In determining the Galois group of the polynomial $p(x) = x^3-x^2-4,$ I concluded that is must be the Klein-$4$ group as follows. First, $p(x) = (x-2)(x^2+x+2)$ and the roots of the irreducible quadratic $x^2+x+2$ are: $$x_{1,2} = \dfrac{-1+\sqrt{-7}}{2}.$$ Therefore, the splitting field of $p(x)$ is $\mathbb{Q}(\sqrt{7}, i).$ Since this is a biquadratic extension and none of $i, \sqrt{7}$ and $\sqrt{7}i$ are squares, the Galois group is then Klein-$4$ group.
However, I found two different answers that disagree with mine. First is from the Dummit and Foote. Specifically, on page 612 it states that:
If the cubic polynomial is reducible and it splits to a linear factor and an irreducible quadratic, it's Galois group is group of order $2.$
The second source is here, where it proceeds to conclude that the polynomial is irreducible and also its Galois group is $S_3,$ on page $5.$
What is the correct answer here?
In the interests of moving this question off the unanswered queue, I'm converting my comment to an answer:
The splitting field is actually $\mathbb Q(\sqrt{7}i)$. Also, the second source is clearly wrong, since $2$ is a root.