Question

How do you write the partial fraction decomposition of the rational expression `(x+6)/(x^3-3x^2-4x+12)`?

Answer 1

Partial fractions are `1/(5(x+2))-2/((x-2))+9/(5(x-3))`

Explanation:

As function in the denominator `f)x)=x^3-3x^2-4x+12` has `2, -2` and `3` as zeros, we have `x^3-3x^2-4x+12=(x-2)(x+2)(x-3)` and we can write partial fractions as

`(x+6)/(x^3-3x^2-4x+12)hArrA/(x+2)+B/(x-2)+C/(x-3)` or

`(x+6)/(x^3-3x^2-4x+12)hArr(A(x-2)(x-3)+B(x+2)(x-3)+C(x+2)(x-3))/((x-2)(x+2)(x-3))` or

`(A(x^2-5x+6)+B(x^2-x-6)+C(x^2-4))/((x-2)(x+2)(x-3))` or

`(x^2(A+B+C)+x(-5A-B)+(6A-6B-4C))/((x-2)(x+2)(x-3))`

Comparing like terms we have three terms `A+B+C=0`, `-5A-B=1` and `6A-6B-4C=6`

From second equation we get `B=-1-5A` and putting this in third equation we get `6A-6(-1-5A)-4C=6` or `6A+6+30A-4C=6` or `4C=36A` or `C=9A`.

Now putting these in first we get `A-1-5A+9A=0` or `5A=1` or `A=1/5`. Hence `B=-2` and `C=9/5`

Hence partial fractions are `1/(5(x+2))-2/((x-2))+9/(5(x-3))`