If polynomial P(x) has a root of multiplicity m at x = a, then P'(x) has a root of multiplicity m-1 at x = a. Write P(x) = (x-a)^m * Q(x) where Q(a) != 0. Then P'(x) = (x-a)^(m-1)[mQ(x) + (x-a)Q'(x)], and the bracketed term equals mQ(a) != 0 at x = a.

Consequence: If P has roots r1 (mult. m1), r2 (mult. m2), ..., rk (mult. mk), then P' has roots at each ri with multiplicity mi-1, PLUS at least k-1 additional roots (one between each consecutive pair by Rolle's).

Total roots of P': sum(mi-1) + (at least k-1) = sum(mi) - k + k - 1 = n - 1 (since degree of P' is n-1 and sum of mi = n). This accounts for all roots.