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 7.5 - Multiplying with More Than One Term and Rationalizing Denominators - Exercise Set: 62

Answer

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

Work Step by Step

Write 16 as $2^4$ to obtain: $=\dfrac{10}{\sqrt[5]{2^4x^2}}$ RECALL: For any real number $a$, $\sqrt[5]{a^5}= a$ Rationalize the denominator by multiplying $\sqrt[5]{2x^3}$ to both the numerator and the denominator. Simplify using the rule above to obtain: $\require{cancel} =\dfrac{10 \cdot \sqrt[5]{2x^3}}{\sqrt[5]{2^4x^2} \cdot \sqrt[5]{2x^3}} \\=\dfrac{10\sqrt[5]{2x^3}}{\sqrt[5]{2^5x^5}} \\=\dfrac{10\sqrt[5]{2x^3}}{2x} \\=\dfrac{5\cancel{10}\sqrt[5]{2x^3}}{\cancel{2}x} \\=\dfrac{5\sqrt[5]{2x^3}}{x}$
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