Question 1069924
If your coefficients must be integers (which are real numbers),
the conjugate of every complex zero must also be a zero.
The conjugate of {{{1+2i}}} is {{{1-2i}}} .
The polynomial in {{{x}}} of least degree whose zeros include
{{{-1}}} , {{{1+2i}}} , and {{{1-2i}}} will be
a polynomial of degree {{{3}}} which can be written as
{{{P(x)=a*(x-(-1))*(x-(1+2i))*(x-(1-2i))}}} for some real number {{{a<>0}}} .
We want to work on that to simplify it and show how we can make the coefficients not only real numbers, but also integers.
{{{P(x)=a*(x+1)*(x-1-2i)*(x-1+2i)}}}
{{{P(x)=a*(x+1)*((x-1)-2i)*((x-1)+2i)}}}
{{{P(x)=a*(x+1)*((x-1)^2-(2i)^2)}}}
{{{P(x)=a*(x+1)*((x^2-2x+1)-2^2*(i)^2)}}}
{{{P(x)=a*(x+1)*(x^2-2x+1+4)}}}
{{{P(x)=a*(x+1)*(x^2-2x+5)}}}
So far the coefficients in this factored form are
{{{1}}} , {{{-2}}} , and {{{5}}} , which are integers,
so it looks like we can just use {{{a=1}}} ,
or any non-zero integer we want for {{{a}}} . 
{{{P(x)=a*(x^3-2x^2+5x+x^2-2x+5)}}}
{{{P(x)=a*(x^3-x^2+3x+5)}}}
Since the coefficients of {{{x^3-x^2+3x+5}}}
are the integers {{{1}}} , {{{-1}}} , {{{3}}} and {{{5}}} ,
the simplest such polynomial, with {{{a=1}}} is
{{{P[1](x)=highlight(x^3-x^2+3x+5)}}} .