\(x^2+2x+6\)\(⋮\)\(x+4\)
\(\Leftrightarrow\)\(\left(x-2\right)\left(x+4\right)+14\)\(⋮\)\(x+4\)
Ta thấy \(\left(x-2\right)\left(x+4\right)\)\(⋮\)\(x+4\)
nên \(14\)\(⋮\)\(x+4\)
hay \(x+4\)\(\inƯ\left(14\right)=\left\{\pm1;\pm2;\pm7\right\}\)
Ta lập bảng sau:
\(x+4\) \(-7\) \(-2\) \(-1\) \(1\) \(2\) \(7\)
\(x\) \(-11\) \(-9\) \(-5\) \(-3\) \(-2\) \(3\)
Vậy....