Let $f(x)=x^3+2x+2 \in \mathbb{Q} [x]$. Find the Galois group of $f$.
I was trying to find the roots of $f$, but I couldn't. Teacher told us it wasn't necessary to find all roots. I know that since the degree of $f$ is odd then it has at least a root in $\mathbb{R}$, say, $r$.
Now, by Eisenstein's criterion with $p=2$ we have that $f$ is irreducible in $ \mathbb{Q} [x]$.
Then I don't know how to continue or how to use that information.
I would appreciate your help, thanks!
Here's a general theorem which fits your problem perfectly:
If $f$ is an irreducible polynomial of prime degree $p$ with rational coefficients and exactly two non-real roots, then the Galois group of $f$ is the full symmetric group $S_p$. [Wikipedia]
$f(x)=x^3+2x+2$ has only one real root because $f'(x)>0$ and so its Galois group is $S_3$.
You can also argue directly. Since $f$ has exactly two nonreal roots, complex conjugation induces an element of order $2$ in the Galois group. Since the Galois group has degree at least $3$ and at most $6=3!$, it must have order $6$. It only remains to prove that it is not abelian.
Cubic:
Cubic is irreducible iff it has no root. By rational root theorem, the only possible roots are $\pm1$, $\pm2$, but none of them are roots, so it has no roots, so it is irreducible (alternatively, Eisenstein with $p=2$) as you asid.
It is a cubic, so the Galois group is a subgroup of $S_3$. It is irreducible, so the Galois group has at least $3$ elements. So it is either $A_3$ or $S_3$.
Its discriminant is $-140$ which is not a square, so it is $S_3$.
