To solve the cubic equation $x^3 - x - 24 = 0$, we can use a combination of methods, starting with trial and error for possible rational solutions, and then using techniques like synthetic division or factoring.

Step 1: Try Rational Roots

Using the Rational Root Theorem, the possible rational roots are the factors of the constant term (-24) divided by the factors of the leading coefficient (1). This gives us possible roots:

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

Now, let's test these values by substituting them into the equation $f(x) = x^3 - x - 24$.

• For $x = 3$: $f(3) = 3^3 - 3 - 24 = 27 - 3 - 24 = 0$

So, $x = 3$ is a root.

Step 2: Factor the Cubic Equation

Now that we know $x = 3$ is a root, we can factor $x^3 - x - 24$ as $(x - 3)(\text{quadratic expression})$.

Perform synthetic division to divide $x^3 - x - 24$ by $x - 3$:

3 & 1 & 0 & -1 & -24 \\ & & 3 & 9 & 24 \\ \hline & 1 & 3 & 8 & 0 \\ \end{array}$$ The quotient is $$x^2 + 3x + 8$$, so we have: $$x^3 - x - 24 = (x - 3)(x^2 + 3x + 8)$$ ### Step 3: Solve the Quadratic Equation Now, solve $$x^2 + 3x + 8 = 0$$ using the quadratic formula: $$x = \frac{-3 \pm \sqrt{3^2 - 4(1)(8)}}{2(1)}$$ $$x = \frac{-3 \pm \sqrt{9 - 32}}{2}$$ $$x = \frac{-3 \pm \sqrt{-23}}{2}$$ $$x = \frac{-3 \pm i\sqrt{23}}{2}$$ Thus, the solutions are: $$x = 3, \quad x = \frac{-3 + i\sqrt{23}}{2}, \quad x = \frac{-3 - i\sqrt{23}}{2}$$ ### Final Answer: The solutions to the equation $$x^3 - x - 24 = 0$$ are: $$x = 3, \quad x = \frac{-3 + i\sqrt{23}}{2}, \quad x = \frac{-3 - i\sqrt{23}}{2}$$ Would you like more details on any step or have any questions? --- Here are 5 related questions: 1. What is the Rational Root Theorem and how does it work? 2. How do you perform synthetic division? 3. How do you use the quadratic formula to solve any quadratic equation? 4. Can all cubic equations be solved by factoring? 5. What are complex numbers and how are they handled in equations? **Tip:** Always check for simple rational solutions first when dealing with cubic equations using the Rational Root Theorem.