Question

How do you express as a partial fraction `(5x-1)/(x^2-x-2)`?

Answer 1

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

Explanation:

`(5x-1)/(x^2-x-2)`

The denominator can be factored using real numbers, so we do that first:

`x^2-x-2 = (x-2)(x+1)`

Now we want:

`A/(x-2) + B/(x+1) = (5x-1)/(x^2-x-2)`

One way of proceeding is to get a single fraction on the left:

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

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

So we want:

`((A+B)x+(A-2B))/(x^2-x-2) = (5x-1)/(x^2-x-2)`

That means we need:

`A+B` `" " =` `" "` `5`
`A-2B` `" "` `=` `-1`

Subtract the second from the first (change the signs and add) to get:

`3B` `=` `6`

So `B=2` and in order to get `A+B=5` we must also have `A=3`

We have:

`3/(x-2)+2/(x+1)`

It is worth taking a moment to check our answer:

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

Yes the top simplifies to `5x-1`, so we should be OK.