Bonsoir.
Developpements :
A = (t + √3)² = t² + 2t√3 + 3.
B = (2 + 3√2)² = 4 + 12√2 + 18 = 22 + 12√2.
C = (1 + u)³ = (u + 1)³ = u³ + 3u² + 3u + 1.
Factorisations :
C = (2a - 3)(a + 5) - 4(a + 5)²
C = (a + 5) [(2a - 3) - 4(a + 5)]
C = (a + 5)(2a - 3 - 4a - 20)
C = (a + 5)(-2a - 23)
C = - (a + 5)(2a + 23).
D = 4x³ - 36x
D = 4x(x² - 9)
D = 4x(x + 3)(x - 3).
E = 1 - z⁴
E = (1 + z²)(1 - z²)
E = (1 + z²)(1 + z)(1 - z).
Exercice avec f(x) :
f(x) = (5x - 3)² - 16 = 25x² - 30x + 9 - 16 = 25x² - 30x - 7
(forme developpee de f)
f(x) = (5x - 3)² - 16 = [(5x - 3) - 4][(5x - 3) + 4] = (5x - 3 - 4)(5x - 3 + 4)
= (5x - 7)(5x + 1) (forme factorisee de f)
a/ f(0) = 25 * 0² - 30 * 0 - 7 = 0 - 0 - 7 = - 7 (via la forme developpee de f)
b/ f(2/5) = (5 * 2/5 - 7)(5 * 2/5 + 1) = (2 - 7)(2 + 1) = - 5 * 3 = - 15
(via la forme factorisee de f)
c/ f(√2) = (5√2 - 3)² - 16 = 50 - 30√2 + 9 - 16 = 43 - 30√2
(via la forme canonique de f, expression initiale)
L equation f(x) = 0 se resout a l aide de la forme factorisee de l expression de f. Equation-produit nul ou au moins un des facteurs est nul.
On a donc :
(5x - 7)(5x + 1) = 0
2 solutions :
5x - 7 = 0 ou 5x + 1 = 0
5x = 7 5x = - 1
x = 7/5 x = - 1/5
S = { - 1/5 ; 7/5 }
Bonne nuit !
