If f(x) = (x-a)^2(x-b)^2(x-c)^2 for a<b<c, prove that f''(x) has four zeroes in (a,c) and that f(v) has one zero in (a,c). Please don't differentiate this more than once!

I took the derivative and got f'(x) = 2(x-a)(x-b)^2(x-c)^2 + 2(x-b)(x-a)^2 (x-c)^2 + 2(x-c)(x-a)^2 (x-b)^2

I dont see how the number of zeroes has decreased from six from the original equation to five in this. Did I do the derivative correctly?

Thanks for the help.