Calculating Galois Group of $x^5+2x^2+2x+5$ over $\mathbb{Q}$

Question

First of all i verified that the quintic is irreducible over $\mathbb{Q}$, since it is irreducibile over $\mathbb{Z}_3$ by Eisenstein's Criterion with 2.

I know that the galois group G should be a transitive subgroup of $S_5$.

The polynomial has only 1 real root.

I tried to apply Dedekind's Theorem by considering the polynomial in $\mathbb{Z}_2$ where we have $x^5+2x^2+2x+5=x^5+1=(x+1)*(x^4+x^3+x^2+x+1)$, so (if i have correctly understood the theorem) there is an element $\sigma \in G$ which is a 4-cycle. For this reason i tend to exclude $A_5$ as the solution, but i don't know how to move further.

Answer 1

All we need to do is to combine the following facts.

1. As Michele observed, modulo $2$ the polynomial $f(x)$ is a product of a linear and a quartic factor.
2. Modulo three we have $$f(x)=x(x^4-1)+2(x^2+1)=(x^2+1)(x^3-x-1),$$ and neither of those degree $\le3$ factors have any zeros in $\Bbb{F}_3$ so they are irreducible.
3. By the first fact the only way $f(x)$ can factor over $\Bbb{Z}$ is linear $\times$ quartic, and by the second fact the only way is quadratic $\times$ cubic. These are incompatible, so $f(x)$ must be irreducible.
4. Because $f(x)$ is irreducible, the Galois group $G$ is a transitive subgroup of $S_5$. Hence $5\mid |G|$. Dedekind's theorem, together with facts 1 and 2, implies that $|G|$ is divisible by both four and three. We can conclude that $60\mid |G|$ so $G$ is either $A_5$ or $S_5$.
5. Michele already observed that $G$ is not a subgroup of $A_5$, so we have eliminated all the alternatives to $G=S_5$.