Yes, the early part of the solution shows that constant term d must be zero so that f(0)=0.

And that leads to the factorisation x(ax^2+bx+c) and this expression=0 from which x=0 (which we are given anyway) or the quadratic ax^2+bx+c=0. There are various ways of solving this. The formula gives us two roots which could be complex. However, we can assume that the factorisation (x-A)(x-B) applies, where A and B are the roots, the zeroes. But, we have to divide through by a first so that we can have this factorised expression: if ax^2+bx+c=0 then x^2+bx/a+c/a=0=x^2-(A+B)x+AB.

We can now equate coefficients: (A+B)=-b/a and AB=c/a. Therefore the product of the zeroes, or roots, is c/a.