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