Rolle's theorem tells us that here there is a zero of the derivative between
each pair or real roots. Hence for our polynomial we have n-1 zeros of the
But the derivative of our polynomial is a polynomial of degree n-1, and we
have shown that it has n-1 zeros.
There is one more issue that needs to be addressed. CaptainBlank is assuming that f(x) has exactly n roots. Then by using Opalg's argument with CaptainBlank's commentary we see there are n-1 real roots for f'(x) and so these are its only roots for deg f'(x) = n-1. But the situation changes if there is not exactly n real roots. But that (this problem) is not really so bad. Say that f(x) = (x-a)^n*(x-b)^m where 'a' is a root of multiplicity n and 'b' of multiplicity m. Then . Thus, 'a' is a zero (of f'(x)) of multiplicity n-1 and 'b' is a zero of multiplicity 'b' is a zero of multiplicity m-1. Now using Opalg's argument there is another zero between 'a' and 'b'. So far we have (n-1)+(m-1)+1 = n+m-2 zeros (counting multiplicity) but since the number of complex zeros must remain even it means the last zero cannot be complex. Thus, there are n+m-1 zeros altogether (counting multiplicity). Thus, f'(x) has only real zeros. Now in general suppose we have then are zeros of multiplicity . Using Opalg's hint we also have zeros between and so there are zeros. Thus in total with multiplicity we have since the number of complex solutions must be even it means the last remaining zero must be real. Thus has only real zeros.
Consider for example this polynomial has real roots only . We want to show has only real roots, we know that there is a root between -1 and 1 but since degree is 6 it can still be that the other roots are complex conjugates in that case f'(x) will have real coefficients but not real roots. So we need to use what I mentioned above.
Now consider it has 3 real roots as much as its degree so its derivative has zeros between 1 and 2 , 2 and 3 hence it is two real solutions equal to its degree and we can stop right there. But this is just a special case because this polynomial has a many roots as its degree. But in the above case this does not work. Thus we need to consider this possibility in the proof.
I agree that multiplicity must be taken into account, for the reasons that ThePerfectHacker gives. However, if a polynomial has a root of multiplicity n, its derivative will have that same root with multiplicity n–1: . So multiplicity ought not to create any problems.