How do you express #(x^2 - 3x + 2) / (4x^3 + 11x^2)# in partial fractions?

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Answer 1 · 2016-12-26T13:27:12

The answer is #=(2/11)/x^2+(-41/121)/x+(285/121)/(4x+11)#

We start the decomposition into partial fractions

#(x^2-3x+2)/(4x^3+11x^2)=(x^2-3x+2)/(x^2(4x+11))#

#=A/x^2+B/x+C/(4x+11)#

#=(A(4x+11)+Bx(4x+11)+Cx^2)/(x^2(4x+11))#

Therefore,

#x^2-3x+2=A(4x+11)+Bx(4x+11)+Cx^2#

Let, #x=0#, #=>#, #2=11A#, #=>#, #A=2/11#

Coefficients of #-3=4A+11B#

#11B=-3-4A=-3-8/11#

#B=-41/121#

Coefficients of #x^2#, #1=4B+C#

#C=1-4B#

#C=1+164/121#

#C=285/121#

So,

#(x^2-3x+2)/(4x^3+11x^2)=(2/11)/x^2+(-41/121)/x+(285/121)/(4x+11)#