The expression you provided is:

\[
\frac{2x^2 - 5x - 42}{2x^2 - 13x + 42}
\]

To simplify this rational expression, we'll attempt to factor both the numerator and the denominator.

### 1. Factor the numerator \( 2x^2 - 5x - 42 \):

We need two numbers whose product is \( 2 \times (-42) = -84 \) and whose sum is \( -5 \). Those numbers are \( -12 \) and \( 7 \).

Thus, we can rewrite the middle term \( -5x \) as \( -12x + 7x \):

\[
2x^2 - 12x + 7x - 42 = 2x(x - 6) + 7(x - 6)
\]

Factoring out the common binomial \( (x - 6) \):

\[
2x^2 - 5x - 42 = (2x + 7)(x - 6)
\]

### 2. Factor the denominator \( 2x^2 - 13x + 42 \):

We need two numbers whose product is \( 2 \times 42 = 84 \) and whose sum is \( -13 \). Those numbers are \( -7 \) and \( -6 \).

Thus, we can rewrite the middle term \( -13x \) as \( -7x - 6x \):

\[
2x^2 - 7x - 6x + 42 = 2x(x - 3) - 7(x - 3)
\]

Factoring out the common binomial \( (x - 3) \):

\[
2x^2 - 13x + 42 = (2x - 7)(x - 3)
\]

### 3. Simplify the rational expression:

Now that we have factored both the numerator and the denominator, the expression becomes:

\[
\frac{(2x + 7)(x - 6)}{(2x - 7)(x - 3)}
\]

This is the simplified form of the given expression.

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Do you want more details, or do you have any questions?

Here are 5 questions that can help expand this concept:
1. How do you determine the correct factors when factoring quadratic expressions?
2. What happens if the numerator and denominator have a common factor that cancels out?
3. Can this expression be simplified further under specific conditions?
4. What is the significance of the roots of the numerator and denominator in this expression?
5. How can this expression be used to solve equations?

**Tip:** Always check for common factors in both the numerator and denominator before simplifying a rational expression.

Explore the step-by-step process of simplifying the rational expression (2x^2 - 5x - 42) / (2x^2 - 13x + 42) by factoring quadratic expressions. This problem demonstrates key algebraic concepts, making it ideal for students in Grades 9-10.

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