Disproving unique factorization of $\mathbb{Q}(\sqrt{-6})$

Question

For homework, we were given a problem that asked to explain why $(\sqrt{-6})(-\sqrt{-6})$ implies $\mathbb{Q}(\sqrt{-6})$ does not have a unique factorization.

I understand the unique part in this case only means unique up to multiplication by a unit, but how does $(\sqrt{-6})(-\sqrt{-6})$ relate to that? Is it that each of the square roots can be factored in more than one way?

Answer 1

We have $$ 10=2\cdot 5=(2-\sqrt{-6})(2+\sqrt{-6}). $$ Now show that these elements are irreducible in $\Bbb Z[\sqrt{-6}]$, so that this ring is not a UFD.