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: 44


$2z^{2}\sqrt[5] (z^{2})$

Work Step by Step

$\sqrt [5] (32z^{12})=\sqrt [5] 32\times\sqrt[5] (z^{10})\times\sqrt[5] (z^{2})=2z^{2}\sqrt[5] (z^{2})$ We know that $\sqrt [5] 32=2$, because $2^{5}=32$. We also know that $\sqrt[5] (z^{10})=z^{2}$, because $(z^{5})^{2}=z^{5\times2}=z^{10}$.
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