How do you write the partial fraction decomposition of the rational expression (x^3-x^2+1) / (x^4-x^3)x3x2+1x4x3?

1 Answer
Dec 23, 2015

(x^3-x^2+1) / (x^4-x^3) = -1/x^3-1/x^2+1/(x-1x3x2+1x4x3=1x31x2+1x1

Explanation:

(x^3-x^2+1) / (x^4-x^3) = (x^3-x^2)/ (x^4-x^3) + 1/(x^4-x^3)x3x2+1x4x3=x3x2x4x3+1x4x3

=(x-1)/(x^2-x) + 1/(x^3(x-1))=x1x2x+1x3(x1)

=(x-1)/(x(x-1)) + 1/(x^3(x-1))=x1x(x1)+1x3(x1)

=1/x + 1/(x^3(x-1))=1x+1x3(x1)

now focus on 1/(x^3(x-1))1x3(x1)

1/(x^3(x-1)) = A/x^3+B/x^2+C/x+D/(x-1)1x3(x1)=Ax3+Bx2+Cx+Dx1

Multiply both side by x^3(x-1)x3(x1)

1 = A(x-1)+Bx(x-1)+Cx^2(x-1)+Dx^31=A(x1)+Bx(x1)+Cx2(x1)+Dx3

1=Ax-A+Bx^2-Bx+Cx^3-Cx^2+Dx^31=AxA+Bx2Bx+Cx3Cx2+Dx3

1=x^3(C+D)+x^2(B-C)+x(A-B)-A1=x3(C+D)+x2(BC)+x(AB)A

C+D = 0C+D=0
B-C = 0BC=0
A-B = 0AB=0
A = -1A=1

Just by looking we have

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

So

(x^3-x^2+1) / (x^4-x^3) = -1/x^3-1/x^2+1/(x-1)x3x2+1x4x3=1x31x2+1x1