Polynomial f (x) = A0 + a1x + a2x ^ 2 +... + anx ^ n, it is proved that f (x) = 0 has n + 1 different roots, then f (x) is equal to 0

Polynomial f (x) = A0 + a1x + a2x ^ 2 +... + anx ^ n, it is proved that f (x) = 0 has n + 1 different roots, then f (x) is equal to 0

F (x) = 0 has n + 1 different roots, let x0, x1, X2 So f (x0) = 0, f (x1) = 0 , f (xn) = 0, that is, A0 + A1 (x0) + A2 (x0) ^ 2 +... + an (x0) ^ n = 0a0 + A1 (x1) + A2 (x1) ^ 2 +... + an (x1) ^ n = 0 a0+a1(xn)+a2(xn)^2+...+an(xn)^n=0...

Let Ao + A1 / 2 + +An / N + 1 = 0, f (x) = Ao + a1x + +Anx ^ n has at least one zero point in (01)

The original function of F (x) is f (x) = AOX + a1x ^ 2 / 2 + a2x ^ 3 / 3 +... Anx ^ (n + 1) / (n + 1) f (0) = 0f (1) = A0 + A1 / 2 +... An / (n + 1) = 0. According to Lagrange mean value theorem, there is a P in (0,1) such that f '(P) = f (1) - f (0) / (1-0) = 0, that is, f (P) = 0, so f (x) has at least

[mathematical analysis] let P (x) be a polynomial, that is, P (x) = anx ^ n +... + a1x + A0
Let P (x) be a polynomial, that is, P (x) = anx ^ n +... + a1x + A0
(1) There exists x0 > 0, such that P (x) is strictly monotone in (- ∞, x0], [XO, + ∞) respectively
(2) If n is even, then when an > 0, P (x) must have a minimum

(1) F (x) = P (x + 1) - P (x) = an (x + 1) ^ n +... + A1 + A0 - [anx ^ n +... + a1x + A0] = an (x + 1) ^ n, when a > 0, x0, (x + 1) ^ n