## Precalculus (6th Edition) Blitzer

$\displaystyle -\frac{5x}{(x-6)(x+4)(x-1)},\qquad x\neq 6,1, -4$
Factor each denominator. Factoring $x^{2}+bx+c$, we search for two factors of c (m and n) such that m+n=b. If they exist, $x^{2}+bx+c =(x+m)(x+n)$ $x^{2}-2x-24=\quad$... we find factors $+4$ and $-6,$ $=(x+4)(x-6)$ $x^{2}-7x+6=\quad$... we find factors $-6$ and $-1,$ $=(x-6)(x-1)$ Build the LCD by listing common factors first. Then list all the "leftover" factors. $LCD=(x-6)(x+4)(x-1)$ $\displaystyle \frac{x}{x^{2}-2x-24}-\frac{x}{x^{2}-7x+6}=\frac{x}{(x-6)(x+4)}-\frac{x}{(x-6)(x-1)}$ $=\displaystyle \frac{x}{(x-6)(x+4)}\times\frac{x-1}{x-1}-\frac{x}{(x-6)(x-1)}\times\frac{x+4}{x+4}$ $=\displaystyle \frac{x^{2}-x}{(x-6)(x+4)(x-1)}-\frac{x^{2}+4x}{(x-6)(x+4)(x-1)}$ $=\displaystyle \frac{x^{2}-x-x^{2}-4x}{(x-6)(x+4)(x-1)}$ $=\displaystyle \frac{-5x}{(x-6)(x+4)(x-1)},\qquad x\neq 6,1, -4$