Réponse :
Bonjour,
1) On remplace x par –1
A(–1) = –3 × (–1)² + 2 × (–1) + 1
A(–1) = –3 × 1 – 2 + 1
A(–1) = –3 – 2 + 1
A(–1) = –4
2) Identité remarquable:
(2x + 1)(2x – 1) ⇔ (2x)² – 1² = 4x² – 1
B(x) = 4(x + 1) + (2x + 1)(2x – 1)
= 4x + 4 + 4x² – 1
= 4x² + 4x + 3
3) C(x) = 3(x + 2) + (x + 2)(2x + 3)
= (x + 2)[3 + (2x + 3)]
= (x + 2)(3 + 2x + 3)
= (x + 2)(2x + 6)
= 2(x + 2)(x + 3)
4) (2x – 3)(x + 1) = 0
Or A × B = 0 ⇔ A = 0 ou B = 0
2x – 3 = 0
2x = 3
x = 3/2
x + 1 = 0
x = –1
Donc S = {–1 ; 3/2}
