How do you write the partial fraction decomposition of the rational expression #(2x^3+2x^2-9x-20)/((x^2-x-6)(x+2))#?

1 Answer
Aug 1, 2017

#(2x^3+2x^2-9x-20)/((x^2-x-6)(x+2))=2+1/(x-3)-3/(x+2)+2/(x+2)^2#

Explanation:

Before we express in partial fractions, ensure that the degree of numerator is less than that of denominator.

#(2x^3+2x^2-9x-20)/((x^2-x-6)(x+2))#

= #(2x^3+2x^2-9x-20)/(x^3+x^2-8x-12)#

= #2+(7x+4)/(x^3+x^2-8x-12)#

= #2+(7x+4)/((x-3)(x+2)(x+2))#

= #2+(7x+4)/((x-3)(x+2)^2)#

Now let #(7x+4)/((x-3)(x+2)^2)-=A/(x-3)+B/(x+2)+C/(x+2)^2#

or #(7x+4)/((x-3)(x+2)^2)=(A(x+2)^2+B(x-3)(x+2)+C(x-3))/((x-3)(x+2)^2)#

if #x=3#, then #25A=25# i.e. #A=1#

if #x=-2#, then #-5C=-10# i.e. #C=2#

and comparing coefficients of #x^2# on both sides

#A+B+C=0# i.e. #B=-3#

Hence

#(2x^3+2x^2-9x-20)/((x^2-x-6)(x+2))=2+1/(x-3)-3/(x+2)+2/(x+2)^2#