Splitting field of a polynomial $f(x) =(x^2-3)(x^2-5)(x^5-1)$ over $\mathbb{Q}$. I was considering the splitting field E of the polynomial $f(x) =(x^2-3)(x^2-5)(x^5-1)$ over $\mathbb{Q}$. 
I expected $E=\mathbb{Q}(\sqrt{5},\sqrt{3},\omega)$,  where $\omega=e^{\frac{2\pi i}{5}}$.
But I saw a textbook that claimed $E=\mathbb{Q}(\sqrt{3},\omega)$; believing that $\pm\sqrt{5}\in \mathbb{Q}(\omega)$.
Can someone please show me how the latter answer is true if it is actually true? Thanks

Note that $$\begin{align}\frac{x^5-1}{x-1}&=x^4+x^3+x^2+x+1=x^2\,\left(t^2+t-1\right)
\\&=x^2\,\left(t-\frac{-1+\sqrt{5}}{2}\right)\left(t-\frac{-1-\sqrt{5}}{2}\right)\,,\end{align}$$
where $t:=x+\dfrac{1}{x}$.  Therefore, $\sqrt{5}$ is already in the field $\mathbb{Q}(\omega)$.
