Intermediate Algebra for College Students (7th Edition)

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

Chapter 6 - Section 6.1 - Rational Expressions and Functions; Multiplying and Dividing - Exercise Set - Page 414: 49


$x^2-3$ and $x\ne-4.$

Work Step by Step

$\frac{x^3+4x^2-3x-12}{x+4}=\frac{x^2(x+4)-3(x+4)}{x+4}=\frac{(x+4)(x^2-3)}{x+4}=x^2-3$ and $x\ne-4.$
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