## Algebra 1: Common Core (15th Edition)

$-\frac{x+3}{x+4}, with$ $x\ne-4,x\ne3$
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 : $\frac{(3+x)(3-x)}{(x+4)(x-3)}=-\frac{x+3}{x+4}$