Are these rings fields? Are the following rings fields?

1) $\Bbb Q[x] /\langle x^2+1\rangle$
Since a polynomial ring taking values on any field is a E.D, and hence a P.I.D, this is a field iff the ideal is prime or maximal.
Any irreducible in this quotient ring is a maximal ideal, and $x^2+1$ is an irreducible polynomial in the quotient ring, since we don't have algebraic closure, hence the quotient ring is a field.

2) $\Bbb F_2[x] / \langle x^2 +1 \rangle$
Not sure how to show if $x^2+1$ is irreducible here, I have a feeling it isn't but no way to expand on that. Some comments, I do know that $\Bbb F_2$ means I am taking values from $\{0,1\}$ and hence $-1=1 \pmod 2$

3) $\Bbb Q[x] / \langle x^4 + 6x^3 + 9x + 6\rangle$
Here I just need to see if $x^4+6x^3+9x+6$ is irreducible.
What I did was a bit strange:
$$x^4+6x^3+9x+6$$
$$=(x+1)x^3+5x^3+9x+6$$
$$=x^3((x+1)+5)+9x+6$$
$$=x(x^2((x+1)+5)+9)+6$$
Which gives us the root $x=-6$ and hence this will be generated by $(x+6)$, so this is not maximal, hence this quotient ring is not a field. This feels sketchy, since perhaps that root is in $\Bbb Q$ but the other ones aren't.

Is my logic correct in 1)&3), how do I do 2)?

It is incorrect. $-6$ is not a root of $x^4+6x^3+9x+6$. Let's look for rational root of that polynomial ${p\over q}$.
We have $q|1$ and $p|6$. (Do you see why?)
So the only rational roots can be $\pm 6$. None of them is and to prove the irreducibility we need more because this quartic could for instance decompose as a product of two irreducible quadratics.
Eisenstein criterion works nicely with $p=3$ in this case
