Ta có: B . (2x2 – 5x + 1)
= (x2 – 4x – 3) . (2x2 – 5x + 1)
= x2 .(2x2 – 5x + 1) – 4x . (2x2 – 5x + 1) – 3.(2x2 – 5x + 1)
= x2 . 2x2 + x2 . (-5x) + x2 . 1 – [4x . 2x2 + 4x . (-5x) + 4x . 1] – [3.2x2 + 3.(-5x) + 3.1]
= 2x4 – 5x3 + x2 – ( 8x3 – 20x2 + 4x) – (6x2 – 15x + 3)
= 2x4 – 5x3 + x2 – 8x3 + 20x2 - 4x – 6x2 + 15x - 3
= 2x4 + (-5x3 – 8x3) + (x2 + 20x2 – 6x2 ) + (-4x + 15x) – 3
= 2x4 - 13x3 + 15x2 + 11x - 3
=A
Vậy ta có phép chia hết A : B = 2x2 – 5x + 1