aa_mod06_les02.ppt

Holt McDougal Algebra 2
Multiplying and Dividing
Rational Expressions
Holt Algebra 2
Warm Up
Lesson Presentation
Lesson Quiz

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

Simplify rational expressions.
Multiply and divide rational
expressions.
Objectives

rational expression
Vocabulary

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

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!
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.

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

Example 1B: Simplifying Rational Expressions
x2 + x – 2
x2 + 2x – 3
(x + 2)(x – 1)
(x – 1)(x + 3)
Factor; then divide out
common factors.
= (x + 2)
(x + 3)
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.

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

Check It Out! Example 1a
Quotient of Powers Property
16x11
8x2
28x11 – 2
18
2x9
=
The expression is undefined at x = 0 because
this value of x makes 8x2 equal 0.

Check It Out! Example 1b
3x + 4
3x2 + x – 4
(3x + 4)
(3x + 4)(x – 1)
common factors.
= 1
(x – 1)
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

3(1) + 4
3(1)2 + (1) – 4
7
0
=
undefined.
Check It Out! Example 1b Continued
Check Substitute x = 1 and x = – into
the original expression.
4
3

Check It Out! Example 1c
6x2 + 7x + 2
6x2 – 5x – 5
(2x + 1)(3x + 2)
(3x + 2)(2x – 3)
common factors.
= (2x + 1)
(2x – 3)
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
2
3

undefined.
Check Substitute x = and x = –
into the original expression.
3
2
2
3
Check It Out! Example 1c Continued

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

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

x is positive, and reorder the terms.
The expression is undefined at x = 5.
10 – 2x
x – 5
–1(2x – 10)
x – 5
–2
1 Simplify.
–1(2)(x – 5)
(x – 5)

Check It Out! Example 2a Continued
= –2 except when x = 5.
10 – 2x
x – 5

x is positive, and reorder the terms.
–x2 + 3x
2x2 – 7x + 3
–1(x2 – 3x)
2x2 – 7x + 3
–x
2x – 1 Simplify.
–1(x)(x – 3)
(x – 3)(2x – 1)
The expression is undefined at x = 3 and x = .
1
2

–x2 + 3x
2x2 – 7x + 3
= except when x =
and x = 3.
–x
2x – 1
1
2

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

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

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

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
=

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

Example 4B: Dividing Rational Expressions
x4 – 9x2
x2 – 4x + 3
÷ x4 + 2x3 – 8x2
x2 – 16
defined.
x4 – 9x2
x2 – 4x + 3
 x2 – 16
x4 + 2x3 – 8x2
Rewrite as
multiplication by
the reciprocal.
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)

x4 y
Rewrite as multiplication
by the reciprocal.
x2
4
÷
12y2
x4y
x2
4

12y2
2
3y
x2
defined.
x2
4

x4y
12y2
3 1

2x2 – 7x – 4
x2 – 9
÷ 4x2– 1
8x2 – 28x +12
defined.
(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)
2x2 – 7x – 4
x2 – 9
 8x2 – 28x +12
4x2– 1

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

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

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

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

Note that x ≠ –4.
x2 + x – 12
x + 4
= –7
(x – 3)(x + 4)
(x + 4)
= –7
x – 3 = –7
x = –4
Because the left side of the original equation is
undefined when x = –4, there is no solution.

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

4x2 – 9
2x + 3
= 5
(2x + 3)(2x – 3)
(2x + 3)
= 5
2x – 3 = 5
x = 4
Note that x ≠ – .
3
2

4x2 – 9
2x + 3
= 5
Check
4(4)2 – 9
2(4) + 3
5
55
11
5
5 5

Lesson Quiz: Part I
1.
2.
x2 – 6x + 5
x2 – 3x – 10
6x – x2
x2 – 7x + 6
x – 1
x + 2
x ≠ –2, 5
–x
x – 1
x ≠ 1, 6

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

aa_mod06_les02.ppt

More Related Content

Similar to aa_mod06_les02.ppt

Recently uploaded

aa_mod06_les02.ppt