Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 10 - Exponents and Radicals - Test: Chapter 10 - Page 695: 9

Answer

$2x^{3}y^{2}\sqrt[5]{x}$

Work Step by Step

Extracting the factors that are perfect powers of the index, the given expression simplifies to \begin{array}{l}\require{cancel} \sqrt[5]{32x^{16}y^{10}} \\\\= \sqrt[5]{32x^{15}y^{10}\cdot x} \\\\= \sqrt[5]{(2x^{3}y^{2})^5\cdot x} \\\\= 2x^{3}y^{2}\sqrt[5]{x} .\end{array}
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