To simplify the expression

$\frac{6x^{2} + x - 2}{2x^{2} - 8x - 5} \times \frac{4x^{2} - 1}{4x^{2} - 4x + 1}$

we need to factor each expression in the numerators and denominators, then simplify by canceling out any common factors.

Step 1: Factor each part

1. Factor $6x^2 + x - 2$:

   To factor $6x^2 + x - 2$, we look for two numbers that multiply to $6 \cdot (-2) = -12$ and add up to $1$. These numbers are $4$ and $-3$.

   So, $6x^2 + x - 2 = (3x - 2)(2x + 1)$

2. Factor $2x^2 - 8x - 5$:

   For $2x^2 - 8x - 5$, we look for two numbers that multiply to $2 \cdot (-5) = -10$ and add up to $-8$. These numbers are $-10$ and $1$.

   So, $2x^2 - 8x - 5 = (2x + 1)(x - 5)$

3. Factor $4x^2 - 1$:

   Notice that $4x^2 - 1$ is a difference of squares: $4x^2 - 1 = (2x - 1)(2x + 1)$

4. Factor $4x^2 - 4x + 1$:

   This is a perfect square trinomial: $4x^2 - 4x + 1 = (2x - 1)^2$

Step 2: Rewrite the expression with factored terms

Substituting these factored forms, the expression becomes:

$\frac{(3x - 2)(2x + 1)}{(2x + 1)(x - 5)} \times \frac{(2x - 1)(2x + 1)}{(2x - 1)^2}$

Step 3: Cancel common factors

Now we can cancel out common factors in the numerator and the denominator:

• Cancel $(2x + 1)$ from the first numerator and first denominator.
• Cancel one $(2x - 1)$ from the second numerator and the second denominator.

This leaves:

$\frac{3x - 2}{x - 5} \times \frac{1}{2x - 1}$

Step 4: Multiply the remaining terms

Multiply across the numerators and the denominators:

$\frac{(3x - 2) \cdot 1}{(x - 5)(2x - 1)} = \frac{3x - 2}{(x - 5)(2x - 1)}$

Final Answer

The simplified form of the expression is:

$\frac{3x - 2}{(x - 5)(2x - 1)}$

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Would you like a breakdown of any particular step or more examples of this type of problem?

1. How do you identify common factors in rational expressions?
2. What are special factorizations, like the difference of squares and perfect square trinomials?
3. Why is it important to factor completely before simplifying?
4. How can you verify your simplifications are correct?
5. What is the role of restrictions in rational expressions (e.g., values that make the denominator zero)?

Tip: Always check for common factors across numerators and denominators before multiplying out terms, as this makes simplification easier.