The expression you provided is:

$\frac{6x^2 + 13 - 5}{9x^2 - 3x - 2} \cdot \frac{3x^2 - 10x - 8}{16x^2 - 1}$

Let’s simplify this step by step.

Step 1: Simplify the Numerators and Denominators

1. First fraction:

   Numerator: $6x^2 + 13 - 5 = 6x^2 + 8$

   Denominator: $9x^2 - 3x - 2$

2. Second fraction:

   Numerator: $3x^2 - 10x - 8$

   Denominator: $16x^2 - 1$

Step 2: Factor each quadratic expression

1. First denominator: $9x^2 - 3x - 2$

   We can factor it as: $9x^2 - 3x - 2 = (3x + 1)(3x - 2)$

2. Second numerator: $3x^2 - 10x - 8$

   We can factor it as: $3x^2 - 10x - 8 = (3x + 2)(x - 4)$

3. Second denominator: $16x^2 - 1$

   This is a difference of squares, so it factors as: $16x^2 - 1 = (4x - 1)(4x + 1)$

Step 3: Rewrite the expression with factored terms

Now the full expression becomes:

$\frac{6x^2 + 8}{(3x + 1)(3x - 2)} \cdot \frac{(3x + 2)(x - 4)}{(4x - 1)(4x + 1)}$

Step 4: Combine and Simplify

Multiply the numerators and the denominators:

$\frac{(6x^2 + 8)(3x + 2)(x - 4)}{(3x + 1)(3x - 2)(4x - 1)(4x + 1)}$

Unfortunately, no common factors cancel out, so this is the simplified form.

Final Answer:

$\frac{(6x^2 + 8)(3x + 2)(x - 4)}{(3x + 1)(3x - 2)(4x - 1)(4x + 1)}$

Would you like more details or have any questions?

Here are five related questions:

1. How can we factor more complex quadratic expressions?
2. What are the properties of difference of squares?
3. How do you handle multiplication of fractions with polynomials?
4. Can common factors always be found between numerators and denominators?
5. What is the importance of factoring in algebraic expressions?

Tip: Always look for common factors when simplifying complex algebraic expressions to reduce them effectively.