Answer
$=1,\qquad$ $x\neq 2, -2, -3, -5$
Work Step by Step
Factor what we can:
$x^{2}-4=(x-2)(x+2)\quad $(diff. of squares)
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}+3x-10=\quad$... we find factors $+5$ and $-2,$
$=(x+5)(x-2)$
$ x^{2}+5x+6=\quad$... we find factors $+3$ and $+2,$
$=(x+3)(x+2)$
$ x^{2}+8x+15=\quad$... we find factors $+3$ and $+5,$
$=(x+3)(x+5)$
Expression $=\displaystyle \frac{(x+2)(x-2)}{(x+5)(x-2)}\div\frac{(x+2)(x+3)}{(x+3)(x+5)}$
... Division with $\displaystyle \frac{a}{b}$ equals multiplication with $\displaystyle \frac{b}{a}.$
(neither a or b can be 0, so $x\neq-2,-3,-5$)
$=\displaystyle \frac{(x+2)(x-2)}{(x+5)(x-2)}\cdot\frac{(x+3)(x+5)}{(x+2)(x+3)}$
...exclude any additional values that yield 0 in the denominator:
$x\neq 2, -2, -3, -5$
... cancel common factors (all cancel)
$=1,\qquad$ $x\neq 2, -2, -3, -5$
