Mathématiques

Question

Faire les factorisations:
(3x-5)² - 4(3x-5)
(7x+3) (x-2) + (7x+3)²
(-x+1) (2x+1) - (2x+1) (x-10)
(3x-8) (x-2) + (5x+7) (3x-8)
3(x-4) - (x+2) (x-4)

1 Réponse

  • 1)=(3x-5)[3x-5)-4]

    (3x-5)(3x-5-4)

    (3x-5)(3x-9)

    2)=(7x+3)[(x-2)+(7x+3)]

    (7x+3)(x-2+7x+3)

    (7x+3)(8x+1)

    3)=2x+1[(-x+1)-(x-10)]

    (2x+1)[-x+1-x+10)

    (2x+1)(-2x+11)

    4)=(3x+8)[(x-2)+(5x+7)

    (3x-8)(x-2+5x+7)

    (3x-8)(6x+5)

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

    (x-4)(3-x-2)

    (x-4)(-x+1)


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