O'Reilly logo

The Humongous Book of Algebra Problems by W. Michael Kelley

Stay ahead with the world's most comprehensive technology and business learning platform.

With Safari, you learn the way you learn best. Get unlimited access to videos, live online training, learning paths, books, tutorials, and more.

Start Free Trial

No credit card required

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 undened. 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 dont
need to include
y 0 because both
of the expressions
would be undened 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 undened when
x = –7 because
substituting –7 into the
denominator gives you
(7) + 7 = 0, and you
cant 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.

With Safari, you learn the way you learn best. Get unlimited access to videos, live online training, learning paths, books, interactive tutorials, and more.

Start Free Trial

No credit card required