a) Ta có:
2x3−5x2+8x−3
=2x3−x2−4x2+2x+6x−3
=x2(2x−1)−2x(2x−1)+3(2x−1)
=(2x−1)(x2−2x+3)
=2(x−21)(x2−2x+3)
b) Ta có:
(x2+x)2+4x2+4x−12
=x4+2x3+x2+4x2+4x−12
=x4+2x3+5x2+4x−12
=x4+2x3+5x2+10x−6x−12
=x3(x+2)+5x(x+2)−6(x+2)
=(x+2)(x3+5x−6)
=(x+2)(x3−x2+x2−x+6x−6)
=(x+2)[x2(x−1)+x(x−1)+6(x−1)]
=(x+2)(x−1)(x2+x+6)