Question

Помогите решить
X^3+3x^2-x-3=0
(x-2)^2(x-3)=12(x-2)
(x-2)(x-3)(x-4)=(x-2)(x-3)(x+5)

Answer 1

$x^3+3x^2-x-3=0 \\ x^2(x+3)-(x+3)=0 \\ (x+3)(x^2-1)=0 \\ \\ x+3=0 \\ x=-3 \\ x=\pm1$

$(x-2)^2(x-3)=12(x-2) \\ (x-2)^2(x-3)-12(x-2)=0 \\ (x-2)((x-2)(x-3)-12)=0 \\ (x-2)(x^2-3x-2x+6-12)=0 \\ (x-2)(x^2-5x-6)=0 \\ \\ x-2=0 \\ x=2 \\ \\ x^2-5x-6=0 \\ x_1=-1; \ x_2=6$

$(x-2)(x-3)(x-4)=(x-2)(x-3)(x+5) \\ (x-2)(x-3)(x-4)-(x-2)(x-3)(x+5)=0 \\ (x-2)(x-3)(x-4-(x+5))=0 \\ (x-2)(x-3)(x-4-x-5)=0 \\ -9(x-2)(x-3)=0 \\ (x-2)(x-3)=0 \\ x_1=2; \ x_2=3$