Find all real and imaginary solutions to the equation x^3 + 3x^2 - 14x - 42 = 0.

Steps to Solve

1. Identify the Cubic Equation

The given equation is

$$ x^3 + 3x^2 - 14x - 42 = 0. $$

2. Use the Rational Root Theorem

Check possible rational roots using factors of the constant term (-42) and the leading coefficient (1). The factors of -42 include ±1, ±2, ±3, ±6, ±7, ±14, ±21, ±42.

3. Test Possible Rational Roots

Let's test $x = 3$:

$$ 3^3 + 3(3^2) - 14(3) - 42 = 27 + 27 - 42 - 42 = -30 ,(\text{not a root}). $$

Now test $x = -6$:

$$ (-6)^3 + 3(-6)^2 - 14(-6) - 42 = -216 + 108 + 84 - 42 = -66 ,(\text{not a root}). $$

Next, test $x = -2$:

$$ (-2)^3 + 3(-2)^2 - 14(-2) - 42 = -8 + 12 + 28 - 42 = -10 ,(\text{not a root}). $$

Test $x = -3$:

$$ (-3)^3 + 3(-3)^2 - 14(-3) - 42 = -27 + 27 + 42 - 42 = 0 ,(\text{is a root}). $$

4. Factor the Polynomial

Since $x = -3$ is a root, we can factor the polynomial using synthetic division by $x + 3$:

Perform the division and we get:

$$ x^3 + 3x^2 - 14x - 42 = (x + 3)(x^2 - 14). $$

5. Solve the Quadratic Equation

Now solve

$$ x^2 - 14 = 0 $$

which gives:

$$ x^2 = 14 \implies x = \pm \sqrt{14}. $$

Thus, the solutions are:

1. Real root: $x = -3$
2. Real roots: $x = \sqrt{14}, -\sqrt{14}$.