Rational polynomial of degree n with n-2 real roots, 2 complex roots and Galois group not being $S_n$

as you may know it is known in Galois theory that if $$f\in\mathbb{Q}[T]$$ and $$\partial f=p$$, with $$p$$ prime, $$f$$ irreducible and $$f$$ has $$p-2$$ real roots and $$2$$ complex roots, then $$\operatorname{Gal}(f)=S_p$$ and therefore, $$f$$ is not resoluble if $$p\ge5$$

So, my question is, is there an $$f\in\mathbb{Q}[T]$$ of degree $$n$$ with $$n-2$$ real roots, $$2$$ complex roots and Galois group not being $$S_n$$ for some composite $$n\in\mathbb{N}$$?

Moreover, is there an example of such polynomial when $$n=4$$?

• This is a really good question to ask. In general, when some theorem has a bunch of conditions, they're all necessary, and finding out exactly what MAKES each one necessary is a really good thing to do to help understanding. Sometimes, of course, they conditions aren't all necessary -- it's just that no one knows how to prove the less-constrained version of the theorem. But those theorems don't tend to come up much in beginning courses. – John Hughes Jul 7 at 23:28

Sure: for instance, $$f(T)=T^4-2$$ works. Its splitting field is $$\mathbb{Q}(\sqrt{2},i)$$ which has degree $$8$$ and thus the Galois group is not $$S_4$$.