Question

How do you divide `(n^3-9n-3)div(n+3)` using synthetic division?

Answer 1

The division gives `(n^2-3n)-3/(n+3)`

Explanation:

Let's do the long division

`n^3` `color(white)(aaaaa)` `-9n-3` `∣` `n+3`
`n^3+3n^2` `color(white)(aaaaaaa)` `∣` `n^2-3n`
`0-3n^2-9n`
`color(white)(aaaaa)` `3n^2-9n`
`color(white)(aaaaaa)` `0+0-3`

So there is a remainder of `-3`

So `(n^3-9n-3)/(n+3)=(n^2-3n)-3/(n+3)`
We can also obtain the remainder
let `f(n)=n^3-9n-3`
then `f(-3)=-27+27-3=-3`