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

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Answer 1 · 2015-12-24T20:16:07

$1/(x-1)-1/x^2-1/x^3$

Factor the denominator.

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

Find a common denominator.

$x^3-x^2+1=Ax^2(x-1)+Bx(x-1)+C(x-1)+Dx^3$

$x^3-x^2+1=Ax^3-Ax^2+Bx^2-Bx+Cx-C+Dx^3$

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

Use this to write the system:

${(A+D=1),(-A+B=-1),(-B+C=0),(-C=1):}$

Solve:

${(A=0),(B=-1),(C=-1),(D=1):}$

Put this back in to the original expression:

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

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