Question

Exercitiul 5, cu polinoame.

Answer 1

$P = (2X^3+X+1)^{2017}\\ \\ Q = X^2-X+1 \\ \\ Q(A) = 0 \Rightarrow A^2-A+1 = 0\Big|\cdot(A+1),\quad A\neq -1 \Rightarrow \\ \Rightarrow A^3+1 = 0 \Rightarrow A^3 = -1\\ \\ \\ P = Q\cdot C+R \\\\ (2X^3+X+1)^{2017}= (X^2-X+1)\cdot C+aX+b\\ \\ \text{Facem }X = A \\ \\ (2A^3+A+1)^{2017}=(A^2-A+1)\cdot C+aA+b\\ \Big(2\cdot (-1)+A+1\Big)^{2017}=aA+b\\ \\(A-1)^{2017} =aA+b$

$A^2-A+1 = 0 \Rightarrow A-1 = A^2 \Rightarrow (A-1)^{2017} = A^{4034} \Rightarrow \\ \Rightarrow (A-1)^{2017} = A^{4032}\cdot A^2 \Rightarrow (A-1)^{2017} = (A^3)^{1344} \cdot A^2 \Rightarrow \\ \Rightarrow (A-1)^{2017} = (-1)^{1344}\cdot A^2 \Rightarrow (A-1)^{2017} = A-1 \\ \\ \Rightarrow A-1 = aA+b\Rightarrow a = 1,\,\,b=-1 \Rightarrow \boxed{R = X-1}$