How do you write the partial fraction decomposition of the rational expression # 1/((x^4)-6(x^3))#?

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Answer 1 · 2015-12-14T16:16:34

#1/(216(x-6))-1/(216x)-1/(36x^2)-1/(6x^3)#

Factor the denominator.

#1/(x^3(x-6))=A/x+B/x^2+C/x^3+D/(x-6)#

#1=A(x^2)(x-6)+B(x)(x-6)+C(x-6)+D(x^3)#

#1=Ax^3-6Ax^2+Bx^2-6Bx+Cx-6C+Dx^3#

#1=x^3(A+D)+x^2(-6A+B)+x(-6B+C)+1(-6C)#

Use this to write the following system:

#{(A+D=0),(-6A+B=0),(-6B+C=0),(-6C=1):}#

Solve to see that:

#{(A=-1/216),(B=-1/36),(C=-1/6),(D=1/216):}#

The fraction decomposes into:

#1/(216(x-6))-1/(216x)-1/(36x^2)-1/(6x^3)#