Let's start by defining the given functions:

1. $f(x) = x^2 - 2x - 8$
2. $g(x) = x - 4$

For $(f + g)(x)$:

$(f + g)(x) = f(x) + g(x) = (x^2 - 2x - 8) + (x - 4)$

Simplifying this:

$(f + g)(x) = x^2 - 2x - 8 + x - 4$

$(f + g)(x) = x^2 - x - 12$

For $(f - g)(x)$:

$(f - g)(x) = f(x) - g(x) = (x^2 - 2x - 8) - (x - 4)$

Simplifying this:

$(f - g)(x) = x^2 - 2x - 8 - x + 4$

$(f - g)(x) = x^2 - 3x - 4$

For $(f \cdot g)(x)$:

$(f \cdot g)(x) = f(x) \cdot g(x) = (x^2 - 2x - 8)(x - 4)$

We'll expand this expression:

$(x^2 - 2x - 8)(x - 4) = x^2(x - 4) - 2x(x - 4) - 8(x - 4)$

Expanding each term:

$= x^3 - 4x^2 - 2x^2 + 8x - 8x + 32$

Simplifying:

$(f \cdot g)(x) = x^3 - 6x^2 + 32$

For $(f / g)(x)$:

$(f / g)(x) = \frac{f(x)}{g(x)} = \frac{x^2 - 2x - 8}{x - 4}$

We can attempt to factor the numerator and simplify. Factor $x^2 - 2x - 8$:

$x^2 - 2x - 8 = (x - 4)(x + 2)$

Thus, we have:

$(f / g)(x) = \frac{(x - 4)(x + 2)}{x - 4}$

As long as $x \neq 4$ (to avoid division by zero), we can cancel $(x - 4)$ from both the numerator and the denominator:

$(f / g)(x) = x + 2 \quad \text{for} \quad x \neq 4$

Summary of results:

1. $(f + g)(x) = x^2 - x - 12$
2. $(f - g)(x) = x^2 - 3x - 4$
3. $(f \cdot g)(x) = x^3 - 6x^2 + 32$
4. $(f / g)(x) = x + 2$ for $x \neq 4$

Would you like further details or explanations on any part of the solution?

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Related Questions:

1. How do you factor quadratic equations like $x^2 - 2x - 8$?
2. What is the difference between polynomial and rational functions?
3. How do you simplify rational expressions with common factors?
4. What happens when you divide by zero in a rational function?
5. How do you expand polynomials in multiplication?

Tip: When simplifying expressions involving polynomials, always try factoring first, especially if there's a common factor you can cancel.