Answer
$=\displaystyle \frac{(x+4)(x+2)}{x-5},\qquad$ $x\neq-6, -3, -1,3,5$
Work Step by Step
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
$=\displaystyle \frac{(x+4)(x+2)}{x-5},\qquad$ $x\neq-6, -3, -1,3,5$
