The equation you've provided is:

$x^3 + 6x^2 + 4x + 24 = 0$

To solve this cubic equation, we can attempt to find at least one real root using techniques like factoring or synthetic division. Let's follow the steps:

Step 1: Try Rational Root Theorem

According to the Rational Root Theorem, possible rational roots could be factors of the constant term (24) divided by the leading coefficient (1). Thus, we check the factors of 24, which are:

$\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 8, \pm 12, \pm 24$

Step 2: Test possible roots

We'll start testing some simple candidates. Let's test $x = -2$.

Substitute $x = -2$ into the equation:

$(-2)^3 + 6(-2)^2 + 4(-2) + 24 = -8 + 24 - 8 + 24 = 32 \neq 0$

Since $x = -2$ doesn't work, let's test $x = -3$.

Substitute $x = -3$ into the equation:

$(-3)^3 + 6(-3)^2 + 4(-3) + 24 = -27 + 54 - 12 + 24 = 0$

Since this gives us 0, $x = -3$ is a root.

Step 3: Factor the cubic equation

Now that we know $x = -3$ is a root, we can divide the cubic polynomial by $(x + 3)$ to factor it further. Using synthetic division or polynomial division:

$x^3 + 6x^2 + 4x + 24 = (x + 3)(x^2 + 3x + 8)$

Step 4: Solve the quadratic

Now solve the quadratic equation $x^2 + 3x + 8 = 0$ using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 1$, $b = 3$, and $c = 8$.

$x = \frac{-3 \pm \sqrt{3^2 - 4(1)(8)}}{2(1)} = \frac{-3 \pm \sqrt{9 - 32}}{2} = \frac{-3 \pm \sqrt{-23}}{2}$

$x = \frac{-3 \pm \sqrt{23}i}{2}$

Thus, the solutions are:

$x = -3, \quad x = \frac{-3 + \sqrt{23}i}{2}, \quad x = \frac{-3 - \sqrt{23}i}{2}$

Final Answer:

The solutions to the equation are: $x = -3, \quad x = \frac{-3 + \sqrt{23}i}{2}, \quad x = \frac{-3 - \sqrt{23}i}{2}$

Would you like more details on any of these steps?

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Here are 5 related questions:

1. How does synthetic division work in solving polynomials?
2. What is the Rational Root Theorem and how is it applied?
3. Can a cubic equation have more than one real root?
4. How do imaginary numbers affect the solutions of quadratic equations?
5. What is the quadratic formula and when is it used?

Tip: Always try rational roots first to simplify cubic or higher-degree polynomials.