Question

How do you write the partial fraction decomposition of the rational expression `(x^3-x^2+1) / (x^4-x^3)`?

Answer 1

`(x^3-x^2+1) / (x^4-x^3) = -1/x^3-1/x^2+1/(x-1`

Explanation:

`(x^3-x^2+1) / (x^4-x^3) = (x^3-x^2)/ (x^4-x^3) + 1/(x^4-x^3)`

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

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

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

now focus on `1/(x^3(x-1))`

`1/(x^3(x-1)) = A/x^3+B/x^2+C/x+D/(x-1)`

Multiply both side by `x^3(x-1)`

`1 = A(x-1)+Bx(x-1)+Cx^2(x-1)+Dx^3`

`1=Ax-A+Bx^2-Bx+Cx^3-Cx^2+Dx^3`

`1=x^3(C+D)+x^2(B-C)+x(A-B)-A`

`C+D = 0`
`B-C = 0`
`A-B = 0`
`A = -1`

Just by looking we have

`A = - 1`
`B = -1`
`C = -1`
`D = 1`

So

`(x^3-x^2+1) / (x^4-x^3) = -1/x^3-1/x^2+1/(x-1)`