We have the following rational expression to combine:

$\frac{5}{x^2 - 4} + \frac{5x + 2}{x + 2}$

Step-by-Step Solution:

1. Factor the Denominators:

   • The first denominator, $x^2 - 4$, is a difference of squares, which can be factored: $x^2 - 4 = (x - 2)(x + 2)$
   • The second denominator is already $x + 2$, so no need to factor it further.

2. Find the Least Common Denominator (LCD):

   • The LCD of $(x - 2)(x + 2)$ and $x + 2$ is $(x - 2)(x + 2)$, since it includes both factors.

3. Rewrite Each Fraction with the LCD:

   • The first fraction $\frac{5}{x^2 - 4}$ already has the denominator $(x - 2)(x + 2)$.
   • The second fraction $\frac{5x + 2}{x + 2}$ needs to be rewritten with the LCD. Multiply both numerator and denominator by $x - 2$: $\frac{5x + 2}{x + 2} = \frac{(5x + 2)(x - 2)}{(x + 2)(x - 2)}$

4. Expand the Numerator of the Second Fraction: $(5x + 2)(x - 2) = 5x(x - 2) + 2(x - 2) = 5x^2 - 10x + 2x - 4 = 5x^2 - 8x - 4$ So, the second fraction becomes: $\frac{5x^2 - 8x - 4}{(x - 2)(x + 2)}$

5. Add the Numerators: Now that both fractions have the same denominator, we can add them: $\frac{5}{(x - 2)(x + 2)} + \frac{5x^2 - 8x - 4}{(x - 2)(x + 2)} = \frac{5 + (5x^2 - 8x - 4)}{(x - 2)(x + 2)}$ Simplify the numerator: $5 + 5x^2 - 8x - 4 = 5x^2 - 8x + 1$

6. Final Combined Expression: The final answer is: $\frac{5x^2 - 8x + 1}{(x - 2)(x + 2)}$

This is the simplified form of the given expression.

Would you like more details or have any questions?

Here are 5 related questions to expand your understanding:

1. How do you multiply and expand binomials like $(5x + 2)(x - 2)$?
2. What is the general process for finding the least common denominator of rational expressions?
3. How can you simplify rational expressions by factoring?
4. What are some common types of factoring techniques, such as factoring by grouping?
5. Why is it important to check for excluded values (i.e., values that make the denominator zero) in rational expressions?

Tip: Always factor denominators first when adding or subtracting rational expressions to find the least common denominator (LCD) efficiently.