Factoriser les expressions algébriques suivantes :

a. C'= (x² + 1)² - 4x²

a²-b²=(a+b) (a-b)

donc C = (x²+1)² - (2x)²

= (x²+1-2x) (x²+1+2x)

= (x²-2x+1) (x²+2x+1)

= (x-1)² (x+1)²

b. D' = (x² - 1)²

= [(x-1) (x+1)]²

c. E' = x² + 2x + 1 + (x + 1)(x + 3)

      = (x+1)² + (x+1) (x+3)

    = (x+1) [(x+1) + (x+3)]

   = (x+1) (2x+4) = 2 (x+1) (x+2)

d. F' = x2 - 2x + 1 + 3x(x - 1)

      = (x-1)² +3x (x-1)

      = (x-1) [(x-1) + 3x] = (x-1) (4x-1)

il faut se rappeler

que a²-2ab+b² = (a-b)²

que a²+2ab+b² = (a+b)²

que a²-b²=(a+b) (a-b)

bonjour ?? de rien :(

Verified answer

Bonsoir !!

Réponse :

a.

[tex]C'= (x^{2} +1)^2-4x^{2} \\C'= (x^{2} +1)^2-(2x)^{2}\\ C'= (x^{2} +1-2x)(x^{2} +1+2x)\\ C'= (x-1)^2(x+1)^2\\C'=[(x-1)(x+1)]^2[/tex]

b.

[tex]D' = (x^{2} - 1)^2\\D'=[(x-1)(x+1)]^2[/tex]

c.

[tex]E' = x^{2} + 2x + 1 + (x + 1)(x + 3)\\E' = (x+1)^2+ (x + 1)(x + 3)\\E'=(x+1)(x+1x+3)\\E'=(x+1)(2x+4)\\E'=2(x+2)(x+1)[/tex]

d.

[tex]F' = x^{2} - 2x + 1 + 3x(x - 1)\\F' = (x-1)^2 + 3x(x - 1)\\F'=(x-1)(x-1+3x)\\F'=(4x-1)(x-1)[/tex]

Bonne chance !