## Precalculus (6th Edition) Blitzer

$=\displaystyle \frac{x-5}{2},\qquad$ $x\neq 1,-5$
$x^{2}-25=(x-5)(x+5)\quad$(diff. of squares) $2x-2=2(x-1)$ $x^{2}+10x+25=(x+5)^{2}\quad$ (square of a sum) $x^{2}+4x-5\quad$ ... find factors of c whose sum is b... ... we find 5 and -1 ... $=(x+5)(x-1)$ Expression $=\displaystyle \frac{(x-5)(x+5)}{2(x-1)}\div\frac{(x+5)^{2}}{(x+5)(x-1)}$ ... Division with $\displaystyle \frac{a}{b}$ equals multiplication with $\displaystyle \frac{b}{a}.$ (neither a or b can be 0, because of the denominator we exclude $-5$ and $+1$, numerator: exclude $-5$ ... we already have) $=\displaystyle \frac{(x-5)(x+5)}{2(x-1)}\cdot\frac{(x+5)(x-1)}{(x+5)^{2}}$ ...exclude any additional values that yield 0 in the denominator: $x\neq 1,-5$ ... cancel common factors $=\displaystyle \frac{x-5}{2},\qquad$ $x\neq 1,-5$