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: 55



Work Step by Step

$x(2x+5)^{2}(x^{2}-4)^{-1/2}+2(x^{2}-4)^{1/2}(2x+5)$ Take out common factor $(2x+5)(x^{2}-4)^{-1/2}$ from the expression: $x(2x+5)^{2}(x^{2}-4)^{-1/2}+2(x^{2}-4)^{1/2}(2x+5)=...$ $...=(2x+5)(x^{2}-4)^{-1/2}[x(2x+5)+2(x^{2}-4)]=...$ Simplify the expression inside brackets: $...=(2x+5)(x^{2}-4)^{-1/2}(2x^{2}+5x+2x^{2}-8)=...$ $...=(2x+5)(x^{2}-4)^{-1/2}(4x^{2}+5x-8)=...$ Change the sign of the exponent of $x^{2}-4$ by putting the factor as a denominator: $...=\dfrac{(2x+5)(4x^{2}+5x-8)}{(x^{2}-4)^{1/2}}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.