Question 940564
If the polynomial must have real coefficients ,
having an imaginary root requires
also having the conjugate imaginary number as a root.
If the polynomial has an irrational root,
to have rational coefficients,
it must also have the conjugate irrational number as a root.
So your polynomial must have as roots
{{{2-i}}} , {{{2+i}}} , {{{sqrt(5)}}} , and {{{-sqrt(5)}}} .
The simplest one would be
{{{f(x)=(x-(2-i))*(x-(2+i))*(x-sqrt(5))*(x+sqrt(5))}}}
{{{f(x)=(x-2+i)*(x-2-i)*(x^2-5)}}}
{{{f(x)=((x-2)^2-i^2)*(x^2-5)}}}
{{{f(x)=(x^2-4x+4+1)*(x^2-5)}}}
{{{f(x)=(x^2-4x+5)*(x^2-5)}}}
{{{f(x)=x^4-4x^3+20x-25}}}