Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 7 - Section 7.3 - Simplifying Radical Expressions - Exercise Set - Page 433: 53


$-4a^{4}b^{3}\sqrt (2b)$

Work Step by Step

$-\sqrt (32a^{8}b^{7})=-\sqrt (16\times2\times a^{8}\times b^{6}\times b)=-(\sqrt 16\times \sqrt (a^{8})\times \sqrt (b^{6})\times \sqrt (2b))=-(4a^{4}b^{3}\sqrt (2b))=-4a^{4}b^{3}\sqrt (2b)$ We know that $\sqrt 16=4$, because $4^{2}=16$. We know that $\sqrt (a^{8})=a^{4}$, because $(a^{4})^{2}=a^{4\times2}=a^{8}$ and also that $\sqrt (b^{6})=b^{3}$, because $(b^{3})^{2}=b^{3\times2}=b^{6}$
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