A degree $4$ polynomial whose Galois group is isomorphic to $S_4$. I am reading an article about Galois groups. The article states that:

It can also be shown that for each degree $d$ there exist polynomials whose Galois group is the fully symmetric group $S_d$.

I think that the Galois group of a quadratic is isomorphic to $S_2$ if the roots are not rational.
I think that the Galois group of $x^3 - a$ is isomorphic to $S_3$ if $a$ is not a perfect cube.
Is it difficult to find an example of a degree $4$ polynomial whose Galois group is isomorphic to $S_4$?  I am reading a text book and so far most of the examples have been for Galois groups of "small" order. 
I have not progressed to the point where I can appreciate a proof of the statement made in the article but I would like to see an example of such a polynomial of degree $4$ or hear from someone knowledgeable that this is a difficult and/or tedious task.

Consider $F=\Bbb C(x_1,x_2,\dots,x_n)$ where the $x_i$'s are indeterminates (variables).  A group of automorphisms can be determined by how they permute the $x_i$'s.  So that group of automorphisms is $S_n$.  Now take the fixed field $K$ of those automorphisms.  Then Gal$(F/K)$ should be $S_n$ by Artin's theorem.
The polynomial is then $P(Y)=(Y-x_1)\cdots(Y-x_n)$ which if you multiply out the coefficients are the symmetric polynomials in the $x_i$'s.
If you want it as an extension of $\Bbb Q$ then that's harder.
