## Precalculus (6th Edition)

Published by Pearson

# Chapter R - Review of Basic Concepts - R.5 Rational Expressions - R.5 Exercises - Page 53: 31

#### Answer

$\color{blue}{x^2-4x+16}$

#### Work Step by Step

The given expression is equivalent to: $=\dfrac{x^3+4^3}{x+4}$ Factor the numerator and the denominator then cancel the common factors. Factor the numerator using the formula $a^3+b^3=(a+b)(a^2-ab+b^2)$ with $a=x$ and $b=4$. : $\require{cancel} =\dfrac{(x+4)(x^2-4x+16)}{x+4} \\=\dfrac{\cancel{(x+4)}(x^2-4x+16)}{\cancel{x+4}} \\=\color{blue}{x^2-4x+16}$

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