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Pikachu972
@Pikachu972
April 2019
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Bonjour , Aidez-moi s'il vous plaît à factoriser .
Factorisez l'expression suivante :
( 2x+1)² - ( 2x+1) (x+3) = ( 2x+1) ( x+3) .
Normalement le facteur est (2x+1) mais je veux factoriser cette expression .
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raymrich
Bonjour,
(2x+1)²-(2x+1)(x+3)-(2x+1)(x+3) = 0⇔
Le facteur commun est (2x+1)
Donc (2x+1)[2x+1-(x+3)-(x+3)] = (2x+1)(2x+1-x-3-x-3) = (2x+1)(-5) = -5(2x+1) = 0
1 votes
Thanks 1
Pikachu972
merci
anomaarieljonathan
J'écris directement la réponse:
( 2x+1 ) ( 2x+1 ) - ( 2x+1 ) ( x+3 ) - ( 2x+1 ) ( x+3 ) = 0
( 2x+1 ) [ ( 2x+1 ) - ( x+3 ) - ( x+3 ) ] = 0
( 2x+1 ) ( 2x + 1 - x - 3 - x - 3 ) = 0
( 2x+1 ) ( 2x - x - x - 3 - 3 + 1 ) = 0
( 2x+1 ) ( 2x - 2x - 6 + 1 ) = 0
( 2x+1 ) ( 0 - 5 ) = 0
( 2x+1 ) ( - 5 ) = 0
- 5 ( 2x+1 ) = 0
( 2x+1 ) = 0 / - 5
( 2x + 1 ) = 0
2x + 1 = 0
2x = - 1
x = - 1 / 2
2 votes
Thanks 1
Pikachu972
merci
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(2x+1)²-(2x+1)(x+3)-(2x+1)(x+3) = 0⇔
Le facteur commun est (2x+1)
Donc (2x+1)[2x+1-(x+3)-(x+3)] = (2x+1)(2x+1-x-3-x-3) = (2x+1)(-5) = -5(2x+1) = 0
( 2x+1 ) ( 2x+1 ) - ( 2x+1 ) ( x+3 ) - ( 2x+1 ) ( x+3 ) = 0
( 2x+1 ) [ ( 2x+1 ) - ( x+3 ) - ( x+3 ) ] = 0
( 2x+1 ) ( 2x + 1 - x - 3 - x - 3 ) = 0
( 2x+1 ) ( 2x - x - x - 3 - 3 + 1 ) = 0
( 2x+1 ) ( 2x - 2x - 6 + 1 ) = 0
( 2x+1 ) ( 0 - 5 ) = 0
( 2x+1 ) ( - 5 ) = 0
- 5 ( 2x+1 ) = 0
( 2x+1 ) = 0 / - 5
( 2x + 1 ) = 0
2x + 1 = 0
2x = - 1
x = - 1 / 2