .
Find each sum or difference.
55. (2x2 − 7) + (8 − 5x
2)
SOLUTION:
56. (3z2 + 2z − 1) + (z
2 − 6)
SOLUTION:
57. (2a − 4a2 + 1) − (5a
2 − 2a − 6)
SOLUTION:
58. (a3 − 3a
2 + 4) − (4a
2 + 7)
SOLUTION:
59. (2ab − 3a + 4b) + (5a + 4ab)
SOLUTION:
60. (8c3 − 3c
2 + c − 2) − (3c
3 + 9)
SOLUTION:
Write a recursive formula for each sequence.61. 16, 2, –12, –26, ...
SOLUTION: Determine if there is a common difference. 2 – 16 = –14 –12 – 2 = –14 –26 – (–12) = –14 The common difference is –14. The first term is 16. To find the next term, subtract 14 from the previous term.
62. –5, 3, 11, 19, ...
SOLUTION: Determine if there is a common difference. 3 – (–5) = 8 11 – 3 = 8 19 – 11 = 8 The common difference is 8. The first term is 3. To find the next term, add 8 to the previous term.
63. 27, 43, 59, 75, ...
SOLUTION: Determine if there is a common difference. 43 – 27 = 16 59 – 43 = 16 75 – 59 = 16 The common difference is 16. The first term is 43. To find the next term, add 16 to the previous term.
64. 80, –200, 500, –1250, ...
SOLUTION: Check for a common difference. –200 – 80 = –280 500 – (–200) = 700 There is no common difference. Check for a common ratio. –200 ÷ 80 = –2.5 500 ÷ (–200) = –2.5 –1250 ÷ 500 = –2.5 The common ratio is –2.5. The first term is 80. To find the next term, multiply the previous term by –2.5.
65. 100, 60, 36, 21.6, ...
SOLUTION: Check for a common difference. 60 – 100 = –40 36 – 60 = –24 There is no common difference. Check for a common ratio. 60 ÷ 100 = 0.6 36 ÷ 60 = 0.6 21.6 ÷ 36 = 0.6 The common ratio is 0.6. The first term is 100. To find the next term, multiply the previous term by 0.6.
66.
SOLUTION: Check for a common difference.
There is no common difference. Check for a common ratio.
4 ÷ 1 = 4 The common ratio is 4.
The first term is . To find the next term, multiply the previous term by 4.
67. TRAVEL In 2003, about 9.5 million people took cruises. Between 2003 and 2008, the number increased by about 740,000 each year. Write the point-slope form of an equation to find the total number of people y taking a cruise for any year x. Estimate the number of people who took a cruise in 2010.
SOLUTION: The statement “increased by 740,000 each year” is a constant rate, or the slope. Now substitute the slope and the point (2003, 9,500,000) into the point slope equation.
To find the number of people who took a cruise in 2010, substitute x = 2010.
So, approximately 14,680,000 people took a cruise in 2010.
Write an equation in function notation for each relation.
68.
SOLUTION: The y-intercept is 0. Find the slope by choosing two points. Choose (0, 0) and (1, 4).
So the slope intercept form is f (x) = 4x + 0 or f (x) = 4x.
69.
SOLUTION:
The y-intercept is 0. Find the slope by choosing two points. Choose (0, 0) and (2, −1).
So in slope intercept form is f (x) = −0.5x + 0 or f (x) = −0.5x.
Simplify.
70. b(b2)(b
3)
SOLUTION:
71. 2y(3y2)
SOLUTION:
72. −y4(−2y
3)
SOLUTION:
73. −3z3(−5z
4 + 2z)
SOLUTION:
74. 2m(−4m4) − 3(−5m
3)
SOLUTION:
75. 4p 2(−2p
3) + 2p
4(5p
6)
SOLUTION:
eSolutions Manual - Powered by Cognero Page 17
8-2 Multiplying a Polynomial by a Monomial
Find each product.
1. 5w( −3w2 + 2w − 4)
SOLUTION:
2. 6g2(3g
3 + 4g2 + 10g − 1)
SOLUTION:
3. 4km2(8km
2 + 2k2m + 5k)
SOLUTION:
4. −3p4r3(2p
2r4 − 6p 6
r3 − 5)
SOLUTION:
5. 2ab(7a4b
2 + a5b − 2a)
SOLUTION:
6. c2d
3(5cd
7 − 3c3d
2 − 4d3)
SOLUTION:
Simplify each expression.
7. t(4t2 + 15t + 4) − 4(3t − 1)
SOLUTION:
8. x(3x2 + 4) + 2(7x − 3)
SOLUTION:
9. −2d(d3c
2 − 4dc2 + 2d
2c) + c
2(dc
2 − 3d4)
SOLUTION:
10. −5w2(8w
2x − 11wx
2) + 6x(9wx
4 − 4w − 3x2)
SOLUTION:
11. GRIDDED RESPONSE Marlene is buying a new plasma television. The height of the screen of the television is one half the width plus 5 inches. The width is 30 inches. Find the height of the screen in inches.
SOLUTION: Let h be the height and let w be the width.
Let w = 30.
The height is 20 inches.
Solve each equation.
12. −6(11 − 2c) = 7(−2 − 2c)
SOLUTION:
13. t(2t + 3) + 20 = 2t(t − 3)
SOLUTION:
14. −2(w + 1) + w = 7 − 4w
SOLUTION:
15. 3(y − 2) + 2y = 4y + 14
SOLUTION:
16. a(a + 3) + a(a − 6) + 35 = a(a − 5) + a(a + 7)
SOLUTION:
17. n(n − 4) + n(n + 8) = n(n − 13) + n(n + 1) + 16
SOLUTION:
Find each product.
18. b(b2 − 12b + 1)
SOLUTION:
19. f (f2 + 2f + 25)
SOLUTION:
20. −3m3(2m
3 − 12m2 + 2m + 25)
SOLUTION:
21. 2j 2(5j
3 − 15j2 + 2j + 2)
SOLUTION:
22. 2pr2(2pr + 5p
2r − 15p )
SOLUTION:
23. 4t3u(2t
2u
2 − 10tu4 + 2)
SOLUTION:
Simplify each expression.
24. −3(5x2 + 2x + 9) + x(2x − 3)
SOLUTION:
25. a(−8a2 + 2a + 4) + 3(6a
2 − 4)
SOLUTION:
26. −4d(5d2 − 12) + 7(d + 5)
SOLUTION:
27. −9g(−2g + g2) + 3(g
2 + 4)
SOLUTION:
28. 2j (7j2k
2 + j k2 + 5k) − 9k(−2j
2k
2 + 2k2 + 3j )
SOLUTION:
29. 4n(2n3p
2 − 3np2 + 5n) + 4p (6n
2p − 2np
2 + 3p )
SOLUTION:
30. DAMS A new dam being built has the shape of a trapezoid. The base at the bottom of the dam is 2 times the
height. The base at the top of the dam is times the height minus 30 feet.
a. Write an expression to find the area of the trapezoidal cross section of the dam. b. If the height of the dam is 180 feet, find the area of this cross section.
SOLUTION:
a. The equation for the area of a trapezoid is .
b1
= 2h
b2 =
b. Let h = 180.
So, the area is 32,940 square feet.
Solve each equation.
31. 7(t2 + 5t − 9) + t = t(7t − 2) + 13
SOLUTION:
32. w(4w + 6) + 2w = 2(2w2 + 7w − 3)
SOLUTION:
33. 5(4z + 6) − 2(z − 4) = 7z(z + 4) − z(7z − 2) − 48
SOLUTION:
34. 9c(c − 11) + 10(5c − 3) = 3c(c + 5) + c(6c − 3) − 30
SOLUTION:
35. 2f (5f − 2) − 10(f2 − 3f + 6) = −8f (f + 4) + 4(2f
2 − 7f )
SOLUTION:
36. 2k(−3k + 4) + 6(k2 + 10) = k(4k + 8) − 2k(2k + 5)
SOLUTION:
Simplify each expression.
37. np2(30p
2 + 9n
2p − 12)
SOLUTION:
38. r2t(10r
3 + 5rt
3 + 15t
2)
SOLUTION:
39. −5q2w
3(4q + 7w) + 4qw
2(7q
2w + 2q) − 3qw(3q
2w
2 + 9)
SOLUTION:
40. −x2z(2z
2 + 4xz3) + xz
2(xz + 5x
3z) + x2
z3(3x
2z + 4xz)
SOLUTION:
41. PARKING A parking garage charges $30 per month plus $0.50 per daytime hour and $0.25 per hour during nights and weekends. Suppose Trent parks in the garage for 47 hours in January and h of those are night and weekend hours. a. Find an expression for Trent’s January bill. b. Find the cost if Trent had 12 hours of night and weekend hours.
SOLUTION: a. First, multiply the price per hour times the number of hours. Let h represent the number of weekend and night hours. For the weekend and night hours, he would spend 0.25h. The number of daytime hours is 47 – h, so he spent 0.50(47 - h) = 23.50 – 0.50h. To find the total cost, add these two values to the cost per month. 30 + (23.50 − 0.50h) + 0.25h = 53.50 − 0.25h. b. If Trent parked 12 night and weekend hours, h = 12
So, the cost for January is $50.50.
42. CCSS MODELING Che is building a dog house for his new puppy. The upper face of the dog house is a trapezoid. If the height of the trapezoid is 12 inches, find the area of the face of this piece of the dog house.
SOLUTION:
The formula for the area of a trapezoid is .
Now substitute h = 12 inches into the equation.
Therefore, the area of the trapezoid is 318 square inches.
43. TENNIS The tennis club is building a new tennis court with a path around it.
a. Write an expression for the area of the tennis court. b. Write an expression for the area of the path. c. What is the perimeter of the outside of the path?
SOLUTION: a. The area of the tennis court can be found by the equation . Substitute the values of the length and width and simplify.
b. The area of the path is the area of the large rectangle minus the area of the smaller rectangle, which is the area ofthe tennis court.
Area of the large rectangle = 2.5x(x + 6) or 2.5x2 + 15x.
Now subtract the area of the tennis court to find the area of just the path.
c. The perimeter of the outside of the path is the perimeter of the larger rectangle.
If x = 36 feet, then the perimeter of the outside of the path is 7(36) + 12 or 264 ft.
44. MULTIPLE REPRESENTATIONS In this problem, you will investigate the degree of the product of a monomialand a polynomial. a. TABULAR Write three monomials of different degrees and three polynomials of different degrees. Determine the degree of each monomial and polynomial. Multiply the monomials by the polynomials. Determine the degree of each product. Record your results in a table like the one below.
b. VERBAL Make a conjecture about the degree of the product of a monomial and a polynomial. What is the degree of the product of a monomial of degree a and a polynomial of degree b?
SOLUTION: a. There possible monomials are 2x, 3x
2, 4x
3. Three polynomials are x2 − 1, x5 + 1, x
6 + 1.
The degree for the monomial is the sum of the exponents. For the example given, case they are 1,2, and 3. The degree for the polynomial is the largest degree of the monomials. For the example given, they are 2, 5, and 6. Use the distributive property to find the product of the monomial and the polynomial. For the example give, they
are2x3 − 2x, 3x
7 + 3x
2, and 4x
9 + 4x. For the example given, the degree of the new polynomials are 3,7 and 9.
b. The degree of the product is the sum of the degree of the monomial and the degree of the polynomial, or a + b.
45. ERROR ANALYSIS Pearl and Ted both worked on this problem. Is either of them correct? Explain your reasoning.
SOLUTION:
Ted’s solution is correct. Pearl made a mistake when distributing on the middle term. She did not add the exponents correctly.
46. CCSS PERSEVERANCE Find p such that 3xp(4x
2p+3 + 2x3p−2
) = 12x12 + 6x
10.
SOLUTION: First simplify the left side of the equation.
Because the coefficients of the terms are the same, the corresponding terms’ exponents can be set equal to each other.
47. CHALLENGE Find 4x−3
y2(2x
5y−4 + 6x
−7y
6 − 4x0y
-2).
SOLUTION:
48. REASONING Is there a value for x that makes the statement (x + 2)2 = x2 + 22
true? If so, find a value for x. Explain your reasoning.
SOLUTION: Solve for x.
When 0 is substituted in for x in the equation, both sides are 22 or 4, which makes the equation true.
49. OPEN ENDED Write a monomial and a polynomial using n as the variable. Find their product.
SOLUTION: Choose a monomial with variable n and a polynomial with variable n. For example, let the monomial be 3n and the polynomial be 4n + 1. Find the product.
50. WRITING IN MATH Describe the steps to multiply a polynomial by a monomial.
SOLUTION: To multiply a polynomial by a monomial, use the Distributive Property. Multiply each term of the polynomial by the monomial. Then simplify by multiplying the coefficients together and using the Product of Powers Property for the variables. Consider the following example:
51. Every week a store sells j jeans and t T-shirts. The store makes $8 for each T-shirt and $12 for each pair of jeans. Which of the following expressions represents the total amount of money, in dollars, the store makes every week? A 8j + 12t B 12j + 8t C 20(j + t) D 96jt
SOLUTION: 12j is the amount made from jeans. 8t is the amount made from T-shirts. The sum, or 12j + 8t, represents the total amount the store makes every week. So, the correct choice is B.
52. If a = 5x + 7y and b = 2y − 3x, what is a + b?F 2x − 9y G 3y + 4x H 2x + 9y J 2x − 5y
SOLUTION:
So, the correct choice is H.
53. GEOMETRY A triangle has sides of length 5 inches and 8.5 inches. Which of the following cannot be the length of the third side? A 3.5 inches B 4 inches C 5.5 inches D 12 inches
SOLUTION: The sum of any two sides of a triangle must be greater than the length of the third side. Choice A: 3.5 + 5 = 8.5 8.5 Choice B: 3.5 + 5 = 8.5 > 8.5 Choice C: 5.5 + 5 = 10.5 > 8.5 Choice D: 12 + 5 = 17 > 8.5 Let x be the third side of the triangle.
3.5 cannot be the third side of the triangle. So, the correct choice is A.
54. SHORT RESPONSE Write an equation in which x varies directly as the cube of y and inversely as the square of z .
SOLUTION: If x varies directly as the cube of y , then both x and y must be in the numerators on either side of the equal sign.
When x varies inversely with the square of z. then the z must be in the denominator. Then . The
combination of the two is .
Find each sum or difference.
55. (2x2 − 7) + (8 − 5x
2)
SOLUTION:
56. (3z2 + 2z − 1) + (z
2 − 6)
SOLUTION:
57. (2a − 4a2 + 1) − (5a
2 − 2a − 6)
SOLUTION:
58. (a3 − 3a
2 + 4) − (4a
2 + 7)
SOLUTION:
59. (2ab − 3a + 4b) + (5a + 4ab)
SOLUTION:
60. (8c3 − 3c
2 + c − 2) − (3c
3 + 9)
SOLUTION:
Write a recursive formula for each sequence.61. 16, 2, –12, –26, ...
SOLUTION: Determine if there is a common difference. 2 – 16 = –14 –12 – 2 = –14 –26 – (–12) = –14 The common difference is –14. The first term is 16. To find the next term, subtract 14 from the previous term.
62. –5, 3, 11, 19, ...
SOLUTION: Determine if there is a common difference. 3 – (–5) = 8 11 – 3 = 8 19 – 11 = 8 The common difference is 8. The first term is 3. To find the next term, add 8 to the previous term.
63. 27, 43, 59, 75, ...
SOLUTION: Determine if there is a common difference. 43 – 27 = 16 59 – 43 = 16 75 – 59 = 16 The common difference is 16. The first term is 43. To find the next term, add 16 to the previous term.
64. 80, –200, 500, –1250, ...
SOLUTION: Check for a common difference. –200 – 80 = –280 500 – (–200) = 700 There is no common difference. Check for a common ratio. –200 ÷ 80 = –2.5 500 ÷ (–200) = –2.5 –1250 ÷ 500 = –2.5 The common ratio is –2.5. The first term is 80. To find the next term, multiply the previous term by –2.5.
65. 100, 60, 36, 21.6, ...
SOLUTION: Check for a common difference. 60 – 100 = –40 36 – 60 = –24 There is no common difference. Check for a common ratio. 60 ÷ 100 = 0.6 36 ÷ 60 = 0.6 21.6 ÷ 36 = 0.6 The common ratio is 0.6. The first term is 100. To find the next term, multiply the previous term by 0.6.
66.
SOLUTION: Check for a common difference.
There is no common difference. Check for a common ratio.
4 ÷ 1 = 4 The common ratio is 4.
The first term is . To find the next term, multiply the previous term by 4.
67. TRAVEL In 2003, about 9.5 million people took cruises. Between 2003 and 2008, the number increased by about 740,000 each year. Write the point-slope form of an equation to find the total number of people y taking a cruise for any year x. Estimate the number of people who took a cruise in 2010.
SOLUTION: The statement “increased by 740,000 each year” is a constant rate, or the slope. Now substitute the slope and the point (2003, 9,500,000) into the point slope equation.
To find the number of people who took a cruise in 2010, substitute x = 2010.
So, approximately 14,680,000 people took a cruise in 2010.
Write an equation in function notation for each relation.
68.
SOLUTION: The y-intercept is 0. Find the slope by choosing two points. Choose (0, 0) and (1, 4).
So the slope intercept form is f (x) = 4x + 0 or f (x) = 4x.
69.
SOLUTION:
The y-intercept is 0. Find the slope by choosing two points. Choose (0, 0) and (2, −1).
So in slope intercept form is f (x) = −0.5x + 0 or f (x) = −0.5x.
Simplify.
70. b(b2)(b
3)
SOLUTION:
71. 2y(3y2)
SOLUTION:
72. −y4(−2y
3)
SOLUTION:
73. −3z3(−5z
4 + 2z)
SOLUTION:
74. 2m(−4m4) − 3(−5m
3)
SOLUTION:
75. 4p 2(−2p
3) + 2p
4(5p
6)
SOLUTION:
eSolutions Manual - Powered by Cognero Page 18
8-2 Multiplying a Polynomial by a Monomial
Find each product.
1. 5w( −3w2 + 2w − 4)
SOLUTION:
2. 6g2(3g
3 + 4g2 + 10g − 1)
SOLUTION:
3. 4km2(8km
2 + 2k2m + 5k)
SOLUTION:
4. −3p4r3(2p
2r4 − 6p 6
r3 − 5)
SOLUTION:
5. 2ab(7a4b
2 + a5b − 2a)
SOLUTION:
6. c2d
3(5cd
7 − 3c3d
2 − 4d3)
SOLUTION:
Simplify each expression.
7. t(4t2 + 15t + 4) − 4(3t − 1)
SOLUTION:
8. x(3x2 + 4) + 2(7x − 3)
SOLUTION:
9. −2d(d3c
2 − 4dc2 + 2d
2c) + c
2(dc
2 − 3d4)
SOLUTION:
10. −5w2(8w
2x − 11wx
2) + 6x(9wx
4 − 4w − 3x2)
SOLUTION:
11. GRIDDED RESPONSE Marlene is buying a new plasma television. The height of the screen of the television is one half the width plus 5 inches. The width is 30 inches. Find the height of the screen in inches.
SOLUTION: Let h be the height and let w be the width.
Let w = 30.
The height is 20 inches.
Solve each equation.
12. −6(11 − 2c) = 7(−2 − 2c)
SOLUTION:
13. t(2t + 3) + 20 = 2t(t − 3)
SOLUTION:
14. −2(w + 1) + w = 7 − 4w
SOLUTION:
15. 3(y − 2) + 2y = 4y + 14
SOLUTION:
16. a(a + 3) + a(a − 6) + 35 = a(a − 5) + a(a + 7)
SOLUTION:
17. n(n − 4) + n(n + 8) = n(n − 13) + n(n + 1) + 16
SOLUTION:
Find each product.
18. b(b2 − 12b + 1)
SOLUTION:
19. f (f2 + 2f + 25)
SOLUTION:
20. −3m3(2m
3 − 12m2 + 2m + 25)
SOLUTION:
21. 2j 2(5j
3 − 15j2 + 2j + 2)
SOLUTION:
22. 2pr2(2pr + 5p
2r − 15p )
SOLUTION:
23. 4t3u(2t
2u
2 − 10tu4 + 2)
SOLUTION:
Simplify each expression.
24. −3(5x2 + 2x + 9) + x(2x − 3)
SOLUTION:
25. a(−8a2 + 2a + 4) + 3(6a
2 − 4)
SOLUTION:
26. −4d(5d2 − 12) + 7(d + 5)
SOLUTION:
27. −9g(−2g + g2) + 3(g
2 + 4)
SOLUTION:
28. 2j (7j2k
2 + j k2 + 5k) − 9k(−2j
2k
2 + 2k2 + 3j )
SOLUTION:
29. 4n(2n3p
2 − 3np2 + 5n) + 4p (6n
2p − 2np
2 + 3p )
SOLUTION:
30. DAMS A new dam being built has the shape of a trapezoid. The base at the bottom of the dam is 2 times the
height. The base at the top of the dam is times the height minus 30 feet.
a. Write an expression to find the area of the trapezoidal cross section of the dam. b. If the height of the dam is 180 feet, find the area of this cross section.
SOLUTION:
a. The equation for the area of a trapezoid is .
b1
= 2h
b2 =
b. Let h = 180.
So, the area is 32,940 square feet.
Solve each equation.
31. 7(t2 + 5t − 9) + t = t(7t − 2) + 13
SOLUTION:
32. w(4w + 6) + 2w = 2(2w2 + 7w − 3)
SOLUTION:
33. 5(4z + 6) − 2(z − 4) = 7z(z + 4) − z(7z − 2) − 48
SOLUTION:
34. 9c(c − 11) + 10(5c − 3) = 3c(c + 5) + c(6c − 3) − 30
SOLUTION:
35. 2f (5f − 2) − 10(f2 − 3f + 6) = −8f (f + 4) + 4(2f
2 − 7f )
SOLUTION:
36. 2k(−3k + 4) + 6(k2 + 10) = k(4k + 8) − 2k(2k + 5)
SOLUTION:
Simplify each expression.
37. np2(30p
2 + 9n
2p − 12)
SOLUTION:
38. r2t(10r
3 + 5rt
3 + 15t
2)
SOLUTION:
39. −5q2w
3(4q + 7w) + 4qw
2(7q
2w + 2q) − 3qw(3q
2w
2 + 9)
SOLUTION:
40. −x2z(2z
2 + 4xz3) + xz
2(xz + 5x
3z) + x2
z3(3x
2z + 4xz)
SOLUTION:
41. PARKING A parking garage charges $30 per month plus $0.50 per daytime hour and $0.25 per hour during nights and weekends. Suppose Trent parks in the garage for 47 hours in January and h of those are night and weekend hours. a. Find an expression for Trent’s January bill. b. Find the cost if Trent had 12 hours of night and weekend hours.
SOLUTION: a. First, multiply the price per hour times the number of hours. Let h represent the number of weekend and night hours. For the weekend and night hours, he would spend 0.25h. The number of daytime hours is 47 – h, so he spent 0.50(47 - h) = 23.50 – 0.50h. To find the total cost, add these two values to the cost per month. 30 + (23.50 − 0.50h) + 0.25h = 53.50 − 0.25h. b. If Trent parked 12 night and weekend hours, h = 12
So, the cost for January is $50.50.
42. CCSS MODELING Che is building a dog house for his new puppy. The upper face of the dog house is a trapezoid. If the height of the trapezoid is 12 inches, find the area of the face of this piece of the dog house.
SOLUTION:
The formula for the area of a trapezoid is .
Now substitute h = 12 inches into the equation.
Therefore, the area of the trapezoid is 318 square inches.
43. TENNIS The tennis club is building a new tennis court with a path around it.
a. Write an expression for the area of the tennis court. b. Write an expression for the area of the path. c. What is the perimeter of the outside of the path?
SOLUTION: a. The area of the tennis court can be found by the equation . Substitute the values of the length and width and simplify.
b. The area of the path is the area of the large rectangle minus the area of the smaller rectangle, which is the area ofthe tennis court.
Area of the large rectangle = 2.5x(x + 6) or 2.5x2 + 15x.
Now subtract the area of the tennis court to find the area of just the path.
c. The perimeter of the outside of the path is the perimeter of the larger rectangle.
If x = 36 feet, then the perimeter of the outside of the path is 7(36) + 12 or 264 ft.
44. MULTIPLE REPRESENTATIONS In this problem, you will investigate the degree of the product of a monomialand a polynomial. a. TABULAR Write three monomials of different degrees and three polynomials of different degrees. Determine the degree of each monomial and polynomial. Multiply the monomials by the polynomials. Determine the degree of each product. Record your results in a table like the one below.
b. VERBAL Make a conjecture about the degree of the product of a monomial and a polynomial. What is the degree of the product of a monomial of degree a and a polynomial of degree b?
SOLUTION: a. There possible monomials are 2x, 3x
2, 4x
3. Three polynomials are x2 − 1, x5 + 1, x
6 + 1.
The degree for the monomial is the sum of the exponents. For the example given, case they are 1,2, and 3. The degree for the polynomial is the largest degree of the monomials. For the example given, they are 2, 5, and 6. Use the distributive property to find the product of the monomial and the polynomial. For the example give, they
are2x3 − 2x, 3x
7 + 3x
2, and 4x
9 + 4x. For the example given, the degree of the new polynomials are 3,7 and 9.
b. The degree of the product is the sum of the degree of the monomial and the degree of the polynomial, or a + b.
45. ERROR ANALYSIS Pearl and Ted both worked on this problem. Is either of them correct? Explain your reasoning.
SOLUTION:
Ted’s solution is correct. Pearl made a mistake when distributing on the middle term. She did not add the exponents correctly.
46. CCSS PERSEVERANCE Find p such that 3xp(4x
2p+3 + 2x3p−2
) = 12x12 + 6x
10.
SOLUTION: First simplify the left side of the equation.
Because the coefficients of the terms are the same, the corresponding terms’ exponents can be set equal to each other.
47. CHALLENGE Find 4x−3
y2(2x
5y−4 + 6x
−7y
6 − 4x0y
-2).
SOLUTION:
48. REASONING Is there a value for x that makes the statement (x + 2)2 = x2 + 22
true? If so, find a value for x. Explain your reasoning.
SOLUTION: Solve for x.
When 0 is substituted in for x in the equation, both sides are 22 or 4, which makes the equation true.
49. OPEN ENDED Write a monomial and a polynomial using n as the variable. Find their product.
SOLUTION: Choose a monomial with variable n and a polynomial with variable n. For example, let the monomial be 3n and the polynomial be 4n + 1. Find the product.
50. WRITING IN MATH Describe the steps to multiply a polynomial by a monomial.
SOLUTION: To multiply a polynomial by a monomial, use the Distributive Property. Multiply each term of the polynomial by the monomial. Then simplify by multiplying the coefficients together and using the Product of Powers Property for the variables. Consider the following example:
51. Every week a store sells j jeans and t T-shirts. The store makes $8 for each T-shirt and $12 for each pair of jeans. Which of the following expressions represents the total amount of money, in dollars, the store makes every week? A 8j + 12t B 12j + 8t C 20(j + t) D 96jt
SOLUTION: 12j is the amount made from jeans. 8t is the amount made from T-shirts. The sum, or 12j + 8t, represents the total amount the store makes every week. So, the correct choice is B.
52. If a = 5x + 7y and b = 2y − 3x, what is a + b?F 2x − 9y G 3y + 4x H 2x + 9y J 2x − 5y
SOLUTION:
So, the correct choice is H.
53. GEOMETRY A triangle has sides of length 5 inches and 8.5 inches. Which of the following cannot be the length of the third side? A 3.5 inches B 4 inches C 5.5 inches D 12 inches
SOLUTION: The sum of any two sides of a triangle must be greater than the length of the third side. Choice A: 3.5 + 5 = 8.5 8.5 Choice B: 3.5 + 5 = 8.5 > 8.5 Choice C: 5.5 + 5 = 10.5 > 8.5 Choice D: 12 + 5 = 17 > 8.5 Let x be the third side of the triangle.
3.5 cannot be the third side of the triangle. So, the correct choice is A.
54. SHORT RESPONSE Write an equation in which x varies directly as the cube of y and inversely as the square of z .
SOLUTION: If x varies directly as the cube of y , then both x and y must be in the numerators on either side of the equal sign.
When x varies inversely with the square of z. then the z must be in the denominator. Then . The
combination of the two is .
Find each sum or difference.
55. (2x2 − 7) + (8 − 5x
2)
SOLUTION:
56. (3z2 + 2z − 1) + (z
2 − 6)
SOLUTION:
57. (2a − 4a2 + 1) − (5a
2 − 2a − 6)
SOLUTION:
58. (a3 − 3a
2 + 4) − (4a
2 + 7)
SOLUTION:
59. (2ab − 3a + 4b) + (5a + 4ab)
SOLUTION:
60. (8c3 − 3c
2 + c − 2) − (3c
3 + 9)
SOLUTION:
Write a recursive formula for each sequence.61. 16, 2, –12, –26, ...
SOLUTION: Determine if there is a common difference. 2 – 16 = –14 –12 – 2 = –14 –26 – (–12) = –14 The common difference is –14. The first term is 16. To find the next term, subtract 14 from the previous term.
62. –5, 3, 11, 19, ...
SOLUTION: Determine if there is a common difference. 3 – (–5) = 8 11 – 3 = 8 19 – 11 = 8 The common difference is 8. The first term is 3. To find the next term, add 8 to the previous term.
63. 27, 43, 59, 75, ...
SOLUTION: Determine if there is a common difference. 43 – 27 = 16 59 – 43 = 16 75 – 59 = 16 The common difference is 16. The first term is 43. To find the next term, add 16 to the previous term.
64. 80, –200, 500, –1250, ...
SOLUTION: Check for a common difference. –200 – 80 = –280 500 – (–200) = 700 There is no common difference. Check for a common ratio. –200 ÷ 80 = –2.5 500 ÷ (–200) = –2.5 –1250 ÷ 500 = –2.5 The common ratio is –2.5. The first term is 80. To find the next term, multiply the previous term by –2.5.
65. 100, 60, 36, 21.6, ...
SOLUTION: Check for a common difference. 60 – 100 = –40 36 – 60 = –24 There is no common difference. Check for a common ratio. 60 ÷ 100 = 0.6 36 ÷ 60 = 0.6 21.6 ÷ 36 = 0.6 The common ratio is 0.6. The first term is 100. To find the next term, multiply the previous term by 0.6.
66.
SOLUTION: Check for a common difference.
There is no common difference. Check for a common ratio.
4 ÷ 1 = 4 The common ratio is 4.
The first term is . To find the next term, multiply the previous term by 4.
67. TRAVEL In 2003, about 9.5 million people took cruises. Between 2003 and 2008, the number increased by about 740,000 each year. Write the point-slope form of an equation to find the total number of people y taking a cruise for any year x. Estimate the number of people who took a cruise in 2010.
SOLUTION: The statement “increased by 740,000 each year” is a constant rate, or the slope. Now substitute the slope and the point (2003, 9,500,000) into the point slope equation.
To find the number of people who took a cruise in 2010, substitute x = 2010.
So, approximately 14,680,000 people took a cruise in 2010.
Write an equation in function notation for each relation.
68.
SOLUTION: The y-intercept is 0. Find the slope by choosing two points. Choose (0, 0) and (1, 4).
So the slope intercept form is f (x) = 4x + 0 or f (x) = 4x.
69.
SOLUTION:
The y-intercept is 0. Find the slope by choosing two points. Choose (0, 0) and (2, −1).
So in slope intercept form is f (x) = −0.5x + 0 or f (x) = −0.5x.
Simplify.
70. b(b2)(b
3)
SOLUTION:
71. 2y(3y2)
SOLUTION:
72. −y4(−2y
3)
SOLUTION:
73. −3z3(−5z
4 + 2z)
SOLUTION:
74. 2m(−4m4) − 3(−5m
3)
SOLUTION:
75. 4p 2(−2p
3) + 2p
4(5p
6)
SOLUTION:
eSolutions Manual - Powered by Cognero Page 19
8-2 Multiplying a Polynomial by a Monomial
Find each product.
1. 5w( −3w2 + 2w − 4)
SOLUTION:
2. 6g2(3g
3 + 4g2 + 10g − 1)
SOLUTION:
3. 4km2(8km
2 + 2k2m + 5k)
SOLUTION:
4. −3p4r3(2p
2r4 − 6p 6
r3 − 5)
SOLUTION:
5. 2ab(7a4b
2 + a5b − 2a)
SOLUTION:
6. c2d
3(5cd
7 − 3c3d
2 − 4d3)
SOLUTION:
Simplify each expression.
7. t(4t2 + 15t + 4) − 4(3t − 1)
SOLUTION:
8. x(3x2 + 4) + 2(7x − 3)
SOLUTION:
9. −2d(d3c
2 − 4dc2 + 2d
2c) + c
2(dc
2 − 3d4)
SOLUTION:
10. −5w2(8w
2x − 11wx
2) + 6x(9wx
4 − 4w − 3x2)
SOLUTION:
11. GRIDDED RESPONSE Marlene is buying a new plasma television. The height of the screen of the television is one half the width plus 5 inches. The width is 30 inches. Find the height of the screen in inches.
SOLUTION: Let h be the height and let w be the width.
Let w = 30.
The height is 20 inches.
Solve each equation.
12. −6(11 − 2c) = 7(−2 − 2c)
SOLUTION:
13. t(2t + 3) + 20 = 2t(t − 3)
SOLUTION:
14. −2(w + 1) + w = 7 − 4w
SOLUTION:
15. 3(y − 2) + 2y = 4y + 14
SOLUTION:
16. a(a + 3) + a(a − 6) + 35 = a(a − 5) + a(a + 7)
SOLUTION:
17. n(n − 4) + n(n + 8) = n(n − 13) + n(n + 1) + 16
SOLUTION:
Find each product.
18. b(b2 − 12b + 1)
SOLUTION:
19. f (f2 + 2f + 25)
SOLUTION:
20. −3m3(2m
3 − 12m2 + 2m + 25)
SOLUTION:
21. 2j 2(5j
3 − 15j2 + 2j + 2)
SOLUTION:
22. 2pr2(2pr + 5p
2r − 15p )
SOLUTION:
23. 4t3u(2t
2u
2 − 10tu4 + 2)
SOLUTION:
Simplify each expression.
24. −3(5x2 + 2x + 9) + x(2x − 3)
SOLUTION:
25. a(−8a2 + 2a + 4) + 3(6a
2 − 4)
SOLUTION:
26. −4d(5d2 − 12) + 7(d + 5)
SOLUTION:
27. −9g(−2g + g2) + 3(g
2 + 4)
SOLUTION:
28. 2j (7j2k
2 + j k2 + 5k) − 9k(−2j
2k
2 + 2k2 + 3j )
SOLUTION:
29. 4n(2n3p
2 − 3np2 + 5n) + 4p (6n
2p − 2np
2 + 3p )
SOLUTION:
30. DAMS A new dam being built has the shape of a trapezoid. The base at the bottom of the dam is 2 times the
height. The base at the top of the dam is times the height minus 30 feet.
a. Write an expression to find the area of the trapezoidal cross section of the dam. b. If the height of the dam is 180 feet, find the area of this cross section.
SOLUTION:
a. The equation for the area of a trapezoid is .
b1
= 2h
b2 =
b. Let h = 180.
So, the area is 32,940 square feet.
Solve each equation.
31. 7(t2 + 5t − 9) + t = t(7t − 2) + 13
SOLUTION:
32. w(4w + 6) + 2w = 2(2w2 + 7w − 3)
SOLUTION:
33. 5(4z + 6) − 2(z − 4) = 7z(z + 4) − z(7z − 2) − 48
SOLUTION:
34. 9c(c − 11) + 10(5c − 3) = 3c(c + 5) + c(6c − 3) − 30
SOLUTION:
35. 2f (5f − 2) − 10(f2 − 3f + 6) = −8f (f + 4) + 4(2f
2 − 7f )
SOLUTION:
36. 2k(−3k + 4) + 6(k2 + 10) = k(4k + 8) − 2k(2k + 5)
SOLUTION:
Simplify each expression.
37. np2(30p
2 + 9n
2p − 12)
SOLUTION:
38. r2t(10r
3 + 5rt
3 + 15t
2)
SOLUTION:
39. −5q2w
3(4q + 7w) + 4qw
2(7q
2w + 2q) − 3qw(3q
2w
2 + 9)
SOLUTION:
40. −x2z(2z
2 + 4xz3) + xz
2(xz + 5x
3z) + x2
z3(3x
2z + 4xz)
SOLUTION:
41. PARKING A parking garage charges $30 per month plus $0.50 per daytime hour and $0.25 per hour during nights and weekends. Suppose Trent parks in the garage for 47 hours in January and h of those are night and weekend hours. a. Find an expression for Trent’s January bill. b. Find the cost if Trent had 12 hours of night and weekend hours.
SOLUTION: a. First, multiply the price per hour times the number of hours. Let h represent the number of weekend and night hours. For the weekend and night hours, he would spend 0.25h. The number of daytime hours is 47 – h, so he spent 0.50(47 - h) = 23.50 – 0.50h. To find the total cost, add these two values to the cost per month. 30 + (23.50 − 0.50h) + 0.25h = 53.50 − 0.25h. b. If Trent parked 12 night and weekend hours, h = 12
So, the cost for January is $50.50.
42. CCSS MODELING Che is building a dog house for his new puppy. The upper face of the dog house is a trapezoid. If the height of the trapezoid is 12 inches, find the area of the face of this piece of the dog house.
SOLUTION:
The formula for the area of a trapezoid is .
Now substitute h = 12 inches into the equation.
Therefore, the area of the trapezoid is 318 square inches.
43. TENNIS The tennis club is building a new tennis court with a path around it.
a. Write an expression for the area of the tennis court. b. Write an expression for the area of the path. c. What is the perimeter of the outside of the path?
SOLUTION: a. The area of the tennis court can be found by the equation . Substitute the values of the length and width and simplify.
b. The area of the path is the area of the large rectangle minus the area of the smaller rectangle, which is the area ofthe tennis court.
Area of the large rectangle = 2.5x(x + 6) or 2.5x2 + 15x.
Now subtract the area of the tennis court to find the area of just the path.
c. The perimeter of the outside of the path is the perimeter of the larger rectangle.
If x = 36 feet, then the perimeter of the outside of the path is 7(36) + 12 or 264 ft.
44. MULTIPLE REPRESENTATIONS In this problem, you will investigate the degree of the product of a monomialand a polynomial. a. TABULAR Write three monomials of different degrees and three polynomials of different degrees. Determine the degree of each monomial and polynomial. Multiply the monomials by the polynomials. Determine the degree of each product. Record your results in a table like the one below.
b. VERBAL Make a conjecture about the degree of the product of a monomial and a polynomial. What is the degree of the product of a monomial of degree a and a polynomial of degree b?
SOLUTION: a. There possible monomials are 2x, 3x
2, 4x
3. Three polynomials are x2 − 1, x5 + 1, x
6 + 1.
The degree for the monomial is the sum of the exponents. For the example given, case they are 1,2, and 3. The degree for the polynomial is the largest degree of the monomials. For the example given, they are 2, 5, and 6. Use the distributive property to find the product of the monomial and the polynomial. For the example give, they
are2x3 − 2x, 3x
7 + 3x
2, and 4x
9 + 4x. For the example given, the degree of the new polynomials are 3,7 and 9.
b. The degree of the product is the sum of the degree of the monomial and the degree of the polynomial, or a + b.
45. ERROR ANALYSIS Pearl and Ted both worked on this problem. Is either of them correct? Explain your reasoning.
SOLUTION:
Ted’s solution is correct. Pearl made a mistake when distributing on the middle term. She did not add the exponents correctly.
46. CCSS PERSEVERANCE Find p such that 3xp(4x
2p+3 + 2x3p−2
) = 12x12 + 6x
10.
SOLUTION: First simplify the left side of the equation.
Because the coefficients of the terms are the same, the corresponding terms’ exponents can be set equal to each other.
47. CHALLENGE Find 4x−3
y2(2x
5y−4 + 6x
−7y
6 − 4x0y
-2).
SOLUTION:
48. REASONING Is there a value for x that makes the statement (x + 2)2 = x2 + 22
true? If so, find a value for x. Explain your reasoning.
SOLUTION: Solve for x.
When 0 is substituted in for x in the equation, both sides are 22 or 4, which makes the equation true.
49. OPEN ENDED Write a monomial and a polynomial using n as the variable. Find their product.
SOLUTION: Choose a monomial with variable n and a polynomial with variable n. For example, let the monomial be 3n and the polynomial be 4n + 1. Find the product.
50. WRITING IN MATH Describe the steps to multiply a polynomial by a monomial.
SOLUTION: To multiply a polynomial by a monomial, use the Distributive Property. Multiply each term of the polynomial by the monomial. Then simplify by multiplying the coefficients together and using the Product of Powers Property for the variables. Consider the following example:
51. Every week a store sells j jeans and t T-shirts. The store makes $8 for each T-shirt and $12 for each pair of jeans. Which of the following expressions represents the total amount of money, in dollars, the store makes every week? A 8j + 12t B 12j + 8t C 20(j + t) D 96jt
SOLUTION: 12j is the amount made from jeans. 8t is the amount made from T-shirts. The sum, or 12j + 8t, represents the total amount the store makes every week. So, the correct choice is B.
52. If a = 5x + 7y and b = 2y − 3x, what is a + b?F 2x − 9y G 3y + 4x H 2x + 9y J 2x − 5y
SOLUTION:
So, the correct choice is H.
53. GEOMETRY A triangle has sides of length 5 inches and 8.5 inches. Which of the following cannot be the length of the third side? A 3.5 inches B 4 inches C 5.5 inches D 12 inches
SOLUTION: The sum of any two sides of a triangle must be greater than the length of the third side. Choice A: 3.5 + 5 = 8.5 8.5 Choice B: 3.5 + 5 = 8.5 > 8.5 Choice C: 5.5 + 5 = 10.5 > 8.5 Choice D: 12 + 5 = 17 > 8.5 Let x be the third side of the triangle.
3.5 cannot be the third side of the triangle. So, the correct choice is A.
54. SHORT RESPONSE Write an equation in which x varies directly as the cube of y and inversely as the square of z .
SOLUTION: If x varies directly as the cube of y , then both x and y must be in the numerators on either side of the equal sign.
When x varies inversely with the square of z. then the z must be in the denominator. Then . The
combination of the two is .
Find each sum or difference.
55. (2x2 − 7) + (8 − 5x
2)
SOLUTION:
56. (3z2 + 2z − 1) + (z
2 − 6)
SOLUTION:
57. (2a − 4a2 + 1) − (5a
2 − 2a − 6)
SOLUTION:
58. (a3 − 3a
2 + 4) − (4a
2 + 7)
SOLUTION:
59. (2ab − 3a + 4b) + (5a + 4ab)
SOLUTION:
60. (8c3 − 3c
2 + c − 2) − (3c
3 + 9)
SOLUTION:
Write a recursive formula for each sequence.61. 16, 2, –12, –26, ...
SOLUTION: Determine if there is a common difference. 2 – 16 = –14 –12 – 2 = –14 –26 – (–12) = –14 The common difference is –14. The first term is 16. To find the next term, subtract 14 from the previous term.
62. –5, 3, 11, 19, ...
SOLUTION: Determine if there is a common difference. 3 – (–5) = 8 11 – 3 = 8 19 – 11 = 8 The common difference is 8. The first term is 3. To find the next term, add 8 to the previous term.
63. 27, 43, 59, 75, ...
SOLUTION: Determine if there is a common difference. 43 – 27 = 16 59 – 43 = 16 75 – 59 = 16 The common difference is 16. The first term is 43. To find the next term, add 16 to the previous term.
64. 80, –200, 500, –1250, ...
SOLUTION: Check for a common difference. –200 – 80 = –280 500 – (–200) = 700 There is no common difference. Check for a common ratio. –200 ÷ 80 = –2.5 500 ÷ (–200) = –2.5 –1250 ÷ 500 = –2.5 The common ratio is –2.5. The first term is 80. To find the next term, multiply the previous term by –2.5.
65. 100, 60, 36, 21.6, ...
SOLUTION: Check for a common difference. 60 – 100 = –40 36 – 60 = –24 There is no common difference. Check for a common ratio. 60 ÷ 100 = 0.6 36 ÷ 60 = 0.6 21.6 ÷ 36 = 0.6 The common ratio is 0.6. The first term is 100. To find the next term, multiply the previous term by 0.6.
66.
SOLUTION: Check for a common difference.
There is no common difference. Check for a common ratio.
4 ÷ 1 = 4 The common ratio is 4.
The first term is . To find the next term, multiply the previous term by 4.
67. TRAVEL In 2003, about 9.5 million people took cruises. Between 2003 and 2008, the number increased by about 740,000 each year. Write the point-slope form of an equation to find the total number of people y taking a cruise for any year x. Estimate the number of people who took a cruise in 2010.
SOLUTION: The statement “increased by 740,000 each year” is a constant rate, or the slope. Now substitute the slope and the point (2003, 9,500,000) into the point slope equation.
To find the number of people who took a cruise in 2010, substitute x = 2010.
So, approximately 14,680,000 people took a cruise in 2010.
Write an equation in function notation for each relation.
68.
SOLUTION: The y-intercept is 0. Find the slope by choosing two points. Choose (0, 0) and (1, 4).
So the slope intercept form is f (x) = 4x + 0 or f (x) = 4x.
69.
SOLUTION:
The y-intercept is 0. Find the slope by choosing two points. Choose (0, 0) and (2, −1).
So in slope intercept form is f (x) = −0.5x + 0 or f (x) = −0.5x.
Simplify.
70. b(b2)(b
3)
SOLUTION:
71. 2y(3y2)
SOLUTION:
72. −y4(−2y
3)
SOLUTION:
73. −3z3(−5z
4 + 2z)
SOLUTION:
74. 2m(−4m4) − 3(−5m
3)
SOLUTION:
75. 4p 2(−2p
3) + 2p
4(5p
6)
SOLUTION:
eSolutions Manual - Powered by Cognero Page 20
8-2 Multiplying a Polynomial by a Monomial
Find each product.
1. 5w( −3w2 + 2w − 4)
SOLUTION:
2. 6g2(3g
3 + 4g2 + 10g − 1)
SOLUTION:
3. 4km2(8km
2 + 2k2m + 5k)
SOLUTION:
4. −3p4r3(2p
2r4 − 6p 6
r3 − 5)
SOLUTION:
5. 2ab(7a4b
2 + a5b − 2a)
SOLUTION:
6. c2d
3(5cd
7 − 3c3d
2 − 4d3)
SOLUTION:
Simplify each expression.
7. t(4t2 + 15t + 4) − 4(3t − 1)
SOLUTION:
8. x(3x2 + 4) + 2(7x − 3)
SOLUTION:
9. −2d(d3c
2 − 4dc2 + 2d
2c) + c
2(dc
2 − 3d4)
SOLUTION:
10. −5w2(8w
2x − 11wx
2) + 6x(9wx
4 − 4w − 3x2)
SOLUTION:
11. GRIDDED RESPONSE Marlene is buying a new plasma television. The height of the screen of the television is one half the width plus 5 inches. The width is 30 inches. Find the height of the screen in inches.
SOLUTION: Let h be the height and let w be the width.
Let w = 30.
The height is 20 inches.
Solve each equation.
12. −6(11 − 2c) = 7(−2 − 2c)
SOLUTION:
13. t(2t + 3) + 20 = 2t(t − 3)
SOLUTION:
14. −2(w + 1) + w = 7 − 4w
SOLUTION:
15. 3(y − 2) + 2y = 4y + 14
SOLUTION:
16. a(a + 3) + a(a − 6) + 35 = a(a − 5) + a(a + 7)
SOLUTION:
17. n(n − 4) + n(n + 8) = n(n − 13) + n(n + 1) + 16
SOLUTION:
Find each product.
18. b(b2 − 12b + 1)
SOLUTION:
19. f (f2 + 2f + 25)
SOLUTION:
20. −3m3(2m
3 − 12m2 + 2m + 25)
SOLUTION:
21. 2j 2(5j
3 − 15j2 + 2j + 2)
SOLUTION:
22. 2pr2(2pr + 5p
2r − 15p )
SOLUTION:
23. 4t3u(2t
2u
2 − 10tu4 + 2)
SOLUTION:
Simplify each expression.
24. −3(5x2 + 2x + 9) + x(2x − 3)
SOLUTION:
25. a(−8a2 + 2a + 4) + 3(6a
2 − 4)
SOLUTION:
26. −4d(5d2 − 12) + 7(d + 5)
SOLUTION:
27. −9g(−2g + g2) + 3(g
2 + 4)
SOLUTION:
28. 2j (7j2k
2 + j k2 + 5k) − 9k(−2j
2k
2 + 2k2 + 3j )
SOLUTION:
29. 4n(2n3p
2 − 3np2 + 5n) + 4p (6n
2p − 2np
2 + 3p )
SOLUTION:
30. DAMS A new dam being built has the shape of a trapezoid. The base at the bottom of the dam is 2 times the
height. The base at the top of the dam is times the height minus 30 feet.
a. Write an expression to find the area of the trapezoidal cross section of the dam. b. If the height of the dam is 180 feet, find the area of this cross section.
SOLUTION:
a. The equation for the area of a trapezoid is .
b1
= 2h
b2 =
b. Let h = 180.
So, the area is 32,940 square feet.
Solve each equation.
31. 7(t2 + 5t − 9) + t = t(7t − 2) + 13
SOLUTION:
32. w(4w + 6) + 2w = 2(2w2 + 7w − 3)
SOLUTION:
33. 5(4z + 6) − 2(z − 4) = 7z(z + 4) − z(7z − 2) − 48
SOLUTION:
34. 9c(c − 11) + 10(5c − 3) = 3c(c + 5) + c(6c − 3) − 30
SOLUTION:
35. 2f (5f − 2) − 10(f2 − 3f + 6) = −8f (f + 4) + 4(2f
2 − 7f )
SOLUTION:
36. 2k(−3k + 4) + 6(k2 + 10) = k(4k + 8) − 2k(2k + 5)
SOLUTION:
Simplify each expression.
37. np2(30p
2 + 9n
2p − 12)
SOLUTION:
38. r2t(10r
3 + 5rt
3 + 15t
2)
SOLUTION:
39. −5q2w
3(4q + 7w) + 4qw
2(7q
2w + 2q) − 3qw(3q
2w
2 + 9)
SOLUTION:
40. −x2z(2z
2 + 4xz3) + xz
2(xz + 5x
3z) + x2
z3(3x
2z + 4xz)
SOLUTION:
41. PARKING A parking garage charges $30 per month plus $0.50 per daytime hour and $0.25 per hour during nights and weekends. Suppose Trent parks in the garage for 47 hours in January and h of those are night and weekend hours. a. Find an expression for Trent’s January bill. b. Find the cost if Trent had 12 hours of night and weekend hours.
SOLUTION: a. First, multiply the price per hour times the number of hours. Let h represent the number of weekend and night hours. For the weekend and night hours, he would spend 0.25h. The number of daytime hours is 47 – h, so he spent 0.50(47 - h) = 23.50 – 0.50h. To find the total cost, add these two values to the cost per month. 30 + (23.50 − 0.50h) + 0.25h = 53.50 − 0.25h. b. If Trent parked 12 night and weekend hours, h = 12
So, the cost for January is $50.50.
42. CCSS MODELING Che is building a dog house for his new puppy. The upper face of the dog house is a trapezoid. If the height of the trapezoid is 12 inches, find the area of the face of this piece of the dog house.
SOLUTION:
The formula for the area of a trapezoid is .
Now substitute h = 12 inches into the equation.
Therefore, the area of the trapezoid is 318 square inches.
43. TENNIS The tennis club is building a new tennis court with a path around it.
a. Write an expression for the area of the tennis court. b. Write an expression for the area of the path. c. What is the perimeter of the outside of the path?
SOLUTION: a. The area of the tennis court can be found by the equation . Substitute the values of the length and width and simplify.
b. The area of the path is the area of the large rectangle minus the area of the smaller rectangle, which is the area ofthe tennis court.
Area of the large rectangle = 2.5x(x + 6) or 2.5x2 + 15x.
Now subtract the area of the tennis court to find the area of just the path.
c. The perimeter of the outside of the path is the perimeter of the larger rectangle.
If x = 36 feet, then the perimeter of the outside of the path is 7(36) + 12 or 264 ft.
44. MULTIPLE REPRESENTATIONS In this problem, you will investigate the degree of the product of a monomialand a polynomial. a. TABULAR Write three monomials of different degrees and three polynomials of different degrees. Determine the degree of each monomial and polynomial. Multiply the monomials by the polynomials. Determine the degree of each product. Record your results in a table like the one below.
b. VERBAL Make a conjecture about the degree of the product of a monomial and a polynomial. What is the degree of the product of a monomial of degree a and a polynomial of degree b?
SOLUTION: a. There possible monomials are 2x, 3x
2, 4x
3. Three polynomials are x2 − 1, x5 + 1, x
6 + 1.
The degree for the monomial is the sum of the exponents. For the example given, case they are 1,2, and 3. The degree for the polynomial is the largest degree of the monomials. For the example given, they are 2, 5, and 6. Use the distributive property to find the product of the monomial and the polynomial. For the example give, they
are2x3 − 2x, 3x
7 + 3x
2, and 4x
9 + 4x. For the example given, the degree of the new polynomials are 3,7 and 9.
b. The degree of the product is the sum of the degree of the monomial and the degree of the polynomial, or a + b.
45. ERROR ANALYSIS Pearl and Ted both worked on this problem. Is either of them correct? Explain your reasoning.
SOLUTION:
Ted’s solution is correct. Pearl made a mistake when distributing on the middle term. She did not add the exponents correctly.
46. CCSS PERSEVERANCE Find p such that 3xp(4x
2p+3 + 2x3p−2
) = 12x12 + 6x
10.
SOLUTION: First simplify the left side of the equation.
Because the coefficients of the terms are the same, the corresponding terms’ exponents can be set equal to each other.
47. CHALLENGE Find 4x−3
y2(2x
5y−4 + 6x
−7y
6 − 4x0y
-2).
SOLUTION:
48. REASONING Is there a value for x that makes the statement (x + 2)2 = x2 + 22
true? If so, find a value for x. Explain your reasoning.
SOLUTION: Solve for x.
When 0 is substituted in for x in the equation, both sides are 22 or 4, which makes the equation true.
49. OPEN ENDED Write a monomial and a polynomial using n as the variable. Find their product.
SOLUTION: Choose a monomial with variable n and a polynomial with variable n. For example, let the monomial be 3n and the polynomial be 4n + 1. Find the product.
50. WRITING IN MATH Describe the steps to multiply a polynomial by a monomial.
SOLUTION: To multiply a polynomial by a monomial, use the Distributive Property. Multiply each term of the polynomial by the monomial. Then simplify by multiplying the coefficients together and using the Product of Powers Property for the variables. Consider the following example:
51. Every week a store sells j jeans and t T-shirts. The store makes $8 for each T-shirt and $12 for each pair of jeans. Which of the following expressions represents the total amount of money, in dollars, the store makes every week? A 8j + 12t B 12j + 8t C 20(j + t) D 96jt
SOLUTION: 12j is the amount made from jeans. 8t is the amount made from T-shirts. The sum, or 12j + 8t, represents the total amount the store makes every week. So, the correct choice is B.
52. If a = 5x + 7y and b = 2y − 3x, what is a + b?F 2x − 9y G 3y + 4x H 2x + 9y J 2x − 5y
SOLUTION:
So, the correct choice is H.
53. GEOMETRY A triangle has sides of length 5 inches and 8.5 inches. Which of the following cannot be the length of the third side? A 3.5 inches B 4 inches C 5.5 inches D 12 inches
SOLUTION: The sum of any two sides of a triangle must be greater than the length of the third side. Choice A: 3.5 + 5 = 8.5 8.5 Choice B: 3.5 + 5 = 8.5 > 8.5 Choice C: 5.5 + 5 = 10.5 > 8.5 Choice D: 12 + 5 = 17 > 8.5 Let x be the third side of the triangle.
3.5 cannot be the third side of the triangle. So, the correct choice is A.
54. SHORT RESPONSE Write an equation in which x varies directly as the cube of y and inversely as the square of z .
SOLUTION: If x varies directly as the cube of y , then both x and y must be in the numerators on either side of the equal sign.
When x varies inversely with the square of z. then the z must be in the denominator. Then . The
combination of the two is .
Find each sum or difference.
55. (2x2 − 7) + (8 − 5x
2)
SOLUTION:
56. (3z2 + 2z − 1) + (z
2 − 6)
SOLUTION:
57. (2a − 4a2 + 1) − (5a
2 − 2a − 6)
SOLUTION:
58. (a3 − 3a
2 + 4) − (4a
2 + 7)
SOLUTION:
59. (2ab − 3a + 4b) + (5a + 4ab)
SOLUTION:
60. (8c3 − 3c
2 + c − 2) − (3c
3 + 9)
SOLUTION:
Write a recursive formula for each sequence.61. 16, 2, –12, –26, ...
SOLUTION: Determine if there is a common difference. 2 – 16 = –14 –12 – 2 = –14 –26 – (–12) = –14 The common difference is –14. The first term is 16. To find the next term, subtract 14 from the previous term.
62. –5, 3, 11, 19, ...
SOLUTION: Determine if there is a common difference. 3 – (–5) = 8 11 – 3 = 8 19 – 11 = 8 The common difference is 8. The first term is 3. To find the next term, add 8 to the previous term.
63. 27, 43, 59, 75, ...
SOLUTION: Determine if there is a common difference. 43 – 27 = 16 59 – 43 = 16 75 – 59 = 16 The common difference is 16. The first term is 43. To find the next term, add 16 to the previous term.
64. 80, –200, 500, –1250, ...
SOLUTION: Check for a common difference. –200 – 80 = –280 500 – (–200) = 700 There is no common difference. Check for a common ratio. –200 ÷ 80 = –2.5 500 ÷ (–200) = –2.5 –1250 ÷ 500 = –2.5 The common ratio is –2.5. The first term is 80. To find the next term, multiply the previous term by –2.5.
65. 100, 60, 36, 21.6, ...
SOLUTION: Check for a common difference. 60 – 100 = –40 36 – 60 = –24 There is no common difference. Check for a common ratio. 60 ÷ 100 = 0.6 36 ÷ 60 = 0.6 21.6 ÷ 36 = 0.6 The common ratio is 0.6. The first term is 100. To find the next term, multiply the previous term by 0.6.
66.
SOLUTION: Check for a common difference.
There is no common difference. Check for a common ratio.
4 ÷ 1 = 4 The common ratio is 4.
The first term is . To find the next term, multiply the previous term by 4.
67. TRAVEL In 2003, about 9.5 million people took cruises. Between 2003 and 2008, the number increased by about 740,000 each year. Write the point-slope form of an equation to find the total number of people y taking a cruise for any year x. Estimate the number of people who took a cruise in 2010.
SOLUTION: The statement “increased by 740,000 each year” is a constant rate, or the slope. Now substitute the slope and the point (2003, 9,500,000) into the point slope equation.
To find the number of people who took a cruise in 2010, substitute x = 2010.
So, approximately 14,680,000 people took a cruise in 2010.
Write an equation in function notation for each relation.
68.
SOLUTION: The y-intercept is 0. Find the slope by choosing two points. Choose (0, 0) and (1, 4).
So the slope intercept form is f (x) = 4x + 0 or f (x) = 4x.
69.
SOLUTION:
The y-intercept is 0. Find the slope by choosing two points. Choose (0, 0) and (2, −1).
So in slope intercept form is f (x) = −0.5x + 0 or f (x) = −0.5x.
Simplify.
70. b(b2)(b
3)
SOLUTION:
71. 2y(3y2)
SOLUTION:
72. −y4(−2y
3)
SOLUTION:
73. −3z3(−5z
4 + 2z)
SOLUTION:
74. 2m(−4m4) − 3(−5m
3)
SOLUTION:
75. 4p 2(−2p
3) + 2p
4(5p
6)
SOLUTION:
eSolutions Manual - Powered by Cognero Page 21
8-2 Multiplying a Polynomial by a Monomial
Find each product.
1. 5w( −3w2 + 2w − 4)
SOLUTION:
2. 6g2(3g
3 + 4g2 + 10g − 1)
SOLUTION:
3. 4km2(8km
2 + 2k2m + 5k)
SOLUTION:
4. −3p4r3(2p
2r4 − 6p 6
r3 − 5)
SOLUTION:
5. 2ab(7a4b
2 + a5b − 2a)
SOLUTION:
6. c2d
3(5cd
7 − 3c3d
2 − 4d3)
SOLUTION:
Simplify each expression.
7. t(4t2 + 15t + 4) − 4(3t − 1)
SOLUTION:
8. x(3x2 + 4) + 2(7x − 3)
SOLUTION:
9. −2d(d3c
2 − 4dc2 + 2d
2c) + c
2(dc
2 − 3d4)
SOLUTION:
10. −5w2(8w
2x − 11wx
2) + 6x(9wx
4 − 4w − 3x2)
SOLUTION:
11. GRIDDED RESPONSE Marlene is buying a new plasma television. The height of the screen of the television is one half the width plus 5 inches. The width is 30 inches. Find the height of the screen in inches.
SOLUTION: Let h be the height and let w be the width.
Let w = 30.
The height is 20 inches.
Solve each equation.
12. −6(11 − 2c) = 7(−2 − 2c)
SOLUTION:
13. t(2t + 3) + 20 = 2t(t − 3)
SOLUTION:
14. −2(w + 1) + w = 7 − 4w
SOLUTION:
15. 3(y − 2) + 2y = 4y + 14
SOLUTION:
16. a(a + 3) + a(a − 6) + 35 = a(a − 5) + a(a + 7)
SOLUTION:
17. n(n − 4) + n(n + 8) = n(n − 13) + n(n + 1) + 16
SOLUTION:
Find each product.
18. b(b2 − 12b + 1)
SOLUTION:
19. f (f2 + 2f + 25)
SOLUTION:
20. −3m3(2m
3 − 12m2 + 2m + 25)
SOLUTION:
21. 2j 2(5j
3 − 15j2 + 2j + 2)
SOLUTION:
22. 2pr2(2pr + 5p
2r − 15p )
SOLUTION:
23. 4t3u(2t
2u
2 − 10tu4 + 2)
SOLUTION:
Simplify each expression.
24. −3(5x2 + 2x + 9) + x(2x − 3)
SOLUTION:
25. a(−8a2 + 2a + 4) + 3(6a
2 − 4)
SOLUTION:
26. −4d(5d2 − 12) + 7(d + 5)
SOLUTION:
27. −9g(−2g + g2) + 3(g
2 + 4)
SOLUTION:
28. 2j (7j2k
2 + j k2 + 5k) − 9k(−2j
2k
2 + 2k2 + 3j )
SOLUTION:
29. 4n(2n3p
2 − 3np2 + 5n) + 4p (6n
2p − 2np
2 + 3p )
SOLUTION:
30. DAMS A new dam being built has the shape of a trapezoid. The base at the bottom of the dam is 2 times the
height. The base at the top of the dam is times the height minus 30 feet.
a. Write an expression to find the area of the trapezoidal cross section of the dam. b. If the height of the dam is 180 feet, find the area of this cross section.
SOLUTION:
a. The equation for the area of a trapezoid is .
b1
= 2h
b2 =
b. Let h = 180.
So, the area is 32,940 square feet.
Solve each equation.
31. 7(t2 + 5t − 9) + t = t(7t − 2) + 13
SOLUTION:
32. w(4w + 6) + 2w = 2(2w2 + 7w − 3)
SOLUTION:
33. 5(4z + 6) − 2(z − 4) = 7z(z + 4) − z(7z − 2) − 48
SOLUTION:
34. 9c(c − 11) + 10(5c − 3) = 3c(c + 5) + c(6c − 3) − 30
SOLUTION:
35. 2f (5f − 2) − 10(f2 − 3f + 6) = −8f (f + 4) + 4(2f
2 − 7f )
SOLUTION:
36. 2k(−3k + 4) + 6(k2 + 10) = k(4k + 8) − 2k(2k + 5)
SOLUTION:
Simplify each expression.
37. np2(30p
2 + 9n
2p − 12)
SOLUTION:
38. r2t(10r
3 + 5rt
3 + 15t
2)
SOLUTION:
39. −5q2w
3(4q + 7w) + 4qw
2(7q
2w + 2q) − 3qw(3q
2w
2 + 9)
SOLUTION:
40. −x2z(2z
2 + 4xz3) + xz
2(xz + 5x
3z) + x2
z3(3x
2z + 4xz)
SOLUTION:
41. PARKING A parking garage charges $30 per month plus $0.50 per daytime hour and $0.25 per hour during nights and weekends. Suppose Trent parks in the garage for 47 hours in January and h of those are night and weekend hours. a. Find an expression for Trent’s January bill. b. Find the cost if Trent had 12 hours of night and weekend hours.
SOLUTION: a. First, multiply the price per hour times the number of hours. Let h represent the number of weekend and night hours. For the weekend and night hours, he would spend 0.25h. The number of daytime hours is 47 – h, so he spent 0.50(47 - h) = 23.50 – 0.50h. To find the total cost, add these two values to the cost per month. 30 + (23.50 − 0.50h) + 0.25h = 53.50 − 0.25h. b. If Trent parked 12 night and weekend hours, h = 12
So, the cost for January is $50.50.
42. CCSS MODELING Che is building a dog house for his new puppy. The upper face of the dog house is a trapezoid. If the height of the trapezoid is 12 inches, find the area of the face of this piece of the dog house.
SOLUTION:
The formula for the area of a trapezoid is .
Now substitute h = 12 inches into the equation.
Therefore, the area of the trapezoid is 318 square inches.
43. TENNIS The tennis club is building a new tennis court with a path around it.
a. Write an expression for the area of the tennis court. b. Write an expression for the area of the path. c. What is the perimeter of the outside of the path?
SOLUTION: a. The area of the tennis court can be found by the equation . Substitute the values of the length and width and simplify.
b. The area of the path is the area of the large rectangle minus the area of the smaller rectangle, which is the area ofthe tennis court.
Area of the large rectangle = 2.5x(x + 6) or 2.5x2 + 15x.
Now subtract the area of the tennis court to find the area of just the path.
c. The perimeter of the outside of the path is the perimeter of the larger rectangle.
If x = 36 feet, then the perimeter of the outside of the path is 7(36) + 12 or 264 ft.
44. MULTIPLE REPRESENTATIONS In this problem, you will investigate the degree of the product of a monomialand a polynomial. a. TABULAR Write three monomials of different degrees and three polynomials of different degrees. Determine the degree of each monomial and polynomial. Multiply the monomials by the polynomials. Determine the degree of each product. Record your results in a table like the one below.
b. VERBAL Make a conjecture about the degree of the product of a monomial and a polynomial. What is the degree of the product of a monomial of degree a and a polynomial of degree b?
SOLUTION: a. There possible monomials are 2x, 3x
2, 4x
3. Three polynomials are x2 − 1, x5 + 1, x
6 + 1.
The degree for the monomial is the sum of the exponents. For the example given, case they are 1,2, and 3. The degree for the polynomial is the largest degree of the monomials. For the example given, they are 2, 5, and 6. Use the distributive property to find the product of the monomial and the polynomial. For the example give, they
are2x3 − 2x, 3x
7 + 3x
2, and 4x
9 + 4x. For the example given, the degree of the new polynomials are 3,7 and 9.
b. The degree of the product is the sum of the degree of the monomial and the degree of the polynomial, or a + b.
45. ERROR ANALYSIS Pearl and Ted both worked on this problem. Is either of them correct? Explain your reasoning.
SOLUTION:
Ted’s solution is correct. Pearl made a mistake when distributing on the middle term. She did not add the exponents correctly.
46. CCSS PERSEVERANCE Find p such that 3xp(4x
2p+3 + 2x3p−2
) = 12x12 + 6x
10.
SOLUTION: First simplify the left side of the equation.
Because the coefficients of the terms are the same, the corresponding terms’ exponents can be set equal to each other.
47. CHALLENGE Find 4x−3
y2(2x
5y−4 + 6x
−7y
6 − 4x0y
-2).
SOLUTION:
48. REASONING Is there a value for x that makes the statement (x + 2)2 = x2 + 22
true? If so, find a value for x. Explain your reasoning.
SOLUTION: Solve for x.
When 0 is substituted in for x in the equation, both sides are 22 or 4, which makes the equation true.
49. OPEN ENDED Write a monomial and a polynomial using n as the variable. Find their product.
SOLUTION: Choose a monomial with variable n and a polynomial with variable n. For example, let the monomial be 3n and the polynomial be 4n + 1. Find the product.
50. WRITING IN MATH Describe the steps to multiply a polynomial by a monomial.
SOLUTION: To multiply a polynomial by a monomial, use the Distributive Property. Multiply each term of the polynomial by the monomial. Then simplify by multiplying the coefficients together and using the Product of Powers Property for the variables. Consider the following example:
51. Every week a store sells j jeans and t T-shirts. The store makes $8 for each T-shirt and $12 for each pair of jeans. Which of the following expressions represents the total amount of money, in dollars, the store makes every week? A 8j + 12t B 12j + 8t C 20(j + t) D 96jt
SOLUTION: 12j is the amount made from jeans. 8t is the amount made from T-shirts. The sum, or 12j + 8t, represents the total amount the store makes every week. So, the correct choice is B.
52. If a = 5x + 7y and b = 2y − 3x, what is a + b?F 2x − 9y G 3y + 4x H 2x + 9y J 2x − 5y
SOLUTION:
So, the correct choice is H.
53. GEOMETRY A triangle has sides of length 5 inches and 8.5 inches. Which of the following cannot be the length of the third side? A 3.5 inches B 4 inches C 5.5 inches D 12 inches
SOLUTION: The sum of any two sides of a triangle must be greater than the length of the third side. Choice A: 3.5 + 5 = 8.5 8.5 Choice B: 3.5 + 5 = 8.5 > 8.5 Choice C: 5.5 + 5 = 10.5 > 8.5 Choice D: 12 + 5 = 17 > 8.5 Let x be the third side of the triangle.
3.5 cannot be the third side of the triangle. So, the correct choice is A.
54. SHORT RESPONSE Write an equation in which x varies directly as the cube of y and inversely as the square of z .
SOLUTION: If x varies directly as the cube of y , then both x and y must be in the numerators on either side of the equal sign.
When x varies inversely with the square of z. then the z must be in the denominator. Then . The
combination of the two is .
Find each sum or difference.
55. (2x2 − 7) + (8 − 5x
2)
SOLUTION:
56. (3z2 + 2z − 1) + (z
2 − 6)
SOLUTION:
57. (2a − 4a2 + 1) − (5a
2 − 2a − 6)
SOLUTION:
58. (a3 − 3a
2 + 4) − (4a
2 + 7)
SOLUTION:
59. (2ab − 3a + 4b) + (5a + 4ab)
SOLUTION:
60. (8c3 − 3c
2 + c − 2) − (3c
3 + 9)
SOLUTION:
Write a recursive formula for each sequence.61. 16, 2, –12, –26, ...
SOLUTION: Determine if there is a common difference. 2 – 16 = –14 –12 – 2 = –14 –26 – (–12) = –14 The common difference is –14. The first term is 16. To find the next term, subtract 14 from the previous term.
62. –5, 3, 11, 19, ...
SOLUTION: Determine if there is a common difference. 3 – (–5) = 8 11 – 3 = 8 19 – 11 = 8 The common difference is 8. The first term is 3. To find the next term, add 8 to the previous term.
63. 27, 43, 59, 75, ...
SOLUTION: Determine if there is a common difference. 43 – 27 = 16 59 – 43 = 16 75 – 59 = 16 The common difference is 16. The first term is 43. To find the next term, add 16 to the previous term.
64. 80, –200, 500, –1250, ...
SOLUTION: Check for a common difference. –200 – 80 = –280 500 – (–200) = 700 There is no common difference. Check for a common ratio. –200 ÷ 80 = –2.5 500 ÷ (–200) = –2.5 –1250 ÷ 500 = –2.5 The common ratio is –2.5. The first term is 80. To find the next term, multiply the previous term by –2.5.
65. 100, 60, 36, 21.6, ...
SOLUTION: Check for a common difference. 60 – 100 = –40 36 – 60 = –24 There is no common difference. Check for a common ratio. 60 ÷ 100 = 0.6 36 ÷ 60 = 0.6 21.6 ÷ 36 = 0.6 The common ratio is 0.6. The first term is 100. To find the next term, multiply the previous term by 0.6.
66.
SOLUTION: Check for a common difference.
There is no common difference. Check for a common ratio.
4 ÷ 1 = 4 The common ratio is 4.
The first term is . To find the next term, multiply the previous term by 4.
67. TRAVEL In 2003, about 9.5 million people took cruises. Between 2003 and 2008, the number increased by about 740,000 each year. Write the point-slope form of an equation to find the total number of people y taking a cruise for any year x. Estimate the number of people who took a cruise in 2010.
SOLUTION: The statement “increased by 740,000 each year” is a constant rate, or the slope. Now substitute the slope and the point (2003, 9,500,000) into the point slope equation.
To find the number of people who took a cruise in 2010, substitute x = 2010.
So, approximately 14,680,000 people took a cruise in 2010.
Write an equation in function notation for each relation.
68.
SOLUTION: The y-intercept is 0. Find the slope by choosing two points. Choose (0, 0) and (1, 4).
So the slope intercept form is f (x) = 4x + 0 or f (x) = 4x.
69.
SOLUTION:
The y-intercept is 0. Find the slope by choosing two points. Choose (0, 0) and (2, −1).
So in slope intercept form is f (x) = −0.5x + 0 or f (x) = −0.5x.
Simplify.
70. b(b2)(b
3)
SOLUTION:
71. 2y(3y2)
SOLUTION:
72. −y4(−2y
3)
SOLUTION:
73. −3z3(−5z
4 + 2z)
SOLUTION:
74. 2m(−4m4) − 3(−5m
3)
SOLUTION:
75. 4p 2(−2p
3) + 2p
4(5p
6)
SOLUTION:
eSolutions Manual - Powered by Cognero Page 22
8-2 Multiplying a Polynomial by a Monomial
Find each product.
1. 5w( −3w2 + 2w − 4)
SOLUTION:
2. 6g2(3g
3 + 4g2 + 10g − 1)
SOLUTION:
3. 4km2(8km
2 + 2k2m + 5k)
SOLUTION:
4. −3p4r3(2p
2r4 − 6p 6
r3 − 5)
SOLUTION:
5. 2ab(7a4b
2 + a5b − 2a)
SOLUTION:
6. c2d
3(5cd
7 − 3c3d
2 − 4d3)
SOLUTION:
Simplify each expression.
7. t(4t2 + 15t + 4) − 4(3t − 1)
SOLUTION:
8. x(3x2 + 4) + 2(7x − 3)
SOLUTION:
9. −2d(d3c
2 − 4dc2 + 2d
2c) + c
2(dc
2 − 3d4)
SOLUTION:
10. −5w2(8w
2x − 11wx
2) + 6x(9wx
4 − 4w − 3x2)
SOLUTION:
11. GRIDDED RESPONSE Marlene is buying a new plasma television. The height of the screen of the television is one half the width plus 5 inches. The width is 30 inches. Find the height of the screen in inches.
SOLUTION: Let h be the height and let w be the width.
Let w = 30.
The height is 20 inches.
Solve each equation.
12. −6(11 − 2c) = 7(−2 − 2c)
SOLUTION:
13. t(2t + 3) + 20 = 2t(t − 3)
SOLUTION:
14. −2(w + 1) + w = 7 − 4w
SOLUTION:
15. 3(y − 2) + 2y = 4y + 14
SOLUTION:
16. a(a + 3) + a(a − 6) + 35 = a(a − 5) + a(a + 7)
SOLUTION:
17. n(n − 4) + n(n + 8) = n(n − 13) + n(n + 1) + 16
SOLUTION:
Find each product.
18. b(b2 − 12b + 1)
SOLUTION:
19. f (f2 + 2f + 25)
SOLUTION:
20. −3m3(2m
3 − 12m2 + 2m + 25)
SOLUTION:
21. 2j 2(5j
3 − 15j2 + 2j + 2)
SOLUTION:
22. 2pr2(2pr + 5p
2r − 15p )
SOLUTION:
23. 4t3u(2t
2u
2 − 10tu4 + 2)
SOLUTION:
Simplify each expression.
24. −3(5x2 + 2x + 9) + x(2x − 3)
SOLUTION:
25. a(−8a2 + 2a + 4) + 3(6a
2 − 4)
SOLUTION:
26. −4d(5d2 − 12) + 7(d + 5)
SOLUTION:
27. −9g(−2g + g2) + 3(g
2 + 4)
SOLUTION:
28. 2j (7j2k
2 + j k2 + 5k) − 9k(−2j
2k
2 + 2k2 + 3j )
SOLUTION:
29. 4n(2n3p
2 − 3np2 + 5n) + 4p (6n
2p − 2np
2 + 3p )
SOLUTION:
30. DAMS A new dam being built has the shape of a trapezoid. The base at the bottom of the dam is 2 times the
height. The base at the top of the dam is times the height minus 30 feet.
a. Write an expression to find the area of the trapezoidal cross section of the dam. b. If the height of the dam is 180 feet, find the area of this cross section.
SOLUTION:
a. The equation for the area of a trapezoid is .
b1
= 2h
b2 =
b. Let h = 180.
So, the area is 32,940 square feet.
Solve each equation.
31. 7(t2 + 5t − 9) + t = t(7t − 2) + 13
SOLUTION:
32. w(4w + 6) + 2w = 2(2w2 + 7w − 3)
SOLUTION:
33. 5(4z + 6) − 2(z − 4) = 7z(z + 4) − z(7z − 2) − 48
SOLUTION:
34. 9c(c − 11) + 10(5c − 3) = 3c(c + 5) + c(6c − 3) − 30
SOLUTION:
35. 2f (5f − 2) − 10(f2 − 3f + 6) = −8f (f + 4) + 4(2f
2 − 7f )
SOLUTION:
36. 2k(−3k + 4) + 6(k2 + 10) = k(4k + 8) − 2k(2k + 5)
SOLUTION:
Simplify each expression.
37. np2(30p
2 + 9n
2p − 12)
SOLUTION:
38. r2t(10r
3 + 5rt
3 + 15t
2)
SOLUTION:
39. −5q2w
3(4q + 7w) + 4qw
2(7q
2w + 2q) − 3qw(3q
2w
2 + 9)
SOLUTION:
40. −x2z(2z
2 + 4xz3) + xz
2(xz + 5x
3z) + x2
z3(3x
2z + 4xz)
SOLUTION:
41. PARKING A parking garage charges $30 per month plus $0.50 per daytime hour and $0.25 per hour during nights and weekends. Suppose Trent parks in the garage for 47 hours in January and h of those are night and weekend hours. a. Find an expression for Trent’s January bill. b. Find the cost if Trent had 12 hours of night and weekend hours.
SOLUTION: a. First, multiply the price per hour times the number of hours. Let h represent the number of weekend and night hours. For the weekend and night hours, he would spend 0.25h. The number of daytime hours is 47 – h, so he spent 0.50(47 - h) = 23.50 – 0.50h. To find the total cost, add these two values to the cost per month. 30 + (23.50 − 0.50h) + 0.25h = 53.50 − 0.25h. b. If Trent parked 12 night and weekend hours, h = 12
So, the cost for January is $50.50.
42. CCSS MODELING Che is building a dog house for his new puppy. The upper face of the dog house is a trapezoid. If the height of the trapezoid is 12 inches, find the area of the face of this piece of the dog house.
SOLUTION:
The formula for the area of a trapezoid is .
Now substitute h = 12 inches into the equation.
Therefore, the area of the trapezoid is 318 square inches.
43. TENNIS The tennis club is building a new tennis court with a path around it.
a. Write an expression for the area of the tennis court. b. Write an expression for the area of the path. c. What is the perimeter of the outside of the path?
SOLUTION: a. The area of the tennis court can be found by the equation . Substitute the values of the length and width and simplify.
b. The area of the path is the area of the large rectangle minus the area of the smaller rectangle, which is the area ofthe tennis court.
Area of the large rectangle = 2.5x(x + 6) or 2.5x2 + 15x.
Now subtract the area of the tennis court to find the area of just the path.
c. The perimeter of the outside of the path is the perimeter of the larger rectangle.
If x = 36 feet, then the perimeter of the outside of the path is 7(36) + 12 or 264 ft.
44. MULTIPLE REPRESENTATIONS In this problem, you will investigate the degree of the product of a monomialand a polynomial. a. TABULAR Write three monomials of different degrees and three polynomials of different degrees. Determine the degree of each monomial and polynomial. Multiply the monomials by the polynomials. Determine the degree of each product. Record your results in a table like the one below.
b. VERBAL Make a conjecture about the degree of the product of a monomial and a polynomial. What is the degree of the product of a monomial of degree a and a polynomial of degree b?
SOLUTION: a. There possible monomials are 2x, 3x
2, 4x
3. Three polynomials are x2 − 1, x5 + 1, x
6 + 1.
The degree for the monomial is the sum of the exponents. For the example given, case they are 1,2, and 3. The degree for the polynomial is the largest degree of the monomials. For the example given, they are 2, 5, and 6. Use the distributive property to find the product of the monomial and the polynomial. For the example give, they
are2x3 − 2x, 3x
7 + 3x
2, and 4x
9 + 4x. For the example given, the degree of the new polynomials are 3,7 and 9.
b. The degree of the product is the sum of the degree of the monomial and the degree of the polynomial, or a + b.
45. ERROR ANALYSIS Pearl and Ted both worked on this problem. Is either of them correct? Explain your reasoning.
SOLUTION:
Ted’s solution is correct. Pearl made a mistake when distributing on the middle term. She did not add the exponents correctly.
46. CCSS PERSEVERANCE Find p such that 3xp(4x
2p+3 + 2x3p−2
) = 12x12 + 6x
10.
SOLUTION: First simplify the left side of the equation.
Because the coefficients of the terms are the same, the corresponding terms’ exponents can be set equal to each other.
47. CHALLENGE Find 4x−3
y2(2x
5y−4 + 6x
−7y
6 − 4x0y
-2).
SOLUTION:
48. REASONING Is there a value for x that makes the statement (x + 2)2 = x2 + 22
true? If so, find a value for x. Explain your reasoning.
SOLUTION: Solve for x.
When 0 is substituted in for x in the equation, both sides are 22 or 4, which makes the equation true.
49. OPEN ENDED Write a monomial and a polynomial using n as the variable. Find their product.
SOLUTION: Choose a monomial with variable n and a polynomial with variable n. For example, let the monomial be 3n and the polynomial be 4n + 1. Find the product.
50. WRITING IN MATH Describe the steps to multiply a polynomial by a monomial.
SOLUTION: To multiply a polynomial by a monomial, use the Distributive Property. Multiply each term of the polynomial by the monomial. Then simplify by multiplying the coefficients together and using the Product of Powers Property for the variables. Consider the following example:
51. Every week a store sells j jeans and t T-shirts. The store makes $8 for each T-shirt and $12 for each pair of jeans. Which of the following expressions represents the total amount of money, in dollars, the store makes every week? A 8j + 12t B 12j + 8t C 20(j + t) D 96jt
SOLUTION: 12j is the amount made from jeans. 8t is the amount made from T-shirts. The sum, or 12j + 8t, represents the total amount the store makes every week. So, the correct choice is B.
52. If a = 5x + 7y and b = 2y − 3x, what is a + b?F 2x − 9y G 3y + 4x H 2x + 9y J 2x − 5y
SOLUTION:
So, the correct choice is H.
53. GEOMETRY A triangle has sides of length 5 inches and 8.5 inches. Which of the following cannot be the length of the third side? A 3.5 inches B 4 inches C 5.5 inches D 12 inches
SOLUTION: The sum of any two sides of a triangle must be greater than the length of the third side. Choice A: 3.5 + 5 = 8.5 8.5 Choice B: 3.5 + 5 = 8.5 > 8.5 Choice C: 5.5 + 5 = 10.5 > 8.5 Choice D: 12 + 5 = 17 > 8.5 Let x be the third side of the triangle.
3.5 cannot be the third side of the triangle. So, the correct choice is A.
54. SHORT RESPONSE Write an equation in which x varies directly as the cube of y and inversely as the square of z .
SOLUTION: If x varies directly as the cube of y , then both x and y must be in the numerators on either side of the equal sign.
When x varies inversely with the square of z. then the z must be in the denominator. Then . The
combination of the two is .
Find each sum or difference.
55. (2x2 − 7) + (8 − 5x
2)
SOLUTION:
56. (3z2 + 2z − 1) + (z
2 − 6)
SOLUTION:
57. (2a − 4a2 + 1) − (5a
2 − 2a − 6)
SOLUTION:
58. (a3 − 3a
2 + 4) − (4a
2 + 7)
SOLUTION:
59. (2ab − 3a + 4b) + (5a + 4ab)
SOLUTION:
60. (8c3 − 3c
2 + c − 2) − (3c
3 + 9)
SOLUTION:
Write a recursive formula for each sequence.61. 16, 2, –12, –26, ...
SOLUTION: Determine if there is a common difference. 2 – 16 = –14 –12 – 2 = –14 –26 – (–12) = –14 The common difference is –14. The first term is 16. To find the next term, subtract 14 from the previous term.
62. –5, 3, 11, 19, ...
SOLUTION: Determine if there is a common difference. 3 – (–5) = 8 11 – 3 = 8 19 – 11 = 8 The common difference is 8. The first term is 3. To find the next term, add 8 to the previous term.
63. 27, 43, 59, 75, ...
SOLUTION: Determine if there is a common difference. 43 – 27 = 16 59 – 43 = 16 75 – 59 = 16 The common difference is 16. The first term is 43. To find the next term, add 16 to the previous term.
64. 80, –200, 500, –1250, ...
SOLUTION: Check for a common difference. –200 – 80 = –280 500 – (–200) = 700 There is no common difference. Check for a common ratio. –200 ÷ 80 = –2.5 500 ÷ (–200) = –2.5 –1250 ÷ 500 = –2.5 The common ratio is –2.5. The first term is 80. To find the next term, multiply the previous term by –2.5.
65. 100, 60, 36, 21.6, ...
SOLUTION: Check for a common difference. 60 – 100 = –40 36 – 60 = –24 There is no common difference. Check for a common ratio. 60 ÷ 100 = 0.6 36 ÷ 60 = 0.6 21.6 ÷ 36 = 0.6 The common ratio is 0.6. The first term is 100. To find the next term, multiply the previous term by 0.6.
66.
SOLUTION: Check for a common difference.
There is no common difference. Check for a common ratio.
4 ÷ 1 = 4 The common ratio is 4.
The first term is . To find the next term, multiply the previous term by 4.
67. TRAVEL In 2003, about 9.5 million people took cruises. Between 2003 and 2008, the number increased by about 740,000 each year. Write the point-slope form of an equation to find the total number of people y taking a cruise for any year x. Estimate the number of people who took a cruise in 2010.
SOLUTION: The statement “increased by 740,000 each year” is a constant rate, or the slope. Now substitute the slope and the point (2003, 9,500,000) into the point slope equation.
To find the number of people who took a cruise in 2010, substitute x = 2010.
So, approximately 14,680,000 people took a cruise in 2010.
Write an equation in function notation for each relation.
68.
SOLUTION: The y-intercept is 0. Find the slope by choosing two points. Choose (0, 0) and (1, 4).
So the slope intercept form is f (x) = 4x + 0 or f (x) = 4x.
69.
SOLUTION:
The y-intercept is 0. Find the slope by choosing two points. Choose (0, 0) and (2, −1).
So in slope intercept form is f (x) = −0.5x + 0 or f (x) = −0.5x.
Simplify.
70. b(b2)(b
3)
SOLUTION:
71. 2y(3y2)
SOLUTION:
72. −y4(−2y
3)
SOLUTION:
73. −3z3(−5z
4 + 2z)
SOLUTION:
74. 2m(−4m4) − 3(−5m
3)
SOLUTION:
75. 4p 2(−2p
3) + 2p
4(5p
6)
SOLUTION:
eSolutions Manual - Powered by Cognero Page 23
8-2 Multiplying a Polynomial by a Monomial
Find each product.
1. 5w( −3w2 + 2w − 4)
SOLUTION:
2. 6g2(3g
3 + 4g2 + 10g − 1)
SOLUTION:
3. 4km2(8km
2 + 2k2m + 5k)
SOLUTION:
4. −3p4r3(2p
2r4 − 6p 6
r3 − 5)
SOLUTION:
5. 2ab(7a4b
2 + a5b − 2a)
SOLUTION:
6. c2d
3(5cd
7 − 3c3d
2 − 4d3)
SOLUTION:
Simplify each expression.
7. t(4t2 + 15t + 4) − 4(3t − 1)
SOLUTION:
8. x(3x2 + 4) + 2(7x − 3)
SOLUTION:
9. −2d(d3c
2 − 4dc2 + 2d
2c) + c
2(dc
2 − 3d4)
SOLUTION:
10. −5w2(8w
2x − 11wx
2) + 6x(9wx
4 − 4w − 3x2)
SOLUTION:
11. GRIDDED RESPONSE Marlene is buying a new plasma television. The height of the screen of the television is one half the width plus 5 inches. The width is 30 inches. Find the height of the screen in inches.
SOLUTION: Let h be the height and let w be the width.
Let w = 30.
The height is 20 inches.
Solve each equation.
12. −6(11 − 2c) = 7(−2 − 2c)
SOLUTION:
13. t(2t + 3) + 20 = 2t(t − 3)
SOLUTION:
14. −2(w + 1) + w = 7 − 4w
SOLUTION:
15. 3(y − 2) + 2y = 4y + 14
SOLUTION:
16. a(a + 3) + a(a − 6) + 35 = a(a − 5) + a(a + 7)
SOLUTION:
17. n(n − 4) + n(n + 8) = n(n − 13) + n(n + 1) + 16
SOLUTION:
Find each product.
18. b(b2 − 12b + 1)
SOLUTION:
19. f (f2 + 2f + 25)
SOLUTION:
20. −3m3(2m
3 − 12m2 + 2m + 25)
SOLUTION:
21. 2j 2(5j
3 − 15j2 + 2j + 2)
SOLUTION:
22. 2pr2(2pr + 5p
2r − 15p )
SOLUTION:
23. 4t3u(2t
2u
2 − 10tu4 + 2)
SOLUTION:
Simplify each expression.
24. −3(5x2 + 2x + 9) + x(2x − 3)
SOLUTION:
25. a(−8a2 + 2a + 4) + 3(6a
2 − 4)
SOLUTION:
26. −4d(5d2 − 12) + 7(d + 5)
SOLUTION:
27. −9g(−2g + g2) + 3(g
2 + 4)
SOLUTION:
28. 2j (7j2k
2 + j k2 + 5k) − 9k(−2j
2k
2 + 2k2 + 3j )
SOLUTION:
29. 4n(2n3p
2 − 3np2 + 5n) + 4p (6n
2p − 2np
2 + 3p )
SOLUTION:
30. DAMS A new dam being built has the shape of a trapezoid. The base at the bottom of the dam is 2 times the
height. The base at the top of the dam is times the height minus 30 feet.
a. Write an expression to find the area of the trapezoidal cross section of the dam. b. If the height of the dam is 180 feet, find the area of this cross section.
SOLUTION:
a. The equation for the area of a trapezoid is .
b1
= 2h
b2 =
b. Let h = 180.
So, the area is 32,940 square feet.
Solve each equation.
31. 7(t2 + 5t − 9) + t = t(7t − 2) + 13
SOLUTION:
32. w(4w + 6) + 2w = 2(2w2 + 7w − 3)
SOLUTION:
33. 5(4z + 6) − 2(z − 4) = 7z(z + 4) − z(7z − 2) − 48
SOLUTION:
34. 9c(c − 11) + 10(5c − 3) = 3c(c + 5) + c(6c − 3) − 30
SOLUTION:
35. 2f (5f − 2) − 10(f2 − 3f + 6) = −8f (f + 4) + 4(2f
2 − 7f )
SOLUTION:
36. 2k(−3k + 4) + 6(k2 + 10) = k(4k + 8) − 2k(2k + 5)
SOLUTION:
Simplify each expression.
37. np2(30p
2 + 9n
2p − 12)
SOLUTION:
38. r2t(10r
3 + 5rt
3 + 15t
2)
SOLUTION:
39. −5q2w
3(4q + 7w) + 4qw
2(7q
2w + 2q) − 3qw(3q
2w
2 + 9)
SOLUTION:
40. −x2z(2z
2 + 4xz3) + xz
2(xz + 5x
3z) + x2
z3(3x
2z + 4xz)
SOLUTION:
41. PARKING A parking garage charges $30 per month plus $0.50 per daytime hour and $0.25 per hour during nights and weekends. Suppose Trent parks in the garage for 47 hours in January and h of those are night and weekend hours. a. Find an expression for Trent’s January bill. b. Find the cost if Trent had 12 hours of night and weekend hours.
SOLUTION: a. First, multiply the price per hour times the number of hours. Let h represent the number of weekend and night hours. For the weekend and night hours, he would spend 0.25h. The number of daytime hours is 47 – h, so he spent 0.50(47 - h) = 23.50 – 0.50h. To find the total cost, add these two values to the cost per month. 30 + (23.50 − 0.50h) + 0.25h = 53.50 − 0.25h. b. If Trent parked 12 night and weekend hours, h = 12
So, the cost for January is $50.50.
42. CCSS MODELING Che is building a dog house for his new puppy. The upper face of the dog house is a trapezoid. If the height of the trapezoid is 12 inches, find the area of the face of this piece of the dog house.
SOLUTION:
The formula for the area of a trapezoid is .
Now substitute h = 12 inches into the equation.
Therefore, the area of the trapezoid is 318 square inches.
43. TENNIS The tennis club is building a new tennis court with a path around it.
a. Write an expression for the area of the tennis court. b. Write an expression for the area of the path. c. What is the perimeter of the outside of the path?
SOLUTION: a. The area of the tennis court can be found by the equation . Substitute the values of the length and width and simplify.
b. The area of the path is the area of the large rectangle minus the area of the smaller rectangle, which is the area ofthe tennis court.
Area of the large rectangle = 2.5x(x + 6) or 2.5x2 + 15x.
Now subtract the area of the tennis court to find the area of just the path.
c. The perimeter of the outside of the path is the perimeter of the larger rectangle.
If x = 36 feet, then the perimeter of the outside of the path is 7(36) + 12 or 264 ft.
44. MULTIPLE REPRESENTATIONS In this problem, you will investigate the degree of the product of a monomialand a polynomial. a. TABULAR Write three monomials of different degrees and three polynomials of different degrees. Determine the degree of each monomial and polynomial. Multiply the monomials by the polynomials. Determine the degree of each product. Record your results in a table like the one below.
b. VERBAL Make a conjecture about the degree of the product of a monomial and a polynomial. What is the degree of the product of a monomial of degree a and a polynomial of degree b?
SOLUTION: a. There possible monomials are 2x, 3x
2, 4x
3. Three polynomials are x2 − 1, x5 + 1, x
6 + 1.
The degree for the monomial is the sum of the exponents. For the example given, case they are 1,2, and 3. The degree for the polynomial is the largest degree of the monomials. For the example given, they are 2, 5, and 6. Use the distributive property to find the product of the monomial and the polynomial. For the example give, they
are2x3 − 2x, 3x
7 + 3x
2, and 4x
9 + 4x. For the example given, the degree of the new polynomials are 3,7 and 9.
b. The degree of the product is the sum of the degree of the monomial and the degree of the polynomial, or a + b.
45. ERROR ANALYSIS Pearl and Ted both worked on this problem. Is either of them correct? Explain your reasoning.
SOLUTION:
Ted’s solution is correct. Pearl made a mistake when distributing on the middle term. She did not add the exponents correctly.
46. CCSS PERSEVERANCE Find p such that 3xp(4x
2p+3 + 2x3p−2
) = 12x12 + 6x
10.
SOLUTION: First simplify the left side of the equation.
Because the coefficients of the terms are the same, the corresponding terms’ exponents can be set equal to each other.
47. CHALLENGE Find 4x−3
y2(2x
5y−4 + 6x
−7y
6 − 4x0y
-2).
SOLUTION:
48. REASONING Is there a value for x that makes the statement (x + 2)2 = x2 + 22
true? If so, find a value for x. Explain your reasoning.
SOLUTION: Solve for x.
When 0 is substituted in for x in the equation, both sides are 22 or 4, which makes the equation true.
49. OPEN ENDED Write a monomial and a polynomial using n as the variable. Find their product.
SOLUTION: Choose a monomial with variable n and a polynomial with variable n. For example, let the monomial be 3n and the polynomial be 4n + 1. Find the product.
50. WRITING IN MATH Describe the steps to multiply a polynomial by a monomial.
SOLUTION: To multiply a polynomial by a monomial, use the Distributive Property. Multiply each term of the polynomial by the monomial. Then simplify by multiplying the coefficients together and using the Product of Powers Property for the variables. Consider the following example:
51. Every week a store sells j jeans and t T-shirts. The store makes $8 for each T-shirt and $12 for each pair of jeans. Which of the following expressions represents the total amount of money, in dollars, the store makes every week? A 8j + 12t B 12j + 8t C 20(j + t) D 96jt
SOLUTION: 12j is the amount made from jeans. 8t is the amount made from T-shirts. The sum, or 12j + 8t, represents the total amount the store makes every week. So, the correct choice is B.
52. If a = 5x + 7y and b = 2y − 3x, what is a + b?F 2x − 9y G 3y + 4x H 2x + 9y J 2x − 5y
SOLUTION:
So, the correct choice is H.
53. GEOMETRY A triangle has sides of length 5 inches and 8.5 inches. Which of the following cannot be the length of the third side? A 3.5 inches B 4 inches C 5.5 inches D 12 inches
SOLUTION: The sum of any two sides of a triangle must be greater than the length of the third side. Choice A: 3.5 + 5 = 8.5 8.5 Choice B: 3.5 + 5 = 8.5 > 8.5 Choice C: 5.5 + 5 = 10.5 > 8.5 Choice D: 12 + 5 = 17 > 8.5 Let x be the third side of the triangle.
3.5 cannot be the third side of the triangle. So, the correct choice is A.
54. SHORT RESPONSE Write an equation in which x varies directly as the cube of y and inversely as the square of z .
SOLUTION: If x varies directly as the cube of y , then both x and y must be in the numerators on either side of the equal sign.
When x varies inversely with the square of z. then the z must be in the denominator. Then . The
combination of the two is .
Find each sum or difference.
55. (2x2 − 7) + (8 − 5x
2)
SOLUTION:
56. (3z2 + 2z − 1) + (z
2 − 6)
SOLUTION:
57. (2a − 4a2 + 1) − (5a
2 − 2a − 6)
SOLUTION:
58. (a3 − 3a
2 + 4) − (4a
2 + 7)
SOLUTION:
59. (2ab − 3a + 4b) + (5a + 4ab)
SOLUTION:
60. (8c3 − 3c
2 + c − 2) − (3c
3 + 9)
SOLUTION:
Write a recursive formula for each sequence.61. 16, 2, –12, –26, ...
SOLUTION: Determine if there is a common difference. 2 – 16 = –14 –12 – 2 = –14 –26 – (–12) = –14 The common difference is –14. The first term is 16. To find the next term, subtract 14 from the previous term.
62. –5, 3, 11, 19, ...
SOLUTION: Determine if there is a common difference. 3 – (–5) = 8 11 – 3 = 8 19 – 11 = 8 The common difference is 8. The first term is 3. To find the next term, add 8 to the previous term.
63. 27, 43, 59, 75, ...
SOLUTION: Determine if there is a common difference. 43 – 27 = 16 59 – 43 = 16 75 – 59 = 16 The common difference is 16. The first term is 43. To find the next term, add 16 to the previous term.
64. 80, –200, 500, –1250, ...
SOLUTION: Check for a common difference. –200 – 80 = –280 500 – (–200) = 700 There is no common difference. Check for a common ratio. –200 ÷ 80 = –2.5 500 ÷ (–200) = –2.5 –1250 ÷ 500 = –2.5 The common ratio is –2.5. The first term is 80. To find the next term, multiply the previous term by –2.5.
65. 100, 60, 36, 21.6, ...
SOLUTION: Check for a common difference. 60 – 100 = –40 36 – 60 = –24 There is no common difference. Check for a common ratio. 60 ÷ 100 = 0.6 36 ÷ 60 = 0.6 21.6 ÷ 36 = 0.6 The common ratio is 0.6. The first term is 100. To find the next term, multiply the previous term by 0.6.
66.
SOLUTION: Check for a common difference.
There is no common difference. Check for a common ratio.
4 ÷ 1 = 4 The common ratio is 4.
The first term is . To find the next term, multiply the previous term by 4.
67. TRAVEL In 2003, about 9.5 million people took cruises. Between 2003 and 2008, the number increased by about 740,000 each year. Write the point-slope form of an equation to find the total number of people y taking a cruise for any year x. Estimate the number of people who took a cruise in 2010.
SOLUTION: The statement “increased by 740,000 each year” is a constant rate, or the slope. Now substitute the slope and the point (2003, 9,500,000) into the point slope equation.
To find the number of people who took a cruise in 2010, substitute x = 2010.
So, approximately 14,680,000 people took a cruise in 2010.
Write an equation in function notation for each relation.
68.
SOLUTION: The y-intercept is 0. Find the slope by choosing two points. Choose (0, 0) and (1, 4).
So the slope intercept form is f (x) = 4x + 0 or f (x) = 4x.
69.
SOLUTION:
The y-intercept is 0. Find the slope by choosing two points. Choose (0, 0) and (2, −1).
So in slope intercept form is f (x) = −0.5x + 0 or f (x) = −0.5x.
Simplify.
70. b(b2)(b
3)
SOLUTION:
71. 2y(3y2)
SOLUTION:
72. −y4(−2y
3)
SOLUTION:
73. −3z3(−5z
4 + 2z)
SOLUTION:
74. 2m(−4m4) − 3(−5m
3)
SOLUTION:
75. 4p 2(−2p
3) + 2p
4(5p
6)
SOLUTION:
eSolutions Manual - Powered by Cognero Page 24
8-2 Multiplying a Polynomial by a Monomial
Find each product.
1. 5w( −3w2 + 2w − 4)
SOLUTION:
2. 6g2(3g
3 + 4g2 + 10g − 1)
SOLUTION:
3. 4km2(8km
2 + 2k2m + 5k)
SOLUTION:
4. −3p4r3(2p
2r4 − 6p 6
r3 − 5)
SOLUTION:
5. 2ab(7a4b
2 + a5b − 2a)
SOLUTION:
6. c2d
3(5cd
7 − 3c3d
2 − 4d3)
SOLUTION:
Simplify each expression.
7. t(4t2 + 15t + 4) − 4(3t − 1)
SOLUTION:
8. x(3x2 + 4) + 2(7x − 3)
SOLUTION:
9. −2d(d3c
2 − 4dc2 + 2d
2c) + c
2(dc
2 − 3d4)
SOLUTION:
10. −5w2(8w
2x − 11wx
2) + 6x(9wx
4 − 4w − 3x2)
SOLUTION:
11. GRIDDED RESPONSE Marlene is buying a new plasma television. The height of the screen of the television is one half the width plus 5 inches. The width is 30 inches. Find the height of the screen in inches.
SOLUTION: Let h be the height and let w be the width.
Let w = 30.
The height is 20 inches.
Solve each equation.
12. −6(11 − 2c) = 7(−2 − 2c)
SOLUTION:
13. t(2t + 3) + 20 = 2t(t − 3)
SOLUTION:
14. −2(w + 1) + w = 7 − 4w
SOLUTION:
15. 3(y − 2) + 2y = 4y + 14
SOLUTION:
16. a(a + 3) + a(a − 6) + 35 = a(a − 5) + a(a + 7)
SOLUTION:
17. n(n − 4) + n(n + 8) = n(n − 13) + n(n + 1) + 16
SOLUTION:
Find each product.
18. b(b2 − 12b + 1)
SOLUTION:
19. f (f2 + 2f + 25)
SOLUTION:
20. −3m3(2m
3 − 12m2 + 2m + 25)
SOLUTION:
21. 2j 2(5j
3 − 15j2 + 2j + 2)
SOLUTION:
22. 2pr2(2pr + 5p
2r − 15p )
SOLUTION:
23. 4t3u(2t
2u
2 − 10tu4 + 2)
SOLUTION:
Simplify each expression.
24. −3(5x2 + 2x + 9) + x(2x − 3)
SOLUTION:
25. a(−8a2 + 2a + 4) + 3(6a
2 − 4)
SOLUTION:
26. −4d(5d2 − 12) + 7(d + 5)
SOLUTION:
27. −9g(−2g + g2) + 3(g
2 + 4)
SOLUTION:
28. 2j (7j2k
2 + j k2 + 5k) − 9k(−2j
2k
2 + 2k2 + 3j )
SOLUTION:
29. 4n(2n3p
2 − 3np2 + 5n) + 4p (6n
2p − 2np
2 + 3p )
SOLUTION:
30. DAMS A new dam being built has the shape of a trapezoid. The base at the bottom of the dam is 2 times the
height. The base at the top of the dam is times the height minus 30 feet.
a. Write an expression to find the area of the trapezoidal cross section of the dam. b. If the height of the dam is 180 feet, find the area of this cross section.
SOLUTION:
a. The equation for the area of a trapezoid is .
b1
= 2h
b2 =
b. Let h = 180.
So, the area is 32,940 square feet.
Solve each equation.
31. 7(t2 + 5t − 9) + t = t(7t − 2) + 13
SOLUTION:
32. w(4w + 6) + 2w = 2(2w2 + 7w − 3)
SOLUTION:
33. 5(4z + 6) − 2(z − 4) = 7z(z + 4) − z(7z − 2) − 48
SOLUTION:
34. 9c(c − 11) + 10(5c − 3) = 3c(c + 5) + c(6c − 3) − 30
SOLUTION:
35. 2f (5f − 2) − 10(f2 − 3f + 6) = −8f (f + 4) + 4(2f
2 − 7f )
SOLUTION:
36. 2k(−3k + 4) + 6(k2 + 10) = k(4k + 8) − 2k(2k + 5)
SOLUTION:
Simplify each expression.
37. np2(30p
2 + 9n
2p − 12)
SOLUTION:
38. r2t(10r
3 + 5rt
3 + 15t
2)
SOLUTION:
39. −5q2w
3(4q + 7w) + 4qw
2(7q
2w + 2q) − 3qw(3q
2w
2 + 9)
SOLUTION:
40. −x2z(2z
2 + 4xz3) + xz
2(xz + 5x
3z) + x2
z3(3x
2z + 4xz)
SOLUTION:
41. PARKING A parking garage charges $30 per month plus $0.50 per daytime hour and $0.25 per hour during nights and weekends. Suppose Trent parks in the garage for 47 hours in January and h of those are night and weekend hours. a. Find an expression for Trent’s January bill. b. Find the cost if Trent had 12 hours of night and weekend hours.
SOLUTION: a. First, multiply the price per hour times the number of hours. Let h represent the number of weekend and night hours. For the weekend and night hours, he would spend 0.25h. The number of daytime hours is 47 – h, so he spent 0.50(47 - h) = 23.50 – 0.50h. To find the total cost, add these two values to the cost per month. 30 + (23.50 − 0.50h) + 0.25h = 53.50 − 0.25h. b. If Trent parked 12 night and weekend hours, h = 12
So, the cost for January is $50.50.
42. CCSS MODELING Che is building a dog house for his new puppy. The upper face of the dog house is a trapezoid. If the height of the trapezoid is 12 inches, find the area of the face of this piece of the dog house.
SOLUTION:
The formula for the area of a trapezoid is .
Now substitute h = 12 inches into the equation.
Therefore, the area of the trapezoid is 318 square inches.
43. TENNIS The tennis club is building a new tennis court with a path around it.
a. Write an expression for the area of the tennis court. b. Write an expression for the area of the path. c. What is the perimeter of the outside of the path?
SOLUTION: a. The area of the tennis court can be found by the equation . Substitute the values of the length and width and simplify.
b. The area of the path is the area of the large rectangle minus the area of the smaller rectangle, which is the area ofthe tennis court.
Area of the large rectangle = 2.5x(x + 6) or 2.5x2 + 15x.
Now subtract the area of the tennis court to find the area of just the path.
c. The perimeter of the outside of the path is the perimeter of the larger rectangle.
If x = 36 feet, then the perimeter of the outside of the path is 7(36) + 12 or 264 ft.
44. MULTIPLE REPRESENTATIONS In this problem, you will investigate the degree of the product of a monomialand a polynomial. a. TABULAR Write three monomials of different degrees and three polynomials of different degrees. Determine the degree of each monomial and polynomial. Multiply the monomials by the polynomials. Determine the degree of each product. Record your results in a table like the one below.
b. VERBAL Make a conjecture about the degree of the product of a monomial and a polynomial. What is the degree of the product of a monomial of degree a and a polynomial of degree b?
SOLUTION: a. There possible monomials are 2x, 3x
2, 4x
3. Three polynomials are x2 − 1, x5 + 1, x
6 + 1.
The degree for the monomial is the sum of the exponents. For the example given, case they are 1,2, and 3. The degree for the polynomial is the largest degree of the monomials. For the example given, they are 2, 5, and 6. Use the distributive property to find the product of the monomial and the polynomial. For the example give, they
are2x3 − 2x, 3x
7 + 3x
2, and 4x
9 + 4x. For the example given, the degree of the new polynomials are 3,7 and 9.
b. The degree of the product is the sum of the degree of the monomial and the degree of the polynomial, or a + b.
45. ERROR ANALYSIS Pearl and Ted both worked on this problem. Is either of them correct? Explain your reasoning.
SOLUTION:
Ted’s solution is correct. Pearl made a mistake when distributing on the middle term. She did not add the exponents correctly.
46. CCSS PERSEVERANCE Find p such that 3xp(4x
2p+3 + 2x3p−2
) = 12x12 + 6x
10.
SOLUTION: First simplify the left side of the equation.
Because the coefficients of the terms are the same, the corresponding terms’ exponents can be set equal to each other.
47. CHALLENGE Find 4x−3
y2(2x
5y−4 + 6x
−7y
6 − 4x0y
-2).
SOLUTION:
48. REASONING Is there a value for x that makes the statement (x + 2)2 = x2 + 22
true? If so, find a value for x. Explain your reasoning.
SOLUTION: Solve for x.
When 0 is substituted in for x in the equation, both sides are 22 or 4, which makes the equation true.
49. OPEN ENDED Write a monomial and a polynomial using n as the variable. Find their product.
SOLUTION: Choose a monomial with variable n and a polynomial with variable n. For example, let the monomial be 3n and the polynomial be 4n + 1. Find the product.
50. WRITING IN MATH Describe the steps to multiply a polynomial by a monomial.
SOLUTION: To multiply a polynomial by a monomial, use the Distributive Property. Multiply each term of the polynomial by the monomial. Then simplify by multiplying the coefficients together and using the Product of Powers Property for the variables. Consider the following example:
51. Every week a store sells j jeans and t T-shirts. The store makes $8 for each T-shirt and $12 for each pair of jeans. Which of the following expressions represents the total amount of money, in dollars, the store makes every week? A 8j + 12t B 12j + 8t C 20(j + t) D 96jt
SOLUTION: 12j is the amount made from jeans. 8t is the amount made from T-shirts. The sum, or 12j + 8t, represents the total amount the store makes every week. So, the correct choice is B.
52. If a = 5x + 7y and b = 2y − 3x, what is a + b?F 2x − 9y G 3y + 4x H 2x + 9y J 2x − 5y
SOLUTION:
So, the correct choice is H.
53. GEOMETRY A triangle has sides of length 5 inches and 8.5 inches. Which of the following cannot be the length of the third side? A 3.5 inches B 4 inches C 5.5 inches D 12 inches
SOLUTION: The sum of any two sides of a triangle must be greater than the length of the third side. Choice A: 3.5 + 5 = 8.5 8.5 Choice B: 3.5 + 5 = 8.5 > 8.5 Choice C: 5.5 + 5 = 10.5 > 8.5 Choice D: 12 + 5 = 17 > 8.5 Let x be the third side of the triangle.
3.5 cannot be the third side of the triangle. So, the correct choice is A.
54. SHORT RESPONSE Write an equation in which x varies directly as the cube of y and inversely as the square of z .
SOLUTION: If x varies directly as the cube of y , then both x and y must be in the numerators on either side of the equal sign.
When x varies inversely with the square of z. then the z must be in the denominator. Then . The
combination of the two is .
Find each sum or difference.
55. (2x2 − 7) + (8 − 5x
2)
SOLUTION:
56. (3z2 + 2z − 1) + (z
2 − 6)
SOLUTION:
57. (2a − 4a2 + 1) − (5a
2 − 2a − 6)
SOLUTION:
58. (a3 − 3a
2 + 4) − (4a
2 + 7)
SOLUTION:
59. (2ab − 3a + 4b) + (5a + 4ab)
SOLUTION:
60. (8c3 − 3c
2 + c − 2) − (3c
3 + 9)
SOLUTION:
Write a recursive formula for each sequence.61. 16, 2, –12, –26, ...
SOLUTION: Determine if there is a common difference. 2 – 16 = –14 –12 – 2 = –14 –26 – (–12) = –14 The common difference is –14. The first term is 16. To find the next term, subtract 14 from the previous term.
62. –5, 3, 11, 19, ...
SOLUTION: Determine if there is a common difference. 3 – (–5) = 8 11 – 3 = 8 19 – 11 = 8 The common difference is 8. The first term is 3. To find the next term, add 8 to the previous term.
63. 27, 43, 59, 75, ...
SOLUTION: Determine if there is a common difference. 43 – 27 = 16 59 – 43 = 16 75 – 59 = 16 The common difference is 16. The first term is 43. To find the next term, add 16 to the previous term.
64. 80, –200, 500, –1250, ...
SOLUTION: Check for a common difference. –200 – 80 = –280 500 – (–200) = 700 There is no common difference. Check for a common ratio. –200 ÷ 80 = –2.5 500 ÷ (–200) = –2.5 –1250 ÷ 500 = –2.5 The common ratio is –2.5. The first term is 80. To find the next term, multiply the previous term by –2.5.
65. 100, 60, 36, 21.6, ...
SOLUTION: Check for a common difference. 60 – 100 = –40 36 – 60 = –24 There is no common difference. Check for a common ratio. 60 ÷ 100 = 0.6 36 ÷ 60 = 0.6 21.6 ÷ 36 = 0.6 The common ratio is 0.6. The first term is 100. To find the next term, multiply the previous term by 0.6.
66.
SOLUTION: Check for a common difference.
There is no common difference. Check for a common ratio.
4 ÷ 1 = 4 The common ratio is 4.
The first term is . To find the next term, multiply the previous term by 4.
67. TRAVEL In 2003, about 9.5 million people took cruises. Between 2003 and 2008, the number increased by about 740,000 each year. Write the point-slope form of an equation to find the total number of people y taking a cruise for any year x. Estimate the number of people who took a cruise in 2010.
SOLUTION: The statement “increased by 740,000 each year” is a constant rate, or the slope. Now substitute the slope and the point (2003, 9,500,000) into the point slope equation.
To find the number of people who took a cruise in 2010, substitute x = 2010.
So, approximately 14,680,000 people took a cruise in 2010.
Write an equation in function notation for each relation.
68.
SOLUTION: The y-intercept is 0. Find the slope by choosing two points. Choose (0, 0) and (1, 4).
So the slope intercept form is f (x) = 4x + 0 or f (x) = 4x.
69.
SOLUTION:
The y-intercept is 0. Find the slope by choosing two points. Choose (0, 0) and (2, −1).
So in slope intercept form is f (x) = −0.5x + 0 or f (x) = −0.5x.
Simplify.
70. b(b2)(b
3)
SOLUTION:
71. 2y(3y2)
SOLUTION:
72. −y4(−2y
3)
SOLUTION:
73. −3z3(−5z
4 + 2z)
SOLUTION:
74. 2m(−4m4) − 3(−5m
3)
SOLUTION:
75. 4p 2(−2p
3) + 2p
4(5p
6)
SOLUTION:
eSolutions Manual - Powered by Cognero Page 25
8-2 Multiplying a Polynomial by a Monomial
Find each product.
1. 5w( −3w2 + 2w − 4)
SOLUTION:
2. 6g2(3g
3 + 4g2 + 10g − 1)
SOLUTION:
3. 4km2(8km
2 + 2k2m + 5k)
SOLUTION:
4. −3p4r3(2p
2r4 − 6p 6
r3 − 5)
SOLUTION:
5. 2ab(7a4b
2 + a5b − 2a)
SOLUTION:
6. c2d
3(5cd
7 − 3c3d
2 − 4d3)
SOLUTION:
Simplify each expression.
7. t(4t2 + 15t + 4) − 4(3t − 1)
SOLUTION:
8. x(3x2 + 4) + 2(7x − 3)
SOLUTION:
9. −2d(d3c
2 − 4dc2 + 2d
2c) + c
2(dc
2 − 3d4)
SOLUTION:
10. −5w2(8w
2x − 11wx
2) + 6x(9wx
4 − 4w − 3x2)
SOLUTION:
11. GRIDDED RESPONSE Marlene is buying a new plasma television. The height of the screen of the television is one half the width plus 5 inches. The width is 30 inches. Find the height of the screen in inches.
SOLUTION: Let h be the height and let w be the width.
Let w = 30.
The height is 20 inches.
Solve each equation.
12. −6(11 − 2c) = 7(−2 − 2c)
SOLUTION:
13. t(2t + 3) + 20 = 2t(t − 3)
SOLUTION:
14. −2(w + 1) + w = 7 − 4w
SOLUTION:
15. 3(y − 2) + 2y = 4y + 14
SOLUTION:
16. a(a + 3) + a(a − 6) + 35 = a(a − 5) + a(a + 7)
SOLUTION:
17. n(n − 4) + n(n + 8) = n(n − 13) + n(n + 1) + 16
SOLUTION:
Find each product.
18. b(b2 − 12b + 1)
SOLUTION:
19. f (f2 + 2f + 25)
SOLUTION:
20. −3m3(2m
3 − 12m2 + 2m + 25)
SOLUTION:
21. 2j 2(5j
3 − 15j2 + 2j + 2)
SOLUTION:
22. 2pr2(2pr + 5p
2r − 15p )
SOLUTION:
23. 4t3u(2t
2u
2 − 10tu4 + 2)
SOLUTION:
Simplify each expression.
24. −3(5x2 + 2x + 9) + x(2x − 3)
SOLUTION:
25. a(−8a2 + 2a + 4) + 3(6a
2 − 4)
SOLUTION:
26. −4d(5d2 − 12) + 7(d + 5)
SOLUTION:
27. −9g(−2g + g2) + 3(g
2 + 4)
SOLUTION:
28. 2j (7j2k
2 + j k2 + 5k) − 9k(−2j
2k
2 + 2k2 + 3j )
SOLUTION:
29. 4n(2n3p
2 − 3np2 + 5n) + 4p (6n
2p − 2np
2 + 3p )
SOLUTION:
30. DAMS A new dam being built has the shape of a trapezoid. The base at the bottom of the dam is 2 times the
height. The base at the top of the dam is times the height minus 30 feet.
a. Write an expression to find the area of the trapezoidal cross section of the dam. b. If the height of the dam is 180 feet, find the area of this cross section.
SOLUTION:
a. The equation for the area of a trapezoid is .
b1
= 2h
b2 =
b. Let h = 180.
So, the area is 32,940 square feet.
Solve each equation.
31. 7(t2 + 5t − 9) + t = t(7t − 2) + 13
SOLUTION:
32. w(4w + 6) + 2w = 2(2w2 + 7w − 3)
SOLUTION:
33. 5(4z + 6) − 2(z − 4) = 7z(z + 4) − z(7z − 2) − 48
SOLUTION:
34. 9c(c − 11) + 10(5c − 3) = 3c(c + 5) + c(6c − 3) − 30
SOLUTION:
35. 2f (5f − 2) − 10(f2 − 3f + 6) = −8f (f + 4) + 4(2f
2 − 7f )
SOLUTION:
36. 2k(−3k + 4) + 6(k2 + 10) = k(4k + 8) − 2k(2k + 5)
SOLUTION:
Simplify each expression.
37. np2(30p
2 + 9n
2p − 12)
SOLUTION:
38. r2t(10r
3 + 5rt
3 + 15t
2)
SOLUTION:
39. −5q2w
3(4q + 7w) + 4qw
2(7q
2w + 2q) − 3qw(3q
2w
2 + 9)
SOLUTION:
40. −x2z(2z
2 + 4xz3) + xz
2(xz + 5x
3z) + x2
z3(3x
2z + 4xz)
SOLUTION:
41. PARKING A parking garage charges $30 per month plus $0.50 per daytime hour and $0.25 per hour during nights and weekends. Suppose Trent parks in the garage for 47 hours in January and h of those are night and weekend hours. a. Find an expression for Trent’s January bill. b. Find the cost if Trent had 12 hours of night and weekend hours.
SOLUTION: a. First, multiply the price per hour times the number of hours. Let h represent the number of weekend and night hours. For the weekend and night hours, he would spend 0.25h. The number of daytime hours is 47 – h, so he spent 0.50(47 - h) = 23.50 – 0.50h. To find the total cost, add these two values to the cost per month. 30 + (23.50 − 0.50h) + 0.25h = 53.50 − 0.25h. b. If Trent parked 12 night and weekend hours, h = 12
So, the cost for January is $50.50.
42. CCSS MODELING Che is building a dog house for his new puppy. The upper face of the dog house is a trapezoid. If the height of the trapezoid is 12 inches, find the area of the face of this piece of the dog house.
SOLUTION:
The formula for the area of a trapezoid is .
Now substitute h = 12 inches into the equation.
Therefore, the area of the trapezoid is 318 square inches.
43. TENNIS The tennis club is building a new tennis court with a path around it.
a. Write an expression for the area of the tennis court. b. Write an expression for the area of the path. c. What is the perimeter of the outside of the path?
SOLUTION: a. The area of the tennis court can be found by the equation . Substitute the values of the length and width and simplify.
b. The area of the path is the area of the large rectangle minus the area of the smaller rectangle, which is the area ofthe tennis court.
Area of the large rectangle = 2.5x(x + 6) or 2.5x2 + 15x.
Now subtract the area of the tennis court to find the area of just the path.
c. The perimeter of the outside of the path is the perimeter of the larger rectangle.
If x = 36 feet, then the perimeter of the outside of the path is 7(36) + 12 or 264 ft.
44. MULTIPLE REPRESENTATIONS In this problem, you will investigate the degree of the product of a monomialand a polynomial. a. TABULAR Write three monomials of different degrees and three polynomials of different degrees. Determine the degree of each monomial and polynomial. Multiply the monomials by the polynomials. Determine the degree of each product. Record your results in a table like the one below.
b. VERBAL Make a conjecture about the degree of the product of a monomial and a polynomial. What is the degree of the product of a monomial of degree a and a polynomial of degree b?
SOLUTION: a. There possible monomials are 2x, 3x
2, 4x
3. Three polynomials are x2 − 1, x5 + 1, x
6 + 1.
The degree for the monomial is the sum of the exponents. For the example given, case they are 1,2, and 3. The degree for the polynomial is the largest degree of the monomials. For the example given, they are 2, 5, and 6. Use the distributive property to find the product of the monomial and the polynomial. For the example give, they
are2x3 − 2x, 3x
7 + 3x
2, and 4x
9 + 4x. For the example given, the degree of the new polynomials are 3,7 and 9.
b. The degree of the product is the sum of the degree of the monomial and the degree of the polynomial, or a + b.
45. ERROR ANALYSIS Pearl and Ted both worked on this problem. Is either of them correct? Explain your reasoning.
SOLUTION:
Ted’s solution is correct. Pearl made a mistake when distributing on the middle term. She did not add the exponents correctly.
46. CCSS PERSEVERANCE Find p such that 3xp(4x
2p+3 + 2x3p−2
) = 12x12 + 6x
10.
SOLUTION: First simplify the left side of the equation.
Because the coefficients of the terms are the same, the corresponding terms’ exponents can be set equal to each other.
47. CHALLENGE Find 4x−3
y2(2x
5y−4 + 6x
−7y
6 − 4x0y
-2).
SOLUTION:
48. REASONING Is there a value for x that makes the statement (x + 2)2 = x2 + 22
true? If so, find a value for x. Explain your reasoning.
SOLUTION: Solve for x.
When 0 is substituted in for x in the equation, both sides are 22 or 4, which makes the equation true.
49. OPEN ENDED Write a monomial and a polynomial using n as the variable. Find their product.
SOLUTION: Choose a monomial with variable n and a polynomial with variable n. For example, let the monomial be 3n and the polynomial be 4n + 1. Find the product.
50. WRITING IN MATH Describe the steps to multiply a polynomial by a monomial.
SOLUTION: To multiply a polynomial by a monomial, use the Distributive Property. Multiply each term of the polynomial by the monomial. Then simplify by multiplying the coefficients together and using the Product of Powers Property for the variables. Consider the following example:
51. Every week a store sells j jeans and t T-shirts. The store makes $8 for each T-shirt and $12 for each pair of jeans. Which of the following expressions represents the total amount of money, in dollars, the store makes every week? A 8j + 12t B 12j + 8t C 20(j + t) D 96jt
SOLUTION: 12j is the amount made from jeans. 8t is the amount made from T-shirts. The sum, or 12j + 8t, represents the total amount the store makes every week. So, the correct choice is B.
52. If a = 5x + 7y and b = 2y − 3x, what is a + b?F 2x − 9y G 3y + 4x H 2x + 9y J 2x − 5y
SOLUTION:
So, the correct choice is H.
53. GEOMETRY A triangle has sides of length 5 inches and 8.5 inches. Which of the following cannot be the length of the third side? A 3.5 inches B 4 inches C 5.5 inches D 12 inches
SOLUTION: The sum of any two sides of a triangle must be greater than the length of the third side. Choice A: 3.5 + 5 = 8.5 8.5 Choice B: 3.5 + 5 = 8.5 > 8.5 Choice C: 5.5 + 5 = 10.5 > 8.5 Choice D: 12 + 5 = 17 > 8.5 Let x be the third side of the triangle.
3.5 cannot be the third side of the triangle. So, the correct choice is A.
54. SHORT RESPONSE Write an equation in which x varies directly as the cube of y and inversely as the square of z .
SOLUTION: If x varies directly as the cube of y , then both x and y must be in the numerators on either side of the equal sign.
When x varies inversely with the square of z. then the z must be in the denominator. Then . The
combination of the two is .
Find each sum or difference.
55. (2x2 − 7) + (8 − 5x
2)
SOLUTION:
56. (3z2 + 2z − 1) + (z
2 − 6)
SOLUTION:
57. (2a − 4a2 + 1) − (5a
2 − 2a − 6)
SOLUTION:
58. (a3 − 3a
2 + 4) − (4a
2 + 7)
SOLUTION:
59. (2ab − 3a + 4b) + (5a + 4ab)
SOLUTION:
60. (8c3 − 3c
2 + c − 2) − (3c
3 + 9)
SOLUTION:
Write a recursive formula for each sequence.61. 16, 2, –12, –26, ...
SOLUTION: Determine if there is a common difference. 2 – 16 = –14 –12 – 2 = –14 –26 – (–12) = –14 The common difference is –14. The first term is 16. To find the next term, subtract 14 from the previous term.
62. –5, 3, 11, 19, ...
SOLUTION: Determine if there is a common difference. 3 – (–5) = 8 11 – 3 = 8 19 – 11 = 8 The common difference is 8. The first term is 3. To find the next term, add 8 to the previous term.
63. 27, 43, 59, 75, ...
SOLUTION: Determine if there is a common difference. 43 – 27 = 16 59 – 43 = 16 75 – 59 = 16 The common difference is 16. The first term is 43. To find the next term, add 16 to the previous term.
64. 80, –200, 500, –1250, ...
SOLUTION: Check for a common difference. –200 – 80 = –280 500 – (–200) = 700 There is no common difference. Check for a common ratio. –200 ÷ 80 = –2.5 500 ÷ (–200) = –2.5 –1250 ÷ 500 = –2.5 The common ratio is –2.5. The first term is 80. To find the next term, multiply the previous term by –2.5.
65. 100, 60, 36, 21.6, ...
SOLUTION: Check for a common difference. 60 – 100 = –40 36 – 60 = –24 There is no common difference. Check for a common ratio. 60 ÷ 100 = 0.6 36 ÷ 60 = 0.6 21.6 ÷ 36 = 0.6 The common ratio is 0.6. The first term is 100. To find the next term, multiply the previous term by 0.6.
66.
SOLUTION: Check for a common difference.
There is no common difference. Check for a common ratio.
4 ÷ 1 = 4 The common ratio is 4.
The first term is . To find the next term, multiply the previous term by 4.
67. TRAVEL In 2003, about 9.5 million people took cruises. Between 2003 and 2008, the number increased by about 740,000 each year. Write the point-slope form of an equation to find the total number of people y taking a cruise for any year x. Estimate the number of people who took a cruise in 2010.
SOLUTION: The statement “increased by 740,000 each year” is a constant rate, or the slope. Now substitute the slope and the point (2003, 9,500,000) into the point slope equation.
To find the number of people who took a cruise in 2010, substitute x = 2010.
So, approximately 14,680,000 people took a cruise in 2010.
Write an equation in function notation for each relation.
68.
SOLUTION: The y-intercept is 0. Find the slope by choosing two points. Choose (0, 0) and (1, 4).
So the slope intercept form is f (x) = 4x + 0 or f (x) = 4x.
69.
SOLUTION:
The y-intercept is 0. Find the slope by choosing two points. Choose (0, 0) and (2, −1).
So in slope intercept form is f (x) = −0.5x + 0 or f (x) = −0.5x.
Simplify.
70. b(b2)(b
3)
SOLUTION:
71. 2y(3y2)
SOLUTION:
72. −y4(−2y
3)
SOLUTION:
73. −3z3(−5z
4 + 2z)
SOLUTION:
74. 2m(−4m4) − 3(−5m
3)
SOLUTION:
75. 4p 2(−2p
3) + 2p
4(5p
6)
SOLUTION:
eSolutions Manual - Powered by Cognero Page 26
8-2 Multiplying a Polynomial by a Monomial
Find each product.
1. 5w( −3w2 + 2w − 4)
SOLUTION:
2. 6g2(3g
3 + 4g2 + 10g − 1)
SOLUTION:
3. 4km2(8km
2 + 2k2m + 5k)
SOLUTION:
4. −3p4r3(2p
2r4 − 6p 6
r3 − 5)
SOLUTION:
5. 2ab(7a4b
2 + a5b − 2a)
SOLUTION:
6. c2d
3(5cd
7 − 3c3d
2 − 4d3)
SOLUTION:
Simplify each expression.
7. t(4t2 + 15t + 4) − 4(3t − 1)
SOLUTION:
8. x(3x2 + 4) + 2(7x − 3)
SOLUTION:
9. −2d(d3c
2 − 4dc2 + 2d
2c) + c
2(dc
2 − 3d4)
SOLUTION:
10. −5w2(8w
2x − 11wx
2) + 6x(9wx
4 − 4w − 3x2)
SOLUTION:
11. GRIDDED RESPONSE Marlene is buying a new plasma television. The height of the screen of the television is one half the width plus 5 inches. The width is 30 inches. Find the height of the screen in inches.
SOLUTION: Let h be the height and let w be the width.
Let w = 30.
The height is 20 inches.
Solve each equation.
12. −6(11 − 2c) = 7(−2 − 2c)
SOLUTION:
13. t(2t + 3) + 20 = 2t(t − 3)
SOLUTION:
14. −2(w + 1) + w = 7 − 4w
SOLUTION:
15. 3(y − 2) + 2y = 4y + 14
SOLUTION:
16. a(a + 3) + a(a − 6) + 35 = a(a − 5) + a(a + 7)
SOLUTION:
17. n(n − 4) + n(n + 8) = n(n − 13) + n(n + 1) + 16
SOLUTION:
Find each product.
18. b(b2 − 12b + 1)
SOLUTION:
19. f (f2 + 2f + 25)
SOLUTION:
20. −3m3(2m
3 − 12m2 + 2m + 25)
SOLUTION:
21. 2j 2(5j
3 − 15j2 + 2j + 2)
SOLUTION:
22. 2pr2(2pr + 5p
2r − 15p )
SOLUTION:
23. 4t3u(2t
2u
2 − 10tu4 + 2)
SOLUTION:
Simplify each expression.
24. −3(5x2 + 2x + 9) + x(2x − 3)
SOLUTION:
25. a(−8a2 + 2a + 4) + 3(6a
2 − 4)
SOLUTION:
26. −4d(5d2 − 12) + 7(d + 5)
SOLUTION:
27. −9g(−2g + g2) + 3(g
2 + 4)
SOLUTION:
28. 2j (7j2k
2 + j k2 + 5k) − 9k(−2j
2k
2 + 2k2 + 3j )
SOLUTION:
29. 4n(2n3p
2 − 3np2 + 5n) + 4p (6n
2p − 2np
2 + 3p )
SOLUTION:
30. DAMS A new dam being built has the shape of a trapezoid. The base at the bottom of the dam is 2 times the
height. The base at the top of the dam is times the height minus 30 feet.
a. Write an expression to find the area of the trapezoidal cross section of the dam. b. If the height of the dam is 180 feet, find the area of this cross section.
SOLUTION:
a. The equation for the area of a trapezoid is .
b1
= 2h
b2 =
b. Let h = 180.
So, the area is 32,940 square feet.
Solve each equation.
31. 7(t2 + 5t − 9) + t = t(7t − 2) + 13
SOLUTION:
32. w(4w + 6) + 2w = 2(2w2 + 7w − 3)
SOLUTION:
33. 5(4z + 6) − 2(z − 4) = 7z(z + 4) − z(7z − 2) − 48
SOLUTION:
34. 9c(c − 11) + 10(5c − 3) = 3c(c + 5) + c(6c − 3) − 30
SOLUTION:
35. 2f (5f − 2) − 10(f2 − 3f + 6) = −8f (f + 4) + 4(2f
2 − 7f )
SOLUTION:
36. 2k(−3k + 4) + 6(k2 + 10) = k(4k + 8) − 2k(2k + 5)
SOLUTION:
Simplify each expression.
37. np2(30p
2 + 9n
2p − 12)
SOLUTION:
38. r2t(10r
3 + 5rt
3 + 15t
2)
SOLUTION:
39. −5q2w
3(4q + 7w) + 4qw
2(7q
2w + 2q) − 3qw(3q
2w
2 + 9)
SOLUTION:
40. −x2z(2z
2 + 4xz3) + xz
2(xz + 5x
3z) + x2
z3(3x
2z + 4xz)
SOLUTION:
41. PARKING A parking garage charges $30 per month plus $0.50 per daytime hour and $0.25 per hour during nights and weekends. Suppose Trent parks in the garage for 47 hours in January and h of those are night and weekend hours. a. Find an expression for Trent’s January bill. b. Find the cost if Trent had 12 hours of night and weekend hours.
SOLUTION: a. First, multiply the price per hour times the number of hours. Let h represent the number of weekend and night hours. For the weekend and night hours, he would spend 0.25h. The number of daytime hours is 47 – h, so he spent 0.50(47 - h) = 23.50 – 0.50h. To find the total cost, add these two values to the cost per month. 30 + (23.50 − 0.50h) + 0.25h = 53.50 − 0.25h. b. If Trent parked 12 night and weekend hours, h = 12
So, the cost for January is $50.50.
42. CCSS MODELING Che is building a dog house for his new puppy. The upper face of the dog house is a trapezoid. If the height of the trapezoid is 12 inches, find the area of the face of this piece of the dog house.
SOLUTION:
The formula for the area of a trapezoid is .
Now substitute h = 12 inches into the equation.
Therefore, the area of the trapezoid is 318 square inches.
43. TENNIS The tennis club is building a new tennis court with a path around it.
a. Write an expression for the area of the tennis court. b. Write an expression for the area of the path. c. What is the perimeter of the outside of the path?
SOLUTION: a. The area of the tennis court can be found by the equation . Substitute the values of the length and width and simplify.
b. The area of the path is the area of the large rectangle minus the area of the smaller rectangle, which is the area ofthe tennis court.
Area of the large rectangle = 2.5x(x + 6) or 2.5x2 + 15x.
Now subtract the area of the tennis court to find the area of just the path.
c. The perimeter of the outside of the path is the perimeter of the larger rectangle.
If x = 36 feet, then the perimeter of the outside of the path is 7(36) + 12 or 264 ft.
44. MULTIPLE REPRESENTATIONS In this problem, you will investigate the degree of the product of a monomialand a polynomial. a. TABULAR Write three monomials of different degrees and three polynomials of different degrees. Determine the degree of each monomial and polynomial. Multiply the monomials by the polynomials. Determine the degree of each product. Record your results in a table like the one below.
b. VERBAL Make a conjecture about the degree of the product of a monomial and a polynomial. What is the degree of the product of a monomial of degree a and a polynomial of degree b?
SOLUTION: a. There possible monomials are 2x, 3x
2, 4x
3. Three polynomials are x2 − 1, x5 + 1, x
6 + 1.
The degree for the monomial is the sum of the exponents. For the example given, case they are 1,2, and 3. The degree for the polynomial is the largest degree of the monomials. For the example given, they are 2, 5, and 6. Use the distributive property to find the product of the monomial and the polynomial. For the example give, they
are2x3 − 2x, 3x
7 + 3x
2, and 4x
9 + 4x. For the example given, the degree of the new polynomials are 3,7 and 9.
b. The degree of the product is the sum of the degree of the monomial and the degree of the polynomial, or a + b.
45. ERROR ANALYSIS Pearl and Ted both worked on this problem. Is either of them correct? Explain your reasoning.
SOLUTION:
Ted’s solution is correct. Pearl made a mistake when distributing on the middle term. She did not add the exponents correctly.
46. CCSS PERSEVERANCE Find p such that 3xp(4x
2p+3 + 2x3p−2
) = 12x12 + 6x
10.
SOLUTION: First simplify the left side of the equation.
Because the coefficients of the terms are the same, the corresponding terms’ exponents can be set equal to each other.
47. CHALLENGE Find 4x−3
y2(2x
5y−4 + 6x
−7y
6 − 4x0y
-2).
SOLUTION:
48. REASONING Is there a value for x that makes the statement (x + 2)2 = x2 + 22
true? If so, find a value for x. Explain your reasoning.
SOLUTION: Solve for x.
When 0 is substituted in for x in the equation, both sides are 22 or 4, which makes the equation true.
49. OPEN ENDED Write a monomial and a polynomial using n as the variable. Find their product.
SOLUTION: Choose a monomial with variable n and a polynomial with variable n. For example, let the monomial be 3n and the polynomial be 4n + 1. Find the product.
50. WRITING IN MATH Describe the steps to multiply a polynomial by a monomial.
SOLUTION: To multiply a polynomial by a monomial, use the Distributive Property. Multiply each term of the polynomial by the monomial. Then simplify by multiplying the coefficients together and using the Product of Powers Property for the variables. Consider the following example:
51. Every week a store sells j jeans and t T-shirts. The store makes $8 for each T-shirt and $12 for each pair of jeans. Which of the following expressions represents the total amount of money, in dollars, the store makes every week? A 8j + 12t B 12j + 8t C 20(j + t) D 96jt
SOLUTION: 12j is the amount made from jeans. 8t is the amount made from T-shirts. The sum, or 12j + 8t, represents the total amount the store makes every week. So, the correct choice is B.
52. If a = 5x + 7y and b = 2y − 3x, what is a + b?F 2x − 9y G 3y + 4x H 2x + 9y J 2x − 5y
SOLUTION:
So, the correct choice is H.
53. GEOMETRY A triangle has sides of length 5 inches and 8.5 inches. Which of the following cannot be the length of the third side? A 3.5 inches B 4 inches C 5.5 inches D 12 inches
SOLUTION: The sum of any two sides of a triangle must be greater than the length of the third side. Choice A: 3.5 + 5 = 8.5 8.5 Choice B: 3.5 + 5 = 8.5 > 8.5 Choice C: 5.5 + 5 = 10.5 > 8.5 Choice D: 12 + 5 = 17 > 8.5 Let x be the third side of the triangle.
3.5 cannot be the third side of the triangle. So, the correct choice is A.
54. SHORT RESPONSE Write an equation in which x varies directly as the cube of y and inversely as the square of z .
SOLUTION: If x varies directly as the cube of y , then both x and y must be in the numerators on either side of the equal sign.
When x varies inversely with the square of z. then the z must be in the denominator. Then . The
combination of the two is .
Find each sum or difference.
55. (2x2 − 7) + (8 − 5x
2)
SOLUTION:
56. (3z2 + 2z − 1) + (z
2 − 6)
SOLUTION:
57. (2a − 4a2 + 1) − (5a
2 − 2a − 6)
SOLUTION:
58. (a3 − 3a
2 + 4) − (4a
2 + 7)
SOLUTION:
59. (2ab − 3a + 4b) + (5a + 4ab)
SOLUTION:
60. (8c3 − 3c
2 + c − 2) − (3c
3 + 9)
SOLUTION:
Write a recursive formula for each sequence.61. 16, 2, –12, –26, ...
SOLUTION: Determine if there is a common difference. 2 – 16 = –14 –12 – 2 = –14 –26 – (–12) = –14 The common difference is –14. The first term is 16. To find the next term, subtract 14 from the previous term.
62. –5, 3, 11, 19, ...
SOLUTION: Determine if there is a common difference. 3 – (–5) = 8 11 – 3 = 8 19 – 11 = 8 The common difference is 8. The first term is 3. To find the next term, add 8 to the previous term.
63. 27, 43, 59, 75, ...
SOLUTION: Determine if there is a common difference. 43 – 27 = 16 59 – 43 = 16 75 – 59 = 16 The common difference is 16. The first term is 43. To find the next term, add 16 to the previous term.
64. 80, –200, 500, –1250, ...
SOLUTION: Check for a common difference. –200 – 80 = –280 500 – (–200) = 700 There is no common difference. Check for a common ratio. –200 ÷ 80 = –2.5 500 ÷ (–200) = –2.5 –1250 ÷ 500 = –2.5 The common ratio is –2.5. The first term is 80. To find the next term, multiply the previous term by –2.5.
65. 100, 60, 36, 21.6, ...
SOLUTION: Check for a common difference. 60 – 100 = –40 36 – 60 = –24 There is no common difference. Check for a common ratio. 60 ÷ 100 = 0.6 36 ÷ 60 = 0.6 21.6 ÷ 36 = 0.6 The common ratio is 0.6. The first term is 100. To find the next term, multiply the previous term by 0.6.
66.
SOLUTION: Check for a common difference.
There is no common difference. Check for a common ratio.
4 ÷ 1 = 4 The common ratio is 4.
The first term is . To find the next term, multiply the previous term by 4.
67. TRAVEL In 2003, about 9.5 million people took cruises. Between 2003 and 2008, the number increased by about 740,000 each year. Write the point-slope form of an equation to find the total number of people y taking a cruise for any year x. Estimate the number of people who took a cruise in 2010.
SOLUTION: The statement “increased by 740,000 each year” is a constant rate, or the slope. Now substitute the slope and the point (2003, 9,500,000) into the point slope equation.
To find the number of people who took a cruise in 2010, substitute x = 2010.
So, approximately 14,680,000 people took a cruise in 2010.
Write an equation in function notation for each relation.
68.
SOLUTION: The y-intercept is 0. Find the slope by choosing two points. Choose (0, 0) and (1, 4).
So the slope intercept form is f (x) = 4x + 0 or f (x) = 4x.
69.
SOLUTION:
The y-intercept is 0. Find the slope by choosing two points. Choose (0, 0) and (2, −1).
So in slope intercept form is f (x) = −0.5x + 0 or f (x) = −0.5x.
Simplify.
70. b(b2)(b
3)
SOLUTION:
71. 2y(3y2)
SOLUTION:
72. −y4(−2y
3)
SOLUTION:
73. −3z3(−5z
4 + 2z)
SOLUTION:
74. 2m(−4m4) − 3(−5m
3)
SOLUTION:
75. 4p 2(−2p
3) + 2p
4(5p
6)
SOLUTION:
eSolutions Manual - Powered by Cognero Page 27
8-2 Multiplying a Polynomial by a Monomial
Find each product.
1. 5w( −3w2 + 2w − 4)
SOLUTION:
2. 6g2(3g
3 + 4g2 + 10g − 1)
SOLUTION:
3. 4km2(8km
2 + 2k2m + 5k)
SOLUTION:
4. −3p4r3(2p
2r4 − 6p 6
r3 − 5)
SOLUTION:
5. 2ab(7a4b
2 + a5b − 2a)
SOLUTION:
6. c2d
3(5cd
7 − 3c3d
2 − 4d3)
SOLUTION:
Simplify each expression.
7. t(4t2 + 15t + 4) − 4(3t − 1)
SOLUTION:
8. x(3x2 + 4) + 2(7x − 3)
SOLUTION:
9. −2d(d3c
2 − 4dc2 + 2d
2c) + c
2(dc
2 − 3d4)
SOLUTION:
10. −5w2(8w
2x − 11wx
2) + 6x(9wx
4 − 4w − 3x2)
SOLUTION:
11. GRIDDED RESPONSE Marlene is buying a new plasma television. The height of the screen of the television is one half the width plus 5 inches. The width is 30 inches. Find the height of the screen in inches.
SOLUTION: Let h be the height and let w be the width.
Let w = 30.
The height is 20 inches.
Solve each equation.
12. −6(11 − 2c) = 7(−2 − 2c)
SOLUTION:
13. t(2t + 3) + 20 = 2t(t − 3)
SOLUTION:
14. −2(w + 1) + w = 7 − 4w
SOLUTION:
15. 3(y − 2) + 2y = 4y + 14
SOLUTION:
16. a(a + 3) + a(a − 6) + 35 = a(a − 5) + a(a + 7)
SOLUTION:
17. n(n − 4) + n(n + 8) = n(n − 13) + n(n + 1) + 16
SOLUTION:
Find each product.
18. b(b2 − 12b + 1)
SOLUTION:
19. f (f2 + 2f + 25)
SOLUTION:
20. −3m3(2m
3 − 12m2 + 2m + 25)
SOLUTION:
21. 2j 2(5j
3 − 15j2 + 2j + 2)
SOLUTION:
22. 2pr2(2pr + 5p
2r − 15p )
SOLUTION:
23. 4t3u(2t
2u
2 − 10tu4 + 2)
SOLUTION:
Simplify each expression.
24. −3(5x2 + 2x + 9) + x(2x − 3)
SOLUTION:
25. a(−8a2 + 2a + 4) + 3(6a
2 − 4)
SOLUTION:
26. −4d(5d2 − 12) + 7(d + 5)
SOLUTION:
27. −9g(−2g + g2) + 3(g
2 + 4)
SOLUTION:
28. 2j (7j2k
2 + j k2 + 5k) − 9k(−2j
2k
2 + 2k2 + 3j )
SOLUTION:
29. 4n(2n3p
2 − 3np2 + 5n) + 4p (6n
2p − 2np
2 + 3p )
SOLUTION:
30. DAMS A new dam being built has the shape of a trapezoid. The base at the bottom of the dam is 2 times the
height. The base at the top of the dam is times the height minus 30 feet.
a. Write an expression to find the area of the trapezoidal cross section of the dam. b. If the height of the dam is 180 feet, find the area of this cross section.
SOLUTION:
a. The equation for the area of a trapezoid is .
b1
= 2h
b2 =
b. Let h = 180.
So, the area is 32,940 square feet.
Solve each equation.
31. 7(t2 + 5t − 9) + t = t(7t − 2) + 13
SOLUTION:
32. w(4w + 6) + 2w = 2(2w2 + 7w − 3)
SOLUTION:
33. 5(4z + 6) − 2(z − 4) = 7z(z + 4) − z(7z − 2) − 48
SOLUTION:
34. 9c(c − 11) + 10(5c − 3) = 3c(c + 5) + c(6c − 3) − 30
SOLUTION:
35. 2f (5f − 2) − 10(f2 − 3f + 6) = −8f (f + 4) + 4(2f
2 − 7f )
SOLUTION:
36. 2k(−3k + 4) + 6(k2 + 10) = k(4k + 8) − 2k(2k + 5)
SOLUTION:
Simplify each expression.
37. np2(30p
2 + 9n
2p − 12)
SOLUTION:
38. r2t(10r
3 + 5rt
3 + 15t
2)
SOLUTION:
39. −5q2w
3(4q + 7w) + 4qw
2(7q
2w + 2q) − 3qw(3q
2w
2 + 9)
SOLUTION:
40. −x2z(2z
2 + 4xz3) + xz
2(xz + 5x
3z) + x2
z3(3x
2z + 4xz)
SOLUTION:
41. PARKING A parking garage charges $30 per month plus $0.50 per daytime hour and $0.25 per hour during nights and weekends. Suppose Trent parks in the garage for 47 hours in January and h of those are night and weekend hours. a. Find an expression for Trent’s January bill. b. Find the cost if Trent had 12 hours of night and weekend hours.
SOLUTION: a. First, multiply the price per hour times the number of hours. Let h represent the number of weekend and night hours. For the weekend and night hours, he would spend 0.25h. The number of daytime hours is 47 – h, so he spent 0.50(47 - h) = 23.50 – 0.50h. To find the total cost, add these two values to the cost per month. 30 + (23.50 − 0.50h) + 0.25h = 53.50 − 0.25h. b. If Trent parked 12 night and weekend hours, h = 12
So, the cost for January is $50.50.
42. CCSS MODELING Che is building a dog house for his new puppy. The upper face of the dog house is a trapezoid. If the height of the trapezoid is 12 inches, find the area of the face of this piece of the dog house.
SOLUTION:
The formula for the area of a trapezoid is .
Now substitute h = 12 inches into the equation.
Therefore, the area of the trapezoid is 318 square inches.
43. TENNIS The tennis club is building a new tennis court with a path around it.
a. Write an expression for the area of the tennis court. b. Write an expression for the area of the path. c. What is the perimeter of the outside of the path?
SOLUTION: a. The area of the tennis court can be found by the equation . Substitute the values of the length and width and simplify.
b. The area of the path is the area of the large rectangle minus the area of the smaller rectangle, which is the area ofthe tennis court.
Area of the large rectangle = 2.5x(x + 6) or 2.5x2 + 15x.
Now subtract the area of the tennis court to find the area of just the path.
c. The perimeter of the outside of the path is the perimeter of the larger rectangle.
If x = 36 feet, then the perimeter of the outside of the path is 7(36) + 12 or 264 ft.
44. MULTIPLE REPRESENTATIONS In this problem, you will investigate the degree of the product of a monomialand a polynomial. a. TABULAR Write three monomials of different degrees and three polynomials of different degrees. Determine the degree of each monomial and polynomial. Multiply the monomials by the polynomials. Determine the degree of each product. Record your results in a table like the one below.
b. VERBAL Make a conjecture about the degree of the product of a monomial and a polynomial. What is the degree of the product of a monomial of degree a and a polynomial of degree b?
SOLUTION: a. There possible monomials are 2x, 3x
2, 4x
3. Three polynomials are x2 − 1, x5 + 1, x
6 + 1.
The degree for the monomial is the sum of the exponents. For the example given, case they are 1,2, and 3. The degree for the polynomial is the largest degree of the monomials. For the example given, they are 2, 5, and 6. Use the distributive property to find the product of the monomial and the polynomial. For the example give, they
are2x3 − 2x, 3x
7 + 3x
2, and 4x
9 + 4x. For the example given, the degree of the new polynomials are 3,7 and 9.
b. The degree of the product is the sum of the degree of the monomial and the degree of the polynomial, or a + b.
45. ERROR ANALYSIS Pearl and Ted both worked on this problem. Is either of them correct? Explain your reasoning.
SOLUTION:
Ted’s solution is correct. Pearl made a mistake when distributing on the middle term. She did not add the exponents correctly.
46. CCSS PERSEVERANCE Find p such that 3xp(4x
2p+3 + 2x3p−2
) = 12x12 + 6x
10.
SOLUTION: First simplify the left side of the equation.
Because the coefficients of the terms are the same, the corresponding terms’ exponents can be set equal to each other.
47. CHALLENGE Find 4x−3
y2(2x
5y−4 + 6x
−7y
6 − 4x0y
-2).
SOLUTION:
48. REASONING Is there a value for x that makes the statement (x + 2)2 = x2 + 22
true? If so, find a value for x. Explain your reasoning.
SOLUTION: Solve for x.
When 0 is substituted in for x in the equation, both sides are 22 or 4, which makes the equation true.
49. OPEN ENDED Write a monomial and a polynomial using n as the variable. Find their product.
SOLUTION: Choose a monomial with variable n and a polynomial with variable n. For example, let the monomial be 3n and the polynomial be 4n + 1. Find the product.
50. WRITING IN MATH Describe the steps to multiply a polynomial by a monomial.
SOLUTION: To multiply a polynomial by a monomial, use the Distributive Property. Multiply each term of the polynomial by the monomial. Then simplify by multiplying the coefficients together and using the Product of Powers Property for the variables. Consider the following example:
51. Every week a store sells j jeans and t T-shirts. The store makes $8 for each T-shirt and $12 for each pair of jeans. Which of the following expressions represents the total amount of money, in dollars, the store makes every week? A 8j + 12t B 12j + 8t C 20(j + t) D 96jt
SOLUTION: 12j is the amount made from jeans. 8t is the amount made from T-shirts. The sum, or 12j + 8t, represents the total amount the store makes every week. So, the correct choice is B.
52. If a = 5x + 7y and b = 2y − 3x, what is a + b?F 2x − 9y G 3y + 4x H 2x + 9y J 2x − 5y
SOLUTION:
So, the correct choice is H.
53. GEOMETRY A triangle has sides of length 5 inches and 8.5 inches. Which of the following cannot be the length of the third side? A 3.5 inches B 4 inches C 5.5 inches D 12 inches
SOLUTION: The sum of any two sides of a triangle must be greater than the length of the third side. Choice A: 3.5 + 5 = 8.5 8.5 Choice B: 3.5 + 5 = 8.5 > 8.5 Choice C: 5.5 + 5 = 10.5 > 8.5 Choice D: 12 + 5 = 17 > 8.5 Let x be the third side of the triangle.
3.5 cannot be the third side of the triangle. So, the correct choice is A.
54. SHORT RESPONSE Write an equation in which x varies directly as the cube of y and inversely as the square of z .
SOLUTION: If x varies directly as the cube of y , then both x and y must be in the numerators on either side of the equal sign.
When x varies inversely with the square of z. then the z must be in the denominator. Then . The
combination of the two is .
Find each sum or difference.
55. (2x2 − 7) + (8 − 5x
2)
SOLUTION:
56. (3z2 + 2z − 1) + (z
2 − 6)
SOLUTION:
57. (2a − 4a2 + 1) − (5a
2 − 2a − 6)
SOLUTION:
58. (a3 − 3a
2 + 4) − (4a
2 + 7)
SOLUTION:
59. (2ab − 3a + 4b) + (5a + 4ab)
SOLUTION:
60. (8c3 − 3c
2 + c − 2) − (3c
3 + 9)
SOLUTION:
Write a recursive formula for each sequence.61. 16, 2, –12, –26, ...
SOLUTION: Determine if there is a common difference. 2 – 16 = –14 –12 – 2 = –14 –26 – (–12) = –14 The common difference is –14. The first term is 16. To find the next term, subtract 14 from the previous term.
62. –5, 3, 11, 19, ...
SOLUTION: Determine if there is a common difference. 3 – (–5) = 8 11 – 3 = 8 19 – 11 = 8 The common difference is 8. The first term is 3. To find the next term, add 8 to the previous term.
63. 27, 43, 59, 75, ...
SOLUTION: Determine if there is a common difference. 43 – 27 = 16 59 – 43 = 16 75 – 59 = 16 The common difference is 16. The first term is 43. To find the next term, add 16 to the previous term.
64. 80, –200, 500, –1250, ...
SOLUTION: Check for a common difference. –200 – 80 = –280 500 – (–200) = 700 There is no common difference. Check for a common ratio. –200 ÷ 80 = –2.5 500 ÷ (–200) = –2.5 –1250 ÷ 500 = –2.5 The common ratio is –2.5. The first term is 80. To find the next term, multiply the previous term by –2.5.
65. 100, 60, 36, 21.6, ...
SOLUTION: Check for a common difference. 60 – 100 = –40 36 – 60 = –24 There is no common difference. Check for a common ratio. 60 ÷ 100 = 0.6 36 ÷ 60 = 0.6 21.6 ÷ 36 = 0.6 The common ratio is 0.6. The first term is 100. To find the next term, multiply the previous term by 0.6.
66.
SOLUTION: Check for a common difference.
There is no common difference. Check for a common ratio.
4 ÷ 1 = 4 The common ratio is 4.
The first term is . To find the next term, multiply the previous term by 4.
67. TRAVEL In 2003, about 9.5 million people took cruises. Between 2003 and 2008, the number increased by about 740,000 each year. Write the point-slope form of an equation to find the total number of people y taking a cruise for any year x. Estimate the number of people who took a cruise in 2010.
SOLUTION: The statement “increased by 740,000 each year” is a constant rate, or the slope. Now substitute the slope and the point (2003, 9,500,000) into the point slope equation.
To find the number of people who took a cruise in 2010, substitute x = 2010.
So, approximately 14,680,000 people took a cruise in 2010.
Write an equation in function notation for each relation.
68.
SOLUTION: The y-intercept is 0. Find the slope by choosing two points. Choose (0, 0) and (1, 4).
So the slope intercept form is f (x) = 4x + 0 or f (x) = 4x.
69.
SOLUTION:
The y-intercept is 0. Find the slope by choosing two points. Choose (0, 0) and (2, −1).
So in slope intercept form is f (x) = −0.5x + 0 or f (x) = −0.5x.
Simplify.
70. b(b2)(b
3)
SOLUTION:
71. 2y(3y2)
SOLUTION:
72. −y4(−2y
3)
SOLUTION:
73. −3z3(−5z
4 + 2z)
SOLUTION:
74. 2m(−4m4) − 3(−5m
3)
SOLUTION:
75. 4p 2(−2p
3) + 2p
4(5p
6)
SOLUTION:
eSolutions Manual - Powered by Cognero Page 28
8-2 Multiplying a Polynomial by a Monomial
Find each product.
1. 5w( −3w2 + 2w − 4)
SOLUTION:
2. 6g2(3g
3 + 4g2 + 10g − 1)
SOLUTION:
3. 4km2(8km
2 + 2k2m + 5k)
SOLUTION:
4. −3p4r3(2p
2r4 − 6p 6
r3 − 5)
SOLUTION:
5. 2ab(7a4b
2 + a5b − 2a)
SOLUTION:
6. c2d
3(5cd
7 − 3c3d
2 − 4d3)
SOLUTION:
Simplify each expression.
7. t(4t2 + 15t + 4) − 4(3t − 1)
SOLUTION:
8. x(3x2 + 4) + 2(7x − 3)
SOLUTION:
9. −2d(d3c
2 − 4dc2 + 2d
2c) + c
2(dc
2 − 3d4)
SOLUTION:
10. −5w2(8w
2x − 11wx
2) + 6x(9wx
4 − 4w − 3x2)
SOLUTION:
11. GRIDDED RESPONSE Marlene is buying a new plasma television. The height of the screen of the television is one half the width plus 5 inches. The width is 30 inches. Find the height of the screen in inches.
SOLUTION: Let h be the height and let w be the width.
Let w = 30.
The height is 20 inches.
Solve each equation.
12. −6(11 − 2c) = 7(−2 − 2c)
SOLUTION:
13. t(2t + 3) + 20 = 2t(t − 3)
SOLUTION:
14. −2(w + 1) + w = 7 − 4w
SOLUTION:
15. 3(y − 2) + 2y = 4y + 14
SOLUTION:
16. a(a + 3) + a(a − 6) + 35 = a(a − 5) + a(a + 7)
SOLUTION:
17. n(n − 4) + n(n + 8) = n(n − 13) + n(n + 1) + 16
SOLUTION:
Find each product.
18. b(b2 − 12b + 1)
SOLUTION:
19. f (f2 + 2f + 25)
SOLUTION:
20. −3m3(2m
3 − 12m2 + 2m + 25)
SOLUTION:
21. 2j 2(5j
3 − 15j2 + 2j + 2)
SOLUTION:
22. 2pr2(2pr + 5p
2r − 15p )
SOLUTION:
23. 4t3u(2t
2u
2 − 10tu4 + 2)
SOLUTION:
Simplify each expression.
24. −3(5x2 + 2x + 9) + x(2x − 3)
SOLUTION:
25. a(−8a2 + 2a + 4) + 3(6a
2 − 4)
SOLUTION:
26. −4d(5d2 − 12) + 7(d + 5)
SOLUTION:
27. −9g(−2g + g2) + 3(g
2 + 4)
SOLUTION:
28. 2j (7j2k
2 + j k2 + 5k) − 9k(−2j
2k
2 + 2k2 + 3j )
SOLUTION:
29. 4n(2n3p
2 − 3np2 + 5n) + 4p (6n
2p − 2np
2 + 3p )
SOLUTION:
30. DAMS A new dam being built has the shape of a trapezoid. The base at the bottom of the dam is 2 times the
height. The base at the top of the dam is times the height minus 30 feet.
a. Write an expression to find the area of the trapezoidal cross section of the dam. b. If the height of the dam is 180 feet, find the area of this cross section.
SOLUTION:
a. The equation for the area of a trapezoid is .
b1
= 2h
b2 =
b. Let h = 180.
So, the area is 32,940 square feet.
Solve each equation.
31. 7(t2 + 5t − 9) + t = t(7t − 2) + 13
SOLUTION:
32. w(4w + 6) + 2w = 2(2w2 + 7w − 3)
SOLUTION:
33. 5(4z + 6) − 2(z − 4) = 7z(z + 4) − z(7z − 2) − 48
SOLUTION:
34. 9c(c − 11) + 10(5c − 3) = 3c(c + 5) + c(6c − 3) − 30
SOLUTION:
35. 2f (5f − 2) − 10(f2 − 3f + 6) = −8f (f + 4) + 4(2f
2 − 7f )
SOLUTION:
36. 2k(−3k + 4) + 6(k2 + 10) = k(4k + 8) − 2k(2k + 5)
SOLUTION:
Simplify each expression.
37. np2(30p
2 + 9n
2p − 12)
SOLUTION:
38. r2t(10r
3 + 5rt
3 + 15t
2)
SOLUTION:
39. −5q2w
3(4q + 7w) + 4qw
2(7q
2w + 2q) − 3qw(3q
2w
2 + 9)
SOLUTION:
40. −x2z(2z
2 + 4xz3) + xz
2(xz + 5x
3z) + x2
z3(3x
2z + 4xz)
SOLUTION:
41. PARKING A parking garage charges $30 per month plus $0.50 per daytime hour and $0.25 per hour during nights and weekends. Suppose Trent parks in the garage for 47 hours in January and h of those are night and weekend hours. a. Find an expression for Trent’s January bill. b. Find the cost if Trent had 12 hours of night and weekend hours.
SOLUTION: a. First, multiply the price per hour times the number of hours. Let h represent the number of weekend and night hours. For the weekend and night hours, he would spend 0.25h. The number of daytime hours is 47 – h, so he spent 0.50(47 - h) = 23.50 – 0.50h. To find the total cost, add these two values to the cost per month. 30 + (23.50 − 0.50h) + 0.25h = 53.50 − 0.25h. b. If Trent parked 12 night and weekend hours, h = 12
So, the cost for January is $50.50.
42. CCSS MODELING Che is building a dog house for his new puppy. The upper face of the dog house is a trapezoid. If the height of the trapezoid is 12 inches, find the area of the face of this piece of the dog house.
SOLUTION:
The formula for the area of a trapezoid is .
Now substitute h = 12 inches into the equation.
Therefore, the area of the trapezoid is 318 square inches.
43. TENNIS The tennis club is building a new tennis court with a path around it.
a. Write an expression for the area of the tennis court. b. Write an expression for the area of the path. c. What is the perimeter of the outside of the path?
SOLUTION: a. The area of the tennis court can be found by the equation . Substitute the values of the length and width and simplify.
b. The area of the path is the area of the large rectangle minus the area of the smaller rectangle, which is the area ofthe tennis court.
Area of the large rectangle = 2.5x(x + 6) or 2.5x2 + 15x.
Now subtract the area of the tennis court to find the area of just the path.
c. The perimeter of the outside of the path is the perimeter of the larger rectangle.
If x = 36 feet, then the perimeter of the outside of the path is 7(36) + 12 or 264 ft.
44. MULTIPLE REPRESENTATIONS In this problem, you will investigate the degree of the product of a monomialand a polynomial. a. TABULAR Write three monomials of different degrees and three polynomials of different degrees. Determine the degree of each monomial and polynomial. Multiply the monomials by the polynomials. Determine the degree of each product. Record your results in a table like the one below.
b. VERBAL Make a conjecture about the degree of the product of a monomial and a polynomial. What is the degree of the product of a monomial of degree a and a polynomial of degree b?
SOLUTION: a. There possible monomials are 2x, 3x
2, 4x
3. Three polynomials are x2 − 1, x5 + 1, x
6 + 1.
The degree for the monomial is the sum of the exponents. For the example given, case they are 1,2, and 3. The degree for the polynomial is the largest degree of the monomials. For the example given, they are 2, 5, and 6. Use the distributive property to find the product of the monomial and the polynomial. For the example give, they
are2x3 − 2x, 3x
7 + 3x
2, and 4x
9 + 4x. For the example given, the degree of the new polynomials are 3,7 and 9.
b. The degree of the product is the sum of the degree of the monomial and the degree of the polynomial, or a + b.
45. ERROR ANALYSIS Pearl and Ted both worked on this problem. Is either of them correct? Explain your reasoning.
SOLUTION:
Ted’s solution is correct. Pearl made a mistake when distributing on the middle term. She did not add the exponents correctly.
46. CCSS PERSEVERANCE Find p such that 3xp(4x
2p+3 + 2x3p−2
) = 12x12 + 6x
10.
SOLUTION: First simplify the left side of the equation.
Because the coefficients of the terms are the same, the corresponding terms’ exponents can be set equal to each other.
47. CHALLENGE Find 4x−3
y2(2x
5y−4 + 6x
−7y
6 − 4x0y
-2).
SOLUTION:
48. REASONING Is there a value for x that makes the statement (x + 2)2 = x2 + 22
true? If so, find a value for x. Explain your reasoning.
SOLUTION: Solve for x.
When 0 is substituted in for x in the equation, both sides are 22 or 4, which makes the equation true.
49. OPEN ENDED Write a monomial and a polynomial using n as the variable. Find their product.
SOLUTION: Choose a monomial with variable n and a polynomial with variable n. For example, let the monomial be 3n and the polynomial be 4n + 1. Find the product.
50. WRITING IN MATH Describe the steps to multiply a polynomial by a monomial.
SOLUTION: To multiply a polynomial by a monomial, use the Distributive Property. Multiply each term of the polynomial by the monomial. Then simplify by multiplying the coefficients together and using the Product of Powers Property for the variables. Consider the following example:
51. Every week a store sells j jeans and t T-shirts. The store makes $8 for each T-shirt and $12 for each pair of jeans. Which of the following expressions represents the total amount of money, in dollars, the store makes every week? A 8j + 12t B 12j + 8t C 20(j + t) D 96jt
SOLUTION: 12j is the amount made from jeans. 8t is the amount made from T-shirts. The sum, or 12j + 8t, represents the total amount the store makes every week. So, the correct choice is B.
52. If a = 5x + 7y and b = 2y − 3x, what is a + b?F 2x − 9y G 3y + 4x H 2x + 9y J 2x − 5y
SOLUTION:
So, the correct choice is H.
53. GEOMETRY A triangle has sides of length 5 inches and 8.5 inches. Which of the following cannot be the length of the third side? A 3.5 inches B 4 inches C 5.5 inches D 12 inches
SOLUTION: The sum of any two sides of a triangle must be greater than the length of the third side. Choice A: 3.5 + 5 = 8.5 8.5 Choice B: 3.5 + 5 = 8.5 > 8.5 Choice C: 5.5 + 5 = 10.5 > 8.5 Choice D: 12 + 5 = 17 > 8.5 Let x be the third side of the triangle.
3.5 cannot be the third side of the triangle. So, the correct choice is A.
54. SHORT RESPONSE Write an equation in which x varies directly as the cube of y and inversely as the square of z .
SOLUTION: If x varies directly as the cube of y , then both x and y must be in the numerators on either side of the equal sign.
When x varies inversely with the square of z. then the z must be in the denominator. Then . The
combination of the two is .
Find each sum or difference.
55. (2x2 − 7) + (8 − 5x
2)
SOLUTION:
56. (3z2 + 2z − 1) + (z
2 − 6)
SOLUTION:
57. (2a − 4a2 + 1) − (5a
2 − 2a − 6)
SOLUTION:
58. (a3 − 3a
2 + 4) − (4a
2 + 7)
SOLUTION:
59. (2ab − 3a + 4b) + (5a + 4ab)
SOLUTION:
60. (8c3 − 3c
2 + c − 2) − (3c
3 + 9)
SOLUTION:
Write a recursive formula for each sequence.61. 16, 2, –12, –26, ...
SOLUTION: Determine if there is a common difference. 2 – 16 = –14 –12 – 2 = –14 –26 – (–12) = –14 The common difference is –14. The first term is 16. To find the next term, subtract 14 from the previous term.
62. –5, 3, 11, 19, ...
SOLUTION: Determine if there is a common difference. 3 – (–5) = 8 11 – 3 = 8 19 – 11 = 8 The common difference is 8. The first term is 3. To find the next term, add 8 to the previous term.
63. 27, 43, 59, 75, ...
SOLUTION: Determine if there is a common difference. 43 – 27 = 16 59 – 43 = 16 75 – 59 = 16 The common difference is 16. The first term is 43. To find the next term, add 16 to the previous term.
64. 80, –200, 500, –1250, ...
SOLUTION: Check for a common difference. –200 – 80 = –280 500 – (–200) = 700 There is no common difference. Check for a common ratio. –200 ÷ 80 = –2.5 500 ÷ (–200) = –2.5 –1250 ÷ 500 = –2.5 The common ratio is –2.5. The first term is 80. To find the next term, multiply the previous term by –2.5.
65. 100, 60, 36, 21.6, ...
SOLUTION: Check for a common difference. 60 – 100 = –40 36 – 60 = –24 There is no common difference. Check for a common ratio. 60 ÷ 100 = 0.6 36 ÷ 60 = 0.6 21.6 ÷ 36 = 0.6 The common ratio is 0.6. The first term is 100. To find the next term, multiply the previous term by 0.6.
66.
SOLUTION: Check for a common difference.
There is no common difference. Check for a common ratio.
4 ÷ 1 = 4 The common ratio is 4.
The first term is . To find the next term, multiply the previous term by 4.
67. TRAVEL In 2003, about 9.5 million people took cruises. Between 2003 and 2008, the number increased by about 740,000 each year. Write the point-slope form of an equation to find the total number of people y taking a cruise for any year x. Estimate the number of people who took a cruise in 2010.
SOLUTION: The statement “increased by 740,000 each year” is a constant rate, or the slope. Now substitute the slope and the point (2003, 9,500,000) into the point slope equation.
To find the number of people who took a cruise in 2010, substitute x = 2010.
So, approximately 14,680,000 people took a cruise in 2010.
Write an equation in function notation for each relation.
68.
SOLUTION: The y-intercept is 0. Find the slope by choosing two points. Choose (0, 0) and (1, 4).
So the slope intercept form is f (x) = 4x + 0 or f (x) = 4x.
69.
SOLUTION:
The y-intercept is 0. Find the slope by choosing two points. Choose (0, 0) and (2, −1).
So in slope intercept form is f (x) = −0.5x + 0 or f (x) = −0.5x.
Simplify.
70. b(b2)(b
3)
SOLUTION:
71. 2y(3y2)
SOLUTION:
72. −y4(−2y
3)
SOLUTION:
73. −3z3(−5z
4 + 2z)
SOLUTION:
74. 2m(−4m4) − 3(−5m
3)
SOLUTION:
75. 4p 2(−2p
3) + 2p
4(5p
6)
SOLUTION:
eSolutions Manual - Powered by Cognero Page 29
8-2 Multiplying a Polynomial by a Monomial
Find each product.
1. 5w( −3w2 + 2w − 4)
SOLUTION:
2. 6g2(3g
3 + 4g2 + 10g − 1)
SOLUTION:
3. 4km2(8km
2 + 2k2m + 5k)
SOLUTION:
4. −3p4r3(2p
2r4 − 6p 6
r3 − 5)
SOLUTION:
5. 2ab(7a4b
2 + a5b − 2a)
SOLUTION:
6. c2d
3(5cd
7 − 3c3d
2 − 4d3)
SOLUTION:
Simplify each expression.
7. t(4t2 + 15t + 4) − 4(3t − 1)
SOLUTION:
8. x(3x2 + 4) + 2(7x − 3)
SOLUTION:
9. −2d(d3c
2 − 4dc2 + 2d
2c) + c
2(dc
2 − 3d4)
SOLUTION:
10. −5w2(8w
2x − 11wx
2) + 6x(9wx
4 − 4w − 3x2)
SOLUTION:
11. GRIDDED RESPONSE Marlene is buying a new plasma television. The height of the screen of the television is one half the width plus 5 inches. The width is 30 inches. Find the height of the screen in inches.
SOLUTION: Let h be the height and let w be the width.
Let w = 30.
The height is 20 inches.
Solve each equation.
12. −6(11 − 2c) = 7(−2 − 2c)
SOLUTION:
13. t(2t + 3) + 20 = 2t(t − 3)
SOLUTION:
14. −2(w + 1) + w = 7 − 4w
SOLUTION:
15. 3(y − 2) + 2y = 4y + 14
SOLUTION:
16. a(a + 3) + a(a − 6) + 35 = a(a − 5) + a(a + 7)
SOLUTION:
17. n(n − 4) + n(n + 8) = n(n − 13) + n(n + 1) + 16
SOLUTION:
Find each product.
18. b(b2 − 12b + 1)
SOLUTION:
19. f (f2 + 2f + 25)
SOLUTION:
20. −3m3(2m
3 − 12m2 + 2m + 25)
SOLUTION:
21. 2j 2(5j
3 − 15j2 + 2j + 2)
SOLUTION:
22. 2pr2(2pr + 5p
2r − 15p )
SOLUTION:
23. 4t3u(2t
2u
2 − 10tu4 + 2)
SOLUTION:
Simplify each expression.
24. −3(5x2 + 2x + 9) + x(2x − 3)
SOLUTION:
25. a(−8a2 + 2a + 4) + 3(6a
2 − 4)
SOLUTION:
26. −4d(5d2 − 12) + 7(d + 5)
SOLUTION:
27. −9g(−2g + g2) + 3(g
2 + 4)
SOLUTION:
28. 2j (7j2k
2 + j k2 + 5k) − 9k(−2j
2k
2 + 2k2 + 3j )
SOLUTION:
29. 4n(2n3p
2 − 3np2 + 5n) + 4p (6n
2p − 2np
2 + 3p )
SOLUTION:
30. DAMS A new dam being built has the shape of a trapezoid. The base at the bottom of the dam is 2 times the
height. The base at the top of the dam is times the height minus 30 feet.
a. Write an expression to find the area of the trapezoidal cross section of the dam. b. If the height of the dam is 180 feet, find the area of this cross section.
SOLUTION:
a. The equation for the area of a trapezoid is .
b1
= 2h
b2 =
b. Let h = 180.
So, the area is 32,940 square feet.
Solve each equation.
31. 7(t2 + 5t − 9) + t = t(7t − 2) + 13
SOLUTION:
32. w(4w + 6) + 2w = 2(2w2 + 7w − 3)
SOLUTION:
33. 5(4z + 6) − 2(z − 4) = 7z(z + 4) − z(7z − 2) − 48
SOLUTION:
34. 9c(c − 11) + 10(5c − 3) = 3c(c + 5) + c(6c − 3) − 30
SOLUTION:
35. 2f (5f − 2) − 10(f2 − 3f + 6) = −8f (f + 4) + 4(2f
2 − 7f )
SOLUTION:
36. 2k(−3k + 4) + 6(k2 + 10) = k(4k + 8) − 2k(2k + 5)
SOLUTION:
Simplify each expression.
37. np2(30p
2 + 9n
2p − 12)
SOLUTION:
38. r2t(10r
3 + 5rt
3 + 15t
2)
SOLUTION:
39. −5q2w
3(4q + 7w) + 4qw
2(7q
2w + 2q) − 3qw(3q
2w
2 + 9)
SOLUTION:
40. −x2z(2z
2 + 4xz3) + xz
2(xz + 5x
3z) + x2
z3(3x
2z + 4xz)
SOLUTION:
41. PARKING A parking garage charges $30 per month plus $0.50 per daytime hour and $0.25 per hour during nights and weekends. Suppose Trent parks in the garage for 47 hours in January and h of those are night and weekend hours. a. Find an expression for Trent’s January bill. b. Find the cost if Trent had 12 hours of night and weekend hours.
SOLUTION: a. First, multiply the price per hour times the number of hours. Let h represent the number of weekend and night hours. For the weekend and night hours, he would spend 0.25h. The number of daytime hours is 47 – h, so he spent 0.50(47 - h) = 23.50 – 0.50h. To find the total cost, add these two values to the cost per month. 30 + (23.50 − 0.50h) + 0.25h = 53.50 − 0.25h. b. If Trent parked 12 night and weekend hours, h = 12
So, the cost for January is $50.50.
42. CCSS MODELING Che is building a dog house for his new puppy. The upper face of the dog house is a trapezoid. If the height of the trapezoid is 12 inches, find the area of the face of this piece of the dog house.
SOLUTION:
The formula for the area of a trapezoid is .
Now substitute h = 12 inches into the equation.
Therefore, the area of the trapezoid is 318 square inches.
43. TENNIS The tennis club is building a new tennis court with a path around it.
a. Write an expression for the area of the tennis court. b. Write an expression for the area of the path. c. What is the perimeter of the outside of the path?
SOLUTION: a. The area of the tennis court can be found by the equation . Substitute the values of the length and width and simplify.
b. The area of the path is the area of the large rectangle minus the area of the smaller rectangle, which is the area ofthe tennis court.
Area of the large rectangle = 2.5x(x + 6) or 2.5x2 + 15x.
Now subtract the area of the tennis court to find the area of just the path.
c. The perimeter of the outside of the path is the perimeter of the larger rectangle.
If x = 36 feet, then the perimeter of the outside of the path is 7(36) + 12 or 264 ft.
44. MULTIPLE REPRESENTATIONS In this problem, you will investigate the degree of the product of a monomialand a polynomial. a. TABULAR Write three monomials of different degrees and three polynomials of different degrees. Determine the degree of each monomial and polynomial. Multiply the monomials by the polynomials. Determine the degree of each product. Record your results in a table like the one below.
b. VERBAL Make a conjecture about the degree of the product of a monomial and a polynomial. What is the degree of the product of a monomial of degree a and a polynomial of degree b?
SOLUTION: a. There possible monomials are 2x, 3x
2, 4x
3. Three polynomials are x2 − 1, x5 + 1, x
6 + 1.
The degree for the monomial is the sum of the exponents. For the example given, case they are 1,2, and 3. The degree for the polynomial is the largest degree of the monomials. For the example given, they are 2, 5, and 6. Use the distributive property to find the product of the monomial and the polynomial. For the example give, they
are2x3 − 2x, 3x
7 + 3x
2, and 4x
9 + 4x. For the example given, the degree of the new polynomials are 3,7 and 9.
b. The degree of the product is the sum of the degree of the monomial and the degree of the polynomial, or a + b.
45. ERROR ANALYSIS Pearl and Ted both worked on this problem. Is either of them correct? Explain your reasoning.
SOLUTION:
Ted’s solution is correct. Pearl made a mistake when distributing on the middle term. She did not add the exponents correctly.
46. CCSS PERSEVERANCE Find p such that 3xp(4x
2p+3 + 2x3p−2
) = 12x12 + 6x
10.
SOLUTION: First simplify the left side of the equation.
Because the coefficients of the terms are the same, the corresponding terms’ exponents can be set equal to each other.
47. CHALLENGE Find 4x−3
y2(2x
5y−4 + 6x
−7y
6 − 4x0y
-2).
SOLUTION:
48. REASONING Is there a value for x that makes the statement (x + 2)2 = x2 + 22
true? If so, find a value for x. Explain your reasoning.
SOLUTION: Solve for x.
When 0 is substituted in for x in the equation, both sides are 22 or 4, which makes the equation true.
49. OPEN ENDED Write a monomial and a polynomial using n as the variable. Find their product.
SOLUTION: Choose a monomial with variable n and a polynomial with variable n. For example, let the monomial be 3n and the polynomial be 4n + 1. Find the product.
50. WRITING IN MATH Describe the steps to multiply a polynomial by a monomial.
SOLUTION: To multiply a polynomial by a monomial, use the Distributive Property. Multiply each term of the polynomial by the monomial. Then simplify by multiplying the coefficients together and using the Product of Powers Property for the variables. Consider the following example:
51. Every week a store sells j jeans and t T-shirts. The store makes $8 for each T-shirt and $12 for each pair of jeans. Which of the following expressions represents the total amount of money, in dollars, the store makes every week? A 8j + 12t B 12j + 8t C 20(j + t) D 96jt
SOLUTION: 12j is the amount made from jeans. 8t is the amount made from T-shirts. The sum, or 12j + 8t, represents the total amount the store makes every week. So, the correct choice is B.
52. If a = 5x + 7y and b = 2y − 3x, what is a + b?F 2x − 9y G 3y + 4x H 2x + 9y J 2x − 5y
SOLUTION:
So, the correct choice is H.
53. GEOMETRY A triangle has sides of length 5 inches and 8.5 inches. Which of the following cannot be the length of the third side? A 3.5 inches B 4 inches C 5.5 inches D 12 inches
SOLUTION: The sum of any two sides of a triangle must be greater than the length of the third side. Choice A: 3.5 + 5 = 8.5 8.5 Choice B: 3.5 + 5 = 8.5 > 8.5 Choice C: 5.5 + 5 = 10.5 > 8.5 Choice D: 12 + 5 = 17 > 8.5 Let x be the third side of the triangle.
3.5 cannot be the third side of the triangle. So, the correct choice is A.
54. SHORT RESPONSE Write an equation in which x varies directly as the cube of y and inversely as the square of z .
SOLUTION: If x varies directly as the cube of y , then both x and y must be in the numerators on either side of the equal sign.
When x varies inversely with the square of z. then the z must be in the denominator. Then . The
combination of the two is .
Find each sum or difference.
55. (2x2 − 7) + (8 − 5x
2)
SOLUTION:
56. (3z2 + 2z − 1) + (z
2 − 6)
SOLUTION:
57. (2a − 4a2 + 1) − (5a
2 − 2a − 6)
SOLUTION:
58. (a3 − 3a
2 + 4) − (4a
2 + 7)
SOLUTION:
59. (2ab − 3a + 4b) + (5a + 4ab)
SOLUTION:
60. (8c3 − 3c
2 + c − 2) − (3c
3 + 9)
SOLUTION:
Write a recursive formula for each sequence.61. 16, 2, –12, –26, ...
SOLUTION: Determine if there is a common difference. 2 – 16 = –14 –12 – 2 = –14 –26 – (–12) = –14 The common difference is –14. The first term is 16. To find the next term, subtract 14 from the previous term.
62. –5, 3, 11, 19, ...
SOLUTION: Determine if there is a common difference. 3 – (–5) = 8 11 – 3 = 8 19 – 11 = 8 The common difference is 8. The first term is 3. To find the next term, add 8 to the previous term.
63. 27, 43, 59, 75, ...
SOLUTION: Determine if there is a common difference. 43 – 27 = 16 59 – 43 = 16 75 – 59 = 16 The common difference is 16. The first term is 43. To find the next term, add 16 to the previous term.
64. 80, –200, 500, –1250, ...
SOLUTION: Check for a common difference. –200 – 80 = –280 500 – (–200) = 700 There is no common difference. Check for a common ratio. –200 ÷ 80 = –2.5 500 ÷ (–200) = –2.5 –1250 ÷ 500 = –2.5 The common ratio is –2.5. The first term is 80. To find the next term, multiply the previous term by –2.5.
65. 100, 60, 36, 21.6, ...
SOLUTION: Check for a common difference. 60 – 100 = –40 36 – 60 = –24 There is no common difference. Check for a common ratio. 60 ÷ 100 = 0.6 36 ÷ 60 = 0.6 21.6 ÷ 36 = 0.6 The common ratio is 0.6. The first term is 100. To find the next term, multiply the previous term by 0.6.
66.
SOLUTION: Check for a common difference.
There is no common difference. Check for a common ratio.
4 ÷ 1 = 4 The common ratio is 4.
The first term is . To find the next term, multiply the previous term by 4.
67. TRAVEL In 2003, about 9.5 million people took cruises. Between 2003 and 2008, the number increased by about 740,000 each year. Write the point-slope form of an equation to find the total number of people y taking a cruise for any year x. Estimate the number of people who took a cruise in 2010.
SOLUTION: The statement “increased by 740,000 each year” is a constant rate, or the slope. Now substitute the slope and the point (2003, 9,500,000) into the point slope equation.
To find the number of people who took a cruise in 2010, substitute x = 2010.
So, approximately 14,680,000 people took a cruise in 2010.
Write an equation in function notation for each relation.
68.
SOLUTION: The y-intercept is 0. Find the slope by choosing two points. Choose (0, 0) and (1, 4).
So the slope intercept form is f (x) = 4x + 0 or f (x) = 4x.
69.
SOLUTION:
The y-intercept is 0. Find the slope by choosing two points. Choose (0, 0) and (2, −1).
So in slope intercept form is f (x) = −0.5x + 0 or f (x) = −0.5x.
Simplify.
70. b(b2)(b
3)
SOLUTION:
71. 2y(3y2)
SOLUTION:
72. −y4(−2y
3)
SOLUTION:
73. −3z3(−5z
4 + 2z)
SOLUTION:
74. 2m(−4m4) − 3(−5m
3)
SOLUTION:
75. 4p 2(−2p
3) + 2p
4(5p
6)
SOLUTION: eSolutions Manual - Powered by Cognero Page 30
8-2 Multiplying a Polynomial by a Monomial
Find each product.
1. 5w( −3w2 + 2w − 4)
SOLUTION:
2. 6g2(3g
3 + 4g2 + 10g − 1)
SOLUTION:
3. 4km2(8km
2 + 2k2m + 5k)
SOLUTION:
4. −3p4r3(2p
2r4 − 6p 6
r3 − 5)
SOLUTION:
5. 2ab(7a4b
2 + a5b − 2a)
SOLUTION:
6. c2d
3(5cd
7 − 3c3d
2 − 4d3)
SOLUTION:
Simplify each expression.
7. t(4t2 + 15t + 4) − 4(3t − 1)
SOLUTION:
8. x(3x2 + 4) + 2(7x − 3)
SOLUTION:
9. −2d(d3c
2 − 4dc2 + 2d
2c) + c
2(dc
2 − 3d4)
SOLUTION:
10. −5w2(8w
2x − 11wx
2) + 6x(9wx
4 − 4w − 3x2)
SOLUTION:
11. GRIDDED RESPONSE Marlene is buying a new plasma television. The height of the screen of the television is one half the width plus 5 inches. The width is 30 inches. Find the height of the screen in inches.
SOLUTION: Let h be the height and let w be the width.
Let w = 30.
The height is 20 inches.
Solve each equation.
12. −6(11 − 2c) = 7(−2 − 2c)
SOLUTION:
13. t(2t + 3) + 20 = 2t(t − 3)
SOLUTION:
14. −2(w + 1) + w = 7 − 4w
SOLUTION:
15. 3(y − 2) + 2y = 4y + 14
SOLUTION:
16. a(a + 3) + a(a − 6) + 35 = a(a − 5) + a(a + 7)
SOLUTION:
17. n(n − 4) + n(n + 8) = n(n − 13) + n(n + 1) + 16
SOLUTION:
Find each product.
18. b(b2 − 12b + 1)
SOLUTION:
19. f (f2 + 2f + 25)
SOLUTION:
20. −3m3(2m
3 − 12m2 + 2m + 25)
SOLUTION:
21. 2j 2(5j
3 − 15j2 + 2j + 2)
SOLUTION:
22. 2pr2(2pr + 5p
2r − 15p )
SOLUTION:
23. 4t3u(2t
2u
2 − 10tu4 + 2)
SOLUTION:
Simplify each expression.
24. −3(5x2 + 2x + 9) + x(2x − 3)
SOLUTION:
25. a(−8a2 + 2a + 4) + 3(6a
2 − 4)
SOLUTION:
26. −4d(5d2 − 12) + 7(d + 5)
SOLUTION:
27. −9g(−2g + g2) + 3(g
2 + 4)
SOLUTION:
28. 2j (7j2k
2 + j k2 + 5k) − 9k(−2j
2k
2 + 2k2 + 3j )
SOLUTION:
29. 4n(2n3p
2 − 3np2 + 5n) + 4p (6n
2p − 2np
2 + 3p )
SOLUTION:
30. DAMS A new dam being built has the shape of a trapezoid. The base at the bottom of the dam is 2 times the
height. The base at the top of the dam is times the height minus 30 feet.
a. Write an expression to find the area of the trapezoidal cross section of the dam. b. If the height of the dam is 180 feet, find the area of this cross section.
SOLUTION:
a. The equation for the area of a trapezoid is .
b1
= 2h
b2 =
b. Let h = 180.
So, the area is 32,940 square feet.
Solve each equation.
31. 7(t2 + 5t − 9) + t = t(7t − 2) + 13
SOLUTION:
32. w(4w + 6) + 2w = 2(2w2 + 7w − 3)
SOLUTION:
33. 5(4z + 6) − 2(z − 4) = 7z(z + 4) − z(7z − 2) − 48
SOLUTION:
34. 9c(c − 11) + 10(5c − 3) = 3c(c + 5) + c(6c − 3) − 30
SOLUTION:
35. 2f (5f − 2) − 10(f2 − 3f + 6) = −8f (f + 4) + 4(2f
2 − 7f )
SOLUTION:
36. 2k(−3k + 4) + 6(k2 + 10) = k(4k + 8) − 2k(2k + 5)
SOLUTION:
Simplify each expression.
37. np2(30p
2 + 9n
2p − 12)
SOLUTION:
38. r2t(10r
3 + 5rt
3 + 15t
2)
SOLUTION:
39. −5q2w
3(4q + 7w) + 4qw
2(7q
2w + 2q) − 3qw(3q
2w
2 + 9)
SOLUTION:
40. −x2z(2z
2 + 4xz3) + xz
2(xz + 5x
3z) + x2
z3(3x
2z + 4xz)
SOLUTION:
41. PARKING A parking garage charges $30 per month plus $0.50 per daytime hour and $0.25 per hour during nights and weekends. Suppose Trent parks in the garage for 47 hours in January and h of those are night and weekend hours. a. Find an expression for Trent’s January bill. b. Find the cost if Trent had 12 hours of night and weekend hours.
SOLUTION: a. First, multiply the price per hour times the number of hours. Let h represent the number of weekend and night hours. For the weekend and night hours, he would spend 0.25h. The number of daytime hours is 47 – h, so he spent 0.50(47 - h) = 23.50 – 0.50h. To find the total cost, add these two values to the cost per month. 30 + (23.50 − 0.50h) + 0.25h = 53.50 − 0.25h. b. If Trent parked 12 night and weekend hours, h = 12
So, the cost for January is $50.50.
42. CCSS MODELING Che is building a dog house for his new puppy. The upper face of the dog house is a trapezoid. If the height of the trapezoid is 12 inches, find the area of the face of this piece of the dog house.
SOLUTION:
The formula for the area of a trapezoid is .
Now substitute h = 12 inches into the equation.
Therefore, the area of the trapezoid is 318 square inches.
43. TENNIS The tennis club is building a new tennis court with a path around it.
a. Write an expression for the area of the tennis court. b. Write an expression for the area of the path. c. What is the perimeter of the outside of the path?
SOLUTION: a. The area of the tennis court can be found by the equation . Substitute the values of the length and width and simplify.
b. The area of the path is the area of the large rectangle minus the area of the smaller rectangle, which is the area ofthe tennis court.
Area of the large rectangle = 2.5x(x + 6) or 2.5x2 + 15x.
Now subtract the area of the tennis court to find the area of just the path.
c. The perimeter of the outside of the path is the perimeter of the larger rectangle.
If x = 36 feet, then the perimeter of the outside of the path is 7(36) + 12 or 264 ft.
44. MULTIPLE REPRESENTATIONS In this problem, you will investigate the degree of the product of a monomialand a polynomial. a. TABULAR Write three monomials of different degrees and three polynomials of different degrees. Determine the degree of each monomial and polynomial. Multiply the monomials by the polynomials. Determine the degree of each product. Record your results in a table like the one below.
b. VERBAL Make a conjecture about the degree of the product of a monomial and a polynomial. What is the degree of the product of a monomial of degree a and a polynomial of degree b?
SOLUTION: a. There possible monomials are 2x, 3x
2, 4x
3. Three polynomials are x2 − 1, x5 + 1, x
6 + 1.
The degree for the monomial is the sum of the exponents. For the example given, case they are 1,2, and 3. The degree for the polynomial is the largest degree of the monomials. For the example given, they are 2, 5, and 6. Use the distributive property to find the product of the monomial and the polynomial. For the example give, they
are2x3 − 2x, 3x
7 + 3x
2, and 4x
9 + 4x. For the example given, the degree of the new polynomials are 3,7 and 9.
b. The degree of the product is the sum of the degree of the monomial and the degree of the polynomial, or a + b.
45. ERROR ANALYSIS Pearl and Ted both worked on this problem. Is either of them correct? Explain your reasoning.
SOLUTION:
Ted’s solution is correct. Pearl made a mistake when distributing on the middle term. She did not add the exponents correctly.
46. CCSS PERSEVERANCE Find p such that 3xp(4x
2p+3 + 2x3p−2
) = 12x12 + 6x
10.
SOLUTION: First simplify the left side of the equation.
Because the coefficients of the terms are the same, the corresponding terms’ exponents can be set equal to each other.
47. CHALLENGE Find 4x−3
y2(2x
5y−4 + 6x
−7y
6 − 4x0y
-2).
SOLUTION:
48. REASONING Is there a value for x that makes the statement (x + 2)2 = x2 + 22
true? If so, find a value for x. Explain your reasoning.
SOLUTION: Solve for x.
When 0 is substituted in for x in the equation, both sides are 22 or 4, which makes the equation true.
49. OPEN ENDED Write a monomial and a polynomial using n as the variable. Find their product.
SOLUTION: Choose a monomial with variable n and a polynomial with variable n. For example, let the monomial be 3n and the polynomial be 4n + 1. Find the product.
50. WRITING IN MATH Describe the steps to multiply a polynomial by a monomial.
SOLUTION: To multiply a polynomial by a monomial, use the Distributive Property. Multiply each term of the polynomial by the monomial. Then simplify by multiplying the coefficients together and using the Product of Powers Property for the variables. Consider the following example:
51. Every week a store sells j jeans and t T-shirts. The store makes $8 for each T-shirt and $12 for each pair of jeans. Which of the following expressions represents the total amount of money, in dollars, the store makes every week? A 8j + 12t B 12j + 8t C 20(j + t) D 96jt
SOLUTION: 12j is the amount made from jeans. 8t is the amount made from T-shirts. The sum, or 12j + 8t, represents the total amount the store makes every week. So, the correct choice is B.
52. If a = 5x + 7y and b = 2y − 3x, what is a + b?F 2x − 9y G 3y + 4x H 2x + 9y J 2x − 5y
SOLUTION:
So, the correct choice is H.
53. GEOMETRY A triangle has sides of length 5 inches and 8.5 inches. Which of the following cannot be the length of the third side? A 3.5 inches B 4 inches C 5.5 inches D 12 inches
SOLUTION: The sum of any two sides of a triangle must be greater than the length of the third side. Choice A: 3.5 + 5 = 8.5 8.5 Choice B: 3.5 + 5 = 8.5 > 8.5 Choice C: 5.5 + 5 = 10.5 > 8.5 Choice D: 12 + 5 = 17 > 8.5 Let x be the third side of the triangle.
3.5 cannot be the third side of the triangle. So, the correct choice is A.
54. SHORT RESPONSE Write an equation in which x varies directly as the cube of y and inversely as the square of z .
SOLUTION: If x varies directly as the cube of y , then both x and y must be in the numerators on either side of the equal sign.
When x varies inversely with the square of z. then the z must be in the denominator. Then . The
combination of the two is .
Find each sum or difference.
55. (2x2 − 7) + (8 − 5x
2)
SOLUTION:
56. (3z2 + 2z − 1) + (z
2 − 6)
SOLUTION:
57. (2a − 4a2 + 1) − (5a
2 − 2a − 6)
SOLUTION:
58. (a3 − 3a
2 + 4) − (4a
2 + 7)
SOLUTION:
59. (2ab − 3a + 4b) + (5a + 4ab)
SOLUTION:
60. (8c3 − 3c
2 + c − 2) − (3c
3 + 9)
SOLUTION:
Write a recursive formula for each sequence.61. 16, 2, –12, –26, ...
SOLUTION: Determine if there is a common difference. 2 – 16 = –14 –12 – 2 = –14 –26 – (–12) = –14 The common difference is –14. The first term is 16. To find the next term, subtract 14 from the previous term.
62. –5, 3, 11, 19, ...
SOLUTION: Determine if there is a common difference. 3 – (–5) = 8 11 – 3 = 8 19 – 11 = 8 The common difference is 8. The first term is 3. To find the next term, add 8 to the previous term.
63. 27, 43, 59, 75, ...
SOLUTION: Determine if there is a common difference. 43 – 27 = 16 59 – 43 = 16 75 – 59 = 16 The common difference is 16. The first term is 43. To find the next term, add 16 to the previous term.
64. 80, –200, 500, –1250, ...
SOLUTION: Check for a common difference. –200 – 80 = –280 500 – (–200) = 700 There is no common difference. Check for a common ratio. –200 ÷ 80 = –2.5 500 ÷ (–200) = –2.5 –1250 ÷ 500 = –2.5 The common ratio is –2.5. The first term is 80. To find the next term, multiply the previous term by –2.5.
65. 100, 60, 36, 21.6, ...
SOLUTION: Check for a common difference. 60 – 100 = –40 36 – 60 = –24 There is no common difference. Check for a common ratio. 60 ÷ 100 = 0.6 36 ÷ 60 = 0.6 21.6 ÷ 36 = 0.6 The common ratio is 0.6. The first term is 100. To find the next term, multiply the previous term by 0.6.
66.
SOLUTION: Check for a common difference.
There is no common difference. Check for a common ratio.
4 ÷ 1 = 4 The common ratio is 4.
The first term is . To find the next term, multiply the previous term by 4.
67. TRAVEL In 2003, about 9.5 million people took cruises. Between 2003 and 2008, the number increased by about 740,000 each year. Write the point-slope form of an equation to find the total number of people y taking a cruise for any year x. Estimate the number of people who took a cruise in 2010.
SOLUTION: The statement “increased by 740,000 each year” is a constant rate, or the slope. Now substitute the slope and the point (2003, 9,500,000) into the point slope equation.
To find the number of people who took a cruise in 2010, substitute x = 2010.
So, approximately 14,680,000 people took a cruise in 2010.
Write an equation in function notation for each relation.
68.
SOLUTION: The y-intercept is 0. Find the slope by choosing two points. Choose (0, 0) and (1, 4).
So the slope intercept form is f (x) = 4x + 0 or f (x) = 4x.
69.
SOLUTION:
The y-intercept is 0. Find the slope by choosing two points. Choose (0, 0) and (2, −1).
So in slope intercept form is f (x) = −0.5x + 0 or f (x) = −0.5x.
Simplify.
70. b(b2)(b
3)
SOLUTION:
71. 2y(3y2)
SOLUTION:
72. −y4(−2y
3)
SOLUTION:
73. −3z3(−5z
4 + 2z)
SOLUTION:
74. 2m(−4m4) − 3(−5m
3)
SOLUTION:
75. 4p 2(−2p
3) + 2p
4(5p
6)
SOLUTION:
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8-2 Multiplying a Polynomial by a Monomial
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