# Thread: Irreducible factorization in number ring.

1. ## Irreducible factorization in number ring.

In the quadratic number ring $Z[\sqrt{2}]$ find an irreducible factorization of the elements $4+\sqrt{2}$ and $6+\sqrt{2}$.

2. Originally Posted by featherbox
In the quadratic number ring $Z[\sqrt{2}]$ find an irreducible factorization of the elements $4+\sqrt{2}$ and $6+\sqrt{2}$.
$4+\sqrt{2}=(2+\sqrt{2})(3-\sqrt{2})$ and $6+\sqrt{2}=(2+\sqrt{2})(5-2\sqrt{2}).$ note that, since the norm of the factors are prime numbers, they are irreducible.