Question

The design for a rectangular solid shipping container has a volume of 30X cubed + 5x minus 15x where x represents inches the box has a height of 5x find the containers cross-sectional area using polynomial division

Answer 1

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:

1. Divide the first term of the numerator \( 30x^3 \) by the first term of the denominator \( 5x \):

\[
\frac{30x^3}{5x} = 6x^2
\]

2. Multiply \( 6x^2 \) by the entire denominator \( 5x \):

\[
6x^2 \cdot 5x = 30x^3
\]

3. Subtract that result from the original numerator:

\[
(30x^3 - 10x) - 30x^3 = -10x
\]

4. Next, take the remaining \( -10x \) and divide it by \( 5x \):

\[
\frac{-10x}{5x} = -2
\]

5. Multiply \( -2 \) by the entire denominator \( 5x \):

\[
-2 \cdot 5x = -10x
\]

6. 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:

\[
\boxed{6x^2 - 2}
\]