Question

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

Answer 1

`(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))`