x(x+1)(x+2)(x+3)=24
(x² + x)(x² + 5x + 6)= 24
x⁴ + 5x³ + 6x² + x³ + 5x² + 6x - 24 = 0
x⁴ + 6x³ + 11x² + 6x -24 = 0
x⁴ - x³ + 7x³ - 7x² + 18x² - 18x + 24x - 24 = 0
x³(x - 1) + 7x²(x - 1) + 18x(x - 1) + 24(x-1) = 0
(x - 1)(x³ + 7x² + 18x + 24) = 0
(x - 1)(x³ + 4x₂ + 3x² + 12x + 6x + 24) = 0
(x - 1)(x²(x+4) + 3x(x + 4) + 6(x+4)) = 0
(x - 1)(x + 4)(x² + 3x + 6) = 0
Am coborat de la o ecuatie de gradul 4 la 3 ecuatii
dintre care 2 ecuatii de gradul 1 si o ecuatie de gradul 2.
Ecuatia astfel descompusa are tot 4 solutii:
(x - 1) = 0 => x₁ = 1
(x + 4) = 0 => x₂ = -4
(x² + 3x + 6) = 0
Rezolvand aceasta ecatie afli solutiile x₃ si x₄
Ecuatia de gradul 2 nu are solutii reale. => x₃ si x₄ sunt solutii complex conjugate.
Nu stiu daca ai invatat la scoala despre numere complexe.
