Factor to find all the zeros x^3-5x^2+16x-30=0

factors of 30:

1,2,3,5,15,30
of which we'd determine that x=3 is a solution such that
x^3-5x^2+16x-30=(x-3)(x^2-2x+10)=0
but the quadratic is not factorable over the reals - so we'd need the quadratic formula to determine that x=1+3i, and x=1-3i are the other two factors. Thus the zeros are:
{3, 1-3i, 1+3i}