Precalculus Skills Builder Workbook with solutions

Skill Builders

(Extra Practice)

1. Factoring Out Monomial Terms

2. Laws of Exponents

3. Function Notation

4. Properties of Lines

5. Multiplying Binomials

6. Special Triangles

7. Simplifying and Combining Radicals

8. Similar Triangles

9. Factoring Standard Forms

10. Inverses

11. Algebraic Fractions

12. Solving Quadratics

13. Inequalities

14. Solving Trigonometric Equations

15. Trigonometric Identities

16.Logarithms

17. Algebra for College Math Courses

18. Completing the Square

19. Complex Numbers

Pre-Calculus with Trigonometry

1. Factoring Out Monomial Terms

Example: The first thing you should look for when factoring expressions is common factors that appear in every term. These common factors could be numbers or variables, so look for common numerical factors and common variables that appear in every term.

3x2y5 ! 5y3 = 3xy3(xy2 ! 2)

Factor out the largest single term possible from the following expressions.

1. 2x3y + 7xy2 2. 42a3b3 + 24a2b5

3. 24!r4h " 20!r2h4 4. 30b11c6d4 +15b11c7d3 ! 35b10c7d4

5. 36x5y5z6 ! 27x3y3z10 +18x6y2z7 6. 6a4b2 ! 7ab4 + 7a2b2

7. 27p6q + 45p5q2 ! 54 p8q 8. 8 j10k2 + 20 j3k4

9. 21x4y3 + 35x2y8 10. 42d3h3 + 63d2h5

11. 24!r2h2 +18!r2h + 54!r2 12. 15mn4 ! 33m3n

13. 2a5b5c ! 6a3b3c6 + 7a2b6c3 14. 15s2t5 ! 25st 7 + 55st5

15. 8u3v ! 32uv

4+ 80uv 16. 6x4y3 + 8x2y5 ! 2x3y2 + 6x2y2

Skill Builders

2. Laws of Exponents

Example: Simplify using the laws of exponents. Solution:

2x!3y2

xy3

"#$

%&'!4

= 2!4 (x!3!1)!4 (y2!3)!4 = 2!4 x16y4 =x16y4

16

Simplify the following expressions.

1. (x3)(x4 ) 2. (y2 )(y3)

3. a9

a6

4. p7

p2

5. (x3)3 6. (y2 )5

7. (z2 )3!" #$2

8. (b3)3!" #$3

9. (z2 )(x4 )!" #$3

10. (y!2 )(y3)"# $%0

11. (a!3)(a5 )"# $%3

12. x4

x( )2

13. (x!3)(x3) + (y4 )(y!4 ) 14. (2x)3(3x)2

15. (5x)2 (2x)2

5x 16. (3x)3(2x)2

4x

!

"#$

%&

2


3. Function Notation

Example: Let f (x) = 2x2 ! 2x + 3 and g(x) = 3x ! 5 . Find the indicated values.

Solutions: a. f (!2) = 2(!2)2 ! 2(!2) + 3 = 15

b. f (x + h) = 2(x + h)2 ! 2(x + h) + 3

= 2(x2 + 2xh + h2 ! 2(x + h) + 3

= 2x2 + 4xh + 2h2 ! 2x ! 2h + 3

c. g( f (x)) = g(2x2 ! 2x + 3) = 3(2x2 ! 2x + 3) ! 5

= 6x2 ! 6x + 4

Find the indicated values.

1. Given f (x) = 3x2 + 5, find f (4).

2. Given f (x) = 5x3 + 3x2, find f (!3).

3. Given f (x) = 5x2 + 3x, find f (x + h).

4. Given f (x) = 21x2 ! 20x, find f (x + h).

5. Given f (x) = 3x2 + 5x and g(x) = 7x + 3, find f (x)g(x).

6. Given f (x) = 6x3 + 7 and g(x) = ex , find f (x)g(x).

7. Given f (x) = 3x2 + 4x + 5 and g(x) = x2 + x +1, find f (x)g(x) if x = 3.

8. Given f (x) = 2x2 + 5x and g(x) = 3x, find f (g(x)).

9. Given f (x) = 2x2 + 5x and g(x) = 3x, find g( f (x)).

10. Given g(x) = 5x2 + 5x + 5 and h(x) = 4x + 4, find g(h(x)).

11. Given g(x) = 5x2 + 5x + 5 and h(x) = 4x + 4, find h(g(x)).

12. Given h(x) = 16 ! 35x , find h(2).

13. Given f (x) = 5 !23x , find f (3).

14. Given f (x) = ex !and! g(x) = 3x2 + 2x + 6, find f (g(x)).

15. Given f (x) = ex and g(x) = 3x2 + 2x + 6, find g( f (x)).

Skill Builders

4. Properties of Lines

Given the points A = (!2, 5) and B = (3, !4) . Example 1: Find the equation of the line that goes through A and B. Write your answer in slope-intercept form. Solution: slope =

(5!(!4))

(!2!3)=

9!5

= !95

Now use the form y = mx + b with m = !9

5 to get the equation y = !

9

5x + b .

Substitute the x and y from point A or point B and solve for b. 5 = !

95(!2) + b

5 =18

5+ b

b =7

5

y = !9

5x + 7

5

Example 2: Find the equation of the line that is perpendicular to line AB and passes through the point (2, 3). Solution: Perpendicular lines have negative reciprocal slopes. Since the slope of AB is ! 9

5, the slope of the line perpendicular to it is 5

9. Point-

slope equations have the form y ! y1 = m(x ! x1) . Substitute (2, 3) for (x1, y1) and 5

9 for m to get: y ! 3 = 5

9(x ! 2) .

Example 3: Find the distance between points A and B. Solution:

d = !2 ! 3( )2+ 5 ! !4( )( )

2= !5( )

2+ 9( )

2= 106


1. Find the equation of the line that is parallel to the line y = 3x + 9 and passes through the point (–4, 3). Write your answer in point-slope form.

2. Find the equation of the line that is parallel to the line y = 2x + 8 and passes through the

point (–4, 3). Write your answer in slope-intercept form. 3. Find the equation of the line that is parallel to the line y = !6x ! 8 and passes through the

point (8, 3). Write your answer in point-slope form. 4. Find the equation of the line that is parallel to the line y = !5x ! 9 and passes through the

point (7, 1). Write your answer in slope-intercept form. 5. Find the equation of the line that contains the points (1, 7) and (–5, –3). Write your answer

in slope-intercept form. 6. Find the equation of the line that contains the points (–5, –10) and (9, 6). Write your

answer in slope-intercept form. 7. Find the equation of the line that contains the points (–1, 4) and (–5, –4). Write your answer

in point-slope form. 8. Find the equation of the line that contains the points (5, 2) and (8, 11). Write your answer in

point-slope form. 9. Find the equation of the line that is perpendicular to the line y = 7x +1 and passes through

the point (3, –1). Write your answer in point-slope form. 10. Find the equation of the line that is perpendicular to the line y = 6x + 2 and passes through

the point (3, 7). Write your answer in slope-intercept form. 11. Find the equation of the line that is perpendicular to the line y = !9x ! 8 and passes through

the point (1,6). Write your answer in slope-intercept form. 12. Find the equation of the line that is perpendicular to the line y = 2x + 5 and passes through

the point (–6, –7). Write your answer in slope-intercept form. 13. Find the distance between the points (–3, –5) and (5, 4). 14. Find the distance between the points (–1, 6) and (3, –4). 15. Find the distance between the points (4, 3) and (18, 14). 16. Find the distance between the points (0, 7) and (4, 16).

Skill Builders

5. Multiplying Binomials

Example: Multiply (x + 2)(x ! 5) . Each term in the first binomial is distributed over the

second binomial, as shown below. Solution: (x + 2)(x ! 5) = x(x ! 5) + 2(x ! 5) = x2 ! 5x + 2x !10 = x2 ! 3x !10

Multiply the following binomials. 1. (x + 3)(x ! 6) 2. (a3 +1)(a2 – 4)

3. (b2 – 3)(b + 5) 4. (z3 +1)(z2 – 1)

5. (p11 – p10 )(p10 – p9 ) 6. (q4 + 4)(q3 + 3)

7. (5h4 +1)(6h – 2) 8. (d5 – d4 )(d6 +1)

9. (x + 3)(x + 6)(x – 7) 10. (x2 + 3)(x – 2)(x +1)

11. (m99 + m98 )(m97 – m96 ) 12. (6n4 + 3n3)(6n3 + 3n2 )

13. (100 j100 – j)(9 j2 +1) 14. (2!k5 + 5k4 )(!k + 3)

15. (c2x + d3y)(2x + c2d) 16. ( f 3y3 + g2x2 )( fy + gx)


6. Special Triangles

Example: The following relationships exist between the sides of the special triangles.

Use these relationships to solve each of the following problems. 1. Solve for angles α and β in the triangle at

right.

1

2

!

"

3

2. Given a 30-60-90 triangle with hypotenuse length 8, find the exact length of the long leg. 3. A 30-60-90 triangle has a hypotenuse 0.5 inches long. How long is the short leg? 4. Given a 30-60-90 triangle whose long leg is 13 cm, how long is the hypotenuse? (Solve

exactly.) 5.

Solve for leg lengths a and b.

6. Given a 45-45-90 triangle with hypotenuse length 13 2 , how long are the legs? 7. If a 45-45-90 triangle’s legs are 17 mm long, exactly how long is its hypotenuse? 8. What is the exact area of the equilateral triangle whose legs are each 6 cm long? 9. Find the exact area of the following isosceles

triangle. 30˚

55

Skill Builders

10. A triangle has angles 30˚, 60˚, and 90˚, and its legs have lengths 7 3 , 7, and 14. How long is the side opposite the 60˚ angle?

11. Solve for α and β in the following triangle.

12. If a 30-60-90 triangle has a hypotenuse length 100, how long is the short leg? 13. A 30-60-90 triangle with long leg length 15 3 has a hypotenuse of what length? 14. Find the area of the triangle at right.

30˚

2020

15. Solve for length x in terms of c.

16. Solve for lengths x and y in terms of b.


7. Simplifying and Combining Radicals

Example: Simplify 288 . When you are asked to write a radical in simpler form, the real task is to reduce the number under the radical sign as much as possible by taking perfect squares outside the radical. Solution: Simplify 288 . Find the factorization of 288, choosing perfect squares as factors whenever possible.

288 = 4 ! 4 !9 !2

288 = 4 4 9 2

288 = 2 !2 ! 3 ! 2

288 = 12 2

Rewrite the following as radicals in simpler form.

1. 1200 2. 845

3. a3b5 4. x7b9

Example: Simplify 3 125 ! 2 80 . Solution: First simplify each radical separately and then combine like terms.

3 125 = 3 25 5 = 15 5

2 80 = 2 16 5 = 8 5

3 125 ! 2 80 = 15 5 ! 8 5 = 7 5


5. 5 + 18 6. 54 + 294

7. 18 + 108 + 50 + 48 8. 99( )2

9. ( 12 )( 75) 10. ( 20 )( 80)

11. ( 50 )( 98)

Skill Builders

Example: Rationalize 4

5.

Solution: To rationalize the denominator you need to multiply both numerator and denominator by the radical in the denominator. This is equivalent to multiplying by 1 so it does not change the value of the expression. 4

5=

4

5

5

5=4 5

5

Rationalize the following expressions. 12. 3

2 13. 5

7 14. a2b4

a5b9

Example: Combine and simplify 2

3+

1

2 3.

Solution: To combine rational expressions you first need to find a common denominator and write the expression as one fraction. Then, you must rationalize the denominator.

2

3+

1

2 3=

2!2

2 3+

1

2 3=

5

2 3=5 3

6

Rewrite the following as single, rationalized expressions. 15. 3

2+

6

2

16. 6

8+

3

8


8. Similar Triangles

Example:

Solution:

Find the length of x.

To find x you need to put corresponding sides in a ratio. In this triangle 17 corresponds to 7 and 12 corresponds to x. So our ratio is: x

12=7

17

17x = 84

x = 4.941

Solve each triangle for the indicated variable. 1. Find the length of x.

2. Find the length of y.

3. Find the length of x.


Skill Builders



7. Find the length of k in terms of x:

8. Find the value of h in terms of y.

9. Find the length of b in terms of x.

10. Find the value of h in terms of x.


11. Kathleen is 6 feet tall. Her shadow is 3 feet long. At the same time of day, a tree casts a shadow that is 21 feet long. How tall is the tree?

12. Kyle is 1.5 meters tall and casts a shadow 4 meters long. A flagpole casts a shadow that is

30 meters long. How tall is the flagpole? 13. Lisa’s dog is 2.5 feet tall and casts a shadow that is 4 feet long. A fire hydrant casts a

shadow that is 6 feet long. How tall is the fire hydrant? 14. Fernando is constructing a scale model of his father’s yacht. The actual yacht has a sail that

is 25 feet tall and the base of the sail is 15 feet long. If the model’s sail is 14 inches tall, how long is its base?

15. Cameron is building a scale model of his sailboat. The right-triangular sail of the actual

boat has a base that is 12 feet long. The height of the actual sail is 20 feet long. The base of the model’s sail is 8 inches long. How long is the hypotenuse of the model’s sail? (The hypotenuse is different from the height!)

16. List all of the similar triangles that are in the

figure at right.

Skill Builders

9. Factoring Standard Forms

Rules for Factoring

Difference of Two Squares: a2 ! b2 = (a + b)(a ! b) Sum of Two Cubes: a3 + b3 = (a + b)(a2 ! ab + b2 ) Difference of Two Cubes: a3 ! b3 = (a ! b)(a2 + ab + b2 )

Special Products (a + b)2 = a2 + 2ab + b2

(a ! b)2 = a2 ! 2ab + b2

(a + b)3 = a3 + 3a2b + 3ab2 + b3

(a ! b)3 = a3 ! 3a2b + 3ab2 ! b3

Factor each of the following expressions.

1. x2!12x + 36 2. z2 ! 6z + 9

3. 9a2 + 24a +16 4. 16x2 + 40xy + 25y2

5. (x + y)2 ! 4(x + y) + 4 6. 4(a ! b)2 ! 8(a ! b) + 4

7. b2 !16 8. 7x4! 7

9. 4x2!1 10. (a ! b)2 ! (a + b)2

11. s6! t

6 12. x3+1

13. a3 ! b3 14. 8p3 + 27q3


10. Inverses

Example: Find the inverse of f (x) = x!2x

Solution: y = x!2

x Start by replacing f(x) with y.

x =y!2

2 Interchange x and y in the equation.

xy = y ! 2 Solve for y.

xy ! y = 2

y(x !1) = 2

y = 2

x!1

f 1(x) = 2x!1

Replace y with f !1(x) .

Find the equation of the inverse for each function below. 1. f (x) = 3x +1 2. f (x) = 5 ! 6x

3. f (x) = x2 + 5 4. f (x) = 4x2

5. f (x) = x!45

6. f (x) = x+1011

7. f (x) = 34x + 8 8. f (x) = 7 ! x2

9. f (x) = 3x 10. f(x) = log x

11. f (x) = x+3x

12. f (x) = x!4x

13. f (x) = 11x!12x2

x 14. f (x) = 3x+4

5x

15. f (x) = 10x!15x2

5x2 16. f (x) = 7x!4

3x

Skill Builders

11. Algebraic Fractions

Facts about Fractions

PKQK

=PQ

where Q ! 0 and K ! 0

Multiplication: PQ!RS=

PRQS

Division: PQ÷RS=

PQ!SR=

PSQR

Addition: PQ+RQ=

P+RQ

Subtraction: PQ!RQ=

P!RQ

Simplify the following fractions.

1. 1 a

b 2. 1 a

1 b

3. x2 –9

x–3 4. x

2 –5x–6

x2 –1

5. x2 –25

5–x 6. a3+b3

a2 –b2

Example:

Solution:

Use Fraction Busters to solve x3+5

x= 2 .

Multiply every term on both sides of the equation by all the terms in the denominator.

Now you have an equation without fractions.

Use the Quadratic Formula to solve.

3 ! x !x

3+ 3 ! x !

5

x= 2 ! 3 ! x

/3 ! x !x

/3+ 3 ! /x !

5

/x= 6x

x2+15 = 6x

x2" 6x +15 = 0

x=3±i 6


Use Fraction Busters to solve for x.

7. 1

x+

3

x2= 4

8. x

3!x+1

5= 2x +

1

5

9. x

3+1

2=

x+2

6!2

3 10. 2

3x+x+2

5= 5x +

1

3


11.

�

1+1

1+1

1+1

12. 1!

1

y

1

y

13. 1!

2

x

1!3

x

14. 1

a+2

b

2

a+1

b

15. x!1

x+1!x+1

x!1

x!1

x+1+x+1

x!1

16. a+b

2a2

a!b

4a

Skill Builders

12. Solving Quadratics

Solve the quadratic equations.

Example 1: 4x2 !12x = 0

Solution: This type of problem is easily factored into 4x(x ! 3) = 0 .

Now you can get the solutions 4x = 0 or x ! 3 = 0 , x = 0 or x ! 3 = 0 .

Example 2: x2! 2x +13 = 0

Solution: This equation cannot be factored so you should use the Quadratic Formula.

x =

2± !48

2=2±4i 3

2= 1± 2i

Example 3: x +1 = x + 2

Solution: When there are square roots in the problem you will need to square both sides.

( x + 2)2 = ( x +1)2

x + 2 = x + 2 x +1

12= x Since there is still a square root, move all

the other terms away from the root.

1

4= x Now square both sides.

0.25 + 2 = 0.25 +1 Make sure you plug the answer back into the original equation because sometimes extra solutions appear when you square equations.

1.5 = 1.5


Solve for the indicated variable.

1. 9x2 !12x = 0 2. 2r +1 = 15r2

3. 1!1

x=12

x2

4. x !4

3x= !

1

3

5. (y + 6)(y ! 2) = !7 6. x + 2 = x ! 4

7. (y ! 5)2 = 9 8. (y + 3)2 = 18

9. x4! x

2= 20 10. x

4+ 20 = 9x2

11. x + 5 = x +1 12. x + 4 = 2 ! 2x

13. x2+ 5x +11 = 0 14. x

2+1 = 0

15. x2! x + 3 = 0 16. x

2= !(5 + x)

Skill Builders

13. Inequalities

Example 1: Solve: 3x ! 5 " 13 .

Solution: Start by writing the inequality. !13 " 3x ! 5 " 13

Add 5 to all three members. !8 " 3x " 18

Divide all three members by 3. ! 8

3" x " 6

The answer written in interval notation is ! 83, 6"# $% .

Remember that if you multiply or divide an inequality by a negative number you must reverse the inequality. For example, solve !2x " 18 , x ! "9 .

Dividing both sides by –2 reverses the inequality.

Example 2: Solve: x2 ! x > 6

Solution: As with equations, we need to have a polynomial equal to zero so that we can factor. Subtract 6 from both sides. x

2! x ! 6 > 0

Factor: (x ! 3)(x + 2) > 0

Find the zero’s. x = 3, x = !2

Test the intervals to find where the quadratic is greater than zero.

-2 3

< >+++ - - - - - - ++++

Here the interval is positive for values less than –2 or greater than 3. It is negative between –2 and 3. Express the solution using inequalities x < !2 or x > 3 .


Solve the inequality. Write your answer in either interval or inequality notation.

1. x ! "2 2. x ! 3 < 7

3. x ! 3 + 2 < 6 4. 2x +1 + 4 ! 7

5. x2+ x ! 6 > 0 6. x

2! x !12 " 0

7. 6x2 < 5x !1 8. (x ! 2)(x ! 3)(x ! 4) > 0 Solve the inequality and graph the solution set.

9. x !1 < !3 10. 3x + 9 ! 6

Example: Graph the solution y + 5 > 2x .

Solution: Begin by treating the expression as an equality and sketch the graph. So begin by graphing y = 2x ! 5 . Since the inequality is y greater than 2x ! 5 the line is dotted because it is not included. Next you should choose a test point on one side of the line and plug it into the inequality. If it makes the inequality true, then shade that side of the graph. If it makes the inequality false, than shade the other side of the graph. A good point to use (as long as it is not on the graph) is (0, 0). If we substitute x = 0 and y = 0 into the inequality, then it becomes 0 + 5 > 10 , which is false so shade the other side of the line.

5–5

5

–5

Graph the solution. 11. x ! 2y < 4 12. x

2+ 5x ! "6

13. 2x + 3y ! 6 14. x ! y > !3

15. 2x + y ! 5 16. 3x + 5y ! 7

x

y

618 Pre-Calculus with Trigonometry

14. Solving Trigonometric Equations

Example 1: Solve sin! = "12!in![0, 2# ]

Solution: Locate where y = !1

2 on the unit circle.

Find the angles (in radians) which will have a y-coordinate of ! 1

2.

! =5"

6!or !! =

11"

6

Example 2: Solve sin 2x = cos x!in![0, 2! ]

Solution: Use the double-angle formula to expand sin 2x : 2 sin x cos x = cos x

Subtract cos x from both sides: 2 sin x cos x ! cos x = 0 Factor: cos x(2 sin x !1) = 0

Solve each part separately: cos x = 0!!!!!!!sin x =1

2

x = 0,!2! !or !x = !

6,! 5!6

Solve the following trigonometric equations by finding approximate values for θ in [0, 2π].

1. sin! = 0.8134 2. cot! = 6.6173

Solve the following trigonometric equations exactly for the indicated variable in [0, 2π].

3. sin x =3

2 4. cos! = 0

5. tan! = 3 6. csc! = 2

7. 6 csc t ! 4 3 = 0 8. sin t cos t = sin t

9. 2 sin! cos! = cos! 10. cos 2t = cos t

11. 2 cos2t = cos t 12. 2 ! sin t = 2 cos2 t

13. tan2 ! " 2 tan! +1 = 0 14. 2 cos2 ! " sin! = 1

15. tan t ! 3 cot t = 0 16. cos t ! sin t = 1

Skill Builders 619

15. Trigonometric Identities

Example: Show that sec2 x + csc2 x = sec2 x csc2 x Solution: Work each side of the equation separately.

Change all trig values to sine and cosine. 1

cos2 x

+1

sin2 x

1

cos2 x sin2 x

Simplify fractions sin2 x+cos2 x

cos2 x sin2 x

1

cos2 x sin2 x

Use the Pythagorean Identity 1

cos2 x sin2 x

1

cos2 x sin2 x

When the left-hand side matches the right hand side, you have verified the identity.

Show that the following equations are identities. 1. sin! csc! = 1

2. (1! sin2 ") sec2 " = 1

3. (sin x !1)(sin x +1) = ! cos2 x

4. cos! + tan! sin! = sec!

5. sin x

2 cos x+

sec x

2 csc x= tan x

6. sin x

sin x!cos x=

1

1!cot x

7. sin2!(1+ cot2!) = 1

8. sin x cot x sec x = 1

9. sin!(csc! " sin! ) = cos2!

10. sin! cos! cot! = 1" sin2!

11. cos2! " sin2! = 2cos2! "1

12. cos4 x ! sin4 x = 1! sin2 x

13. 1

cos2 !+1+

sin2 !

cos2 != 2 sec2 !

14. sin!

csc!"cot != 1+ cos!

15. 1!cos2 "

1+sin"= sin"

16. 1+sin!

cos!=

cos!

1"sin!


16. Logarithms

Example 1: Write as a sum or difference of multiples of logarithms: logbx

y3

Solution: Use log properties to break down expression.

Quotient Rule: logb x – logb y3

Power Rule: 12logb x – 3 logb y

Example 2: Solve: ln(x + 3) ! ln(x ! 2) = ln 4

Solution: Use quotient rule to combine terms. lnx+3

x!2= ln 4

Take ex on both sides. x+3

x!2= 4

Solve the expression. x + 3 = 4x ! 8 11 = 3x

x =11

3

Make sure the solution is in the domain of the expression. In this case the domain is x > 2 . Since 11

3> 2 , it is a valid solution.

Write each expression as a sum or difference of multiples of logarithms. 1. logb x

2y3 2. logbx9

y7

3. logbx5

y26 4. ln

x5y2

z5

5. ln4x+3

x4

3 6. ln1

6x2!7x!37

Write each expression as a single logarithm.

7. log557 + log5

4025

8. log23211

+ log212116

! log245

9. log 4x2 ! 9( ) ! ln 8x3 ! 27( ) 10. 3 lna2b

c2+ 2 ln

b2c

a4+ 2 ln

abc

2

Solve each equation without using a calculator.

11. log3(x +1) = 2 12. log5(5x !1) = !2

13. log2x = log 3+ log(x !1) 14. ln x + ln(x ! 2) = ln(x + 4)

15. 2 ln(x +1) ! ln(x + 4) = ln(x !1) 16. log4 x + log4 (6x +10) = 1

Skill Builders 621

17. Algebra for College Mathematics

Example 1: Rationalize the numerator: 2x+2h! 2x

h

Solution: Multiply numerator and denominator by the conjugate of the numerator.

2x+2h! 2x( ) 2x+2h+ 2x( )

h 2x+2h+ 2x( )

Simplify: 2x+2h!2x

h 2x+2h+ 2x( )= 2h

h 2x+2h+ 2x( )= 2

2x+2h+ 2x

Example 2: Factor x3 4 from 2x3 4 ! 5x7 4 Solution: Subtract the exponent of the factor from each term.

x3 4 2x3 4!3 4 ! 5x7 4!3 4( )

Simplify: x3 4 2 – 5x( )

Example 3: Simplify: 4x!1 3+3x2 35x! 4 3

Solution: Multiply the numerator and the denominator by x4 3 to clear negative exponents.

4x!1 3+3x2 3( )x4 3

5x! 4 3( )x4 3

Distribute and simplify: 4x+3x2

5

Rationalize the numerator of the following expressions.

1. x+h! x

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

h

3. (5x+5h)1 2 !(5x)1 2

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

h


Factor out the specified term from the expression.

5. x1 2 :!!5x3 2 – 2x1 2 6. x

2 3 : !!x2 3+ 3x5 3

7. 2x!3

: !!8x!2

! 2x!3

+ 4 8. 3x!2 : !!9x!2+ 2x!1

+ 6

9. x!1 3 : !!4x5 3

! 7x!1 3+ x

2 3 10. 2x!2 3

: !!2x!2 3

+ 6x1 3

+ 8x4 3

Simplify.

11. x!2

+x!3

x!1

12. 4x!3+5x2

x!4

+1

13. 2x!1 2

x+1 2

x1 2

!x!1 2

14. x2 3

+x!1 3

x5 3

!x!4 3

15. x+2( )

!3+ x+2( )

!2

x+2( )!1 16. (x+x!1)!1 2 +(x+x!1)! 3 2

(x+x!1)1 2

Skill Builders 623

18. Completing the Square

Example: Solve 2x2 !12x ! 7 = 0 .

Solution: x2! 6x !

7

2= 0 Divide both sides by the coefficient of x2

so that x2 will have a coefficient of 1. x

2! 6x =

7

2 Subtract the constant term from both sides.

x2! 6x + (!3)2 = 7

2+ (!3)2 Complete the square. Add the square of

one half the coefficient of x to both sides. (x ! 3)2 = 25

2 Factor the left and combine the

right. x ! 3 = ±5 2

2 Take the square root of both sides and

solve for x. The solutions are 3+ 5 2

2 and

�

3!5 2

2.

Solve each of the following exactly by completing the square. Note: Some solutions will be in

complex form.

1. x2+ 4x = 12 2. a

2! 2a = 8

3. d2! 6d = 16 4. p2 +12p = !27

5. f 2 +10 f ! 22 = 0 6. z2 ! 8z +1 = 0

7. y2 + 6y !1 = 0 8. w2! 5w ! 3 = 0

9. c2! 5c ! 2 = 0 10. 2b2 ! 4b ! 8 = 0

11. 3n2 ! 9n !12 = 0 12. 3t2! 8t +1 = 0

13. 5m2 +12m !1 = 0 14. h2! 2h + 5 = 0

15. 4r2 ! 3r + 5 = 0 16. 6v2! 3v +1 = 0


19. Complex Numbers

Complex numbers behave just like binomials when they are added or subtracted. Example 1: If z1 = 4 ! 5i and z2 = 1+10i , find: a. z1 + z2

= (4 ! 5i) + (1+10i)

= 4 ! 5i +1+10i

= 5 + 5i

b. z1 ! z2

= (4 ! 5i) ! (1+10i)

= 4 ! 5i !1!10i

= 3!15i

Complex numbers also behave like binomials when you multiply them. The only

thing you need to remember is that i2 = !1 . Example 2: Multiply: (4 ! 5i)(1+10i) Solution: = 4 + 35i ! 50i2 Multiply each term in one binomial by each term

in the other. = 4 + 35i + 50 = 54 + 35i When dividing one complex number by another you rationalize the denominator by

multiplying the numerator and denominator by the complex conjugate. Example 3: Divide: 4+3i

52i

Solution: =4 + 3i

5 ! 2i"5 + 2i

5 + 2i=20 + 23i + 6i2

25 ! 4i2=14 + 23i

29=14

29+23

29i

Express in a + bi form. 1. 2i + 3i 2. 6i + 4i ! 3i

3. 1

2 + 3i 4. 5i

2

5. 4i2! 3i + 5 6. i

2(a ! b)

7. – i i4 ! i3 + i2 ! i 8. (i + 2)(i ! 3)

9. i(i2+ i ! 4) 10. 4i5 +3i3

i2

11. (i2+ i)(i +1) 12. (3i + 4)(2i

2! 5)

13. 5+i

5i 14. 32i

3+2i

Skill Builder Answers Pre-Calculus with Trigonometry

1. Factoring Out Monomial Terms 1. xy(2x2 + 7y)

2. 6a2b3(7a + 4b2)

3. 4π r2h(6r2 − 5h3)

4. 5b10c6d 3(6bd + 3bc − 7cd)

5. 9x3y2z6(4x2y3 − 3yz4 + 2x3z)

6. ab2(6a3 − 7b2 + 7a)

7. 9p5q(3p + 5p − 6p3)

8. 4 j 3k2(2 j 7 + 5k2)

9. 7x2y3(3x2 + 5y5) 10. 21d 2h3(2d + 3h2)

11. 6π r2(4h2 + 3h + 9)

12. 3mn(5n3 −11m2)

13. a2b3c(2a3b2 − 6ac5 + 7b3c2) 14. 5st5(3s − 5t2 +11)

15. 8uv(u2 − 4v3 +10)

16. 2x2y2(3x2y + 4y3 − x + 3)

2. Law of Exponents 1. x7

2. y5

3. a3

4. p5

5. x9

6. y10

7. z12

8. b27

9. z6x12 10. 1

11. a6

12. x6

13. 2

14. 72x5

15. 20x3 16. 729x8


3. Function Notation 1. 53

2. –108

3. 5x2 +10xh + 5h2 + 3x + 3h

4. 21x2 + 42xh + 21h2 − 20x − 20h

5. 21x3 + 44x2 +15x

6. 6x3ex + 7ex

7. 572

8. 18x2 +15x

9. 6x2 +15x 10. 80x2 +180x +105

11. 20x2 + 20x + 24

12. 944784

13. 2560

14. e(3x2+2x+6)

15. 3e2x + 2ex + 6 16. 729x8

4. Properties of Lines 1. y − 3= 3(x + 4)

2. y = 2x +11

3. y − 3 = −6(x −8)

4. y = −5x + 36

5. y = 53 x +

163

6. y = 8

7 x –307

7. y − 4 = 2(x +1)or y + 4 = 2(x + 5)

8. y − 2 = 3(x − 5)or y –11= 3(x – 8)

9. y +1= – 17 (x – 3) 10. y = – 16 x +152

11. y = 1

9 x +539

12. y = – 12 x –10

13. 145

14. 2 29

15. 317 16. 97


5. Multiplying Binomials 1. x2 − 3x −18

2. a5 − 4a3 + a2 − 4

3. b3 + 5b2 − 3b −15

4. z5 − z3 + z2 −1

5. p21 − 2p20 + p19

6. q7 + 3q4 + 4q3 +12

7. 30h5 −10h4 + 6h − 2

8. d11 − d10 + d 5 − d 4

9. x3 + 2x2 − 45x −126

10. x4 − x3 + x2 − 3x − 6

11. m196 −m194

12. 36n7 + 36n6 + 9n5

13. {900 j102 +100 j100 − 9 j3 − j}

14. 2π 2k6 +11π k5 +15k 4

15. 2c2x2 + 2d 3xy + c4dx + c2d 4y 16. f 4y4 + f 3gxy3 + fg2x2y + g3x3

6. Special Triangles 1. α = 30º , β = 60º

2. 4 3

3. 0.25 inches

4. 26 33 cm

5. Both a and b = 5 .

6. 13 each

7. 17 2 mm

8. 9 3 cm2

9. 25 34 units2

10. 7 3

11. α = 30º , β = 60º

12. 50

13. 30

14. 100 3

15. x = c 3 16. x = b 3, y = 2b


7. Simplifying and Combining Radicals 1. 20 3

2. 13 5

3. ab2 ab

4. x3b4 xb

5. 5 + 3 2

6. 10 6

7. 8 2 +10 3

8. 99

9. 30

10. 40

11. 70

12. 3 22

13. 5 77

14. ab

ab

15. 9 22 16. 9 2

4

8. Similar Triangles 1. x = 48

9

2. x =16.5

3. x =11.2

4. x = 12821

5. x = 215

6. x = 22

3

7. k = 2x

8. h = 4y10−y

9. b = x2

10. h = 13x

x+5

11. 42 feet

12. 11.25 meters

13. 3.75 feet

14. 8.4 inches

15. 8 343 16. ∆ABC, ∆APB, ∆CPB


9. Factoring Standard Forms 1. (x − 6)2

2. (z − 3)2

3. (3a + 4)2

4. (4x + 5y)2

5. (x + y − 2)2

6. 4(a − b −1)2

7. (b − 4)(b + 4) 8. 7(x +1)(x −1)(x2 +1)

9. (2x −1)(2x +1) 10. −4ab

11. (s3 − t3)(s3 + t3)

12. (x +1)(x2 − x +1)

13. (a − b)(a2 + ab + b2)

14. (2p + 3q)(4p2 − 6pq + 9q2)

10. Inverses 1. f −1(x) = x−1

3 2. f −1(x) = 5−x6

3. f −1(x) = ± x − 5 4. f −1(x) = ± x2

5. f −1(x) = 5x + 4 6. f −1(x) = 11x −10

7. f −1(x) = 43 (x − 8) 8. f −1(x) = 7− x

9. f −1(x) = log3 x 10. f -1(x) = 10x

11. f −1(x) = 3x−1 12. f −1(x) = 4

1−x

13. f −1(x) = 11−x12 14. f −1(x) = 45x−3

15. f −1(x) = 2x+3 16. f −1(x) = −4

3x−7


11. Algebraic Functions 1. 1

ab 2. ba

3. x + 3 4. x−6x−1

5. −(x + 5) 6. a2−ab+b2a−b

7. x =1, − 34 8. − 3

14

9. x = −5 10. x = 1± 2881144

11. 53 12. y−1

y

13. x−2x−3 14. 2a+b

2b+a

15. −2x1+x2

16. 2(a+b)a(a−b)

12. Solving Quadratics 1. x = 0, 43 2. r = − 1

5 ,13

3. x = −3, 4

4. x =1, − 43

5. y = −5,1

6. x = 7

7. y = 2, 8

8. y = −3± 3 2

9. x = ± 5

10. x = −2, 2, ± 5

11. x = 4

12. x = 0

13. x = −5±i 192

14. x = ±i

15. x = 1±i 112

16. x = −1±i 19

2


13. Inequalities 1. all real numbers 2. −4 < x <10

3. −1< x < 7

4. x ≤ −2 and x ≥1

5. x < −3, x > 2

6. −3≤ x ≤ 4

7. 13 << x << 1

2

8. 2 < x < 3, x > 4

9. no solution

10. x ≤ −5 and x ≥ −1

11. x = 4

12.

13.

14.

15.

16.

x

y

0 –2 –4 4 2

x

y

x

y


14. Solving Trigonometric Equations 1. θ ≈ 2, 4.28 radians 2. θ ≈ 0.15, 3.29 radians

3. x = π3 ,

2π3

4. θ = π

2 ,3π2

5. θ = π3 ,

4π3

6. θ = π

4 ,3π4

7. t = π3 ,

2π3

8. t = π, 2π

9. θ = π6 ,

π2 ,

3π2 ,

5π6

10. t = 0, 2π3 ,

4π3

11. t = π3 ,

π2 ,

3π2 ,

5π3

12. t = 0, π6 ,

5π6 ,π

13. θ = π4 ,

5π4

14. θ = π

6 ,5π63π2

15. t = π3 ,

2π3 ,

4π3 ,

5π3 16. t = 0, 3π2

15. Trigonometric Identities Answers will vary for this section, because students will need to prove the given identities.

16. Logarithms 1. 2 logb x + 3logb y 2. 9 logb x − 7logb y

3. 56 logb x −

13 logb y

4. 5 ln x + 2 ln y − 1

5 ln z

5. 13 ln(4x + 3)−

43 ln y

6. − 1

7 ln(3x +1)−17 (2x − 3)

7. log5 87

8. log2 552

9. ln 2x+34x2+6x+9

10. ln b74

11. 8

12. 26125

13. 3

14. 4

15. 5 16. 13


17. Algebra for College Mathematics 1. 1

x+h+ x 2. 3

3(x+h)+ 3x

3. 5(5x+5h)1/2+(5x)1/2

4. 2x+h(x+h)1/2+(x)1/2

5. x1/2(5x − 2)

6. x2/3(1+ 3x)

7. 2x −3(8x −1+ 4x3)

8. 3x−2(3+ 23 x + 2x

2 )

9. x−1/2(4x2 − 7 + x)

10. 2x−2/3(1+ 6x + 8x2 )

11. x+1x2

12. 4x+5x61+x4

13. 2+xx−1

14. x2+x

x3−1

15. 1+ x+2x−2

16. x3+x2+x

x4+2x2+1

18. Completing the Square 1. x = −6, 2 2. a = −2, 4

3. d = −2, 8 4. p = −3, −9

5. f = −5± 47 6. z = 4 ± 15

7. y = −3± 10 8. w = 5± 372

9. c = 5± 332 10. b =1± 5

11. n = 4, −1 12. t = 4± 133

13. m = −6± 415 14. h =1± 2i

15. r = 3±i 718 16. v = 3±i 15

12


19. Complex Numbers 1. 5i 2. 7i

3. 213 −313 i 4. −5

5. 1− 3i 6. b − a

7. 0 8. −7− i

9. −1− 5i 10. −i

11. −2 12. −28− 21i

13. 1213 +

513 i 14. 5

13 −1213 i

Education

Precalculus Skills Builder Workbook with solutions