Odpowiedź:
a)
w(x) = 2x³ - 5x² + 2x + 1 = (x - 1)(2x² - 3x - 1)
(x - 1)(2x² - 3x - 1) = 0
x - 1 = 0∨ 2x² - 3x - 1=0
x = 1
2x² - 3x - 1 = 0
a = 2 , b = - 3 , c = - 1
Δ = b² - 4ac = (- 3)² - 4 * 2 * (- 1) = 9 + 8 = 17
√Δ = √17
x₁ = ( - b - √Δ)/2a = (3 - √17)/4
x₂ = ( - b + √Δ)/2a = (3 + √17)/4
x₀ = { 1 , (3 - √17)/4 , (3 + √17)/4}
b)
w(x) = x³ + 3x² - x - 6 = (x + 2)(x² + x - 3)
(x + 2)(x² + x - 3) = 0
x + 2 = 0 ∨ x² + x - 3 = 0
x = - 2
x² + x - 3 = 0
a = 1, b = 1, c = - 3
Δ = b² - 4ac = 1² - 4 * 1 * (- 3) = 1 + 12 = 13
√Δ = √13
x₁ = ( - b - √Δ)/2a = (- 1 - √13)/1 = 1 + √13
x₂ = ( - b + √Δ)/2a = (- 1 + √13)/1 = √13 - 1
x₀ = { - 2 , (1 - √13) , (√13 - 1)}
