Find the possible length and width of a rectangular box with a volume of 5×²+60×+160 centimeter if its height is 5cm

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).

Explore how to find the dimensions (length and width) of a rectangular box with a given volume expression. This problem involves solving a quadratic equation to determine possible dimensions based on the polynomial volume. Suitable for Grades 10-12.

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