1) f(x) = (x-3)² - 4 → On reconnaît l'identité remarquable a² - b²
= (√((x-3)²) + √4)(√((x-3)²) - √4)
= ((x-3) + 2)((x-3) - 2)
= (x-1)(x-5)
2) f(0) = (0-1)(0-5)
= -1 * (-5)
= 5
f(3) = (3-3)² - 4
= 0² - 4
= -4
f(x) = 0
(x-1)(x-5) = 0 → si (x-1) = 0 ou (x-5) = 0
x - 1 = 0
x = 1
ou
x - 5 = 0
x = 5
f(x) = 5
(x-3)² - 4 = 5
(x-3)² = 5 + 4
(x-3)² = 9 → Si (x-3) = 3 ou (x-3) = -3
x - 3 = √9
x - 3 = 3
x = 3 + 3 = 6
ou (0-3)² = 9
(-3)² = 9
x = 0
