One manifestation involves primitive solutiins for $x^2+5y^2$, where $x$ and $y$ are whole numbers. "Primitive" means $x$ amd $y$ are relatively prime.
We cannot in general factor such numbers into prime factors having the form above. For instance, we have
$1189=41×29\implies8^2+(5×15^2)=[6^2+(5×1^2)][3^2+(5×2^2)]$
but
$6=3×2\implies1^2+(5×1^2)=\text{no }x^2+5y^2\text{ factors}$
This shows up as $1189$ actually having a unique factorization in $\mathbb{Z}[\sqrt{-5}]$ but $6$ famously does not. The nonunique factorization of $6$ goes along with it having prime factors that fail to have the form $x^2+5y^2$.
In class field theiry we learn that $\mathbb{Z}[\sqrt{-5}]$ has class number $2$. With domains corresponding to positive definite quadratic forms this implies that full factorability of numbers having the given quadratic form ($x^2+5y^2$) can be gained by introducing a second quadratic form. In the case of $x^2+5y^2$ we may choose $2x^2+2xy+3y^2$ as the second form. Then we can factor such numbers as $6$:
$6=3×2\implies1^2+(5×1^2)=[(2×0^2)+(2×0×1)+(3×1^2)][(2×1^2)+(2×1×0)+(3×0^2)]$
Introducing the second quadratic form to repair the prime factorization of $x^2+5y^2$ numbers corresponds to adding another set of algebraic integers to $\mathbb{Z}[\sqrt{-5}]$ such that the second set has norms with this alternate form. One such set is given by $\frac12(a\sqrt2+b\sqrt{-10})$. With this augmentation $\mathbb{Z}[\sqrt{-5}]$ is converted to a UFD.