1.
P (1) = 2×1^3 -3×1^2 + 1 = 2×1 - 3×1 + 1 = 2 - 3 +1 = 0
Donc 1 est une racine de ce polynôme :
P (x) = (x-1)(ax^2 +bx + c) = ax^3 + bx^2 + cx - ax^2 - bx - c = ax^3 + (b-a)x^2 + (c-b)x - c
On remarque que :
■a = 2
■b-a = -3
■c-b = 0
■c = -1
==> cest à dire :
●a = 2
●b = -1
●c = -1
donc: P (x) = (x-1)(2x^2 -1x -1)
2.
a)
P (x) = 0
(x-1)(2x^2 -1x -1) = 0
x-1 = 0 ou 2x^2-x-1 = 0
x=1 ou x = -1/2
b)
P (x) = 1
2x^3 -3x^2 +1 = 1
2x^3 -3x^2 = 0
x^2 ( 2x-3) = 0
x^2 = 0 ou 2x-3 = 0
x = 0 ou x = 3/2
c)
P (x) = 2x+1
2x^3 -3x^2 +1 = 2x+1
2x^3 -3x^2 -2x = 0
x (2x^2 -3x -2) = 0
x = 0 ou 2x^2 -3x -2 = 0
x = 0 ou x = -1/4 ou x = 2
