Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter P - Section P.6 - Rational Expressions - Exercise Set - Page 85: 45


$=\displaystyle \frac{3x^{2}+4}{(x+2)(x-2)}, \quad x\neq-2,2$

Work Step by Step

The denominators are different. Find a common denominator. LCD=$(x+2)(x-2)$ $\displaystyle \frac{2x}{x+2}+\frac{x+2}{x-2}= \displaystyle \frac{2x(x-2)}{(x+2)(x-2)}+\frac{(x+2)(x+2)}{(x-2)(x+2)}$ $=\displaystyle \frac{2x^{2}-4x+x^{2}+4x+4}{(x+2)(x-2)}$ $=\displaystyle \frac{3x^{2}+4}{(x+2)(x-2)}$ ... exclude values that yield 0 in the denominator: $x\neq-2,2$ ... and, cancel common factors (here, there are none) $=\displaystyle \frac{3x^{2}+4}{(x+2)(x-2)}, \quad x\neq-2,2$
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