Algebra and Trigonometry 10th Edition

$-\frac{x+1}{5+x}$ with $x\ne-5$ and $x\ne2$
$-x=x-2x$ and $-3x=-5x+2x$ $\frac{x^2-x-2}{10-3x-x^2}=\frac{x^2+x-2x-2}{10-5x+2x-x^2}=\frac{x(x+1)-2(x+1)}{5(2-x)+x(2-x)}=\frac{(x-2)(x+1)}{(5+x)(2-x)}=\frac{-(2-x)(x+1)}{(5+x)(2-x)}$ Notice, at this point, that we must have that $x\ne-5$ and $x\ne2$ because the denominator can not be equal to $0$. $\frac{-(2-x)(x+1)}{(5+x)(2-x)}=-\frac{x+1}{5+x}$