To find the cross-sectional area of the rectangular solid shipping container using polynomial division, we start with the given volume and height.
The volume \( V \) of the container is given by:
\[ V = 30x^3 + 5x - 15x \]
First, simplify the volume expression:
\[ V = 30x^3 - 10x \]
The height of the container is given as \( h = 5x \).
To find the cross-sectional area \( A \), we can use the formula for volume:
\[ V = \text{Base Area} \times \text{Height} \]
Given that the volume can be expressed in terms of base area \( A \) and height \( h \):
\[ A \times h = V \]
Substituting the expressions we have:
\[ A \times 5x = 30x^3 - 10x \]
To find the base area \( A \), we will divide the volume by the height:
\[ A = \frac{V}{h} = \frac{30x^3 - 10x}{5x} \]
Now, simplify the expression using polynomial division:
- Divide the first term of the numerator \( 30x^3 \) by the first term of the denominator \( 5x \):
\[ \frac{30x^3}{5x} = 6x^2 \]
- Multiply \( 6x^2 \) by the entire denominator \( 5x \):
\[ 6x^2 \cdot 5x = 30x^3 \]
- Subtract that result from the original numerator:
\[ (30x^3 - 10x) - 30x^3 = -10x \]
- Next, take the remaining \( -10x \) and divide it by \( 5x \):
\[ \frac{-10x}{5x} = -2 \]
- Multiply \( -2 \) by the entire denominator \( 5x \):
\[ -2 \cdot 5x = -10x \]
- Finally, subtract that from the remaining term:
\[ -10x - (-10x) = 0 \]
Now, we have completed the division. Therefore, the cross-sectional area \( A \) is:
\[ A = 6x^2 - 2 \]
Final Result:
The cross-sectional area of the container is:
\[ \boxed{6x^2 - 2} \]
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