## Precalculus (6th Edition) Blitzer

$=\displaystyle \frac{(x-3)(x+3)}{x(x+4)},\qquad x\neq 0,-4,3$
Factor what you can: $x^{2}-9=$ difference of squares $= (x-3)(x+3)$ $x^{2}$ is factored $x^{2}-3x=x(x-3)$ Factoring $x^{2}+bx+c$ = $x^{2}+x-12$ we search for two factors of c (m and n) such that m+n=b. If they exist, $x^{2}+bx+c =(x+m)(x+n)$ $x^{2}+x-12 = \quad$... we find factors $-3$ and $+4,$ $=(x+4)(x-3)$ Expression $=\displaystyle \frac{(x-3)(x+3)}{x^{2}}\cdot\frac{x(x-3)}{(x+4)(x-3)}\qquad$ ...exclude the values that yield 0 in the denominator: $x\neq 0,-4,3$ ... cancel common factors: $x$, and $(x-3)$ $=\displaystyle \frac{(x-3)(x+3)}{x(x+4)},\qquad x\neq 0,-4,3$