## Intermediate Algebra for College Students (7th Edition)

Published by Pearson

# Chapter 7 - Section 7.1 - Radical Expressions and Functions - Exercise Set - Page 512: 90

#### Answer

$-2(x-2)$ or, $4-2x$

#### Work Step by Step

Simplify $\sqrt[5] {(-32)(x-2)^5}$ For $n$ is even, we have $\sqrt [n] {a^n}=|a|$ For $n$ is odd, we have $\sqrt [n] {a^n}=a$ Thus, $\sqrt[5] {(-32)(x-2)^5}=\sqrt[5] {(-2)^5(x-2)^5}$ or, $=\sqrt[5] {(-2(x-2))^5}$ or, $=-2(x-2)$ or, $=-2x+4$ or, $=4-2x$

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