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: 21


$=\displaystyle \frac{x^{2}+2x+4}{3x},\qquad x\neq-2,0,2$

Work Step by Step

Factor what you can: $ x^{3}-8= \quad$...difference of cubes = $x^{3}-2^{3},$ $=(x-2)(x^{2}+2x+4)$ $x^{2}-4 =$ difference of squares $= (x-2)(x+2)$ $x+2$ is factored $3x$ is factored Expression $ =\displaystyle \frac{(x-2)(x^{2}+2x+4)}{(x-2)(x+2)}\cdot\frac{(x+2)}{3x} $ $\qquad$ ...exclude the values that yield 0 in the denominator: $ x\neq -2,0,2$ ... cancel common factors: $=\displaystyle \frac{x^{2}+2x+4}{3x},\qquad x\neq-2,0,2$
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