How do you express # (5x^2+3x-2)/(x^3+2x^2) # in partial fractions?

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Answer 1 · 2016-10-27T08:54:51

The result is
#(5x^2+3x-2)/(x^3+2x^2)=-1/x^2+2/x+3/(x+2)#

We need to simplify the denominator
#x^3+2x^2=x^2(x+2)#

Let #(5x^2+3x-2)/(x^3+2x^2)=(5x^2+3x-2)/(x^2(x+2))=A/x^2+B/x+C/(x+2)#
#=(A(x+2)+Bx+Cx^2)/(x^2(x+2))#
So #5x^2+3x-2=A(x+2)+Bx(x+2)+Cx^2#

Let #x=0# then #-2=2A# #=>##A=-1#

Coefficients of #x^2#
#5=B+C#

Coefficients of #x#
#3=A+2B# #=># #3=-1+2B# #=>##B=2#

and #5=2+C# #=>##C=3#

so the result is
#(5x^2+3x-2)/(x^3+2x^2)=-1/x^2+2/x+3/(x+2)#