Answer
$\displaystyle \frac{(x+3)^{2}}{(x-3)^{2}}$
Work Step by Step
$=\displaystyle \frac{x^{2}+7x+12}{x^{2}-7x+12}\div\frac{x^{2}+x-12}{x^{2}-x-12}=\frac{x^{2}+7x+12}{x^{2}-7x+12}\cdot\frac{x^{2}-x-12}{x^{2}+x-12}$
In each trinomial $x^{2}+bx+c$, we search for factors of c whose sum is b:
$x^{2}+7x+12$ ... $,\quad $(4 and 3)$\quad $factors to $(x+4)(x+3)$
$x^{2}-7x+12$ ... $,\quad $($-4$ and $-3$)$\quad $factors to $(x-4)(x-3)$
$x^{2}-x-12$... $,\quad $($-4$ and $+3$)$\quad $factors to $(x-4)(x+3)$
$x^{2}+x-12$... $,\quad $($+4$ and $-3$)$\quad $factors to $(x+4)(x-3)$
$=\displaystyle \frac{(x+4)(x+3)(x-4)(x+3)}{(x-4)(x-3)(x+4)(x-3)}$
after canceling $(x+4)$ and $(x-4),$
$=\displaystyle \frac{(x+3)^{2}}{(x-3)^{2}}$
