Odpowiedź:
a) 2 x³ + x² - 3 = 0
x = 1 , bo 2*1³ + 1² - 3 = 2 + 1 - 3 = 0
więc
(2 x³ + x² - 3 ) : ( x - 1) = 2 x² + 3 x + 3
2 x² + 3 x + 3 = 0
Δ = 3² - 4*2*3 = 9 - 24 < 0 - brak rozwiązań.
Odp. x = 1
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b) x³ + 5 x² + 6 x + 2 = 0
x = - 1
bo (-1)³ + 5*( -1)² + 6*( -1) + 2 = - 1 + 5 - 6 + 2 = 0
więc
( x³ + 5 x² + 6 x + 2) : ( x + 1) = x² + 4 x + 2
x² + 4 x + 2 = 0
Δ = 4² - 4*1*2 = 16 - 8 = 8 = 4*2
√Δ = 2 √2
x = [tex]\frac{- 4 - 2\sqrt{2} }{2*1} = -2 - \sqrt{2}[/tex] lub x = - 2 + [tex]\sqrt{2}[/tex]
Odp. x = - 2 - [tex]\sqrt{2}[/tex] lub x = - 1 lub x = - 2 + [tex]\sqrt{2}[/tex]
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Szczegółowe wyjaśnienie:
