Let $f(x)\in \mathbb{Q}[x]$ be an irreducible polynomial of degree five with exactly three real roots and let $K$ be the splitting field of $f$. Prove that Gal$(K/Q) \cong S_5$.
My attempt
The splitting field of an irreducible polynomial of degree $5$ has Galois group which is a transitive subgroup of $S_5$. Hence, if we can show that the Galois group has elements of cycle type $[2][1][1][1]$ and $[3][1][1][1]$ then $S_5$ is the only such subgroup possible (we are eliminating the possible subgroups of $F_{20}$ with the cycle of order three since $3 \nmid 20$ and $A_5$ since the transposition cannot be an even permutation).
Conjugation is a field automorphism that fixes $\mathbb{Q}$. It is in thus in the Galois Group and has cycle type $[2][1][1][1]$.
My question
How do I show that there is an element of order three in the Galois group?