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: 111

Answer

$\frac{7(2x-1)^2(2x+5)}{2\sqrt{x+3}}$

Work Step by Step

Factor out $2x-1$ and $x+3$: $3(2x-1)^{2}(2)(x+3)^{1/2}+(2x-1)^{3}(\displaystyle \frac{1}{2})(x+3)^{-1/2}=(2x-1)^{2}(x+3)^{-1/2}[6(x+3)+(2x-1)(\frac{1}{2})] =(2x-1)^{2}(x+3)^{-1/2}(6x+18+x-\frac{1}{2})=(2x-1)^{2}(x+3)^{-1/2}(7x+\frac{35}{2})=\frac{(2x-1)^2(7x+35/2)}{\sqrt{x+3}}=\frac{(2x-1)^2(14x+35)}{2\sqrt{x+3}}=\frac{7(2x-1)^2(2x+5)}{2\sqrt{x+3}}$
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