2x2+3x+2=2x2+2x+x+2=2x(x+1)+(x+2)
Vì 2x(x+1) chia hết cho x+1
=> x+2 chia hết cho x+1
Ta có: x+2=x+1+1
x nguyên => x+1 nguyên => x+1 thuộc Ư (1)={-1;1}
Với x+1=1 => x=0
Với x+1=-1 => x=-2
Vậy x={0;-2} thì 2x2+3x+2 chia hết cho x+1
Ta có : 2.x2+3x+2 \(⋮\)x+1
=) [ 2.x2+3x+2 - ( x + 1 ) ] \(⋮\)x+1
=) [ 2.x2+3x+2 - 3( x + 1 ) ] \(⋮\)x+1
=) [ 2.x2+3x+2 - (3x + 3 ) ] \(⋮\)x+1
=) 2.x2+3x+2 - 3x - 3 \(⋮\)x+1
=) 2.x2 - 1 \(⋮\)x+1=) [(2.x2 - 1-(x+1)] \(⋮\)x+1=) [(2.x2 - 1-x(x+1)] \(⋮\)x+1=) [(2.x2 - 1-(x2+x)] \(⋮\)x+1=) [(2.x2 - 1-2(x2+x)] \(⋮\)x+1=) [(2.x2 - 1-(2x2+2x)] \(⋮\)x+1=) [(2.x2 - 1-(2x2+2x)] \(⋮\)x+1=) 2.x2 - 1-2x2-2x \(⋮\)x+1=) -1 - 2x \(⋮\)x+1=) [(-1 - 2x+(x+1)] \(⋮\)x+1=) [(-1 - 2x+2(x+1)] \(⋮\)x+1=) [(-1 - 2x+(2x+2)] \(⋮\)x+1=) -1 - 2x+2x+2 \(⋮\)x+1=) 1 \(⋮\)x+1sau đó bạn tìm x