How do you write the partial fraction decomposition of the rational expression 1/(x^3-5x^2)1x35x2?

1 Answer
Dec 14, 2015

1/(x^3-5x^2)=1/(5x)-1/(5x^2)-1/(5(x-5))1x35x2=15x15x215(x5)

Explanation:

Factor the denominator.

1/(x^2(x-5))=A/x+B/x^2+C/(x-5)1x2(x5)=Ax+Bx2+Cx5

1=Ax(x-5)+B(x-5)+Cx^21=Ax(x5)+B(x5)+Cx2

1=Ax^2-5Ax+Bx-5B+Cx^21=Ax25Ax+Bx5B+Cx2

1=x^2(A+C)+x(-5A-5B)+1(-5B)1=x2(A+C)+x(5A5B)+1(5B)

Thus, {(A+C=0),(-5A-5B=0),(-5B=1):}

Solve to find that {(A=1/5),(B=-1/5),(C=-1/5):}

Plug these back in:

1/(x^2(x-5))=1/(5x)-1/(5x^2)-1/(5(x-5))