How do you express #(x+13)/(x^3+2x^2-5x-6)# in partial fractions?

1 Answer
May 8, 2016

#(x+3)/(x^3+2x^2-5x-6)=1/(3(x-2))-1/(3(x+1))#

Explanation:

Let us factorize #x^3+2x^2-5x-6#. It is apparent that #2# is a zero of the polynomial and hence #(x-2)# is a factor of #x^3+2x^2-5x-6#. Dividing by #(x-2)#,

#x^3+2x^2-5x-6=(x-2)(x^2+4x+3)# and hence #x^3+2x^2-5x-6=(x-2)(x+3)(x+1)#.

Now let the partial fractions be given by

#(x+3)/(x^3+2x^2-5x-6)=(x+3)/((x-2)(x+3)(x+1))=1/((x-2)(x+1))hArrA/(x-2)+B/(x+1)#

or #1/((x-2)(x+1))hArr(A(x+1)+B(x-2))/((x-2)(x+1))#

or #1/((x-2)(x+1))hArr((A+B)x+(A-2B))/((x-2)(x+1))#

Hence #A+B=0# and #A-2B=1# or #A=1/3# and #B=-1/3#

Hence #(x+3)/(x^3+2x^2-5x-6)=1/(3(x-2))-1/(3(x+1))#