Question

$Fie -. x- . apartine. -Z . Aratati : a= x^{5} -5 x^{3} +4x+120 , divide -120$

Answer 1

$a=x^5-5x^3+4x+120 \\ \\ a=x^5-x^3-4x^3+4x+120 \\ \\ a=x^3(x^2-1)-4x(x^2-1)+120 \\ \\ a=x(x^2-1)(x^2-4)+120 \\ \\a=x(x-1)(x+1)(x-2)(x+2) +120 \\ \\ a=(x-2)(x-1)x(x+1)(x+2)+120$

$(x-2)(x-1)x(x+1)(x+2)~este~produsul~a~5~numere~intregi \\ \\ consecutive \Rightarrow~sunt~cel~putin~doua~numere~pare,~dintre~care \\ \\ unul~este~divizibil~doar~cu~2,~iar~altul~este~divizibil~cu~4,~astfel \\ \\ ~produsul~lor~este~divizibil~cu~8.~Exact~un~numar~este~divizibil~cu~3. \\ \\ Exact~un~numar~este~divizibil~cu~5.~ \\ \\ Deci~(x-2)(x-1)x(x+1)(x+2) ~ \vdots~(8 \cdot 3 \cdot 5)~(adica~120...) \\ \\ Deci~a=M_{120}+120 ~ \vdots ~120.$