 Solution

For a purely imaginary candidate $z=it$, $t\in \mathbb{R}$, the imaginary part of the polynomial is $2t(t^2-4)$. The candidate $t=0$ is not a root, while $t=\pm 2$ both make the real part zero. The resulting factorization into quadratic polynomials givesThe roots of a polynomial sum to $0+2=2$ and have product $4\cdot 4=16$, as required by Vieta's formulas for a monic quartic with cubic coefficient $-2$ and constant coefficient $16$.