Question 868265
If a polynomial with rational coefficients has a non-real zero, it must also have the conjugate non-real zero.
That is to say that if {{{3+i}}} is a zero, {{{3-i}}} must be a zero too.
That means that the factorization of the polynomial must contain the factors
{{{x-(3+i)=x-3-i=(x-3)-i}}} and {{{x-(3-i)=(x-3)+i}}} .
So the polynomial must have as a factor
{{{((x-3)+i)((x-3)-i)=(x-3)^2-i^2=x^2-6x+9-(-1)=x^2-6x+9+1=x^2-6x+10}}} .
The same applied to the zero {{{2-i}}} means that
{{{x-(2-i)=x-2+i=(x-2)+i}}} and {{{x-(2+i)=x-2-i=(x-2)-i}}} area factors,
and the polynomial must have as a factor
{{{((x-2)+i)((x-2)-i)=(x-2)^2-i^2=x^2-4x+4-(-1)=x^2-4x+4+1=x^2-4x+5}}} .
In sum, the polynomial must have for a factor
{{{(x^2-6x+10)(x^2-4x+5)=x^4-10x^3+39x^2-70x+50}}} .
The simplest answer would be
{{{highlight(f(x)=x^4-10x^3+39x^2-70x+50)}}} .
Of course, multiplying that polynomial by any rational number we can get another answer,
like {{{g(x)=3x^4-30x^3+117x^2-210x+150}}} ,
or {{{g(x)=(2/3)x^4-(20/3)x^3+26x^2-(140/3)x+100/3}}} .