2i is a zero so (x-2i) is a factor
Since 2i is a zero, so is its conjugate -2i, so (x+2i) is a factor
3i is a zero so (x-3i) is a factor
Since 3i is a zero, so is its conjugate -3i, so (x-i) is a factor

There also could be a constant factor k

So

f(x) = k(x-2i)(x+2i)(x-3i)(x+3i)

f(x) = k[(x-2i)(x+2i)][(x-3i)(x+3i)]

f(x) = k[x²+2ix-2ix-4i²][x²+3i-3i-9i²]

f(x) = k[x²-4i²][x²-9i²]

Replace i² by (-1)

f(x) = k[x²-4(-1)][x²-9(-1)]

f(x) = k[x²+4][x²+9]

f(x) = k[x4+9x2+4x2+36]

f(x) = k[x4+13x2+36]

Since we are given 

f(-2) = 104, so we substitute (-2) for x and set it equal to 104:

f(-2) = k[(-2)4+13(-2)2+36] = 104

             k[16+13(4)+36] = 104

                k[16+52+36] = 104

                     k[104] = 104

                          k = 1

So we substitute 1 for k

f(x) = k[x4+13x2+36]

f(x) = 1[x4+13x2+36]

f(x) = x4+13x2+36

Edwin