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

1 Answer
Dec 23, 2015

(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)