Polynomial arithmetic uses coefficient ring arithmetic How is it works ?
What is different between divide with remainders in a ring and without ?
e.g I have this question:
Calculate $\frac{6x^5+2x^4+5x^3+x+2}{5x^3+x^2+6}$ in the ring $\mathbb{F}_{7}[x]$.
So I how do I do this ?
Because the way I calculated, the result is: 
$1.2x^2+0.16x+0.968$ with the reminder: $-9.128x^2 +x -3.808$
So what is the differences ?
I know I probably wrong but this is what I know
 A: Hint: You must calculate in $F_7$, where $6/5=4$ because $4 \cdot 5 = 20 = 6 \mod 7$
A: Your polynomial ring $\Bbb F_7[x]$ has coefficient ring being the field $\,\Bbb F_7 = $ integers mod $7$, so coefficient arithmetic is not real number arithmetic but, instead, integer arithmetic modulo $7,\,$  where fractions are calculated as $\,a/b = ab^{-1},$ and inverses $\,b^{-1}$may be calculated by the extended Euclidean algorithm (or inspection). However, we can eliminate all fractional arithmetic by first scaling $\,f,g\,$ so that the divisor $\,g\,$ is monic, i.e. has lead coeff $= 1.\,$ To do so you need to multiply both by by $\,\color{blue}{1}/\color{#0a0}5\equiv \color{blue}8/\color{#0a0}{{-}2}\equiv -4\equiv \color{#c00}3\pmod 7.\,$ Then division yields $\,f/5\, =\, q\ g/5 + r,\,$ so scaling by $5$ yields the division of $f$ by $g$.
Remark $\ $ You can also obtain the correct result by reducing your real-number result modulo $7$ using, by above, $\,1/10 \equiv 1/\color{#c00}3\equiv \color{#0a0}5,\,$ so $\,a/10^k\equiv a\,5^k.\,$ This works because the real numbers produced by the division will all be rationals with (lowest-terms) denominators that are a power of the lead coeff $5$, which is invertible mod $\,7.\,$ Indeed writing your coefficients as reduced fractions we find
$$ 0.2 = \frac{1}5,\ \ 0.16 = \frac{4}{5^2},\ \ 0.968 = \frac{121}{5^3},\ \ 0.128 = \frac{16}{5^3},\ \ 0.808 = \frac{101}{5^3}$$
But doing it that way ends up being much more work than the mentioned scaling method mod $7$.
