Prove that the fields $\mathbb Z_{11}[x]/\langle x^2+1\rangle$ and $\mathbb Z_{11}[x]/\langle x^2+x+4 \rangle$ are isomorphic I have been stuck in this problem for some time now.

Prove that $x^2+2$ and $x^2+x+4$ are irreducible over $\mathbb{Z}_{11}$. Also, prove further $\mathbb Z_{11}[x]/\langle x^2+1\rangle$
  and $\mathbb Z_{11}[x]/\langle x^2+x+4\rangle$ are isomorphic, each
  having $121$ elements.

The first part is easy to prove since there is no element of $\mathbb Z_{11}$ that satisfies either of the polynomials given in the question. However, proving that  $\mathbb Z_{11}[x]/\langle x^2+1\rangle$ and $\mathbb Z_{11}[x]/\langle x^2+x+4\rangle$ are isomorphic has been a challenge for me. How do I proceed? Moreover, how do I show that the the fields $\mathbb {Z}_{11}[x]/\langle x^2+1\rangle$ and $\mathbb {Z}_{11}[x]/\langle x^2+x+4\rangle$ each have $121$ elements?

Here's an explicit isomorphism $f:\mathbb{Z}_{11}[x]/\langle x^2 + 1\rangle\rightarrow \mathbb{Z}_{11}[x]/\langle x^2+x+4\rangle$ between the 2 fields.  Abusing notation, I'm going to refer to an element in the domain by it's natural preimage in $\mathbb{Z}_{11}[x]$, and likewise in the range.
First note that $1$ is uniquely determined in a field, so we must send $1$ to $1$.  Using additivitiy, $f(n) = n$ for any $n\in\mathbb{Z}/11\mathbb{Z}$.  So, the only remaining question is what $f(x)$ will be.  Notice that $x^2 = -1$, so $f(x)$ must be a squareroot of $-1$ in the other field.
Writing $f(x) = ax+b$, we get $$-1 = f(x)^2 = a^2 x^2 +2abx + b^2 = a^2(-x-4)+2abx + b^2 = (2ab-a^2)x + b^2-4.$$
Now, $a\neq 0$ (else $f$ is not injective), so we learn that $2b=a$  and $b^2-4 = -1$.
The second equation, $b^2-4 = -1$ has 2 solutions (mod 11), $b= 5$ and $b = 6 (=-5)$.  We'll pick $b=5$ (the choice doesn't matter).  Then since $2b=a$, we get $a = 10 = -1$.
Thus, we have $f(x) = -x + 5$.
Putting this altogether, we have $f(ax+b) = -ax + 5a + b$ defining our isomorphism.
To see $f$ is injective, assume $f(ax+b) = 0$.  Then, since $-ax + 5a+b = 0$, we must have $-a = 0$, so $a=0$.  Once we know $a=0$, $b=0$ follows.  So $f$ is injective.  To see it's surjective, notice $f(-ax + 5a+b) = ax + b$, so $f$ is surjective.
Finally, we check it's a homomorphism.  We have \begin{align*}f((ax+b) + (cx+d)) &= -(a+c)x + 5(a+c) + (b+d)\\\ &= -ax + 5a +b + -cx + 5c + d\\\ &= f(ax+b) + f(cx+d).\end{align*}
We also have \begin{align*} f((ax+b)(cx+d)) &= f(acx^2 + (ad+bc)x + bd)\\\ &= f((ad+bc)x + bd-ac)\\\ &=-(ad+cb)x + 5(ad+bc) + bd-ac\end{align*}
while \begin{align*} f(ax+b)f(cx+d) &= (-ax + 5a +b)(-cx+5c+d)\\\ &=acx^2 -5acx -adx -5acx +25ac+5ad-bcx+5bc+bd \\\ &= ac(-x-4) +x(-10ac-ad-bc) + 25ac+5ad+5bc+bd \\\ &= x(-11ac-ad-bc) + (21 ac+5ad+5bc+bd) \\\ &= x(-ad-bc) + 5ad+5bc +bd-ac\end{align*} 
so $f((ax+b)(cx+d)) = f(ax+b)f(cx+d)$.  Thus, $f$ is the desired isomorphism.
