Precalculus (6th Edition) Blitzer

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

Chapter P - Section P.5 - Factoring Polynomials - Exercise Set - Page 69: 100


$\displaystyle \frac{x^{2}+4}{(x^{2}+3)^{5/3}}$

Work Step by Step

$\left[(x^{2}+3)^{-2/3}=(x^{2}+3)^{-5/3+1}=(x^{2}+3)(x^{2}+3)^{-5/3}\right]$ $(x^{2}+3)^{-2/3}+(x^{2}+3)^{-5/3}=(x^{2}+3)(x^{2}+3)^{-5/3}+(x^{2}+3)^{-5/3}$ ... factor out $(x^{2}+3)^{-5/3}$ $=(x^{2}+3)^{-5/3}[(x^{2}+3)+1]$ $=(x^{2}+3)^{-5/3}[x^{2}+4] \qquad $... apply $a^{-n}=\displaystyle \frac{1}{a^{n}}$ $=\displaystyle \frac{x^{2}+4}{(x^{2}+3)^{5/3}}$
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