Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter R - Algebra Reference - R.6 Exponents - R.6 Exercises - Page R-25: 56



Work Step by Step

$(4x^{2}+1)^{2}(2x-1)^{-1/2}+16x(4x^{2}+1)(2x-1)^{1/2}$ Take out common factor $(4x^{2}-1)(2x-1)^{-1/2}$ from the expression: $(4x^{2}+1)^{2}(2x-1)^{-1/2}+16x(4x^{2}+1)(2x-1)^{1/2}=...$ $...=(4x^{2}+1)(2x-1)^{-1/2}[(4x^{2}+1)+16x(2x-1)]=...$ Simplify the expression inside brackets: $...=(4x^{2}+1)(2x-1)^{-1/2}(4x^{2}+1+32x^{2}-16x)=...$ $...=(4x^{2}+1)(2x-1)^{-1/2}(36x^{2}-16x+1)=...$ Change the sign of the exponent of $2x-1$ by putting the factor as a denominator: $...=\dfrac{(4x^{2}+1)(36x^{2}-16x+1)}{(2x-1)^{1/2}}$
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