Question

How do you long divide `n^3` by `n^2-4` ?

Answer 1

`n^3/(n^2-4) = n + (4n)/(n^2-4)`

Explanation:

The slightly tricky thing with long dividing such polynomials is that you need to include all relevant powers of `n`.

So in our example, the divisor should be treated as `n^2+0n-4`.

Then all of the terms should line up correctly and the result will be correct:

`color(white)(n^2 +0n - 4 " ") underline(color(white)(" "0) n color(white)(n^3 + 0n^2 + 0n + 0))`
`n^2 +0n -4 " " ) " " n^3 + 0n^2 + 0n + 0`
`color(white)(n^2 +0n -4 " " x " ") underline(n^3 + 0n^2 - 4n)`
`color(white)(n^2 +0n -4 " " x " " n^3 + 0n^2 +) 4n + 0`

So the skew asymptote is the function `f(n) = n`

and we can express the division as:

`n^3/(n^2-4) = n + (4n)/(n^2-4)`

graph{(y - (x^3/(x^2-4)))(y-x) = 0 [-20, 20, -10, 10]}