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The Humongous Book of Algebra Problems by W. Michael Kelley

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Chapter Twenty — Rational Expressions
The Humongous Book of Algebra Problems
449
20.17 Simplify the expression:
.
Factor the denominators.
The denominators consist of two unique factors: (x + 6) and (x – 6). The least
common denominator is (x + 6)
2
(x – 6)
2
.
Write each fraction using the least common denominator.
Expand the numerator of each fraction.
Combine the numerators to simplify the expression.
Therefore
Both factors
are raised to
the rst power and
the second power
somewhere in the
denominators of the
expression. Choose
the higher power
for the LCD.
(x + 6)
2
(x – 6)
2
= (x
2
+ 12x + 36)
(x
2
– 12x + 36)
= x
4
– 72x
2
+ 1,296
Chapter Twenty — Rational Expressions
The Humongous Book of Algebra Problems
450
Note: Problems 20.18–20.19 refer to the function f(x) defined below.
20.18 Identify the least common denominator d(x) of the function f(x) and express
f(x) using the least common denominator.
Factor the denominators in each of the fractions.
The denominators consist of three unique factors, and the least common
denominator is the product of all three: (x + 9)(x – 8)(x + 1). Express f(x) using
the least common denominator.
Combine the numerators to simplify f(x).
Group the terms of the numerator according to the power to which x is raised
in each.
Factor x
2
and x out of the corresponding quantities.
Chapter Twenty — Rational Expressions
The Humongous Book of Algebra Problems
451
Note: Problems 20.18–20.19 refer to the function f(x) defined below.
20.19 If , given function d(x) defined in Problem 20.18,
calculate the values of a, b, and c.
According to Problem 20.18, .
The problem states that .
Set the rational functions equal.
Two equivalent fractions with equivalent denominators must have equivalent
numerators.
4x
3
+ 7x
2
+ 37x – 103 = ax
3
+ (a + b)x
2
+ (1 + 9bc)x + (–31 + 8c)
For the cubic polynomials to be equal, each of the corresponding terms must be
equal.
Set the x
3
-terms equal and solve for a.
Set the x
2
-terms equal, substitute a = 4 into the equation, and solve for b.
Set the x-terms equal, substitute b = 3 into the equation, and solve for c.
|Because f(x)
equals both of the
fractions (the one
that starts with ax
3
in
the numerator and the
one that starts with
4x
3
), those fractions are
equal to each other.
You can drop the
denominators and
set the numerators
equal.
In other
words, the x
3
-
terms are equal
(4x
3
= ax
3
), the x
2
-
terms are equal
(7x
2
= [a + b]x
2
), the
x–terms are equal,
and the constants
are equal.

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