Proving a Galois group is not contained in $A_4$

Question

I have been asked to prove that the Galois group for $f(x)=x^4-6x+2$ over $\mathbb{Q}$ is not contained in $A_4$. I have already proven that $f(x)$ is irreducible and that it contains a 3-cycle. How would I go about proving this?

Answer 1

An alternative to @Lord Shark's answer. I suspect that this one is more useful in contrived examples rather than real-life problems. But if you had to produce a polynomial with Galois group $S_{31}$ (say) this might be the way to go.

$f(1)=-3$ and $f(\pm\infty)=\infty$ so $f$ has at least two real roots.

$f'(x)=4x^3-6$ is non-decreasing, so $f$ has exactly two real roots.

Complex conjugation therefore gives a non-trivial element of the Galois group, which acts as a transposition. Hence the Galois group is not in $A_4$.

Answer 2

An irreducible polynomial $f(x)$ of degree $n$ over $\Bbb Q$ has its Galois group a subgroup of $A_n$ (in the action on zeros of $f$) if and only if the discriminant of $f$ is a square of a rational number.

See Discriminant of a trinomial $x^n+ax^m+b$ for more about evaluating discriminants of polynomials with only three terms.