## Algebra and Trigonometry 10th Edition

$-\frac{x+1}{2x+1}$ with $x\ne-\frac{1}{2}$ and $x\ne4$
$3x=4x-x$ and $-7x=-8x+x$ $\frac{4+3x-x^2}{2x^2-7x-4}=\frac{4+4x-x-x^2}{2x^2-8x+x-4}=\frac{4(1+x)-x(1+x)}{2x(x-4)+1(x-4)}=\frac{(4-x)(1+x)}{(2x+1)(x-4)}=\frac{-(x-4)(x+1)}{(2x+1)(x-4)}$ Notice at this point that we must have that $x\ne-\frac{1}{2}$ and $x\ne4$ because the denominator can not be equal to $0$. $\frac{-(x-4)(x+1)}{(2x+1)(x-4)}=-\frac{x+1}{2x+1}$