(x³ + 3) divided by (x - 1) First put placeholder terms for the 2nd and 1st powers of x: x³ + 0x² + 0x - 1 Then 1 | 1 0 0 3 | 1 1 1 1 1 1 4 answer: 1x² + 1x + 1 + or just x² + x + 1 + Edwin