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Factorisation |
Utilisation d’un facteur commun : on sait que a (b + c) = ab + ac (Distributivité)
Inversement : ab + ac = a (b + c) ; on a mis " a " en facteur commun
Trucs utiles : a = 1xa = 1a et (a + 5) = 1 x (a + 5) = 1(a + 5)I)Factorisation par reconnaissance d’un facteur commun :
. 4x + 4y = 4(x + y)
. 8y + y = 8y + 1y = y(8 + 1) = y x 9 = 9y
. 5a + 10b = 5 x a + 5 x 2b = 5 x (a + 2b) = 5(a + 2b)
. (3x + 4)(2x - 7) - (3x + 4) = (3x + 4)(2x - 7) - 1(3x + 4) = (3x + 4)[(2x - 7) - 1]
= (3x + 4)(2x - 7 - 1) = (3x + 4)(2x - 8) = (3x + 4)2(x - 4) = 2(3x + 4)(x - 4)
. (3x + 5)
(2x - 3) + (3x + 5) (6x - 5) = (3x + 5) [(2x - 3) + (6x - 5)]= (3x + 5)(2x - 3 + 6x - 5) = (3x +5) (8x - 8) = (3x + 5)8(x - 1) = 8(3x + 5)(x - 1)
. (3x +2)(5x - 4) - (3x + 2)(2x - 1) = (3x + 2)[(5x - 4) - (2x - 1)] =(3x + 2)(5x - 4 -2x +1)
= (3x + 2)(3x - 3) = (3x + 2)3(x - 1) = 3(3x + 2)(x - 1)
. (5x + 4)² - (5x + 4)(6x - 7) = (5x + 4)(5x + 4) - (5x + 4)(6x - 7) = (5x + 4)[(5x + 4) - (6x - 7)]
= (5x + 4)(5x + 4 - 6x + 7) = (5x + 4)(- x + 11)
II) Factorisation par reconnaissance d'une identité remarquable :
(a + b)² = a² + b² + 2ab Ex : (3x + 4)² = (3x)² + (4)² + 2(3x)(4) = 9x² + 16 + 24x
(a - b )² = a² + b² - 2ab Ex : (5x - 3)² = (5x)² + (3)² - 2(5x)(3) = 25x² + 9 - 30x
(a + b)(a - b) = a² - b² Ex : (6x + 5)(6x - 5) = (6x)² - (5)² = 36x² - 25
Exemples :
. 9x² + 16 + 24x =(3x)² + (4)² + 2(3x)(4) = (3x + 4)² = (3x + 4)(3x + 4)
. 25x² - 49 = (5x)² - (7)² = (5x +7)(5x - 7)
. 36x² + 25 - 60x = (6x)² + (5)² - 2(6x)(5) = (6x - 5)² = (6x - 5)(6x - 5)
. 36 - (x + 3)² = (6)² - (x + 3)² = [6 + (x + 3)][6 - (x + 3)] = (6 + x + 3)(6 - x - 3)
= (x + 9)(- x + 3)
. (3x - 5)² - 64 = (3x - 5)² - (8)² = [(3x - 5) + 8][(3x - 5) - 8] = (3x - 5 + 8)(3x - 5 - 8)
= (3x + 3)(3x - 13) = 3(x + 1)(3x - 13)
III) Utilisation des deux méthodes précédentes :
. x² - 4 + (x + 2)(5x - 3) = (x + 2)(x - 2) + (x + 2)(5x - 3) = (x + 2)[(x - 2) + (5x - 3)]
= (x + 2)(x - 2 + 5x - 3) = (x + 2)(6x - 5)
. 4x² + 16 - 16x + (2x + 4)(5x - 9) = (2x)² + (4)² - 2(2x)(4) + (2x - 4)(5x - 9)
= (2x - 4)² + (2x - 4)(5x - 9) = (2x - 4)(2x - 4) + (2x - 4)(5x - 9) = (2x - 4)[(2x - 4) + (5x - 9)]
= (2x - 4)(2x - 4 + 5x - 9) = (2x - 4)(7x - 13)
. (3x + 4)(5x - 2) - (25x² - 4) = (3x + 4)(5x - 2) - ((5x)² - (2)²)
= (3x + 4)(5x - 2) - (5x + 2)(5x - 2) = (5x -2)[(3x + 4) - (5x + 2)] = (5x - 2)(3x + 4 - 5x - 2)
= (5x - 2)(- 2x + 2) = (5x - 2)2(- x + 1) = 2(5x - 2)(- x + 1)