Question 819539
There is one more root; include 6-i as one of them.


{{{P(x)=0=(x-3)(x-(6+i))(x-(6-i))}}}
{{{P(x)=(x-3)(x-6-i)(x-6+i)}}}
{{{(x-3)((x-6)-i)((x-6)+i)}}}, recognize the difference of two squares for the result, and that i*i=-1,
{{{(x-3)((x-6)^2+1)}}}
{{{(x-3)(x^2-12x+36+1)}}}
{{{(x-3)(x^2-12x+37)}}}
and you can finish the multiplication for the two polynomial factors.