Chapter Twenty — Rational Expressions

The Humongous Book of Algebra Problems

440

Simplifying Rational Expressions

Reducing fractions by factoring

Note: Problems 20.1–20.2 refer to the expression .

20.1 Simplify the rational expression and identify the restriction on the answer.

The quotient of exponential expressions with the same base is equal to the base

raised to the difference of the exponents: .

Therefore, when x ≠ 0.

Note: Problems 20.1–20.2 refer to the expression .

20.2 Explain why the restriction stated in Problem 20.1 is necessary.

The expression is not deﬁned for x = 0; substituting x = 0 into the

expression results in , which is indeterminate. However, 5x is deﬁned when

x = 0. The expressions and 5x are truly not equivalent if their values differ

when x = 0, so stating they are equal must include the disclaimer that the

equality is not guaranteed at x = 0.

20.3 Simplify the expression and indicate restrictions, if any, on the result.

Write the rational expression as a product of three fractions, the quotient of the

coefﬁcients, the quotient of the x-terms, and the quotient of the y-terms.

Express the coefﬁcients as prime factorizations to reduce the corresponding

fraction.

The numerator

contains x

2

and the

denominator contains

x (which equals x

1

),

so subtract the

powers of x.

This is the

restriction the problem

mentioned. The fraction

is equal to 5x only

when x ≠ 0. Why? See

Problem 20.2.

Avoid

dividing by

0 at all costs.

0 ÷ 0 is called an

indeterminate value,

and any other number

divided by 0 is said

to be undened. In

either case, 0 in the

denominator is

bad.

Restrictions

are the x-values

that make the

denominator zero.

See Problems

2.30–2.32 for more

information about prime

factorizations.

Chapter Twenty — Rational Expressions

The Humongous Book of Algebra Problems

441

Apply the exponential property referenced in Problem 20.1.

Place variable expressions with positive exponents in the numerator and

expressions with negative exponents in the denominator.

Therefore, when x ≠ 0.

20.4 Simplify the expression: and indicate restrictions, if any, on the result.

Factor the numerator, a difference of perfect squares.

The numerator and denominator contain the common factor (x + 7).

To reduce the fraction, eliminate the shared factor.

Therefore, when x ≠ –7.

20.5 Simplify the expression and indicate restrictions, if any,

on the result.

Factor the numerator and denominator; each contains a greatest common

factor.

Whereas 5 is a prime number, 10 is not. Express 10 as a prime factorization:

10 = 2 · 5.

After you

place the y

–4

in

the denominator

where it belongs,

change the negative

exponent to a

positive.

You don’t

need to include

y ≠ 0 because both

of the expressions

would be undened in

that case. You need

to include restrictions

only on variables that

would make one

of the fraction

denominators

equal 0 but not

the other.

The difference

of perfect squares

formula (from Problem

12.26) is (a

2

– b

2

) =

(a + b)(a – b).

The fraction

is undened when

x = –7 because

substituting –7 into the

denominator gives you

(–7) + 7 = 0, and you

can’t divide by 0.

Chapter Twenty — Rational Expressions

The Humongous Book of Algebra Problems

442

Factor the quadratic expression in the denominator: x

2

+ 6x – 27 = (x + 9)(x – 3).

The numerator and denominator contain common factors 5 and (x – 3).

Eliminate the common factors to reduce the rational expression.

Therefore, when x ≠ 3.

20.6 Simplify the expression and indicate restrictions, if any, on the result.

The terms of the numerator have the greatest common factor y. Factoring it out

of the expression results in a difference of perfect squares: y

3

– 4y = y(y

2

– 4).

Factor the difference of perfect squares in the numerator and the difference of

perfect cubes in the denominator.

Reduce the fraction to lowest terms.

Therefore, when y ≠ 2.

Either

of these

fractions is an

acceptable nal

answer.

You eliminate

the factor (x – 3)

from the denominator.

Set it equal to 0 and

solve to get the

restriction:

x = 3.

The formula

is a

3

– b

3

=

(a – b)(a

2

+ ab + b

2

).

Check out Problems

12.31 and 12.33 for

more information

about factoring

differences of

perfect cubes.

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