f(x) = (2x+1)(x-3) + (2x+1)(3x+2)
= 2x^2-6x+x-3 + 6x^2+4x+3x+2
=8x^2+2x-1
f(x) = (2x+1)(x-3) + (2x+1)(3x+2)
= (2x+1)(x-3+3x+2)
= (2x+1)(4x-1)
f(0) = -1
f(-1/2) = 0
f(V2) = 8*(V2)^2+2V2-1 = 16+2V2-1 = 2v2+15
f(x) = 0
donc (2x+1)(4x-1) = 0
donc 2x+1=0ou 4x-1=0
donc x=-1/2 ou x=1/4
f(x)=8x^2
donc 8x^2+2x-1 = 8x^2
donc 2x-1=0
donc x=1/2
voila, bon courage....