How do you express #(x^3-x^2+1) / (x^4-x^3)# in partial fractions?

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Answer 1 · 2017-05-22T17:45:37

The answer is #=-1/x^3-1/(x^2)+1/(x-1)#

The denominator is

#x^4-x^3=x^3(x-1)=x(x-1)#

Let's perform the decomposition into partial fractions

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

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

#=(A(x-1)+Bx(x-1)+Cx^2(x-1)+Dx^3)/(x^3(x-1))#

The denominators are the same, we compare the numerators

#x^3-x^2+1=A(x-1)+Bx(x-1)+Cx^2(x-1)+Dx^3#

Let #x=0#, #=>#, #1=-A#, #=>#, #A=-1#

Let #x=1#, #=>#, #1=D#

Coefficients of #x^2#,

#-1=B-C#

Coefficients of #x^3#,

#1=C+D#, #=>#, #C=1-D=1-1=0#

#B=C-1=0-1=-1#

Therefore,

#(x^3-x^2+1)/(x^4-x^3)=-1/x^3-1/(x^2)+(0)/x+1/(x-1)#