To determine the length of the rectangle, we can use the formula for the area of a rectangle:

$\text{Area} = \text{Length} \times \text{Width}$

Step 1: Substitute the given expressions for $A(x)$ and $W(x)$:

• $A(x) = x^2 - x - 6$
• $W(x) = x^2 - 9$

Thus: $\text{Length} = \frac{\text{Area}}{\text{Width}} = \frac{A(x)}{W(x)}$

Step 2: Perform the division:

$\text{Length} = \frac{x^2 - x - 6}{x^2 - 9}$

Both the numerator and denominator are factorizable:

• $x^2 - x - 6 = (x - 3)(x + 2)$
• $x^2 - 9 = (x - 3)(x + 3)$

So: $\text{Length} = \frac{(x - 3)(x + 2)}{(x - 3)(x + 3)}$

Cancel the common factor $(x - 3)$, provided $x \neq 3$: $\text{Length} = \frac{x + 2}{x + 3}$

Step 3: Interpret the result:

The length of the rectangle corresponds to the factorization result: $\text{Length} = x + 3 \quad \text{and} \quad x + 2$

This matches Option B: $\boxed{B. \, x + 3x + 2}$

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Would you like further details or steps? Here are 5 related questions to explore:

1. How do we determine if two polynomials can be factored?
2. Why can $x - 3$ be canceled in the fraction?
3. What happens to the dimensions of the rectangle when $x = 3$?
4. Can $A(x)$ or $W(x)$ ever result in a negative value for $x$?
5. What would the area of the rectangle be for a specific value, like $x = 5$?

Tip: Always double-check for restrictions in the domain (e.g., values where the denominator becomes zero).