## Precalculus (6th Edition) Blitzer

$\displaystyle \frac{ x+3}{x-4},\qquad x\neq-3, 2,\ 4,\ 8$
Division = Multiplication, with the reciprocal of the divisor, $... =\displaystyle \frac{x^{2}-5x-24}{x^{2}-x-12}\cdot\frac{x^{2}+x-6}{x^{2}-10x+16}$ Before multiplying, we factor what we can, so we can cancel common factors. Factoring $x^{2}+bx+c$, we search for factors (m and n) of $c$ whose sum is $b.$ If they (m and n) exist, we obtain $(x+m)(x+n)$. $x^{2}-5x-24= (x-8)(x+3)$ $-8$ and $+3$ are factors of $-24$ whose sum is $-5$ $x^{2}-x-12=(x-4)(x+3)$ $-4$ and $+3$ are factors of $-12$ whose sum is $-1$ $x^{2}-10x+16=(x-8)(x-2)$ $-8$ and $-2$ are factors of $+16$ whose sum is $-10$ $x^{2}+x-6=$ $-2$ and $+3$ are factors of $-6$ whose sum is $+10$ Problem = $\displaystyle \frac{ (x-8)(x+3)\cdot(x-2)(x+3)}{(x-4)(x+3)\cdot(x-8)(x-2)},\qquad x\neq-3, 2,\ 4,\ 8$ All numbers producing zero in the denominator are excluded from the domain. There are common factors; reduce the expression: = $\displaystyle \frac{ x+3}{x-4},\qquad x\neq-3, 2,\ 4,\ 8$