Answer Key 53

Chapter 8

Lesson 8.1.1

8-1. a. (x + 4)(y + x + 2) = xy + x2 + 6x + 4y + 8 b. 18x2 + 9x ! 2

8-2. a. (2x + 3)(x + 2) b. (2x +1)(3x + 2) c. no solution

d. (2x + y)(y + 3) ; Conclusion. Not every expression can be factored.

8-3. a. (3x +1)(2x + 5) = 6x2 +17x + 5

b. (5x ! 2)(y + 3) = 5xy +15x ! 2y ! 6

c. (4x ! 3)(3x + 4) = 12x2 + 7x !12

8-4. The product of each diagonal is equal. 6x2 !5 = 30x2 and 2x !15x = 30x2 .

8-5. Diagonals: part (a) are both 30x2 , part (b) are both !30xy , part (c) are both !144x2 .

Typical response: “The product of one diagonal always equals the product of the

other diagonal.”

8-6. (2x ! 3)(x + 2y ! 4) = 2x2 + 4xy !11x ! 6y +12

8-7. a. 12x2 +17x ! 5

b. 4x2 ! 28x + 49

8-8.

8-9. a. m = 2 , (0,!! 12)

b. m = !3 , (0,!!7)

c. m = !2

3, (0,!8)

d. m = 0 , (0,!!2)

8-10. a. (0, –8); It is the constant in the equation.

b. (–2, 0) and (4, 0); The product of the x-intercepts equals the constant term.

c. (1, –9); Its x-coordinate is midway between the x-intercepts.

8-11. a. !1 b. ! 7.24 c. ! " 4.24

–8

2

–80

10 – 4

–7

12

–3 0

7

0

7 –9

0

–81

9 3x

5x

6x2

2x x

–6x

!7x2

–7x

54 Algebra Connections

Lesson 8.1.2

8-12. a. (5x ! 2)(2x ! 7) = 10x2 ! 39x +14 b. !35x " ! 4x = 10x2 "14 = 140x2

8-13. a. (2x + 3)(x +1)

b. One corner should contain 4x , while the other should contain 6x ; (3x + 4)(x + 2) .

c. Their sum is 7x, and their product is 12x2 .

d. The product 12x2 should be placed at the top of the diamond problem, 7x at the

bottom, and terms 3x and 4x should be in the middle.

e. (2x + 3)(x + 2)

8-14. a. One corner contains 6x2 , and the opposite corner contains 12.

b. The product of the x2 and units terms (in this case, 72x2 ) goes on top, while the

x-term (17x ) goes on bottom.

d. (2x + 3)(3x + 4)

8-15. a. (x + 3)(x + 6) b. (4x ! 3)(x + 5) c. (2x ! 3)(2x !1)

d. not factorable because there are no integers that multiply to get !9x2 (the diagonal

of the generic rectangle) and add to get 5x.

8-16. a. (x ! 6)(x + 2) b. (2x +1)2 c. (x ! 5)(2x +1)

d. (x + 4)(3x ! 2)

8-17. a. x-intercepts (–1, 0) and (3, 0), y-intercept. (0, –3)

b. x-intercept (2, 0), no y-intercept

c. x-intercepts (–3, 0), (–1, 0), and (1, 0), y-intercept (0, 2)

d. x-intercept (8, 0), y-intercept (0, –20)

8-18. a. (0, –9); It is the constant in the equation.

b. (3, 0) and (–3, 0)

8-19. a. (6, 9) b. (0, 2)

8-20. a. x = ! 1023

b. all numbers c. c = 0

8-21. y = 1

4x + 400

Answer Key 55

Lesson 8.1.3

8-22. a. (x + 3)2

b. (2x + 3)(x +1)

c. not factorable

d. (3m + 7)(m ! 2)

8-23. a. (3x ! 2)(3x + 2)

b. 4x(3x ! 4)

c. (4k ! 3)(2k !1)

d. 20(2 ! 5m)

8-24. (2x ! 6)(2x +1) or

(x ! 3)(4x + 2)

8-25. See table below

Multiply x ! 2 2x + 1

x + 7 x2+ 5x ! 14 2x

2+ 15x + 7

3x + 1 3x2! 5x ! 2 6x

2+ 5x +1

8-27. s + 2s + s + 3 = 51 ; 12, 24, and

15 cm

8-28. a. 9 units b. 15 units

c. 10 units d. 121 un 2

8-29. a. (k ! 2)(k !10)

b. (2x + 7)(3x ! 2)

c. (x ! 4)2

d. (3m +1)(3m !1)

8-30. (2, 5)

8-31. y = !x + 8

8-32. a. 5 b. –6

c. 5 or –6 d. ! 14

e. 8 f. ! 14

or 8

Lesson 8.1.4

8-33. a. (3x ! 2)2 b. (9m +1)(9m !1)

c. (x ! 4)(x ! 7)

d. (3n + 3)(n + 2) or (n +1)(3n + 6) .

8-34. a. Yes, because there are two

different arrangements of tiles that

build a rectangle.

b. Because there is a common factor

of 3 in each of the terms of the

original expression and in one of

the two binomials in either of the

two partially factored forms.

c. (i) and (iii) both have common

factors, so they could have more

than one factored form.

8-35. a. 5 b. 5(2x2 + 5x ! 3)

c. Yes; 5(2x !1)(x + 3) .

8-36. a. 5(x + 4)(x !1)

b. 3x(x + 3)(x ! 5)

c. 2(x + 5)(x ! 5)

d. y(x ! 5)(x + 2)

8-37. a. (2x + 5)(x !1)

b. (x ! 3)(x + 2)

c. (3x +1)(x + 4)

d. It is not factorable because no

integers have a product of 14 and

a sum of 5.

8-38. y = 3

4x ! 3

8-39. a. in 7 weeks

b. Joman will score more with 1170

points, while Jhalil will have 970.

8-40. a. Michelle is correct. b. (– 4, 0)

8-41. 45, 46, 47; x + (x +1) + (x + 2) = 138

8-42. a. 2 b. 3 c. 1

56 Algebra Connections

Lesson 8.2.1

8-45. a. 2 b. –3 c. " –6.1

8-46. y = !3x + 25

8-47. y = 3x ! 5 ; m = 3 and b = 5

8-48. There is only one line of symmetry. horizontal through the middle.

8-49. a. x-intercepts (–2, 0) and (0, 0), y-intercept (0, 0)

b. x-intercepts (–3, 0) and (5, 0), y-intercept (0, 3)

c. x-intercepts (–1, 0) and (1, 0), y-intercept (0, –1)

d. x-intercept (9, 0), y-intercept (0, 6)

8-50. a. 6x2 + x !12 b. 25x2 ! 20x + 4

Lesson 8.2.2

8-51. a. Longest: Maggie, Highest: Jen

b. Jen. (0, 0) and (8, 0), Maggie. (3, 0) and

(14, 0), Imp. (2, 0) and (12, 0), Al. (10, 0) and

(16, 0); the x-intercepts tell where the balloon

was launched and where it landed.

c. Jen. (4, 32), Maggie. (8.5, 30.25), Imp. (7, 25),

and Al. (13, 27); maximum height.

8-52. You should be able to connect rule ! table, table " graph, graph " situation, and

table " situation.

8-53. a. One way to write the rule is y = (x +1)(x + 2) + 2 . b. Yes

8-54. vertex. (4, –9), x-intercepts. (1, 0) and (7, 0), y-intercept. (0, 7)

8-55. a. 3 –7 6 –2 b. …it does not change the value of the number.

c. It tells us that a = 0 . d. 0 for all e. …the result is always 0.

8-56. a. x-intercepts (2, 0), (– 4, 0), and (3, 0), y-intercept. (0, 18)

b. x-intercepts (3, 0) and (8, 0), y-intercept. (0, –3)

c. x-intercept (1, 0) and y-intercept (0, – 4)

8-57. a. (–3, 0) b. ! 12

8-58. a. no solution b. (7, 2)

2 4 6 8 10 12 14 16

5

10

15

20

25

30

Jen Maggie

Al Imp

Solution to part (a)

Answer Key 57

Lesson 8.2.3

8-59. a. No; the y-intercept is not enough information.

b. No; the parabola could vary in width and direction.

c. Yes; solution shown at right.

8-60. a. y = 0 for all x-intercepts and x = 0 for all y-intercepts.

b. (0, –12)

c. 0 = 2x2 + 5x !12

d. Not yet, because it has an x2 term.

8-61. a. At least one of the two numbers must be zero.

b. At least one of the three numbers must be zero.

c. Typical response: “If the product of two or more numbers is zero, then you know

that one of the numbers must be zero.”

8-62. a. 0 = (2x ! 3)(x + 4)

b. 2x ! 3 = 0 or x + 4 = 0 , so x = 3

2 or x = ! 4 .

c. The roots are at ( 32, !0) and (– 4, 0).

d. The solution graph is shown at right.

8-63. This parabola should have roots (–3, 0) and (2, 0) and y-intercept (0, –6).

8-64. roots: (–1, 0) and (–2, 0), y-intercept: (0, 4)

8-65. a. One is a product and the other is a sum.

b. first: x = !2 or x = 1 ; second: x = ! 12

8-66. a. x = 2 or x = !8 b. x = 3 or x = 1

c. x = !10 or x = 2.5 d. x = 7

8-67. a. The line x = 0 is the y-axis, so this system is actually finding where the line

5x ! 2y = 4 crosses the y-axis.

b. (0, –2)

8-68. a. 4; Since the vertex lies on the line of symmetry, it must lie halfway between the

x-intercepts.

b. (4, –2)

8-69. a. 2(x ! 2)(x +1) b. 4(x ! 3)2

8-70. a. The symbol “#” represents “greater than or equal to” and the symbol “>”

represents “greater than.”

b. 5 > 3 c. x ! 9 d. –2 is less than 7.

58 Algebra Connections

Lesson 8.2.4

8-71. The parabola should have y-intercept (0, 2) and roots (–1, 0) and (–2, 0).

8-72. a. x = –2 or x = – 4 b. x = 1 or x = 43

c. x = –5 or x = 32

d. x = 0 or x = –6 e. x = 5 or x = –1.5 f. x = 2 or x = –6

8-73. a. y = (x + 3)(x ! 2) = x2+ x ! 6

b. y = (x + 5)(x !1) = x2 + 4x ! 5

8-74. By symmetry, (12, 0) is also a root. Thus, the quadratic must be of the form

y = a(x ! 2)(x !12) . Since the parabola points down, a must be negative. Testing a

point shows that y = !(x ! 2)(x !12) = !x2 +14x ! 24 is correct.

8-76. The result must be the original expression because multiplying and factoring are

opposite processes; 65x2 + 212x !133 .

8-77. a. x = 3 or x = ! 23

b. x = 2 or x = 5

c. x = !3 or x = 2

d. x = 12

or x = !1

2

8-78.

8-79. a. true b. false c. true d. true e. false f. false

8-80. a. –1 b. " 1.6 c. –3

8-81. y = !4

3x

a. ! 43

b. Yes; it makes the equation true and lies on the graph of the line.

Answer Key 59

Lesson 8.2.5

8-82. (1) " b, (2) " e, (3) " a, (4) " g, (5) " d, (6) " i

8-83. Letter A. The client should order the parabola y = (x !1)(x + 6) .

Letter B. The parabola y = (x ! 5)2 should be recommended.

Letter C. The parabola y = !(x + 3)(x ! 2) should be recommended.

8-84. a. y = !2x(x ! 8) = !2x2 +16x b. y = !3(x !10)(x !16)

8-85. a. y = x2 + 2x ! 8 b. y = x2 ! 6x + 9 c. y = x2 ! 7x

d. y = !x2 ! 4x + 5

8-86. m =1

2, (0, 4) 8-87. a. " –1.4 and " 0.3 b. The quadratic is not factorable.

8-88. a. x = 4 or x = –10 b. x = –8 or x = 1.5

8-89. a. 4 b. –10 c. –8 d. 1.5; They are the same.

8-90. a. (1, –1) b. (–2, 12

)

Lesson 8.3.1

8-91. a. The quadratic is not factorable. b. There are two roots (x-intercepts).

c. The intercepts are " –1.5 and 4.5.

8-92. a. a = 1 , b = !3 , c = !7 b. 3± 37

2 " 4.5 and –1.5; yes

8-93. a. Graphing and factoring with the Zero Product Property

8-94. a. x = –2 or ! 13

b. x = 7 or –2.5 c. x = –0.5 or –0.75 d. no solution

8-96. a. x = 6 or 7 b. x = 23

or – 4 c. x = 0 or 5 d. x = 3 or –5

8-97. x = 6 or 7; yes

8-98. no

a. The parabola should be tangent to the x-axis.

b. Answers vary, but the parabola should not cross the x-axis.

8-99. y = 1

2x + 9 8-100. line. (a) and (c); parabola. (b) and (d)

8-101. A and D

8-102. a. false b. true c. true d. true e. true f. false

g. true h. false

60 Algebra Connections

Lesson 8.3.2

8-103. a. (3x ! 2)(2x + 5) = 0 ,

x =2

3 or ! 5

2

b. a = 6 , b = 11, c = !10 , x = 2

3

or ! 52

c. Yes

8-104. a. x = 5.5 or x = !5.5

b. x = 2 or x = !1

2

c. x = !3 or x = 14 d. x = 5

6

8-105. a. " 315 and –315 feet; The bases

of the arch are 315 feet from

the center.

b. " 630 feet

c. 630 feet; y-intercept

8-106. a. x = 5 b. x = 13

or –6

c. x = –1 or 53

d. x = ± 3

4

8-107. x = 13

or –6; yes

8-108. a. y = (x + 3)(x !1) = x2 + 2x ! 3

b. y = (x ! 2)(x + 2) = x2 ! 4

8-109. If x = width, x(2x + 5) = 403 ;

width = 13 cm.

8-110. (b) and (c) are solutions.

8-111. a. She solved for x when y = 0 .

b. The y-intercept is (0, –5), and

a shortcut is to solve for y

when x = 0; y = 3

5x ! 5 .

c. x. (8, 0), y. (0, –12)

Lesson 8.3.3

8-112. a. x = !3 or x = !9

b. x = !17.6 or x = !0.36

c. x = ! 43

or x = 12

d. x = 4

e. no solution

f. x = 2.5 or x = !0.9

8-114. a. The Zero Product Property

only works when a product

equals zero.

b. x2 ! 3x ! 4 = 0

c. x = 4 or x = –1; no

8-115. a. x = 0 seconds and x = 2.4

seconds, so it is in the air for

2.4 seconds.

b. at x = 1.2 secs c. 2.88 feet

d. The sketch should have roots

(0, 0) and (2.4, 0) and vertex

(1.2, –2.88).

8-116. If n = # nickels and q = # of

quarters, 0.05n + 0.25q = 1.90 ,

n = 2q + 3 , and n = 13, so Daria

has 13 nickels.

8-117. a. x = ±0.08 b. x = 29

or – 4

c. no solution

d. x " 1.4 or –17.4

8-118. While the expressions may vary,

each should be equivalent to

y = x2 + 4x + 3 .

8-119. a. x = 2 b. x = 15

c. x = –2 d. all numbers

8-120. Line L has slope 4, while line M

has slope 3. Therefore, line L is

steeper.

8-121. D