## Precalculus (6th Edition) Blitzer

$=\displaystyle \frac{(x+4)(x+2)}{x-5},\qquad$ $x\neq-6, -3, -1,3,5$
Factor what we can: Factoring $x^{2}+bx+c$ 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 $+$4 and $-3,$ $=(x+4)(x-3)$ $x^{2}+x-30=\quad$... we find factors $+6$ and $-5,$ $=(x+6)(x-5)$ $x^{2}+5x+6=\quad$... we find factors $+2$ and $+3,$ $=(x+2)(x+3)$ $x^{2}+7x+6=\quad$... we find factors $+6$ and $+1,$ $=(x+6)(x+1)$ Expression $=\displaystyle \frac{(x+4)(x-3)}{(x+6)(x-5)}\cdot\frac{(x+2)(x+3)}{(x+1)(x-3)}\div\frac{(x+3)}{(x+6)(x+1)}$ ... Division with $\displaystyle \frac{a}{b}$ equals multiplication with $\displaystyle \frac{b}{a}.$ (neither a or b can be 0, so $x\neq-6,-3,-1$) $=\displaystyle \frac{(x+4)(x-3)}{(x+6)(x-5)}\cdot\frac{(x+2)(x+3)}{(x+1)(x-3)}\cdot\frac{(x+6)(x+1)}{(x+3)}$ ...exclude any additional values that yield 0 in the denominator: $x\neq-6, -3, -1,3,5$ ... cancel common factors $=\displaystyle \frac{(x+4)(x+2)}{x-5},\qquad$ $x\neq-6, -3, -1,3,5$