I've made some progress finding the Galois group of $f = x^5 - 2x^3 - x^2 + 2$ but I am having some difficulties.
I've factorised it over $\mathbb{Q}$ as $f = (x^2 - 2)(x-1)(x^2 + x + 1)$ and so I can see the splitting field must be $\mathbb{Q}(\sqrt(2), w)$ where $w$ is a primitive cube root of unity. Now I am trying to calculate the degree of this extension. I know that it must be divisible by 2 because it has $\mathbb{Q}(\sqrt(2))$ as a subfield and I know it must be less than 4 (because adjoining the primitive root is a degree 2 extension). Now how do I distinguish between the degree being 2 or 4? I.e. How do we know there is no quadratic polynomial which has a root generating the whole extension?
I also know that $f$ has distinct roots and so it is a subgroup of $S5$ but if the extension is 4, how do I distinguish between the possibilities $V4$ or $C4$?
Thanks
As you say, it has $\mathbb{Q}(\sqrt{2})$ as a subfield. So if the degree over $\mathbb{Q}$ were $2$, it would be equal to that subfield (by multiplicativity of degrees and the observation that $[F: L] = 1$ iff $F = L$.) If that were so, in particular, $\mathbb{Q}(\sqrt{2})$ would contain a non-trivial third root of unity, which is absurd (proof?)
For calculating the Galois group -- think about where each root of the polynomial can go under an automorphism. Do you have an automorphism of order 4, or not?