49.
f(x) = (x-4-1)(x-4+1) = (x-5)(x-3)
il est plus facile de développer d’après cette forme: f(x) = x² - 5x - 3x + 15
= x² - 8x + 15
f(5) = 0 ; f(rac(2) = 2+15 - 8rac(2) = 17 - 8rac(2) ; f(0) = 15
racines (x-5)(x-3) = 0 => x = 3 et x = 5
50.
f(x) = 9x² + 6x + 1 - 49 = 9x² + 6x - 48
f(x) = (3x+1-7)(3x+1 + 7) = ((3x-6)(3x+8)
f(0) = -48 ; f(rac(2) = 18 + 6rac(2) - 48 = -30 + 6rac(2) ; f(1/3) = -5.9 = -45
((3x-6)(3x+8) = 0 => x = 2 ou x = -8/3
9x² + 6x - 48 = -48 = 9x² + 6x = 0 => 3x(3x + 2) = 0 => x = 0 ou x = -2/3
