${\displaystyle {}P(2)=8-4-10+6=0\,,}$

somit ist ${\displaystyle {}2}$ eine Nullstelle von ${\displaystyle {}P}$.

${\displaystyle {}X^{3}-X^{2}-5X+6=(X-2)(X^{2}+X-3)\,.}$

Es geht also noch um die Nullstellen von ${\displaystyle {}X^{2}+X-3}$. Diese sind

${\displaystyle {}x_{1,2}={\frac {\pm {\sqrt {1+12}}-1}{2}}={\frac {\pm {\sqrt {13}}-1}{2}}\,.}$

${\displaystyle {}P=(X-2){\left(X-{\frac {{\sqrt {13}}-1}{2}}\right)}{\left(X+{\frac {{\sqrt {13}}+1}{2}}\right)}\,.}$