Question 850065
That includes zero at 2-i.


{{{p=(x+1)(x-(2+i))(x-(2-i))}}}
{{{p=(x+1)((x-2)-i)((x-2)+i)}}}
{{{p=(x+1)((x-2)^2+1)}}}--------Do you understand that step?
{{{p=(x+1)(x^2-4x+4+1)}}}
{{{highlight(p=(x+1)(x^2-4x+5))}}}
{{{(x+1)x^2-4x(x+1)+5x+5}}}
{{{x^3+x^2-4x^2-4x+5x+5}}}
{{{highlight(p=x^3-2x^2+x+5)}}}