the solutions for the first equation are x = 2, x = -5, and x = 3. the solutions for the second equation are x = 2, x = -2, x = -√3, and x = √3.
1. x^3 + 3x^2 - 4x - 12 = 0
a) Factoring:
First, notice that the constant term (-12) can be factored as 223. We can try to group the terms to find factors that multiply to -12 and add up to -4.
We can group the equation as: ((x^3 + 3x^2) - 4x) - 12 = 0.
Factor out x from the first group: x(x^2 + 3x) - 12 = 0.
Now, try to factor x^2 + 3x. Remember that the constant term is -12 and the coefficient of x is 3. We need to find two numbers that multiply to -12 and add up to 3. These numbers are -2 and 5.
Factor x^2 + 3x as (x - 2)(x + 5).
Substitute this back into the equation: x(x - 2)(x + 5) - 12 = 0.
Factor out -1: (x - 2)(x + 5)(x - 3) = 0.
Set each factor equal to zero and solve for x:
x - 2 = 0 --> x = 2
x + 5 = 0 --> x = -5
x - 3 = 0 --> x = 3
Therefore, the solutions for the first equation are x = 2, x = -5, and x = 3.
b) Cardano formula:
2. 2x^5 + 24x = 14x^3
First, factor out 2x from both sides: 2x(x^4 + 12) = 14x^3.
Divide both sides by 2x (assuming x ≠ 0): x^4 + 12 = 7x^2.
Move all terms to one side: x^4 - 7x^2 + 12 = 0.
This equation can be factored as (x^2 - 4)(x^2 - 3) = 0.
Factor each quadratic further: (x + 2)(x - 2)(x + √3)(x - √3) = 0.
Set each factor equal to zero and solve for x:
x + 2 = 0 --> x = 2
x - 2 = 0 --> x = -2
x + √3 = 0 --> x = -√3
x - √3 = 0 --> x = √3
Therefore, the solutions for the second equation are x = 2, x = -2, x = -√3, and x = √3.