Z twierdzenia Bezout otrzymujemy: W(5) = 1 W(-3) = -7 Wielomian W(x) możemy zapisać też jako: W(x) = (x - 5) * (x + 3) * Q(x) + ax + b, gdzie Q(x) to pewien wielomian. W(5) = 1 => (5 - 5) * (5 + 3) * Q(5) + a * 5 + b = 1 => 5a + b = 1 W(-3) = -7 => (-3 - 5) * (-3 + 3) * Q(-3) + a * (-3) + b = -7 => -3 a + b = -7 Rozwiązujemy układ równań: 5a + b = 1 -3a + b = -7 5a + b = 1 b = 3a - 7 5a + 3a - 7 = 1 b = 3a - 7 a = 1 b = -4 Zatem W(x) = (x - 5) * (x + 3) * Q(x) + (x - 4) W(x) = (x² - 2x - 15) * Q(x) + (x - 4) Reszta z dzielenia wielomianu W(x) przez wielomian P(x) wynosi x - 4.
W(5) = 1 W(-3) = -7 albo tak: W(x) = (x - 5) × (x + 3) × Q(x) + ax + b, W(5) = 1 => (5 - 5) × (5 + 3) × Q(5) + a × 5 + b = 1 => 5a + b = 1 W(-3) = -7 => (-3 - 5) × (-3 + 3) × Q(-3) + a × (-3) + b = -7 => -3 a + b = -7 _____________________________ 5a + b = 1 -3a + b = -7 ___________________ 5a + b = 1 b = 3a - 7 ___________________ 5a + 3a - 7 = 1 b = 3a - 7 ___________________ a = 1 b = -4 ____________________________ W(x) = (x - 5) × (x + 3) × Q(x) + (x - 4) W(x) = (x² - 2x - 15) × Q(x) + (x - 4) Reszta z dzielenia wielomianu wynosi x - 4.
