Galois group of $x^5-5x+10$ I was illustrating the theorem on solvability by radicals through some examples of degree $5$ polynomials. One I chose was $x^5-5x+10$. I was (perhaps wrongly) going to prove that the Galos group is $S_5$, with expectation that it has exactly three real roots. 
However, it has exactly one real root (which I saw by looking graph of this polynomial in $\mathbb{R}^2$ using some online software). Then I get stucked, and couldn't completely find the Galois group of this polynomial. 
How do we proceed to determine the Galois group of this polynomial over $\mathbb{Q}$?
 A: Observing discriminant of polynomial is very helpful to find its Galois group. It is known that if $D$ is discriminant of polynomial $f(x)$, then Galois group $G=G_{f}$ of $f(x)$ is contained in $A_{5}$ iff $D\in \mathbb{Q}^{2}$. (This holds for general fields with $char\neq 2$, not only $\mathbb{Q}$. Proof is not hard and you can find proof in "Abstract Algebra", Chap 14, Proposition 33 of Dummit-Foote.) As Wolfram says, its discriminant is 30450000 which is not a square. So $G$ is not contained in $A_{5}$. 
Also, by Eisenstein criterion with $p=5$, $f(x)$ is irreducible and $[\mathbb{Q}(\alpha):\mathbb{Q}]=5$ for any $f(\alpha)=0$. So order of $|G|$ has to be divided by $5$. 
Then if you see here, the only candidates are $S_{5}$ and general Affine group $GA(1,5)$. If we reduce the polynomial modulo $3$, then $\overline{f}(x)=x^{5}+x+1=(x-1)^{2}(x^{3}-x^{2}+1)$ in $\mathbb{F}_{3}[x]$. Since $x^{3}-x^{2}+1$ is irreducible over $\mathbb{F}_{3}$, order of $G_{\overline{f}}$ has to be divided by $3$. Since $G_{\overline{f}}\leq G_{f}$ (this is non-trivial result, and see here for the Tate's proof), $|G_{f}|$ is also divided by $3$ and contains a $3$-cycle. We can show that $5$-cycle and $3$-cycle in $S_{5}$ generates $A_{5}$, so the answer is $G_{f}=S_{5}$. 
I think it will be possible to find a prime $p$ s.t. $\overline{f}(x)\in\mathbb{F}_{p}[x]$ has an irreducible factor of degree $2$, and then you don't have to calculate discriminant of $f$. 
