Question

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

Answer 1

`{-2*x-3}/{x^2-x}={-5}/{x-1}+3/x`

Explanation:

We begin with
`{-2*x-3}/{x^2-x}`
First we factor the bottom to get
`{-2*x-3}/{x(x-1)}`.

We have a quadratic on the bottom and a linear on the top this means we are looking for something of the form
`A/{x-1}+B/x`, where `A` and `B` are real numbers.

Starting with
`A/{x-1}+B/x`, we use fraction addition rules to get
`{A*x}/{x(x-1)}+{B*(x-1)}/{x(x-1)}={A*x+Bx-B}/{x(x-1)}`

We set this equal to our equation

`{(A+B)x-B}/{x(x-1)}={-2*x-3}/{x(x-1)}`.

From this we can see that
`A+B=-2` and `-B=-3`.

We end up with
`B=3` and `A+3=-2` or `A=-5`.

So we have
`{-5}/{x-1}+3/x={-2*x-3}/{x^2-x}`