Question

How do you write the partial fraction decomposition of the rational expression `(2x-6)/((x-1)(x-2)^2)`?

Answer 1

`color(blue)(4/(x-2)-2/((x-2)^2)-4/(x-1))`

Explanation:

`(2x-6)/((x-1)(x-2)^2)`

We expect the form of the partial fractions to be:

`(2x-6)/((x-1)(x-2)^2)-=A/(x-1)+B/((x-2)^2)+C/(x-2)`

Adding `RHS`

`2x-6-=A(x-2)^2+B(x-1)+C(x-1)(x-2)`

Because this is an identity numerators will be equal:

We are looking to find the values of A B and C.

We now set `x` to values that will eliminate most of the unknowns we are looking for.

`2x-6-=A(x-2)^2+B(x-1)+C(x-1)(x-2)`

Setting `x=2`

`2(2)-6-=A((2)-2)^2+B((2)-1)+C((2)-1)((2)-2)`

`-2-=B`

`B=-2`

Setting `x=1`

`2(1)-6-=A((1)-2)^2+B((1)-1)+C((1)-1)((1)-2)`

`-4-=A`

`A=-4`

Setting `x=0`

`2(0)-6-=A((0)-2)^2+B((0)-1)+C((0)-1)((0)-2)`

`-6-=4A-B+2C`

`4A-B+2C=-6`

We already know A and B:

`4(-4)-(-2)+2C=-6`

`C=4`

So our partial fractions are:

`color(blue)(4/(x-2)-2/((x-2)^2)-4/(x-1))`