College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter R - Section R.7 - Rational Expressions - R.7 Assess Your Understanding: 21

Answer

$\dfrac{2x(x^2+4x+16)}{x+4};x \ne -4, 0, 4$

Work Step by Step

Factor each polynomial completely to obtain: $=\dfrac{2(2)(x)(x)}{(x-4)(x+4)} \cdot \dfrac{(x-4)(x^2+4x+16)}{2(x)}$ Cancel the common factors to obtain: $\require{cancel} \\=\dfrac{2\cancel{(2)}\cancel{(x)}(x)}{\cancel{(x-4)}(x+4)} \cdot \dfrac{\cancel{(x-4)}(x^2+4x+16)}{\cancel{2}\cancel{(x)}} \\=\dfrac{2x(x^2+4x+16)}{x+4};x \ne -4, 0, 4$ ($x$ cannot be $-4$, $0$, and $4$ because they make the denominator equal to zero, which in turn makes the rational expression/s undefined.)
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