x⁴ - 3x³- 14x² - 20x - 24 = 0
x⁴ + 2x³ - 5x³ - 10x² - 4x² - 8x - 12x - 24 = 0
x³ (x+2) - 5x²(x+2) - 4x(x+2) - 12(x+2) = 0
(x+2)(x³-5x²-4x-12) = 0
(x+2)(x³-6x²+x²-6x+2x-12) = 0
(x+2)[x²(x-6)+x(x-6)+2(x-6)] = 0
(x+2)(x-6)(x²+x+2) = 0
x+2 = 0 → x = -2
lub
x-6 = 0 → x = 6
lub
x²+x+2 = 0
Δ < 0
zatem x²+x+2>0 dla każdego x
Odp. Pierwiastkami wielomianu x⁴ - 3x³- 14x² - 20x - 24 są x = -2 i x = 6.
x⁴ + 2x³ - 3x² - 4x + 4 = 0
x³(x+2) - 3x² - 6x + 2x + 4 = 0
x³(x+2) - 3x(x+2) + 2(x+2) = 0
(x+2)(x³-3x+2) = 0
(x+2)(x³+2x²-2x²-4x+x+2) = 0
(x+2)[x²(x+2)-2x(x+2)+(x+2)] = 0
(x+2)(x+2)(x²-2x+1) = 0
(x+2)²(x-1)² = 0
x₁ = x₂ = -2
x₃ = x₄ = 1
Odp. Wielomian x⁴ + 2x³ - 3x² - 4x + 4 ma cztery pierwiastki, w tym dwa podwójne: x₁ = x₂ = -2 i x₃ = x₄ = 1.
