Question

[Nguyễn Nguyên Trung]

Tìm n nguyên để 6n² + n - 8 chia cho 2n + 1

Answer 1

[Pham Van Hung]

\(6n^2+n-8=6n^2+3n-2n-1-7\)

                        \(=3n\left(2n+1\right)-\left(2n+1\right)-7\)

                        \(=\left(2n+1\right)\left(3n-1\right)-7\)

Để \(6n^2+n-8⋮\left(2n+1\right)\Rightarrow-7⋮\left(2n+1\right)\)

\(\Rightarrow2n+1\inƯ\left(-7\right)=\left\{-7;-1;1;7\right\}\)

\(\Rightarrow n\in\left\{-4;-1;0;3\right\}\)