Để A chia hết cho 2n+1 thì \(2n^2+n-6⋮2n+1\)
=>\(n\left(2n+1\right)-6⋮2n+1\)
=>\(-6⋮2n+1\)
=>\(2n+1\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\)
mà 2n+1 lẻ(do n nguyên)
nên \(2n+1\in\left\{1;-1;3;-3\right\}\)
=>\(2n\in\left\{0;-2;2;-4\right\}\)
=>\(n\in\left\{0;-1;1;-2\right\}\)