P(X) = X¹⁰⁰+X⁵⁰-2X⁴-X³+X+1
P(X) = Q(X)•(X³+X)+(aX²+bX+c)
Considerăm A rădăcina lui X³+X
A³+A = 0 => A(A²+1) = 0 =>
A = 0
sau
A²+1 = 0
(1) A = 0
P(A) = aA²+bA+c
=> 0+0-0-0+0+1 = 0+0+c = c = 1
(2) A²+1 = 0 => A² = -1|^(50) => A¹⁰⁰ = 1
A² = -1|^25 => A⁵⁰= -1
A² = -1|•A => A³ = -A
P(A) = aA²+bA+1
=> A¹⁰⁰+A⁵⁰-2A⁴-A³+A+1=
= aA²+bA+1
=> 1-1-2+A+A+1 = -a+bA+1
=> 2A-1 = bA+1-a
=> b = 2, 1-a = -1 => a = 2
=> R(X) = 2X²+2X+1
