Answer
$=\displaystyle \frac{x+3}{x-2},\qquad x\neq-3, -2,2,3$
Work Step by Step
Factoring $x^{2}+bx+c $
we search for two factors of c (m and n) such that m+n=b.
If they exist, $x^{2}+bx+c =(x+m)(x+n)$
Factor what you can:
$ x^{2}+5x+6= \quad$... we find factors $+3$ and $+2,$
$=(x+3)(x+2)$
$ x^{2}+x-6\quad$... we find factors $+3$ and $-2,$
$=(x+3)(x-2)$
$x^{2}-x-6= \quad$... we find factors $-3$ and $+2,$
$=(x-3)(x+2)$
$x^{2}-9 =$ difference of squares $= (x-3)(x+3)$
Expression $=\displaystyle \frac{(x+3)(x+2)}{(x+3)(x-2)}\cdot\frac{(x-3)(x+3)}{(x-3)(x+2)}$
$\qquad$
...exclude the values that yield 0 in the denominator:
$x\neq-3, -2,2,3$
... cancel common factors:
$=\displaystyle \frac{x+3}{x-2},\qquad x\neq-3, -2,2,3$
