Ta có:
`(ax+b)(x^2 + cx + 1) = x^3 - 3x +2`
`ax.x^2 + ax . cx +ax + bx^2 + bcx + b = x^3- 3x+2`
`ax^3 + acx^2 +ax + bx^2 + bcx + b= x^3 - 3x+ 2`
`ax^3 + (acx^2 + bx^2) + (ax + bcx) + b = x^3- 3x+2`
`ax^3 + x^2(ac + b) + x(a+bc) = x^3 - 3x +2`
`=> ax^3 = x^3 ; x^2(ac+b) =0; x(a+bc) = -3x ; b= 2`
`=>a =1 ; ac+b =0; a+bc = -3 ; b= 2`
Thay `a = 1; b= 2` vào `ac +b =0`
`=> c + 2 =0`
`=> c= 0-2`
`=> c= -2`
Vậy `a = 1 ; b= 2; c=-2`
