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

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Answer 1 · 2015-12-14T21:17:43

#1/(x^3-5x^2)=1/(5x)-1/(5x^2)-1/(5(x-5))#

Factor the denominator.

#1/(x^2(x-5))=A/x+B/x^2+C/(x-5)#

#1=Ax(x-5)+B(x-5)+Cx^2#

#1=Ax^2-5Ax+Bx-5B+Cx^2#

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