To find the real-number solutions of the equation:

$x^3 - 8x^2 - 3x + 24 = 0$

we will use the Rational Root Theorem to test possible rational roots, followed by synthetic division and factoring.

Step 1: Rational Root Theorem

The Rational Root Theorem states that any rational root of the polynomial $ax^3 + bx^2 + cx + d = 0$ is of the form $\frac{p}{q}$, where $p$ is a factor of the constant term (24), and $q$ is a factor of the leading coefficient (1).

Thus, the possible rational roots are $\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 8, \pm 12, \pm 24$.

Step 2: Testing the possible roots

We will check these values by substituting them into the equation to see if any of them satisfy it.

Checking $x = 2$:

$2^3 - 8(2)^2 - 3(2) + 24 = 8 - 32 - 6 + 24 = -30 + 24 = 0$

Thus, $x = 2$ is a root of the polynomial.

Step 3: Factor the polynomial

Since $x = 2$ is a root, we can factor $(x - 2)$ out of the polynomial. Using synthetic division:

2 & 1 & -8 & -3 & 24 \\ & & 2 & -12 & -30 \\ \hline & 1 & -6 & -15 & 0 \\ \end{array}$$ The quotient is $$x^2 - 6x - 15$$, so the original polynomial can be written as: $$(x - 2)(x^2 - 6x - 15) = 0$$ ### Step 4: Solve the quadratic equation Solve $$x^2 - 6x - 15 = 0$$ using the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $$a = 1$$, $$b = -6$$, and $$c = -15$$. $$x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(-15)}}{2(1)}$$ $$x = \frac{6 \pm \sqrt{36 + 60}}{2}$$ $$x = \frac{6 \pm \sqrt{96}}{2}$$ $$x = \frac{6 \pm 4\sqrt{6}}{2}$$ $$x = 3 \pm 2\sqrt{6}$$ ### Step 5: List all solutions The real-number solutions are: $$x = 2, \quad x = 3 + 2\sqrt{6}, \quad x = 3 - 2\sqrt{6}$$ Would you like more details on any steps or have any questions? ### Related Questions: 1. How do you factor higher-degree polynomials using synthetic division? 2. What is the Rational Root Theorem, and how is it applied? 3. How does the quadratic formula work with irrational roots? 4. Can you explain the process of synthetic division in more detail? 5. How do we determine if a polynomial has complex roots? ### Tip: For polynomials of degree 3 or higher, always start by looking for rational roots using the Rational Root Theorem before attempting more advanced factoring techniques.