Question 759889
{{{x[1]=1-3i}}}

then
{{{x[2]=1+3i}}}


{{{
(x-1-3i)(x-1+3i)=x^2-x-3ix-x+1+3i+3ix-3i+9
}}}

so one factor must be

{{{
x^2-2x+10
}}}

after long division...

{{{P(x)/(x^2-2x+10)=x+1}}}


zeros:


{1-3i, 1+3i, -1}}}