on doit passe de ka+kb à k(a+b) quand on factorise

A = 3 * x - 3 * 2 = 3 (x-2)

B = x * 7x + x * 1 = x (7x+1)

C = 9x² * (-2x) + 9x² * 1 = 9x² (-2x + 1)

on sait que

(a+b)² = a² + 2ab+ b² , que (a-b)² = a² -2ab+ b² et que a²-b² = (a+b) (a-b)

D = (x - 2)²

E = (3x + 5)²

F = (2x)² - 4² = (2x+4) (2x-4) = 4 (x+2) (x-2)

G = (9)² - (√5x)² = (9+√5x) (9-√5x)