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: 27


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

Work Step by Step

Factor what we can. $\displaystyle \frac{4x^{2}+10}{x-3}\div\frac{6x^{2}+15}{x^{2}-9}=\frac{2(2x^{2}+5)}{(x-3)}\div\frac{3(2x^{2}+5)}{(x-3)(x+3)}$ Division with $\displaystyle \frac{a}{b}$ equals multiplication with $\displaystyle \frac{b}{a}.$ (neither a or b can be 0, because of the denominator we exclude $-3$ and $3$, $2x^{2}+5$ can never be 0 so there is nothing to exclude there) $=\displaystyle \frac{2(2x^{2}+5)}{(x-3)}\cdot\frac{(x-3)(x+3)}{3(2x^{2}+5)}$ ...exclude any additional values that yield 0 in the denominator: $x\neq 3, -3$ ... cancel common factors $=\displaystyle \frac{2(x+3)}{3},\qquad x\neq 3, -3$
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