Question 1099340
{{{P(z)=(z^2-6z+10)(z-r)}}} where {{{r}}} is the other root.
The independent term is {{{-10r=-20}}} ,
so {{{r=2}}} or we were deceived about {{{3+i}}} being a root.
Let's multiply.
{{{P(z)=(z^2-6z+10)(z-2)=z^3-6z^2+10z-2z^2+12z-20=z^3-8z^2+22z-20}}}
If that polynomial is the same function as {{{P(z)=z^3-kz^2+22z-20 }}}
(same values for every complex value of {{{z}}} ),
then {{{k=8}}} .