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: 59


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

Work Step by Step

$\sqrt [4](32x^{12}y^{5})=\sqrt[4] (16\times x^{12}\times y^{4}\times 2y)=\sqrt[4] 16\times \sqrt[4] (x^{12})\times \sqrt [4](y^{4})\times \sqrt[4] (2y)=2x^{3}y\sqrt[4] (2y)$ We know that $\sqrt[4] 16=2$, because $2^{4}=16$. We also know that $\sqrt[4] (x^{12})=x^{3}$, because $(x^{3})^{4}=x^{3\times4}=x^{12}$ and that $\sqrt[4] (y^{4})=y$, because $(y)^{4}=y^{4}$.
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.