Find the quotient and the remainder of $(n^6-7)/(n^2+1)$ 
Given that $n$ belong to $\mathbb{N}$.
Find the quotent and the remainder of $(n^6-7)/(n^2+1)$.

So I tried to divide them up and got a negative expression $(-n^4-7)$.
How to continue?
Or what can be done differently?
How to find the quotent and the remainder?

I assume that by "dish" you mean "quotient"; that's not terminology I'm familiar with, but it seems to make sense in context.
When you're dividing polynomials, remember that you are not done until the remainder is of lower degree than the denominator. So you're correct that the first step of dividing $n^2 + 1$ into $n^6 - 7$ gives you a remainder of $-n^4 - 7$ (with $n^4$ in the quotient); but since $-n^4 - 7$ is of higher degree than $n^2 + 1$, we have to keep going.
To go into $-n^4 - 7$, we need to multiply by $-n^2$; so now our quotient so far is $n^4 - n^2$, and our remainder is $(-n^4 - 7) - (-n^2 (n^2 + 1)) = n^2 - 7$.
Since $n^2 - 7$ is still at least as high in degree as $n^2 + 1$, we have to divide one more time; this gives us a quotient of $n^4 - n^2 + 1$ and a remainder of $(n^2 - 7) - (n^2 + 1) = -8$.
Since the remainder now has degree $0$, which is less than the degree of our divisor, we can stop now; so the quotient is $n^4 - n^2 + 1$ and the remainder is $-8$.
