Question 40800
Yes.  In fact 0 is a "tripple zero" of the function.
You can continue factoring using difference of two squares to get linear factors as required in the problem:
{{{x^3(4x^2-3) = x^3(2x+sqrt(3))(2x-sqrt(3))}}}.  To get the zeros, set the whole expression equal to zero.  Since the product equals zero, one of the factors equals zero, so set each factor equal to zero and solve for x.  The zeros are:
0, {{{-sqrt(3)/2}}}, and {{{sqrt(3)/2}}}