Question

How do you express `(x²+2) / (x+3)` in partial fractions?

Answer 1

`x/1 + {-3x+2}/{x+3}`

Explanation:

because the top quadratic and the bottom is linear you're looking for something or the form

`A/1 + B/(x+3)`, were `A` and `B` will both be linear functions of `x` (like 2x+4 or similar).

We know one bottom must be one because x+3 is linear.

We're starting with
`A/1 + B/(x+3)`.
We then apply standard fraction addition rules. We need to get then to a common base.

This is just like numerical fractions `1/3+1/4=3/12+4/12=7/12.`

`A/1 + B/(x+3)=> {A*(x+3)}/{1*(x+3)} + B/(x+3)={A*(x+3)+B}/{x+3}`.
So we get the bottom automatically.

Now we set `A*(x+3)+B=x^2+2`
`Ax + 3A + B=x^2+2`
`A` and `B` are linear terms so the `x^2` must come from `Ax`.
let `Ax=x^2` `=>` `A=x`
Then
`3A + B=2`
substituting `A=x`, gives
`3x + B=2`
or
`B=2-3x`
in standard from this is `B=-3x+2`.
Putting it all together we have
`x/1 + {-3x+2}/{x+3}`