How do you simplify $\frac{8 - x}{x^{2} - 7 x - 8}$ $(8-x)/(x^2-7x-8)$ ?

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Answer 1 · 2018-03-22T12:08:45

$- \frac{1}{x + 1}$ $-1/(x+1)$

Lets begin by factoring the quadratic in the denominator:

$x^{2} - 7 x - 8 = (x - 8) (x + 1)$ $x^2-7x-8=(x-8)(x+1)$

The first factor is, in fact, the negative of the numerator:

$8 - x = - (x - 8)$ $8-x = -(x-8)$

So we can rewrite our fraction as:

$\frac{- (x - 8)}{(x - 8) (x + 1)}$ $(-(x-8))/((x-8)(x+1))$

Finally

$- \frac{1}{x + 1}$ $-1/(x+1)$

How do you simplify 8−xx2−7x−8(8-x)/(x^2-7x-8)?