## College Algebra (10th Edition)

$\displaystyle \frac{(x+1)(x+2) }{(x-2)(x-1)}$
$=\displaystyle \frac{x^{2}+7x+6}{x^{2}+x-6}\div\frac{x^{2}+5x-6}{x^{2}+5x+6}=\frac{x^{2}+7x+6}{x^{2}+x-6}\cdot\frac{x^{2}+5x+6}{x^{2}+5x-6}$ 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 $(x+3),$ $=\displaystyle \frac{(x+1)(x+2) }{(x-2)(x-1)}$