How do you express 3/((x-2)(x+1))3(x−2)(x+1) in partial fractions?

2 Answers
Cem Sentin
Mar 16, 2018

1/(x-2)-1/(x+1)1x−2−1x+1

Explanation:

3/[(x-2)(x+1)]=[(x+1)-(x-2)]/[(x-2)*(x+1)]=1/(x-2)-1/(x+1)3(x−2)(x+1)=(x+1)−(x−2)(x−2)⋅(x+1)=1x−2−1x+1

maganbhai P.
Mar 16, 2018

1/(x-2)-1/(x+1)1x−2−1x+1

Explanation:

Here,
3/((x-2)(x+1))3(x−2)(x+1)
we have two factors in denominator: (x-2) and (x+1)(x−2)and(x+1).
We take, color(red)((x+1)-(x-2))=x+1-x+2=color(red)(3(x+1)−(x−2)=x+1−x+2=3
:.color(red)(3)/((x-2)(x+1))=(color(red)((x+1)-(x-2)))/((x-2)(x+1))
=cancel((x+1))/((x-2)cancel((x+1)))-cancel((x-2))/(cancel((x-2))(x+1))=1/(x-2)-1/(x+1)

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