Exercice n°3 : Factoriser en utilisant l'identité remarquable : a - b = (a + b)(a - b) A=100x²-1 B=4x²-9 C=144-16x D=(3x+2)²-9 E=(3-4x)²-25 aidez moi svp​

Bonjour,

100x² - 1 = (10x)²-1² = (10x+1)(10x-1)

4x²-9 = (2x)²-3² = (2x+3)(2x-3)

144-16x² = 12²-(4x)² = (12+4x)(12-4x)

(3x+2)²-9 = (3x+2)²-3² = (3x+2+3)(3x+2-3) = (3x+5)(3x-1)

(3-4x)²-25 = (3-4x)²-5² = (3-4x+5)(3-4x-5) = (8-4x)(-4x-2)

Réponse :

Explications étape par étape

Bonsoir

Factoriser en utilisant l'identité remarquable : a - b = (a + b)(a - b)

A=100x²-1

A = (10x)^2 - 1^2

A = (10x - 1)(10x + 1)

B=4x²-9

B = (2x)^2 - 3^2

B = (2x - 3)(2x + 3)

C=144-16x^2 (il manque un carré ?)

C = 12^2 - (4x)^2

C = (12 - 4x)(12 + 4x)

C = 4(3 - x) * 4(3 + x)

C = 16(3 - x)(3 + x)

D=(3x+2)²-9

D = (3x + 2)^2 - 3^2

D = (3x + 2 - 3)(3x + 2 + 3)

D = (3x - 1)(3x + 5)

E=(3-4x)²-25

E = (3 - 4x)^2 - 5^2

E = (3 - 4x - 5)(3 - 4x + 5)

E = (-4x - 2)(-4x + 8)

E = -2(2x + 1) * -4(x - 2)

E = (-2) * (-4)(2x + 1)(x - 2)

E = 8(2x + 1)(x - 2)