## Precalculus (6th Edition) Blitzer

$=1,\qquad$ $x\neq 2, -2, -3, -5$
Factor what we can: $x^{2}-4=(x-2)(x+2)\quad$(diff. of squares) Factoring $x^{2}+bx+c$ we search for two factors of c (m and n) such that m+n=b. If they exist, $x^{2}+bx+c =(x+m)(x+n)$ $x^{2}+3x-10=\quad$... we find factors $+5$ and $-2,$ $=(x+5)(x-2)$ $x^{2}+5x+6=\quad$... we find factors $+3$ and $+2,$ $=(x+3)(x+2)$ $x^{2}+8x+15=\quad$... we find factors $+3$ and $+5,$ $=(x+3)(x+5)$ Expression $=\displaystyle \frac{(x+2)(x-2)}{(x+5)(x-2)}\div\frac{(x+2)(x+3)}{(x+3)(x+5)}$ ... Division with $\displaystyle \frac{a}{b}$ equals multiplication with $\displaystyle \frac{b}{a}.$ (neither a or b can be 0, so $x\neq-2,-3,-5$) $=\displaystyle \frac{(x+2)(x-2)}{(x+5)(x-2)}\cdot\frac{(x+3)(x+5)}{(x+2)(x+3)}$ ...exclude any additional values that yield 0 in the denominator: $x\neq 2, -2, -3, -5$ ... cancel common factors (all cancel) $=1,\qquad$ $x\neq 2, -2, -3, -5$