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

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Answer 1 · 2018-08-07T07:30:03

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

Perform the decomposition into partial fractions as follows

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

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

The denominators are the same, compare the numerators

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

Let #x=2#, #=>#, #4-4+1=A#, #=>#, #A=1#

Coefficients of #x^2#

#1=C#

Coefficients of #x#

#-2=B-4C#

#B=4C-2=2#

Finally,

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