4. Пусть (х) = х² + x + 1. Найдите ф(0) + (1) + φ(2) + φ(3).

φ(0) = 0² + 0 + 1 = 0 + 0 + 1 = 1

φ(1) = 1² + 1 + 1 = 1 + 1 + 1 = 3

φ(2) = 2² + 2 + 1 = 4 + 2 + 1 = 7

φ(3) = 3² + 3 + 1 = 9 + 3 + 1 = 13

φ(0) + φ(1) + φ(2) + φ(3) = 1 + 3 + 7 + 13 = 24

Ответ: 24