Polynomial Division in a field

Question

I am asked to find the quotient and remainder of $x^3+[2]x^2+[5]$ divided by $x^2+[3]$ in $\mathbb{Z_7}[x]$

Answer 1

Because the polynomial $x^2+[3]$ that we are dividing by is monic (lead coefficient $1$) the division goes exactly like ordinary polynomial division. Except at the end you put square brackets around the coefficients. And you may for example wish to replace $[47]$, say, by $[5]$, or $-[2]$.

If we are not dividing by a monic polynomial, again it is much like ordinary polynomial division, except that where you would use division, you use division modulo $7$.

Remark: The issue does not come up in your problem, but let us find the quotient and remainder when $[3]x^3$ is divided by $[2]x^2+x$. From now on, for convenience, we leave out square brackets. "Divide" $3x^3$ by $2x^2+x$. The inverse of $2$ modulo $7$ is $4$, so the "quotient" is $12x$, or preferably $5x$. Multiply $2x^2+x$ by $5x$. We get $2x^3+5x^2$.

But importantly, there is a shortcut. We want to divide $3x^3$ by $2x^2+x$. Replace the $3$ by $10$. Now we can divide "normally."

Subtract. We get $-5x^2$, or preferably $2x^2$. Note that $2x^2+x$ goes into $2x^2$ once. After we do the usual arithmetic, we get remainder $-x$, or if you prefer, $6x$, and quotient $5x+1$.