How do you simplify $\frac{8 p + 8}{2 p^{2} - 4 p - 6}$ $(8p+8)/(2p^2-4p-6)$ ?

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Answer 1 · 2016-08-15T20:17:55

$= \frac{4}{p - 3}$ $= 4/(p-3)$

Factor numerator and denominator first.

$\frac{8 (p + 1)}{2 (p^{2} - 2 p - 3)} = \frac{8 (p + 1)}{2 (p - 3) (p + 1)}$ $(8(p+1))/(2(p^2 -2p-3)) = (8(p+1))/(2(p-3)(p+1)$

$((cancel8^4)cancel(p+1))/(cancel2(p-3)cancel(p+1)) =$

$= 4/(p-3)$

How do you simplify (8p+8)/(2p^2-4p-6)8p+82p2−4p−6(8p+8)/(2p^2-4p-6)?