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 - Section 7.4 - Adding, Subtracting, and Dividing Radical Expressions - Exercise Set: 61

Answer

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

Work Step by Step

RECALL: (1) The quotient rule: $\dfrac{\sqrt[n]{a}}{\sqrt[n]{b}}=\sqrt[n]{\dfrac{a}{b}}$ where $\sqrt[n]{a}$ and $\sqrt[n]{b}$ are real numbers and $b\ne0$ (2) $\dfrac{a^m}{a^n} = a^{m-n}, a \ne =0$ Use the quotient rule above to obtain: $\require{cancel}=\sqrt[4]{\dfrac{32x^{10}y^{8}}{2x^2y^{-2}}} \\=\sqrt[4]{\dfrac{16\cancel{32}\cancel{x^{10}}x^8y^8}{\cancel{2x^2}y^{-2}}} \\=\sqrt[4]{\dfrac{16x^8y^8}{y^{-2}}}$ Use rule (2) above to obtain: $=\sqrt[4]{16x^8y^{8-(-2)}} \\=\sqrt[4]{16x^8y^{8+2}} \\=\sqrt[4]{16x^8y^{10}}$ Factor the radicand so that at least one factor is a perfect fourth power to obtain: $=\sqrt[4]{16x^8y^8(y^2)} \\=\sqrt[4]{2^4(x^2)^4(y^2)^4(y^2)} \\=\sqrt[4]{(2x^2y^2)^4(y^2)}$ Simplify to obtain: $=2x^2y^2\sqrt[4]{y^2}$
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