1. Multiplying and Dividing Rational Expressions
8-2
Warm Up
Lesson Presentation
Lesson Quiz
Holt McDougal Algebra 2
Holt Algebra 2

2. Simplify each expression. Assume all variables are nonzero.
Warm Up
Simplify each expression. Assume all variables are nonzero.
1. x5  x2 x7
2. y3  y3 y6 3. x6 x2 4. y2 y5 1 y3 x4
Factor each expression.
5. x2 – 2x – 8
(x – 4)(x + 2)
6. x2 – 5x
x(x – 5)
7. x5 – 9x3
x3(x – 3)(x + 3)

3. Objectives Simplify rational expressions.
Multiply and divide rational expressions.

4. Vocabulary
rational expression

5. In Lesson 8-1, you worked with inverse variation functions such as y =
In Lesson 8-1, you worked with inverse variation functions such as y = . The expression on the right side of this equation is a rational expression. A rational expression is a quotient of two polynomials. Other examples of rational expressions include the following: 5 x

6. Because rational expressions are ratios of polynomials, you can simplify them the same way as you simplify fractions. Recall that to write a fraction in simplest form, you can divide out common factors in the numerator and denominator.
When identifying values for which a rational expression is undefined, identify the values of the variable that make the original denominator equal to 0.
Caution!

7. Example 1A: Simplifying Rational Expressions
Simplify. Identify any x-values for which the expression is undefined.
10x8
6x4
510x8 – 4 36 5 3 x4 =
Quotient of Powers Property
The expression is undefined at x = 0 because this value of x makes 6x4 equal 0.

8. Example 1B: Simplifying Rational Expressions
Simplify. Identify any x-values for which the expression is undefined.
x2 + x – 2
x2 + 2x – 3
(x + 2)(x – 1)
(x – 1)(x + 3)
(x + 2)
(x + 3)
Factor; then divide out common factors. =
The expression is undefined at x = 1 and x = –3 because these values of x make the factors (x – 1) and (x + 3) equal 0.

9. Example 1B Continued
Check Substitute x = 1 and x = –3 into the original expression.
(1)2 + (1) – 2
(1)2 + 2(1) – 3
(–3)2 + (–3) – 2
(–3)2 + 2(–3) – 3 4 = =
Both values of x result in division by 0, which is undefined.

10. Check It Out! Example 1a
Simplify. Identify any x-values for which the expression is undefined.
16x11
8x2
28x11 – 2 18
2x9 =
Quotient of Powers Property
The expression is undefined at x = 0 because this value of x makes 8x2 equal 0.

11. Check It Out! Example 1b
Simplify. Identify any x-values for which the expression is undefined.
3x + 4
3x2 + x – 4
(3x + 4)
(3x + 4)(x – 1) 1
(x – 1)
Factor; then divide out common factors. =
The expression is undefined at x = 1 and x = –because these values of x make the factors (x – 1) and (3x + 4) equal 0. 4 3

12. Check It Out! Example 1b Continued
Check Substitute x = 1 and x = – into the original expression. 4 3
3(1) + 4
3(1)2 + (1) – 4 7 =
Both values of x result in division by 0, which is undefined.

13. Check It Out! Example 1c
Simplify. Identify any x-values for which the expression is undefined.
6x2 + 7x + 2
6x2 – 5x – 5
(2x + 1)(3x + 2)
(3x + 2)(2x – 3)
(2x + 1)
(2x – 3)
Factor; then divide out common factors. =
The expression is undefined at x =– and x = because these values of x make the factors (3x + 2) and (2x – 3) equal 0. 3 2

14. Check It Out! Example 1c Continued
Check Substitute x = and x = – into the original expression. 3 2
Both values of x result in division by 0, which is undefined.

15. Example 2: Simplifying by Factoring by –1
Simplify Identify any x values for which the expression is undefined.
4x – x2
x2 – 2x – 8
–1(x2 – 4x)
x2 – 2x – 8
Factor out –1 in the numerator so that x2 is positive, and reorder the terms.
–1(x)(x – 4)
(x – 4)(x + 2)
Factor the numerator and denominator. Divide out common factors. –x
(x + 2 )
Simplify.
The expression is undefined at x = –2 and x = 4.

16. Example 2 Continued
Check The calculator screens suggest that
= except when x = – or x = 4.
4x – x2
x2 – 2x – 8 –x
(x + 2)

17. Check It Out! Example 2a
Simplify Identify any x values for which the expression is undefined.
10 – 2x
x – 5
–1(2x – 10)
x – 5
Factor out –1 in the numerator so that x is positive, and reorder the terms.
–1(2)(x – 5)
(x – 5)
Factor the numerator and denominator. Divide out common factors. –2 1
Simplify.
The expression is undefined at x = 5.

18. Check It Out! Example 2a Continued
Check The calculator screens suggest that
= –2 except when x = 5.
10 – 2x
x – 5

19. Check It Out! Example 2b
Simplify Identify any x values for which the expression is undefined.
–x2 + 3x
2x2 – 7x + 3
–1(x2 – 3x)
2x2 – 7x + 3
Factor out –1 in the numerator so that x is positive, and reorder the terms.
–1(x)(x – 3)
(x – 3)(2x – 1)
Factor the numerator and denominator. Divide out common factors. –x
2x – 1
Simplify.
The expression is undefined at x = 3 and x = . 1 2

20. Check It Out! Example 2b Continued
Check The calculator screens suggest that
= except when x = and x = 3. –x
2x – 1 1 2
–x2 + 3x
2x2 – 7x + 3

21. You can multiply rational expressions the same way that you multiply fractions.

22. Example 3: Multiplying Rational Expressions
Multiply. Assume that all expressions are defined.
3x5y3
2x3y7 
10x3y4
9x2y5
x – 3
4x + 20 
x + 5
x2 – 9 A. B.
3x5y3
2x3y7 
10x3y4
9x2y5 3 5
x – 3
4(x + 5) 
x + 5
(x – 3)(x + 3) 3
5x3
3y5 1
4(x + 3)

23. Check It Out! Example 3
Multiply. Assume that all expressions are defined. x 15  20 x4 2x x7
10x – 40
x2 – 6x + 8 
x + 3
5x + 15 A. B. x 15  20 x4 2x x7 2
10(x – 4)
(x – 4)(x – 2) 
x + 3
5(x + 3) 2 2 3
2x3 3 2
(x – 2)

24. You can also divide rational expressions
You can also divide rational expressions. Recall that to divide by a fraction, you multiply by its reciprocal. 1 2 3 4 ÷ 1 2 4 3  2 2 3 = =

25. Example 4A: Dividing Rational Expressions
Divide. Assume that all expressions are defined.
5x4
8x2y2 ÷
8y5 15
5x4
8x2y2  15
8y5
Rewrite as multiplication by the reciprocal.
5x4
8x2y2  15
8y5 2 3 3
x2y3 3

26. Example 4B: Dividing Rational Expressions
Divide. Assume that all expressions are defined.
x4 – 9x2
x2 – 4x + 3 ÷
x4 + 2x3 – 8x2
x2 – 16
Rewrite as multiplication by the reciprocal.
x4 – 9x2
x2 – 4x + 3 
x2 – 16
x4 + 2x3 – 8x2
x2 (x2 – 9)
x2 – 4x + 3 
x2 – 16
x2(x2 + 2x – 8)
x2(x – 3)(x + 3)
(x – 3)(x – 1) 
(x + 4)(x – 4)
x2(x – 2)(x + 4)
(x + 3)(x – 4)
(x – 1)(x – 2)

27. Check It Out! Example 4a
Divide. Assume that all expressions are defined. x2 4 ÷
12y2
x4y
Rewrite as multiplication by the reciprocal. x2 4 
x4y
12y2 x2 4 
12y2 3 1
x4 y 2 3y x2

28. Check It Out! Example 4b
Divide. Assume that all expressions are defined.
2x2 – 7x – 4
x2 – 9 ÷
4x2– 1
8x2 – 28x +12
2x2 – 7x – 4
x2 – 9 
8x2 – 28x +12
4x2– 1
(2x + 1)(x – 4)
(x + 3)(x – 3) 
4(2x2 – 7x + 3)
(2x + 1)(2x – 1)
(2x + 1)(x – 4)
(x + 3)(x – 3) 
4(2x – 1)(x – 3)
(2x + 1)(2x – 1)
4(x – 4)
(x +3)

29. Example 5A: Solving Simple Rational Equations
Solve. Check your solution.
x2 – 25
x – 5
= 14
(x + 5)(x – 5)
(x – 5)
= 14
Note that x ≠ 5.
x + 5 = 14
x = 9

30. Example 5A Continued
x2 – 25
x – 5
= 14
Check
(9)2 – 25
9 – 5 14 56 4 14

31. Example 5B: Solving Simple Rational Equations
Solve. Check your solution.
x2 – 3x – 10
x – 2
= 7
(x + 5)(x – 2)
(x – 2)
= 7
Note that x ≠ 2.
x + 5 = 7
x = 2
Because the left side of the original equation is undefined when x = 2, there is no solution.

32. Example 5B Continued
Check A graphing calculator shows that 2 is not a solution.

33. Check It Out! Example 5a
Solve. Check your solution.
x2 + x – 12
x + 4
= –7
(x – 3)(x + 4)
(x + 4)
= –7
Note that x ≠ –4.
x – 3 = –7
x = –4
Because the left side of the original equation is undefined when x = –4, there is no solution.

34. Check It Out! Example 5a Continued
Check A graphing calculator shows that –4 is not a solution.

35. Check It Out! Example 5b
Solve. Check your solution.
4x2 – 9
2x + 3
= 5
(2x + 3)(2x – 3)
(2x + 3)
= 5
Note that x ≠ – 3 2
2x – 3 = 5
x = 4

36. Check It Out! Example 5b Continued
= 5
Check
4(4)2 – 9
2(4) + 3 5 55 11 5

37. Lesson Quiz: Part I
Simplify. Identify any x-values for which the expression is undefined. 1.
x2 – 6x + 5
x2 – 3x – 10
x – 1
x + 2
x ≠ –2, 5
6x – x2
x2 – 7x + 6 –x
x – 1 2.
x ≠ 1, 6

38. Lesson Quiz: Part II
Multiply or divide. Assume that all expressions are defined. 3.
x + 1
3x + 6 
6x + 12
x2 – 1 2
x – 1 4.
x2 + 4x + 3
x2 – 4 ÷
x2 + 2x – 3
x2 – 6x + 8
(x + 1)(x – 4)
(x + 2)(x – 1)
Solve. Check your solution.
4x2 – 1
2x – 1
= 9 5.
x = 4