Bonjour voila ma question. Factorisez si possible les expressions suivantes: [tex](x + {1})^{2} - 2(x + {1})^{2} [/tex] [tex](2x + {3})^{2} + (x - 2)(2x +

Bonjour,☕️

A= (x+1)²×(x-2).

B=(2x+3)[(2x+3)+(x-2)]=(2x+3)[3x+1].

C=(3x-5)[(3x-5)-(x+4)]=(3x-5)[2x-9].

D=[(2t+1)-(t+1)].[(2t+1)+(t+1)]= t.[3t+2].

E=[(3t-7)-5)].[(3t-7)+5]=[3t-12].[3t-2]=3(t-4)(3t-2).

La dernière est à vous, même manière fait en D


:)

Bonjour,

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

(2x+3)²+(x-2)(2x+3)
=(2x+3)(2x+3+x-2)
=(2x+3)(3x+1)

(3x-5)²-(x+4)(3x-5)
=(3x-5)(3x-5-x-4)
=(3x-5)(2x-9)

(2t+1)²-(t+1)²
=((2t+1)+(t+1))((2t+1)-(t+1)) car (a²-b²)=(a-b)(a+b)
=(2t+1+t+1)(2t+1-t-1)
=t(3t+2)

(3t+1)²-25
=(3t+1)²-5²
=(3t+1-5)(3t+1+5) car a²-b²=(a-b)(a+b)
=(3t-4)(3t+6)
=3(3t-4)(t+2)

(3t+4)²-(2t+3)²
=((3t+4)+(2t+3))((3t+4)-(2t+3)) car a²-b²=(a+b)(a-b)
=(3t+4+2t+3)(3t+4-2t-3)
=(5t+7)(t+1)