This is a biquadratic polynomial with roots $\pm \sqrt{3}$ and $\pm\sqrt{5}$. So, the splitting field is $\Bbb Q(\pm\sqrt{3}, \pm\sqrt{5})=\Bbb Q(\sqrt{3}, \sqrt{5})$. For the degree part, can I conclude that since this polynomial is irreducible in $\Bbb Q$ that the degree is $4$?
Say for example, that I didn't know about tha result, then how would I go about proving it? I am aware of the tower law, but not sure how to utilise it properly in this case.
Thanks in advance.
The degree is indeed $4$, but not because the polynomial is irreducible: $$ x^4-8x^2+15=(x^2-3)(x^2-5) $$ The degree can be computed by noting that $$ [\mathbb{Q}(\sqrt{3},\sqrt{5}):\mathbb{Q}(\sqrt{3})]=2 $$ because $\sqrt{5}\notin\mathbb{Q}(\sqrt{3})$ (prove it).