I am trying to find the splitting field of the polynomial $p(x)=4 x^{8} - 8x^{6} - 12x^{4} - 40x^{2} - 160$ over $\mathbb{Q}$.
Here’s my attempt:
Let $u=x^{2}$. We can factor out a $4$ so we get $4(u^{4} - 2u^{3} - 3u^{2} - 10u - 40)$. Using the Rational Roots Theorem, we can divide $p(x)$ by some obvious factor. One of the factors is $u + 2$.
$$\frac{u^{4}-2u^{3}-3u^{2}-10u-40}{u+2} = u^{3}-4u^{2}+5u-20$$
So we have $(u+2)(u^{3}-4u^{2}+5u-20)$. We can factor by grouping and get $4(u+2)(u-4)(u^{2}+5)$. Setting this equal to zero, we get our quasi solutions $x= \pm i \sqrt{2},\pm 2, \pm (-5)^{1/4}$.
Letting $\zeta_{4}^{k}=e^{2ki\pi/4}$, we have
$$\begin{split} \zeta_4&=e^{2i\pi/4}=i\\ \zeta_4^{2}&=e^{4i\pi/4}=-1\\ \zeta_4^3&=e^{6i\pi/4}=i\\ \zeta_4^4&=1 \end{split}$$
So our final solutions are $x= \pm i\sqrt{2},\pm 2, \pm \sqrt[4]{-5},\zeta_{4} \left(\pm (-5)^{1/4}\right)$.
Does this mean our splitting field is $\text{Split}_{\mathbb{Q}}\big(p(x)\big)= \mathbb{Q}\left(2,\sqrt{2},(-5)^{1/4},\zeta_{4}\right)?$
As you have shown we have $$ p(x)=4(x^4 + 5)(x^2 + 2)(x + 2)(x - 2). $$ Now we can read off the roots and find the splitting field. We can write $\Bbb Q(i,\sqrt{2},\sqrt[4]{5})$. The degree is $16$ over $\Bbb Q$.