Intermediate Algebra (6th Edition)

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

Chapter 7 - Sections 7.1-7.5 - Integrated Review - Radicals and Rational Exponents: 22


$2xy^{2}\sqrt[4] (x^{3}y^{2})$

Work Step by Step

$\sqrt[4] (16x^{7}y^{10})=\sqrt[4] (16\times x^{4}\times y^{8}\times x^{3}y^{2})=\sqrt[4] 16\times \sqrt[4] (x^{4})\times \sqrt[4] (y^{8})\times \sqrt[4] (x^{3}y^{2})=2xy^{2}\sqrt[4] (x^{3}y^{2})$ We know that $\sqrt[4] 16=2$, because $2^{4}=16$. We know that $\sqrt[4] (x^{4})=x$, because $(x)^{4}=x^{4}$. We know that $\sqrt[4] (y^{8})=y^{2}$, because $(y^{2})^{4}=y^{2\times4}=y^{8}$.
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