Question

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

Answer 1

`(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`