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

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Answer 1 · 2017-02-05T08:14:16

The answer is #=(1/4)/(x^2)+(-15/16)/(x)+(15/16)/(x-4)#

Let's perform the decomposition into partial fractions

#(4x-1)/(x^2(x-4))=A/(x^2)+B/(x)+C/(x-4)#

#=(A(x-4)+B(x(x-4))+C(x^2))/(x^2(x-4))#

As the denominators are the same, we compare the numerators

#4x-1=A(x-4)+B(x(x-4))+C(x^2)#

Let #x=0#, #=>#, #-1=-4A#, #=>#, #A=1/4#

Let #x=4#, #=>#, #15=16C#, #=>#, #C=15/16#

Coefficients of #x^2#

#0=B+C#, #=>#, #B=-15/16#

Therefore

#(4x-1)/(x^2(x-4))=(1/4)/(x^2)+(-15/16)/(x)+(15/16)/(x-4)#