Odpowiedź:
Szczegółowe wyjaśnienie:
Zad 1.
a) W(x) = 15x³ -6x² +5x - 2
W(x) = 3x²(5x-2) + 1(5x-2)
W(x) = (5x-2)(3x²+1) = 0
x = 2/5, (w drugim nawiasie wartości zawsze będą ≥ 1)
b) W(x) = -9x⁶ + 18x⁵ -12x⁴
W(x) = -3x⁴(3x²-6x+4) = 0
x1 = 0, Δ = 36 - 48 < 0 brak pierwiastków
c) W(x) = -3x⁴ -7x³ + 6x² + 14x
W(x) = x(-3x³ -7x² + 6x + 14)
W(x) = x(x²(-3x-7) -2(-3x-7))
W(x) = x(-3x-7)(x²-2)
x₁ = 0, x₂ = -7/3, x₃ = √2, x₄ = -√2
d) W(x) = x⁴ -6x² + 9
W(x) = (x²-3)²
x₁ = -√3, x₂ = √3
Zad 2.
a) W(x) = x³ - 7x - 6
Dzielniki wyrazu wolnego: {±6, ±3, ±2, ±1}
W(-1) = 0, W(-2) = 0, W(3) = 0
x ∈ {-2, -1, 3}
b) W(x) = 3x³ + 11x² -3x + 4
Dzielniki wyrazu wolnego: {±4, ±2, ±1}
W(-4) = 0
x = -4
Zad 3.
a) (x+5)(x²-7) = 0
x₁ = -5, x₂ = -√7, x₃ = √7
b) x⁴ - 3x³ = 10x²
x²(x²-3x-10) = 0
x₁ = 0, Δ = 9 + 40 = 49 => √Δ = 7
x₂ = (3-7)/2 = -2
x₃ = (3+7)/2 = 5
c) 2x⁵ -18x³ + x² - 9 = 0
2x³(x²-9) + 1(x²-9) = 0
(x²-9)(2x³+1) = 0
x₁ = -3, x₂ = 3, x₃ = -1/∛2
