Answer
$\displaystyle \frac{ x+3}{x-4},\qquad x\neq-3, 2,\ 4,\ 8$
Work Step by Step
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:
= $\displaystyle \frac{ x+3}{x-4},\qquad x\neq-3, 2,\ 4,\ 8$
