\(2x^2+x-7=2x^2-8+x-2+3=2\left(x^2-4\right)+\left(x-2\right)+3\)
\(=2\left(x-2\right)\left(x+2\right)+\left(x-2\right)+3=\left(x-2\right)\left(2x+5\right)+3\)chia hết cho x-2
mà (x-2)(2x+5) chia hết cho x-2 => 3 chia hết cho x-2
=> \(x-2\inƯ\left(3\right)=\left\{-3;-1;3\right\}\Leftrightarrow x\in\left\{-1;1;5\right\}\)