Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.3 - Algebraic Expressions - 1.3 Exercises - Page 34: 128


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

Work Step by Step

$3(2x-1)^{2}(2)(x+3)^{1/2}+(2x-1)^{3}(\frac{1}{2})(x+3)^{-1/2}$ Take out common factor $(2x-1)^{2}(x+3)^{-1/2}$: $(2x-1)^{2}(x+3)^{-1/2}[3(2)(x+3)+(\frac{1}{2})(2x-1)]=...$ $...=(2x-1)^{2}(x+3)^{-1/2}[6(x+3)+(\frac{1}{2})(2x-1)]=...$ Simplify the expression inside brackets: $...=(2x-1)^{2}(x+3)^{-1/2}\Big(6x+18+x-\dfrac{1}{2}\Big)=...$ $...=(2x-1)^{2}(x+3)^{-1/2}\Big(7x+\dfrac{35}{2}\Big)=...$ Take out common factor $\dfrac{7}{2}$ from $\Big(7x+\dfrac{35}{2}\Big)$: $...=\dfrac{7}{2}(2x-1)^{2}(x+3)^{-1/2}(2x+5)=...$ Organize the expression: $...=\dfrac{7(2x-1)^{2}(2x+5)}{2(x+3)^{1/2}}$ or $\dfrac{7(2x-1)^{2}(2x+5)}{2\sqrt{x+3}}$
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