## Precalculus (6th Edition) Blitzer

$\displaystyle \frac{x^{2}+40x-25}{(x+5)(x-4)},\qquad x\neq-5,4$
Factor each denominator. $x^{2}+x-20= (x+5)(x-4)$ ... two factors of $-20$ with sum $+1$ are $+5$ and $-4.$ $LCD=(x+5)(x-4)$ $...= \displaystyle \frac{6x^{2}+17x-40}{(x+5)(x-4)}+\frac{3}{x-4}-\frac{5x}{x+5}$ $= \displaystyle \frac{6x^{2}+17x-40}{(x+5)(x-4)}+\frac{3}{x-4}\times\frac{x+5}{x+5}-\frac{5x}{x+5}\times\frac{x-4}{x-4}$ $=\displaystyle \frac{6x^{2}+17x-40}{(x+5)(x-4)} +\frac{3x+15}{(x+5)(x-4)}-\frac{5x^{2}-4x}{(x+5)(x-4)}$ $=\displaystyle \frac{6x^{2}+17x-40+3x+15-5x^{2}+20x}{(x+5)(x-4)}$ = $\displaystyle \frac{x^{2}+40x-25}{(x+5)(x-4)},\qquad x\neq-5,4$