Question

How do you express `(4x^3) / (x^3 + 2x^2 - x - 2)` in partial fractions?

Answer 1

`(4x^3)/(x^3+2x^2-x-2) = 4+2/(3(x-1))+2/(x+1)-32/(3(x+2)`

Explanation:

Note that:

`x^3+2x^2-x-2 = (x^3+2x^2)-(x+2)`

`color(white)(x^3+2x^2-x-2) = x^2(x+2)-1(x+2)`

`color(white)(x^3+2x^2-x-2) = (x^2-1)(x+2)`

`color(white)(x^3+2x^2-x-2) = (x-1)(x+1)(x+2)`

So:

`(4x^3)/(x^3+2x^2-x-2) = 4+A/(x-1)+B/(x+1)+C/(x+2)`

So:

`4x^3 = 4(x^3+2x^2-x-2)+A(x+1)(x+2)+B(x-1)(x+2)+C(x-1)(x+1)`

Putting `x=1` we get:

`4 = 6A" "` so `" "A=2/3`

Putting `x=-1` we get:

`-4 = -2B" "` so `" "B=2`

Putting `x=-2` we get:

`-32 = 3C" "` so `" "C = -32/3`

So:

`(4x^3)/(x^3+2x^2-x-2) = 4+2/(3(x-1))+2/(x+1)-32/(3(x+2)`