Réponse :
factoriser à l'aide d'une identité remarquable
les expressions A , B , C , D , E , F , G , H et I sont des identités remarquables de la forme: a² - b² = (a+b)(a-b)
A = x² - 1 ⇔ A = x² - 1² = (x+1)(x - 1)
B = 4 x² - 9 ⇔ B = (2 x)² - 3² = (2 x +3)(2 x - 3)
C = (3 x + 1)² - 25 ⇔ C = (3 x + 1)² - 5² = (3 x + 1 + 5)(3 x + 1 - 5)
= (3 x + 6)(3 x - 4)
= 3(x + 2)(3 x - 4)
D = (4 x - 3)² - x² = (4 x - 3 + x)(4 x - 3 - x)
= (5 x - 3)(3 x - 3)
= 3(x - 1)(5 x - 3)
E = (2 - 5 x)² - 9 x² ⇔ E = (2 - 5 x)² - (3 x)² = (2 - 5 x + 3 x)(2 - 5 x - 3 x)
= (2 - 2 x)(2 - 8 x)
= 4(1 - x)(1 - 4 x)
F = (4 x - 5)² - (3 x + 2)² = (4 x - 5 + 3 x +2)(4 x - 5 - 3 x - 2)
= (7 x - 3)(x - 7)
G = (1 - x)² - (4 x + 3)² = (1 - x + 4 x + 3)(1 - x - 4 x - 3)
= (3 x + 4)(- 5 x - 2)
H = (2 x + 5)² - 4(x - 5)² ⇔ H = (2 x + 5)² - (2(x - 5))²
= (2 x + 5 + 2(x-5))(2 x + 5 - 2(x - 5))
= (2 x + 5 + 2 x - 10)(2 x + 5 - 2 x + 10)
= 15(4 x - 5)
I = 9(7 x - 1)² - 25(3 x + 1)² ⇔ I = (3(7 x - 1))² - (5(3 x + 1))²
= (9(7 x - 1) + 5(3 x + 1))(9(7 x - 1) - 5(3 x + 1))
= 63 x - 9 + 15 x + 5)(63 x - 9 - 15 x - 5)
= (78 x - 4)(48 x - 14)
= 4(39 x - 1)(24 x - 7)
Explications étape par étape
