Answer
Zeros: $\ \ \pm 3i, \ \ \pm 2i,$
$f(x)=(x-3i)(x+3i)(x-2i)(x+2i)$
Work Step by Step
Degree $4$: there are $4$ complex zeros.
Substituting $t=x^{2}$ try to factor $t^{2}+13t+36$
Two factors of 36 with sum 13 ... are 9 and 4.
$t^{2}+13t+36=(t+9)(t+4)$
$x^{4}+13x^{2}+36=(x^{2}+9)(x^{2}+4)$
Zeros: $\pm 3i, \pm 2i,$
$f(x)=(x-3i)(x+3i)(x-2i)(x+2i)$
