Such a polynomial couldn't exist: given a polynomial [tex]p(x)[/tex] with real coefficients, if a complex number [tex]z[/tex] is a root of the polynomial, i.e. [tex]p(z)=0[/tex], then also the conjugate [tex]\overline{z}[/tex] must be a solution: [tex]p(\overbar{z})=0[/tex]
The closest we can get to your question is a polynomial with real coefficients and roots
[tex]x=2,\quad x=\pm 3i,\quad x=-1[/tex]
This polynomial would be
[tex](x-2)(x-3i)(x+3i)(x+1) = (x-2)(x^2+9)(x+1)=x^4 - x^3 + 7 x^2 - 9 x - 18[/tex]
