Answer
$=\displaystyle \frac{x-1}{x+2},\qquad x\neq-2, -1,2,3$
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 $+4,$
$=(x-3)(x-2)$
$ x^{2}-2x-3\quad$... we find factors $-3$ and $+1,$
$=(x-3)(x+1)$
$x^{2}-1 =$ difference of squares $= (x-1)(x+1)$
$x^{2}-4 =$ difference of squares $= (x-2)(x+2)$
Expression $=\displaystyle \frac{(x-3)(x-2)}{(x-3)(x+1)}\cdot\frac{(x+1)(x-1)}{(x-2)(x+2)}$
$\qquad$
...exclude the values that yield 0 in the denominator:
$x\neq-2, -1,2,3$
... cancel common factors:
$=\displaystyle \frac{x-1}{x+2},\qquad x\neq-2, -1,2,3$
