Odpowiedź:
a)
(x³ + 125)(x²- 64) = 0
x³ + 125 = 0 ∨ (x - 8)(x + 8) = 0
(x + 5)(x² - 5x + 25) = 0 ∨ (x - 8)(x + 8) = 0
x = - 5 ∨ x² - 5x + 25 = 0 ∨ x = 8 ∨ x = - 8
x² - 5x + 25 = 0
a = 1, b = - 5 ,c = 25
Δ = b² - 4ac = (- 5)² - 4 * 1 * 25 = 25 - 100 = - 75
Δ < 0 więc brak miejsc zerowych
x = - 5 , x = 8 , x = - 8
b)
(x² - 4x + 4)(x³-5x² + 6x)= 0
x(x²-4x+4)(x²-5x + 6)= 0
x = 0 ∨ x² - 4x + 4 = 0 ∨ x² - 5x + 6 = 0
x² - 4x + 4 = 0
a = 1 , b = - 4 , c = 4
Δ = b² - 4ac = (- 4)² - 4 * 1 * 4 = 16 - 16 = 0
x₁ = x₂ = - b/2a = 4/2 = 2
x² - 5x + 6 = 0
a = 1 , b = - 5 , c = 6
Δ = b² - 4ac = (- 5)² - 4 * 1 * 6 = 25 - 24 = 1
√Δ = √1 = 1
x₁ = ( - b - √Δ)/2a = ( 5 - 1)/2 = 4/2 = 2
x₂ = ( - b + √Δ)/2a = (5 + 1)/2 = 6/2 = 3
x = 0 , x = 2 , x = 3
c)
x³ - 2x² - 6x - 8 = 0
Schemat Hornera (dzielenie wielomianu przez jednomian )
1 - 2 - 6 - 8
4 4 8 8
1 2 2 0
x³ -2x² - 6x - 8 = (x - 4)(x² + 2x + 2) = 0
x - 4 = 0 ∨ x² + 2x+ 2 = 0
x²+ 2x+2 =0
a = 1 , b = 2 , c = 2
Δ = b² - 4ac = 2² - 4 * 1 * 2 = 4 - 8 = - 4
Δ < 0 , więc brak miejsc zerowych
x - 4 = 0
x = 4
