Ta có:
(x2−2x−3)x=x3−2x2−3x
(x2+x)(x−3)=x3−3x2+x2−3x=x3−2x2−3x
nên (x2–2x–3)x=(x2+x)(x–3)
do đó: x2−2x−3x2+x = x−3x
(x−3)(x2−x)=x3−x2−3x2+3x=x3−4x2+3x
x(x2−4x+3)=x3−4x2+3x
nên (x−3)(x2−x)=x(x2−4x+3)
do đó x−3x = x2−4x+3x2−x
Vậy: x2−2x−3x2+x=x−3x=x2−4x+3x2−x