Réponse :
f(x) = 2 x² + 4 x - 16
1) donner la forme canonique de f(x)
f(x) = 2 x² + 4 x - 16
= 2(x² + 2 x - 8)
= 2(x² + 2 x - 8 + 1 - 1)
= 2(x²+2 x + 1 - 9)
= 2((x + 1)² - 9)
donc f(x) = 2(x + 1)² - 18
2) donner la forme factoriser de f(x)
f(x) = 2((x + 1)² - 9)
= 2((x + 1)² - 3²)
= 2((x + 1 + 3)(x + 1 - 3)
f(x) = 2(x + 4)(x - 2)
3) a) résoudre f(x) = 0
f(x) = 2(x + 4)(x - 2) = 0 ⇔ x + 4 = 0 ⇔ x = - 4 ou x = 2
b) f(x) = - 16 ⇔ f(x) = 2 x² + 4 x - 16 ⇔ 2 x² + 4 x = 0 ⇔ 2 x (x + 2) = 0
⇔ 2 x = 0 ⇔ x = 0 ou x + 2 = 0 ⇔ x = - 2
c) f(x) = - 18
f(x) = 2 x² + 4 x - 16 = - 18 ⇔ 2 x² + 4 x + 2 = 0 ⇔ 2 (x² + 2 x + 1) = 0
⇔ 2 (x + 1)² = 0 ⇔ x + 1 = 0 ⇔ x = - 1 racine double
Explications étape par étape
