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: 112

Answer

$\frac{1}{3}(x+6)^{-2/3}(2x-3)(14x+69)$

Work Step by Step

Factor out $x+6$ and $2x-3$: $\displaystyle \frac{1}{3}(x+6)^{-2/3}(2x-3)^{2}+(x+6)^{1/3}(2)(2x-3)(2)=\frac{1}{3}(x+6)^{-2/3}(2x-3)[(2x-3)+(3)(x+6)(4)] =\frac{1}{3}(x+6)^{-2/3}(2x-3)[2x-3+12x+72]=\frac{1}{3}(x+6)^{-2/3}(2x-3)(14x+69)$
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