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 84: 25


$= \displaystyle \frac{(x-2)^{2}}{x},\qquad x\neq 0,-2,2$

Work Step by Step

Factor what we can. Recognize a difference of squares. $\displaystyle \frac{x^{2}-4}{x}\div\frac{x+2}{x-2}=\frac{(x-2)(x+2)}{x}\div\frac{x+2}{x-2}$ Division with $\displaystyle \frac{a}{b}$ equals multiplication with $\displaystyle \frac{b}{a}.$ (neither a or b can be 0, we exclude $-2$ and $2$) $=\displaystyle \frac{(x-2)(x+2)}{x}\cdot\frac{(x-2)}{(x+2)}$ ...exclude any additional values that yield 0 in the denominator: $x\neq 0,-2,2$ ... cancel common factors $= \displaystyle \frac{(x-2)^{2}}{x},\qquad x\neq 0,-2,2$
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