Precalculus (6th Edition) Blitzer

$=\displaystyle \frac{2x^{2}+5x+12}{(x-3)(x+2)}, \quad x\neq-2,3$
The denominators are different. Find a common denominator. LCD=$(x+2)(x-3)$ $\displaystyle \frac{3x}{x-3}-\frac{x+4}{x+2}= \displaystyle \frac{3x(x-3)}{(x+2)(x-3)}-\frac{(x+4)(x+2)}{(x+2)(x-3)}$ $=\displaystyle \frac{3x^{2}+6x-(x^{2}+x-12)}{(x-3)(x+2)}$ $=\displaystyle \frac{3x^{2}+6x-x^{2}-x+12}{(x-3)(x+2)}$ $=\displaystyle \frac{2x^{2}+5x+12}{(x-3)(x+2)}$ ... exclude values that yield 0 in the denominator: $x\neq-2,3$ ... and, cancel common factors to factor $2x^{2}+5x+12$, we search for factors of ac=24 whose sum is b=5 and here we can't find any. So here, there are no common factors to cancel. $=\displaystyle \frac{2x^{2}+5x+12}{(x-3)(x+2)}, \quad x\neq-2,3$