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

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

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Answer 1 · 2016-09-03T12:23:18

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

Explanation:

Well you have to do something so factorise the denominator
$\frac{2 - p}{(p - 2) (p + 1)}$ $(2-p)/((p-2)(p+1))$

And it helps
$- \frac{(p - 2)}{(p - 2) (p + 1)}$ $-((p-2))/((p-2)(p+1))$

Cancel
$- \frac{1}{p + 1}$ $-1/(p+1$

Answer 2 · 2016-09-03T12:37:55

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

Explanation:

The first step is to factorise the denominator.

$\Rightarrow p^{2} - p - 2$ $rArrp^2-p-2$

Consider the factors of - 2 which sum to give the coefficient of the middle term, that is - 1.

These are - 2 and + 1 , since.

$(- 2 \times + 1) = - 2 and - 2 + 1 = - 1$ $(-2xx+1)=-2" and " -2+1=-1$

$\Rightarrow p^{2} - p - 2 = (p - 2) (p + 1)$ $rArrp^2-p-2=(p-2)(p+1)$

Fraction can now be expressed as $\frac{2 - p}{(p - 2) (p + 1)}$ $(2-p)/((p-2)(p+1))$

Take out a common factor of - 1 in the numerator.

$\Rightarrow 2 - p = - 1 (p - 2)$ $rArr2-p=-1(p-2)$

and now we have a fraction which may be simplified.

$rArr(-1cancel((p-2)))/(cancel((p-2))(p+1))=(-1)/(p+1)=-1/(p+1)$

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

2 Answers

Explanation:

Explanation:

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How do you simplify (2-p)/(p^2-p-2)2−pp2−p−2(2-p)/(p^2-p-2)?

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