Question 668545
We know that all imaginary roots come in conjugate pairs. 3i comes in the form 0 + 3i and so 0-3i is its conjugate.  Similarly 2-i has conjugate pair 2+i.

This means we can form a product of linear factors like so:

(x+3i)(x-3i)(x-(2-i)(x-(2+i)

FOIL out each pair of binomials by their conjugates to get two nice easy quadratics.

(x^2+9)(x^2-4x+5)  = x^4 - 4x^3 + 14x^2 - 36x + 45