## Precalculus (6th Edition) Blitzer

$=\displaystyle \frac{3}{x-3}, \quad x\neq-4,3$
The fractions have a common denominator, we subtract the numerators... $\displaystyle \frac{x^{2}+3x}{x^{2}+x-12}-\frac{x^{2}-12}{x^{2}+x-12}=\frac{x^{2}+3x-(x^{2}-12)}{x^{2}+x-12}$ $=\displaystyle \frac{3x+12}{x^{2}+x-12}$ factor what we can; for the denominator, we search for two factors of c=-12 whose sum is b=1 $=\displaystyle \frac{3(x+4)}{(x+4)(x-3)}$ ... exclude values that yield 0 in the denominator: $x\neq-4,3$ ... and, cancel the common factor $=\displaystyle \frac{3}{x-3}, \quad x\neq-4,3$