how do I find a 4th degree polynomial that has zeros; i,3i with f(-1)=60

Question

I know the conjucants are -i and -3i

(x-i)(x-3i) (x+i)(x=3i)

Answer 1

So, multiply it out to get

f(x) = a(x^2+1)(x^2+9) = a(x^4+10x^2+9)

Now, f(-1) = a(1+10+9) = 20a, so we need a=3 and thus

f(x) = 3(x^4+10x^2+9)

BTW, that's "conjugates"