Lesson 11-3
Example 1 Find Excluded Values
State the excluded value of each expression.
2y
a.
3y +1
Exclude the values for which 3y + 1 = 0.
3y + 1 = 0
The denominator cannot be zero.
3y = -1
Subtract 1 from each side.
y=-
1
3
Divide each side by 3.
1
3
Therefore, y cannot equal - .
b.
-b
b 2 -9
Exclude the values for which b 2 – 9 = 0.
b2 – 9 = 0
The denominator cannot be zero.
(b – 3)(b + 3) = 0
Factor.
b – 3 = 0 or b + 3 = 0
b=3
b = -3
Zero Product Property
Therefore, b cannot equal 3 or -3.
Real-World Example 2 Use Rational Expressions
GEOMETRY The area of the rectangle at the right
is 15a4b2c. Write an expression that gives the length
of the rectangle.
Understand
Plan
Solve
2
3a b
?
You know the area and the width of the rectangle.
A
Use the formula for length of a rectangle,  = . Substitute 15a4b2c for A and
w
3a2b for w.
A
=w
15a4b2c
= 3a2b
Write the equation.
4 2
2
Replace A with 15a b c and w with 3a b.
821
Simplify the denominator.
49
 5.3
Simplify. Use a calculator.
=
Check
So, the height of the cylinder is about 5.3 inches.
Use estimation to determine whether the answer is reasonable.
821
 5.6 or 5  The solution is reasonable.
49(3)
Test Example 3 Expressions Involving Monomials
30x 2 y 3
Which expression is equivalent to
?
25xz
6xy 2
xy 3
63
A
B
C
z
5
5z
6 xy 3
D
5z
Read the Test Item
The expression
30 x 2 y 3
is a monomial divided by a monomial.
25 xz
Solve the Test Item
Step 1 The GCF of the numerator and denominator is 5x
Step 2 Divide the numerator and denominator by the GCF.
Step 3 Simplify. The correct answer is D.
5 x 6 xy 3 
5 x 5 z 
5 x 6 xy 3 
5 x 5 z 
6 xy 3
5z
Example 4 Simplify Rational Expressions
x2 – 2x – 8
Simplify
. State the excluded values of x.
x+2
x2 – 2x – 8 (x + 2)(x – 4)
x+2 =
x+2
Factor.
1
Divide the numerator and denominator by
the GCF, x + 2.
(x + 2)(x – 4)
=
x+2
1
=x–4
Simplify.
Exclude the values for which x + 2 equals 0.
x+2=0
The denominator cannot equal zero.
x = –2
Subtract 2 from each side.
So, x  –2.
Example 5 Recognize Opposites
45 – 9x
Simplify 2
. State the excluded values of x.
x – 3x – 10
45 – 9x
9(5 – x)
=
x2 – 3x – 10 (x – 5)(x + 2)
9(–1)(x – 5)
=
(x – 5)(x + 2)
Factor.
Rewrite 5 – x as –1(x – 5).
1
9(–1)(x – 5)
=
(x – 5)(x + 2)
Divide out the common factor, x – 5.
1
9
=–x+2
Simplify.
Exclude the values for which x2 – 3x – 10 equals 0.
x2 – 3x – 10 = 0
The denominator cannot equal zero.
(x – 5)(x + 2) = 0
Factor.
x = 5 or x = –2
Zero Product Property
So, x  –5 and x  –2.
Example 6 Rational Functions
x2 – 8x – 7
Find the roots of f(x) =
.
x+2
x2 – 8x – 7
f(x) = x + 2
0=
x2 – 8x – 7
x+2
Original function
f(x) = 0
(x – 1)(x – 7)
Factor.
x+2
When x = 1 or x = 7, the numerator becomes 0, so f(x) = 0. Therefore, the roots of the function are
1 and 7.
0=