Answer
$\pm3 i,\pm2 i$
$f(x)=(x+3i)(x-3i)(x+2i)(x-2i)$
Work Step by Step
Step 1. Factor the function to get a zero $f(x)=(x^2+9)(x^2+4)$
Step 2. Solve $x^2+9=0$ to get $x=\pm3 i$
Step 3. Solve $x^2+4=0$ to get $x=\pm2 i$
Step 4. Thus $f(x)=(x+3i)(x-3i)(x+2i)(x-2i)$
