ring isomorphism question Could any one tell me how to solve this two?

$1.$ As a ring $\mathbb{Z}[i]/(3-i)\cong\ ?$
$2.$ $L=\mathbb{R}[x]/(x^2-x+1),\ M=\mathbb{R}[x]/(x^2+x+1),\ N=\mathbb{R}[x]/(x^2+2x+1)$. Who is isomorphic to whom as a ring?

I know the definition of $\mathbb{Z}[i]/(3-i)= x(=a+ib)+(3-i)=c+id$
please help.
 A: Sketch: $\mathbb{Z}[i]\cong \mathbb{Z}[x]/(x^2+1)$. So, $\mathbb{Z}[i]/(3-i)\cong \mathbb{Z}[x]/(x^2+1,3-x)$. But, using the same trick
$$\mathbb{Z}[x]/(x^2+1,x-3)\cong (\mathbb{Z}[x]/(x-3))/(x^2+1)\cong \mathbb{Z}/10\mathbb{Z}$$
Hint: Modding out by a quadratic in $\mathbb{R}[x]$ always either gives you $\mathbb{R}^2$ or $\mathbb{C}$. What does this depend on?
EDIT: As YACP points out below, the above should have obviously said a squarefree quadratic. I didn't notice that one of the polynomials wasn't squarefree. For that, you have $N\cong \mathbb{R}[\varepsilon]$ with $\varepsilon^2=0$. This isn't an integral domain, and so shouldn't bother you too much.
A: Well, my algebra teacher's favorite way of explaining the quotients is to say that, by the quotient $\mathbb F[x]/(f(x))$ we understand the ring $S\subset \mathbb F[x]$ where $f(x)$ is regarded as $0$.
Hence, $\mathbb Z[x]/(x^2+1)$ is just $\mathbb Z$ adjoined by an element $x$ which satisfies $x^2+1=0,$ i.e. $i$.  So $\mathbb Z[i]\cong \mathbb Z[x]/(x^2+1).$ Moreover, the image of $3-x$ in the quotient $\mathbb Z[x]/(x^2+1)$ is precisely $3-i$, so that our desired quotient is just $\mathbb Z[x]/(x^2+1,3-x).$ As explained by Alex, this quotient is exactly isomorphic to $\mathbb Z/10\mathbb Z.$
Now, for your second question, by forming quotients you just add some roots of the polynomials in question into $\mathbb R$. If the roots are complex, then the resulting quotient is $\mathbb C$.  

In fact $L\cong M\cong \mathbb C$, and $N$ is isomorphic to an algebra over $\mathbb R$, of dimension $2$, and generated by an element $x$ satisfying $(x+1)^2=0$, while $x\not=-1$.  

P.S. As this is only a supplement to the answer by Alex, I made it CW.
