Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 7 - Review Exercises - Page 577: 39

Answer

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

Work Step by Step

Simplify . $\sqrt[4]{32x^3y^{11}z^5}$ As per the product rule, we have $\sqrt[n] {pq}=\sqrt[n] {p}\sqrt[n] {q}$ $\sqrt[4]{32x^3y^{11}z^5}=\sqrt[4] {32}\sqrt[4] {x^3}\sqrt[4] {y^{11}} \sqrt[4] {z^5}$ Thus, $=\sqrt[4] {2^4}\sqrt[4] {(y^2)^4}\sqrt[4] {z^4} \sqrt[4]{2x^3y^3z}$ or, $=\sqrt[4]{(2y^2z)^4} \sqrt[4]{2x^3y^3z}$ Hence, the above exponent in radical form can be written as: or, $2y^2z \sqrt[4]{2x^3y^3z}$
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