bonjour

je te montre

A (x) = x² + 6 x + 9

d =  36 - 4 ( 1 * 9 ) = 36 - 36 = 0

d = 0 donc une seule solution

x 0  =   - 6 / 2 = - 3

A (x) = ( x + 3 ) ( x + 3 )

B  ( x) = - x² + 2 x + 15

d =  4 - 4 ( - 1 * 15 )=  4  + 60 = 64

x 1 = (  2 - 8 ) / - 2 = - 6 / - 2 = 3

x 2 = ( 2 + 8 ) / - 2 = - 5

b (x) = (  x - 3 ) ( x + 5 )  

continue

32

a)

A(x) = x² + 6x + 9

A(x) = (x + 3)²

b)

B(x) = -x² + 2x + 15

      = - (x² - 2x) + 15

     = - (x² - 2x + 1 - 1) + 15

    = - [(x - 1)² -1] + 15

     = -(x -1)² + 1 + 15

    = 16 - (x - 1)²

    = 4² - (x -1)²  

   = [4 - (x - 1)][ 4 + (x - 1)]

   = (-x + 5)(x + 3)

    = - (x - 5)(x + 3)

45

cas général

f(x) = ax² + bx + c

f(x) = a(x - x1)(x - x2)

1)

f(x) = a(x -2)(x + 8)     a non nul

2)

f(x) est de la forme   f(x) = a(x + 1/2)(x - 5)

on calcule a en écrivant que f(0) = -10

f(0) = a*(1/2) *(-5) = (-5/2)a

(-5/2)a = -10

a = -10 /(-5/2)

a = 4

f(x) = 4(x + 1/2)(x - 5)

f(x) = 2(2x + 1)(x - 5)

3)

idem