Question 420134
This problem can be solved in at least a couple of ways.  I show the most direct.  Note that if 1+i is a zero of the polynomial, then so is 1 - i.

{{{p(x) = a(x + 2)(x - (1 + i))(x - (1 - i)) = a(x+2)((x-1)-i)((x-1)+i)

=a(x+2)((x-1)^2  + 1) = a(x+2)(x^2 - 2x  + 2) = a(x^3 - 2x + 4)}}}.

Here, a is any non-zero constant.