Question 1058087
You are pretty much on the right track.

{{{ x^3 - 3x^2 = (x^2)(x-3) }}}

So the zeros are:
0  (multiplicity 2)
3  (multiplicity 1)


The higher the multiplicity of a zero, the flatter the function is near that zero.   This is easy to see using Calculus:  for multiplicity of  2 or more, the derivative of f(x) (= rate of change of f(x)) will also have that zero.