To do this, you need to factor the polynomial to find its zeros.
The 1st Fundamental Theorem of Algebra says there should be 3 zeros (real or imaginary) as the polynomial is a degree of 3;
Descartes' rule of signs indicates that there should be two positive roots or two imaginary roots, as there are 2 changes in sign, and 1 negative root, as there is one change in sign when you replace x with -x.
So we should see roots in one of these two forms:
(ex ± a + bi)(ex ± a - bi)(cx + d) or
(dx - a)(ex - b)(fx + c)
Using the rational root theorem, if there are rational roots, they have to be in the form of the factors of 36 the last coefficient) divided by the factors of 3 (the first coefficient). So either 1, 2, 3, 4, 6, 9, 12, 18, or 36 divided by either 1 or 3. I would look for the negative root first because it has to be rational, so the possibilities are x= to -1/3, -2/3, -1, -4/3, -2, -3, -4, -6, -9, -12, -18, or -36. If you plug in each these values into the polynomial for x, you will find one of them will produce a zero for the value of the polynomial; then, you can use either long division of polynomials or synthetic division to find the remainder of the polynomial, after factoring out the negative root.
Once you pull out the real, negative root, the remainder will be a polynomial of degree 2 (i.e. a quadratic), which you can then apply the quadratic formula to in order to locate the last two roots (positive or imaginary).
Using a graphing calculator, you can have it graph the function and you can see which negative value of x is where the function crosses the x-axis to find the negative root; you can also check or setup a table to locate values where the function equals zero.
Multiplicity means that a root appears at least twice (it could be more); in this case it would mean that the positive root would be a zero twice, e.g. factors of the polynomial would be (x-6)(x-6) or x^2-12x+36.
Feel free to follow up with me or ask another question if you do not understand any of these steps. And I can email you more specifics about any step.