Let me solve this step by step. The given expression is:

$\frac{2}{x - 5} + \frac{x + 5}{3x + 6} \times \frac{x^2 - x - 6}{x^2 - 25}.$

Step 1: Factorize all components

1. $x^2 - x - 6$ can be factorized as: $x^2 - x - 6 = (x - 3)(x + 2).$

2. $x^2 - 25$ is a difference of squares: $x^2 - 25 = (x - 5)(x + 5).$

3. $3x + 6$ can be factorized as: $3x + 6 = 3(x + 2).$

Substituting these factorizations, the expression becomes:

$\frac{2}{x - 5} + \frac{x + 5}{3(x + 2)} \times \frac{(x - 3)(x + 2)}{(x - 5)(x + 5)}.$

Step 2: Simplify the multiplication

The second term involves multiplication. Simplify it:

$\frac{x + 5}{3(x + 2)} \times \frac{(x - 3)(x + 2)}{(x - 5)(x + 5)}.$

Cancel out $x + 2$ and $x + 5$ (as they are common terms):

$= \frac{x - 3}{3(x - 5)}.$

Thus, the expression becomes:

$\frac{2}{x - 5} + \frac{x - 3}{3(x - 5)}.$

Step 3: Combine into a single fraction

Find a common denominator for $\frac{2}{x - 5}$ and $\frac{x - 3}{3(x - 5)}$. The common denominator is $3(x - 5)$. Rewrite both fractions with this denominator:

$\frac{2}{x - 5} = \frac{6}{3(x - 5)}, \quad \frac{x - 3}{3(x - 5)} = \frac{x - 3}{3(x - 5)}.$

Combine them:

$\frac{6}{3(x - 5)} + \frac{x - 3}{3(x - 5)} = \frac{6 + (x - 3)}{3(x - 5)}.$

Simplify the numerator:

$6 + (x - 3) = x + 3.$

So the fraction becomes:

$\frac{x + 3}{3(x - 5)}.$

Final Answer:

The simplified and fully factorized form of the expression is:

$\frac{x + 3}{3(x - 5)}.$

Would you like me to explain any specific step or solve a related problem? Here are some questions to explore:

1. How do you determine the common denominator for fractions?
2. Why do terms like $x + 2$ and $x + 5$ cancel during simplification?
3. How do you factorize quadratic expressions like $x^2 - x - 6$?
4. What are the restrictions on $x$ for this expression (e.g., values that make the denominator zero)?
5. How do you simplify expressions involving mixed operations?

Tip: Always check for restrictions (e.g., where the denominator becomes zero) in rational expressions to avoid undefined values!