Here's a straightforward problem involving rational functions:

Problem:

Simplify the rational function:

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

Solution:

1. Factor the Numerator and Denominator:

   Start by factoring both the numerator and the denominator.

   • Numerator: $2x^2 - 8 = 2(x^2 - 4)$
   • Denominator: $x^2 - 4$

   Notice that $x^2 - 4$ is a difference of squares, which can be factored as:

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

   Therefore, the numerator becomes:

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

   So, the rational function is now:

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

2. Cancel Common Factors:

   We can cancel out the common factors in the numerator and the denominator, which are $(x - 2)(x + 2)$. This results in:

   $f(x) = 2$

   However, note that the factors $x - 2$ and $x + 2$ cannot be zero. Therefore, $x \neq 2$ and $x \neq -2$. These are restrictions that come from the original denominator.

3. Final Answer:

   The simplified form of the rational function is:

   $f(x) = 2, \quad \text{where } x \neq 2 \text{ and } x \neq -2.$

Conclusion:

The simplified function is a constant, but with restrictions on the domain where the original function was undefined.

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Would you like more details on any step, or do you have any questions?

Here are 5 related questions to explore more:

1. How do you factor more complex polynomials in rational functions?
2. What happens if there are no common factors to cancel in a rational function?
3. How do you find vertical and horizontal asymptotes of rational functions?
4. What are the steps to perform polynomial long division in rational expressions?
5. How can we determine the holes in the graph of a rational function?

Tip: Always check for restrictions in the domain when simplifying rational functions!