Defining multiplication of quotient ring with respect to polynomial Let $R=\mathbb{Z}_{11}[x]/(x^2+2)$. I would like to define multiplication in $R$. 
To my understanding, it is required that if $[ax+b][cx+d]=[rx+s]$ we want to find $r$ and $s$ in terms of $a,b,c,d$. I am having a hard time getting started with this. How do I begin thinking about this?
I understand that we must satisfy the requirements of closure, associativity, identity, distributivity, and commutativity, I just don't know how to begin.
Also, can one help me show that $[3x+1]$ is a unit in $R$? 
Thank you for your help. 
 A: Use the fact that $x^2\equiv -2$ in the quotient:
\begin{align}
(ax+b)(cx+d)
&=acx^2 + (ad+bc)x + bd
\\&=\equiv ac(-2)+ (ad+bc)x + bd
\\&=(ad+bc)x+(bd-2ac)
\end{align}
Therefore, $r=ad+bc$ and $s=bd-2ac$.
This trick applies more generally to defining multiplication in $R[x]/(p(x))$, where $p(x)=x^n+a_{n-1}x^{n-1}+\dots+a_0$. Use the fact that $x^n\equiv  -a_{n-1}x^{n-1}-\dots-a_0$. 
To show $[3x+1]$, it suffices to find an inverse. You can compute the inverse of $[3x+1]$ by writing the fraction $\frac1{3x+1}$ and multiplying the top and bottom by the conjugate $-3x+1$:
$$
\frac{1}{3x+1}=\frac{1-3x}{(1+3x)(1-3x)}=\frac{1-3x}{1-9x^2}\equiv\frac{1-3x}{1-9(-2)}=\frac{1-3x}{8}=7(1-3x)=x+7
$$
To verify this is the inverse, use the multiplication rule we derived:
$$
(3x+1)(x+7)=(3\cdot 7+1\cdot 1)x+(1\cdot 7-2\cdot 1\cdot 3)=\dots
$$
A: There are several ways to think about quotienting out by polynomials.
If you are familiar with quotienting out by normal subgroups, here elements are cosets of the form $ax+b+\langle x^2+2\rangle$. And any time you see an $x^2+2$ you may absorb it into the ideal $\angle x^2+2\rangle$. 
This is kind of equivalent to saying "modulo $x^2+2$". Just like when you do modular arithmetic with the integers, if you see $x^2+2$ , you may reduce it down to $0$.
Example: Let us see what happens when we multiply by $x+\langle x^+2\rangle$  to $3x+1+\langle x^+2\rangle$
So we get $3x^2+x+\langle x^+2\rangle=3(x^2+2)+(x-6)+\langle x^+2\rangle=x-6+\langle x^+2\rangle$
If this is clumsy notation, you may think about it as 
$x\times(3x+1)\equiv 3x^+x\equiv x-6 (\textrm{mod } x^2+2$
Here, we notice that I can get rid of all the $x$ terms if I do $(-3x+1)(3x+1)\equiv -3(x-6)+(3x+1)\equiv 19\equiv 8(\textrm{mod } x^2+2$
You should know that numbers in $\mathbb{Z}_11=\mathbb{Z}/11\mathbb{Z}$ have multiplicatie inverses except for $0$.  
So I will leave it as an exercise for you to figure out what number you have to multiply to make that $8$ into a $1$.
