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Michonne15
@Michonne15
January 2021
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Bonsoir, je bloque sur une question, la 3. Je n'y arrive pas du tout, pouvez-vous m'aider s'il vous plaît?
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scoladan
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Bonjour,
f(x) = 2x³ - x² - 9x - 6
1) f(-1) = -2 - 1 + 9 - 6 = 0
2) (x + 1)(ax² + bx + c)
= ax³ + (a + b)x² + (c + b)x + c
Par analogie des termes de même degré :
a = 2
a + b = -1
b + c = -9
c = -6
soit a = 2, b = -3 et c = -6
⇒ f(x) = (x+1)(2x² - 3x - 6)
3) f(x) = 0
⇔ x = -1 ou 2x² - 3x - 6 = 0
Δ = (-3)² - 4x2x(-6) = 9 + 48 = 57
donc 2 autres solutions :
x = (3 - √57)/4 soit environ -1,137
et x = (3 + √57)/4 soit environ 2,637
Et donc f(x) = (x+1)(x - (3 - √57)/4)(x - (3 + √57)/4)
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Verified answer
Bonjour,f(x) = 2x³ - x² - 9x - 6
1) f(-1) = -2 - 1 + 9 - 6 = 0
2) (x + 1)(ax² + bx + c)
= ax³ + (a + b)x² + (c + b)x + c
Par analogie des termes de même degré :
a = 2
a + b = -1
b + c = -9
c = -6
soit a = 2, b = -3 et c = -6
⇒ f(x) = (x+1)(2x² - 3x - 6)
3) f(x) = 0
⇔ x = -1 ou 2x² - 3x - 6 = 0
Δ = (-3)² - 4x2x(-6) = 9 + 48 = 57
donc 2 autres solutions :
x = (3 - √57)/4 soit environ -1,137
et x = (3 + √57)/4 soit environ 2,637
Et donc f(x) = (x+1)(x - (3 - √57)/4)(x - (3 + √57)/4)