## Precalculus (6th Edition) Blitzer

$\displaystyle \frac{x-1}{x+2},\ \qquad x\neq-2,-1$
Factor each denominator. $x^{2}+3x+2= (x+1)(x+2)$ ... two factors of $2$ with sum $3$ are $+1$ and $+2.$ $LCD=(x+1)(x+2)$ $\displaystyle \frac{4x^{2}+x-6}{x^{2}+3x+2}-\frac{3x}{x+1}+\frac{5}{x+2}= \displaystyle \frac{4x^{2}+x-6}{(x+1)(x+2)}-\frac{3x}{x+1}\times\frac{x+2}{x+2}+\frac{5}{x+2}\times\frac{x+1}{x+1}$ $= \displaystyle \frac{4x^{2}+x-6}{(x+1)(x+2)}- \displaystyle \frac{3x^{2}+6x}{(x+1)(x+2)}+ \displaystyle \frac{5x+5}{(x+1)(x+2)}$ $=\displaystyle \frac{4x^{2}+x-6-3x^{2}-6x+5x+5}{(x+1)(x+2)}$ $=\displaystyle \frac{x^{2}-1}{(x+1)(x+2)}\qquad$... recognize a difference of squares $=\displaystyle \frac{(x-1)(x+1)}{(x+1)(x+2)}\qquad$... We have a common factor. Reduce. $=\displaystyle \frac{x-1}{x+2},\ \qquad x\neq-2,-1$ (-1 is also excluded as $(x+1)$ was a part of the expression before reducing)