To find the possible length and width of a rectangular box with a given volume, we can use the formula for the volume of a rectangular box:

$\text{Volume} = \text{Length} \times \text{Width} \times \text{Height}$

Given:

• Volume = $5x^2 + 60x + 160$ cm³
• Height = 5 cm

Let's denote the length by $L$ and the width by $W$. The formula for volume becomes:

$L \times W \times 5 = 5x^2 + 60x + 160$

To isolate the product of length and width:

$L \times W = \frac{5x^2 + 60x + 160}{5}$

Simplifying the right side:

$L \times W = x^2 + 12x + 32$

So, the possible product of the length and width is given by $L \times W = x^2 + 12x + 32$.

Now, we can factorize the quadratic expression $x^2 + 12x + 32$ to find the possible values of $L$ and $W$:

$x^2 + 12x + 32 = (x + 4)(x + 8)$

Thus, the possible dimensions (length $L$ and width $W$) could be $L = x + 4$ and $W = x + 8$, or vice versa.

Summary:

• Length = $x + 4$ cm
• Width = $x + 8$ cm

These represent possible dimensions of the rectangular box based on the given polynomial expression.

Would you like further details on how to choose specific values for $x$ or have any other questions?

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Here are some related questions:

1. What values of $x$ make the dimensions positive?
2. How would changing the height affect the dimensions?
3. What happens if the height is a variable instead of a constant?
4. How can you verify that the factorization is correct?
5. How does this problem change if the volume is given as a different polynomial?

Tip: Always double-check the factorization to ensure that the dimensions are physically meaningful (e.g., non-negative).