## Introductory Algebra for College Students (7th Edition)

$\displaystyle \frac{x(x+3)}{(x-2)(x-1)}$
Dividing with $\displaystyle \frac{P}{Q}$ equals multiplying with the reciprocal, $\displaystyle \frac{Q}{P}.$ $\displaystyle \frac{x^{2}+x}{x^{2}-4}\div\frac{x^{2}-1}{x^{2}+5x+6}=\frac{x^{2}+x}{x^{2}-4}\cdot\frac{x^{2}+5x+6}{x^{2}-1}\qquad$... factor what you can $x^{2}+x=x(x+1)$ $x^{2}-4=(x+2)(x-2)$ $x^{2}-1=(x+1)(x-1)$ $x^{2}+5x+6=(x+2)(x+3)$ $=\displaystyle \frac{x(x+1)}{(x+2)(x-2)}\cdot\frac{(x+2)(x+3)}{(x+1)(x-1)}\qquad$... divide out the common factors $=\displaystyle \frac{x(1)}{(1)(x-2)}\cdot\frac{(1)(x+3)}{(1)(x-1)}\qquad$ = $\displaystyle \frac{x(x+3)}{(x-2)(x-1)}$