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bjr , j'ai besoin d'aide dans ce exercice
factoriser
A = ( 2x - 1)² - ( 5x + 1 )( 6x - 3 ) + ( 8x² - 2 )
E = ( 3x + 1 )² - 2( 9x² - 1 ) - ( 3x + 1 )( 5x + 3 )
F=( 3x - 4 )( x - 2 ) - ( 6x - 8 )( x - 3 )
B= 2( x - 2 )( x + 1 ) + ( x² - 4 ) - 3( 1 - x )( 4 - 2x )

Sagot :

Explications étape par étape :

A = ( 2x - 1 )² - ( 5x + 1 )( 6x - 3 ) + ( 8x² - 2 )

⇔ A = ( 2x - 1 )² - ( 5x + 1 )( 6x - 3 ) + ( 4x + 2 ) ( 2x - 1 )

⇔ A = ( 2x - 1 )² - 3( 5x + 1 ) ( 2x - 1 ) + ( 4x + 2 ) ( 2x - 1 )

⇔ A = ( 2x - 1 ) [ ( 2x - 1 ) -3( 5x + 1 ) +  ( 4x + 2 ) ]

⇔ A = ( 2x - 1 ) ( 2x - 1 - 15x - 3 + 4x + 2 )

⇔ A = ( 2x - 1 ) ( -9x - 2 )

E =  ( 3x + 1 )² - 2( 9x² - 1 ) - ( 3x + 1 )( 5x + 3 )

⇔ E = ( 3x + 1 )²- 2( 3x - 1 ) ( 3x + 1 ) - ( 3x + 1 )( 5x + 3 )

⇔ E = ( 3x + 1 ) [ ( 3x + 1 ) - 2 (3x - 1 ) - ( 5x + 3 ) ]

⇔ E =  ( 3x + 1 ) ( 3x + 1 - 6x + 2 - 5x - 3 )

⇔ E =  ( 3x + 1 ) ( -8x )

F = ( 3x - 4 )( x - 2 ) - ( 6x - 8 )( x - 3 )

⇔ F = ( 3x - 4 ) ( x - 2 ) - 2( 3x - 4 ) ( x - 3 )

⇔ F = ( 3x - 4 ) [ ( x - 2 ) - 2 ( x - 3 ) ]

⇔ F =  ( 3x - 4 ) ( x - 2 - 2x + 6 )

⇔ F =  ( 3x - 4 ) ( -x + 4 )

B= 2( x - 2 )( x + 1 ) + ( x² - 4 ) - 3( 1 - x )( 4 - 2x )

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

⇔ B = ( x - 2 ) [ 2( x + 1 ) + ( x + 2 ) + 6 ( 1 - x ) ]

⇔ B = ( x - 2 ) ( 2x + 2 + x + 2 + 6 -6x )

⇔ B = ( x - 2 ) ( -3x + 10 )

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