Verified answer

Bonsoir

Une identité remarquable qu’il faut résoudre :

(x2-3)(x2+3x-10)=(x+5)(x2-5x+6)

(x^2 - 3)(x^2 + 3x - 10) = (x + 5)(x^2 - 5x + 6)

x^4 + 3x^3 - 10x^2 - 3x^2 - 9x + 30 = x^3 - 5x^2 + 6x + 5x^2 - 25x + 30

x^4 + 3x^3 - 13x^2 - 9x + 30 = x^3 - 19x + 30

x^4 + 3x^3 - x^3 - 13x^2 - 9x + 19x + 30 - 30 = 0

x^4 + 2x^3 - 13x^2 + 10x = 0

x(x^3 + 2x^2 - 13x + 10) = 0

x = 0

Ou

x^3 + 2x^2 - 13x + 10 = 0

On remarque que pour x = 1 c’est vrai :

= (x - 1)(ax^2 + bx + c)

= ax^3 + bx^2 + cx - ax^2 - bx - c

a = 1

b - a = 2

b = 2 + a = 2 + 1 = 3

c - b = -13

c = -13 + b = -13 + 3 = -10

(x - 1)(x^2 + 3x - 10) = 0

x = 1 ou

x^2 + 3x - 10 = 0

X1 = (-3 - 7)/2 = -10/2 = -5

X2 = (-3 + 7)/2 = 4/2 = 2

x(x - 1)(x + 5)(x - 2) = 0