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The design for a rectangular solid shipping container has a volume of 30X cubed + 5x minus 15x where x represents inches the bo...Asked by Honeybeee
The design for a rectangular solid shipping container has a volume of 24x3+6x2−9x where x represents inches. The box has a height of 3x . Find the container’s cross-sectional area using polynomial division.(1 point) cross-sectional area = $$
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GPT-4o mini
To find the cross-sectional area of the rectangular solid shipping container given its volume \( V(x) = 24x^3 + 6x^2 - 9x \) and height \( h = 3x \), we will perform polynomial division to divide the volume by the height.
1. **Write down the expressions**:
- Volume: \( V(x) = 24x^3 + 6x^2 - 9x \)
- Height: \( h(x) = 3x \)
2. **Perform polynomial division**:
We want to divide \( V(x) \) by \( h(x) \):
\[
\frac{V(x)}{h(x)} = \frac{24x^3 + 6x^2 - 9x}{3x}
\]
To perform the division, we can factor out \( 3x \):
- Divide the leading term of the numerator by the leading term of the denominator:
\[
\frac{24x^3}{3x} = 8x^2
\]
- Multiply \( 8x^2 \) by \( 3x \):
\[
8x^2 \cdot 3x = 24x^3
\]
- Subtract this from the original polynomial:
\[
(24x^3 + 6x^2 - 9x) - 24x^3 = 6x^2 - 9x
\]
Now divide the next term:
- Divide the leading term of the modified numerator by the leading term of the denominator:
\[
\frac{6x^2}{3x} = 2x
\]
- Multiply \( 2x \) by \( 3x \):
\[
2x \cdot 3x = 6x^2
\]
- Subtract:
\[
(6x^2 - 9x) - 6x^2 = -9x
\]
Now divide the next term:
- Divide the leading term:
\[
\frac{-9x}{3x} = -3
\]
- Multiply \( -3 \) by \( 3x \):
\[
-3 \cdot 3x = -9x
\]
- Subtract:
\[
-9x - (-9x) = 0
\]
After the division, we've found:
\[
\frac{V(x)}{h(x)} = 8x^2 + 2x - 3
\]
3. **Cross-sectional area**:
Therefore, the cross-sectional area \( A(x) \) of the container is given by:
\[
A(x) = 8x^2 + 2x - 3
\]
Thus, the container's cross-sectional area is:
\[
\boxed{8x^2 + 2x - 3} \text{ square inches.}
\]
1. **Write down the expressions**:
- Volume: \( V(x) = 24x^3 + 6x^2 - 9x \)
- Height: \( h(x) = 3x \)
2. **Perform polynomial division**:
We want to divide \( V(x) \) by \( h(x) \):
\[
\frac{V(x)}{h(x)} = \frac{24x^3 + 6x^2 - 9x}{3x}
\]
To perform the division, we can factor out \( 3x \):
- Divide the leading term of the numerator by the leading term of the denominator:
\[
\frac{24x^3}{3x} = 8x^2
\]
- Multiply \( 8x^2 \) by \( 3x \):
\[
8x^2 \cdot 3x = 24x^3
\]
- Subtract this from the original polynomial:
\[
(24x^3 + 6x^2 - 9x) - 24x^3 = 6x^2 - 9x
\]
Now divide the next term:
- Divide the leading term of the modified numerator by the leading term of the denominator:
\[
\frac{6x^2}{3x} = 2x
\]
- Multiply \( 2x \) by \( 3x \):
\[
2x \cdot 3x = 6x^2
\]
- Subtract:
\[
(6x^2 - 9x) - 6x^2 = -9x
\]
Now divide the next term:
- Divide the leading term:
\[
\frac{-9x}{3x} = -3
\]
- Multiply \( -3 \) by \( 3x \):
\[
-3 \cdot 3x = -9x
\]
- Subtract:
\[
-9x - (-9x) = 0
\]
After the division, we've found:
\[
\frac{V(x)}{h(x)} = 8x^2 + 2x - 3
\]
3. **Cross-sectional area**:
Therefore, the cross-sectional area \( A(x) \) of the container is given by:
\[
A(x) = 8x^2 + 2x - 3
\]
Thus, the container's cross-sectional area is:
\[
\boxed{8x^2 + 2x - 3} \text{ square inches.}
\]
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