Question

How do you express `(3x^2 - 4x - 2) / [(x-1)(x-2)]` in partial fractions?

Answer 1

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

Explanation:

`(3x^2-4x-2)/((x-1)(x-2))`

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

`=(3x^2-9x+6+5x-8)/(x^2-3x+2)`

`=(3(x^2-3x+2)+5x-8)/(x^2-3x+2)`

`=3 + (5x-8)/(x^2-3x+2)`

`=3 + (5x-8)/((x-1)(x-2))`

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

Use Heaviside's cover-up method to find:

`A=(5(1)-8)/((1)-2) = (-3)/(-1) = 3`

`B=(5(2)-8)/((2)-1) = 2/1 = 2`

So:

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