Question

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

Answer 1

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

Explanation:

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