Dividing polynomials in $\mathbb{F}_7[x]$

Question

I am trying to divide $4x^4+3x^3+2x^2+x+1$ by $2x^2+x+1$ in $\mathbb{F}_7[x]$.

Normally outside of $\mathbb{F}_7[x]$ I know that the answer would be $2x^2+(1/2)x-(1/4)$ with a remainder of $(3/4)x+(5/4)$. But because this is in $\mathbb{F}_7[x]$ and the coefficients must be in $\mathbb{F}_7$, I am confused on how to deal with the fractions. Can anyone help?

Answer 1

$\frac{1}{2}=\frac{1×4}{2×4}=\frac{4}{8}=\frac{4}{1}$ as 8 mod 7 is 1.Here you need to make denominator 1 by selecting suitable number.
$\frac{-1}{4}=\frac{6×2}{8}=\frac{6×2}{1}=\frac{5}{1}$. Since $-1=6 mod 7$. Similarly you can try other fraction mod 7.

Answer 2

$\mathbb{F}_7$ is a field so everything you have written is valid. Notice that $2\cdot 4\equiv_7 1$, so $2,4$ are inverses of each other. From this you get: $$2x^2+4x-2$$ with a remainder of $6x+3$.