# Prerequisites - P.5 - Rational Expressions - Exercises - Page 48: 28

$-\frac{x+1}{2x+1}$ with $x\ne-\frac{1}{2}$ and $x\ne4$

#### Work Step by Step

$3x=4x-x$ 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 $0$. $\frac{-(x-4)(x+1)}{(2x+1)(x-4)}=-\frac{x+1}{2x+1}$

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