Question

How do you decompose `(x-6)/(x^2-2x)` into partial fractions?

Answer 1

Begin with this equation:

`(x - 6)/(x^2 - 2x) = A/x + B/(x - 2)`

Explanation:

Multiply both sides of the equation by `x(x - 2)`:

`x - 6 = A(x - 2) + Bx`

Let x = 0:

`0 - 6 = A(0 - 2) + B(0)`

`-6 = A(-2)`

`A = 3`

Let x = 2:

`2 - 6 = A(2 - 2) + B(2)`

`-4 = 2B`

`B = -2`

`A = 3 and B = -2`

Check:

`3/x - 2/(x - 2) =`

`3/x(x - 2)/(x - 2) - 2/(x - 2)x/x =`

`(3x - 6)/(x(x - 2)) - (2x)/(x(x - 2)) =`

`(3x - 6 - 2x)/(x(x - 2)) =`

`(x - 6)/(x^2 - 2x)`

The checks.

The answer is:

`(x - 6)/(x^2 - 2x) = 3/x - 2/(x - 2)`