How do you write 2/(x^3-x^2) 2x3−x2 as a partial fraction decomposition?

1 Answer
Narad T.
Nov 8, 2016

The result is =-2/x^2-2/x+2/(x-1)=−2x2−2x+2x−1

Explanation:

Let's factorise the denominator, x^3-x^2=x^2(x-1)x3−x2=x2(x−1)
So, 2/(x^3-x^2)=2/(x^2(x-1))=A/x^2+B/x+C/(x-1)2x3−x2=2x2(x−1)=Ax2+Bx+Cx−1
=(A(x-1)+Bx(x-1)+Cx^2)/(x^2(x-1))=A(x−1)+Bx(x−1)+Cx2x2(x−1)

:. 2=A(x-1)+Bx(x-1)+Cx^2
If x=0=>2=-A=>A=-2
Coefficients of x, 0=A-B=>B=-2
Coefficients of x^2, 0=B+C=>C=2

2/(x^3-x^2)=2/(x^2(x-1))=-2/x^2-2/x+2/(x-1)

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