Answer
$P(x)=4x^{4}+52x^{2}+36$
Work Step by Step
Since $2i, 3i$ are zeros, then so $-2i, -3i$ by the Conjugate Zeros Theorem. This mean that P(x) have the following form:
$P(x)=a[(x-2i)(x-(-2i))][(x-3i)(x-(-3i))]$
$P(x)=a(x-2i)(x+2i)(x-3i)(x+3i)$
$P(x)=a(x^{2}-4i^{2})(x^{2}-9i^{2})$
$P(x)=a(x^{4}-9x^{2}i^{2}-4x^{2}i^{2}+36i^{4})$
$P(x)=a(x^{4}+13x^{2}+36)$
To make all all coefficients interger, we set $a=4$ and get
$P(x)=4x^{4}+52x^{2}+36$
