Use the Rational Zero Test to list all possible rational zeros of f. f(x)=x^3-4x^2-4x+16

since the leading coefficient is 1, we only look at the constant terms factors
16:
1, -1, 2, -2, 4, -4, 8, -8, 16, -16
in fact toe actual zeros are -2, 2, and 4

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}

List all possible rational zeros f(x)= x^3-5x^2-4x+20

p/q: 20/1 -> 
+-1, +-2, +-4, +-5, +-10, +-20