Question 313486
Complex roots only come in complex conjugate pairs.
So if {{{x=-2i}}} is a root so is {{{x=2i}}}.
Same for {{{x=-3i}}}.
So the equation looks like,
{{{f(x)=A(x-2i)(x+2i)(x-3i)(x+3i)}}} where A is any non-zero constant.
{{{f(x)=A(x^2+2ix-2ix-4i^2)(x^2+3ix-3ix-9i^2)}}}
{{{f(x)=A(x^2+4)(x^2+9)}}}
{{{f(x)=A((x^2+4)x^2+(x^2+4)9)}}}
{{{f(x)=A((x^4+4x^2)+(9x^2+36))}}}
{{{f(x)=A(x^4+13x^2+36)}}}