College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter P, Prerequisites - Section P.6 - Factoring - P.6 Exercises: 113

Answer

$\frac{\frac{1}{3}x^{2}+3}{(x^{2}+3)^{4/3}}$

Work Step by Step

Factor out $x^2+3$: $(x^{2}+3)^{-1/3}-\displaystyle \frac{2}{3}x^{2}(x^{2}+3)^{-4/3}=(x^{2}+3)^{-4/3}[(x^{2}+3)-\frac{2}{3}x^{2}]=(x^{2}+3)^{-4/3}(\frac{1}{3}x^{2}+3)=\frac{\frac{1}{3}x^{2}+3}{(x^{2}+3)^{4/3}}$
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