a) (x3 – 7x + 3 – x2): (x – 3)
ta có: x3 – 7x + 3 – x2 = x3 – x2 – 7x + 3
Vậy: (x3 – 7x + 3 – x2): (x – 3)
= ( x3 – x2 – 7x + 3) : (x – 3) = x2 + 2x - 1
b) (2x4 – 3x3 – 2 + 6x): (x2 – 2)
ta có: 2x4 – 3x3 – 2 + 6x = 2x4 – 3x3 + 6x – 2
Vậy: (2x4 – 3x3 – 2 + 6x) = (x2 – 2)( 2x2 – 3x + 4) + 6