Answer
$-\frac{x+3}{x+4}, with$ $x\ne-4,x\ne3$
Work Step by Step
Given : $\frac{9-x^{2}}{x^{2}+x-12}=\frac{(3+x)(3-x)}{x^{2}+x-12}$
(Since $(a^{2}-b^{2})=(a+b)(a-b)$)
This becomes : $\frac{(3+x)(3-x)}{(x+4)(x-3)}=-\frac{x+3}{x+4}$
Algebra 1: Common Core (15th Edition)
by Charles, Randall I.
Published by
Prentice Hall
ISBN 10:
0133281140
ISBN 13:
978-0-13328-114-9
Chapter 11 - Rational Expressions and Functions - Mid-Chapter Quiz - Page 690: 5
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