Let a be a root of a polynomial P.
a is a multiple root of P if and only if a is also a root of P' (the first derivative of P)
(1) Since a is a root of P, there exists a polynomial Q such that (see Theorem, here):
P = (x - a)Q
(2) Using the Product Rule of Derivatives (see Lemma 4, here), we know that:
P' = Q + (x-a)Q'
(3) But then (x-a) only divides P' if and only if it also divides Q.
If a polynomial has only simple roots
Its first derivative does not share any of those roots.
(1) Assume that a polynomial P has only simple roots.
(2) Assume that a root a divides both P and P'
(3) Then by Lemma 1 above, a is a multiple root of P.
(4) But by step #1 this is impossible so we reject our assumption in step #2.
If a polynomial P has only simple roots
Then P,P' are relatively prime.
(1) Assume that P,P' are not relatively prime.
(2) Then, they have a common irreducible factor of degree 1.
[Since by definition, two polynomials are relatively prime if their only common factor is of degree 0]
(3) Then there exists a polynomial of the form X-a that divides both P and P' (See Thereom, here)
(4) Then from Lemma 1 above, a is a multiple root of P
(5) But this is impossible, so we reject our assumption in step #1.
- Jean-Pierre Tignol, Galois' Theory of Algebraic Equations, World Scientific, 2001