Answer
$ \displaystyle \frac{(x+1)(x+2) }{(x-2)(x-1)}$
Work Step by Step
$=\displaystyle \frac{x^{2}+7x+6}{x^{2}+x-6}\div\frac{x^{2}+5x-6}{x^{2}+5x+6}=\frac{x^{2}+7x+6}{x^{2}+x-6}\cdot\frac{x^{2}+5x+6}{x^{2}+5x-6}$
In each trinomial $x^{2}+bx+c$, we search for factors of c whose sum is b:
$x^{2}+7x+6$ ... $,\quad $($6$ and $1$)$\quad $factors to $(x+6)(x+1)$
$x^{2}+5x+6$ ... $,\quad $($+2$ and $+3$)$\quad $factors to $(x+2)(x+3)$
$x^{2}+x-6$... $,\quad $($-2$ and $+3$)$\quad $factors to $(x-2)(x+3)$
$x^{2}+5x-6$... $,\quad $($+6$ and $-1$)$\quad $factors to $(x+6)(x-1)$
$=\displaystyle \frac{(x+6)(x+1)(x+2)(x+3)}{(x-2)(x+3)(x+6)(x-1)}$
after canceling $(x+6)$ and $(x+3),$
$=\displaystyle \frac{(x+1)(x+2) }{(x-2)(x-1)}$
