To find the cross-sectional area of the rectangular solid shipping container, we start with the given volume polynomial \( V = 24x^3 + 6x^2 - 9x \) and the height of the container \( h = 3x \).
The volume of a rectangular solid is given by the formula:
\[ V = \text{Base Area} \times \text{Height} \]
We can express this in terms of the cross-sectional area \( A \):
\[ A = \frac{V}{h} \]
Substituting the volume and height:
\[ A = \frac{24x^3 + 6x^2 - 9x}{3x} \]
To perform the division \( \frac{24x^3 + 6x^2 - 9x}{3x} \), we divide each term in the numerator by \( 3x \):
-
For the term \( 24x^3 \): \[ \frac{24x^3}{3x} = 8x^2 \]
-
For the term \( 6x^2 \): \[ \frac{6x^2}{3x} = 2x \]
-
For the term \( -9x \): \[ \frac{-9x}{3x} = -3 \]
Now, combining these results, we have:
\[ A = 8x^2 + 2x - 3 \]
Thus, the cross-sectional area of the container is:
\[ \boxed{8x^2 + 2x - 3} \]