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: 101


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

Work Step by Step

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