bonjour,
1) developper f(x) =
x-3+3(x²-6x+9)+x²-9 =
x-3+3x²-18x+27+x²-9 =
4x²-17x+15
2) f(x) = 1(x-3)+3(x-3)²+(x-3)(x+3)
(x-3)[1+3(x-3)+(x+3)]=
(x-3)(1+3x-9+x+3) =
(x-3)(4x+5)
3) f(0) : 4x²-17x+15 =
4*0²-17*0+15 = 15
f(4:5) =
4(4/5)² - 17*4/5 + 15 =
4*16/25 - 68/5 + 15 =
64/ 25- 340/25 + 375/25 = 99/25
f(x) = 0 forme facto
(x-3)(4x+5) = 0
x-3 =0
x = 3
4x+5 =0
4x = 5
x = 5/4
f(x) = 15
4x²-17x+15 = 15
4x²-17x+15-15 =0
4x²-17x = 0
x(4x-17) = 0
4x-17 =0
4x =17
x =17/4
x = 0
la suite ? illisible
