2nde – Lycée Lafayette Brioude – http://cecbertrandmath.free.fr/
Savoir-Faire : Factoriser une expression littérale
Définition : Factoriser, c’est transformer une somme en produit.
Méthode : Pour factoriser une expression, on repère le nombre de termes et on cherche s’il y a un facteur commun à chaque terme :
- s’il y a un facteur commun : on utilise la règle de distributivité simple Remarque : le nombre de termes du 2ème facteur doit être égal au nombre de termes de l’expression initiale
- s’il n’y a pas de facteur commun, il faut utiliser une identité remarquable
Outils forme factorisée forme développée
► Distributivité simple k×(a + b) = k×a + k×b
(2 termes : 1 facteur commun k)
► Identités remarquables (a + b)2 = a2 + 2ab + b2 (a - b)2 = a2 − 2ab + b2 (a + b) (a – b) = a2 − b2
Exemples :
A = 8x2 – 10x 2 termes B = 27x2 – 18x – 9 3 termes
A = 2x×4x – 2x×5 B = 9×3x2 – 9×2x – 9×1
A = 2x (4x – 5) 2 termes B = 9 (3x2 – 2x – 1) 3 termes
C = 3 (x + 2) + (x + 2) (1 – 3x) 2 termes
C = 3×(x + 2) + (x + 2)×(1 – 3x) D = (x – 1)2 – (x – 1) (2x – 5) 2 termes C = (x + 2)×[3 + (1 – 3x)] 2 termes D = (x – 1)×(x – 1) – (x – 1)× (2x – 5) C = (x + 2) (3 + 1 – 3x) D = (x – 1) [(x – 1) – (2x – 5)] 2 termes
C = (x + 2) (4 – 3x) D = (x – 1) (x – 1 – 2x + 5)
D = (x – 1) (-x + 4)
E = x2 + 4x + 4 F = 16x2 – 9
E = (x + 2)2 F = (4x + 3) (4x – 3)
G = (x – 3)2 – 25 H = (x – 1)2 – (3 – 2x)2
G = [(x – 3) + 5] [(x – 3) – 5] H = [(x – 1) + (3 – 2x)] [(x – 1) – (3 – 2x)]
G = (x – 3 + 5) (x – 3 – 5) H = (x – 1 + 3 – 2x) (x – 1 – 3 + 2x) G = (x + 2) (x – 8) H = (-x + 2) (3x – 4)
On factorise
2nde – Lycée Lafayette Brioude – http://cecbertrandmath.free.fr/
Exercice : Factoriser chacune des expressions suivantes : A = 25x2 – 10x
B = (x + 3) (3 – x) + 7 (x + 3) + 2x (x + 3) C = (x + 1)² + (x + 1) (3x + 1)
D = (x – 3)² – (x – 3) (4x + 1)
E = (x + 1) (2x – 5) + (2x – 5)² – 3 (2x – 5) F = (3x – 4) (2 – x) – (3x – 4)²
G = (x – 1)2 – 16
H = (7 – x)2 – (2x + 1)² I = x2 – 14x + 49
J = 16x2 – 81
2nde – Lycée Lafayette Brioude – http://cecbertrandmath.free.fr/
Correction :
A = 25x2 – 10x A=5 (5x x−2)
B = (x + 3) (3 – x) + 7 (x + 3) + 2x (x + 3) B=(x+3)(x+10)
C = (x + 1)² + (x + 1) (3x + 1) C=2(x+1)(2x+1)
D = (x – 3)² – (x – 3) (4x + 1) D=(x−3)( 3− −x 4)
E = (x + 1) (2x – 5) + (2x – 5)² – 3 (2x – 5) E=(2x−5)(3x−7)
F = (3x – 4) (2 – x) – (3x – 4)² F = −2(3x−4)(2x−3)
G = (x – 1)2 – 16 G=(x+3)(x−5)
H = (7 – x)2 – (2x + 1)² H = −3(x+8)(x−2)
I = x2 – 14x + 49 I =(x−7)2
J = 16x2 – 81 J =(4x+9)(4x−9)
Correction détaillée : A = 25x2 – 10x
A = 5x×5x – 2×5x A = 5x (5x – 2)
B = (x + 3) (3 – x) + 7 (x + 3) + 2x (x + 3) B = (x + 3)×(3 – x) + 7×(x + 3) + 2x×(x + 3) B = (x + 3) × [(3 – x) + 7 + 2x ]
B = (x + 3) (x + 10)
C = (x + 1)² + (x + 1) (3x + 1)
C = (x + 1)×(x + 1) + (x + 1)× (3x + 1) C = (x + 1) [(x + 1) + (3x + 1)]
C = (x + 1) (x + 1 + 3x + 1) C = (x + 1) (4x + 2)
C = 2 (x + 1) (2x + 1)
2nde – Lycée Lafayette Brioude – http://cecbertrandmath.free.fr/
D = (x – 3)² – (x – 3) (4x + 1)
D = (x – 3)×(x – 3) – (x – 3)× (4x + 1) D = (x – 3) [(x – 3) – (4x + 1)]
D = (x – 3) (x – 3 – 4x – 1) D = (x – 3) (-3x – 4)
E = (x + 1) (2x – 5) + (2x – 5)² – 3 (2x – 5)
E = (x + 1)×(2x – 5) + (2x – 5)× (2x – 5) – 3×(2x – 5) E = (2x – 5) [(x + 1) + (2x – 5) – 3]
E = (2x – 5) (x + 1 + 2x – 5 – 3) E = (2x – 5) (3x – 7)
F = (3x – 4) (2 – x) – (3x – 4)²
F = (3x – 4)× (2 – x) – (3x – 4)×(3x – 4) F = (3x – 4) [(2 – x) – (3x – 4)]
F = (3x – 4) (2 – x – 3x + 4) F = (3x – 4) (-4x + 6)
F = -2 (3x – 4) (2x – 3)
G = (x – 1)2 – 16
G = [(x – 1) + 4] [(x – 1) – 4]
G = (x – 1 + 4) (x – 1 – 4)
G = (x + 3) (x – 5)
H = (7 – x)2 – (2x + 1)²
H = [(7 – x) + (2x + 1)] [(7 – x) – (2x + 1)]
H = (7 – x + 2x + 1) (7 – x – 2x – 1) H = (x + 8) (-3x + 6)
H = -3 (x + 8) (x – 2)
I = x2 – 14x + 49 I = (x – 7)2
J = 16x2 – 81
J = (4x + 9) (4x – 9)
