Question

Решите уравнение x^3+3x^2=4x+12

Answer 1

x³+3x² = 4x+12;

x²(x + 3) = 4(x + 3);

x²(x + 3) - 4(x + 3) = 0;

(x + 3)(x² - 4) = 0;

(x + 3)(x - 2)(x + 2) = 0;

x₁ = -3; x₂ = 2; x₃ = -2.

Ответ: -3; ±2.

Answer 2

Ответ:

х^3 + 3x^2 - 4x - 12 = 0

x^2(x+3) - 4(x+3) = 0

(x+3)(x^2 - 4) = 0

(x+3)(x-2)(x+2) = 0

x=-3

x=2

x=-2

Объяснение: