How do you solve this?

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How do you solve this?

A warehouse is packing materials to ship out to a customer. A small box has the dimensions of x inches, (x+ 2) inches and (2x) inches. Where x is the length in inches of width. A large box has the dimensions of (x+5) inches, (x + 7) inches and (3x) inches.

*Recall that Volume = (Length)(Width)(Height)

Part 1: Write an expression that represents the dimensions of the small box.

Part 2: Write an expression that represents the dimensions of the large box.

Part 3: What is the difference in the volumes of the two boxes? Show or explain or work.

(I think I understand parts 1 and 2 but I'm definitely struggling with part 3)

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Answer 1 · 2017-02-10T15:29:32

The difference in volume is $x (x^{2} + 32 x + 105)$ $x(x^2 + 32x + 105)$ cubic inches.

Explanation:

I'm going to do all three parts just so that you can compare your work.

$V_{small} = x (2 x) (x + 2) = x (2 x^{2} + 4 x) = 2 x^{3} + 4 x^{2}$ $V_"small" = x(2x)(x +2) = x(2x^2 + 4x) = 2x^3 + 4x^2$

$V_{large} = (x + 5) (x + 7) (3 x) = (x^{2} + 12 x + 35) (3 x) = 3 x^{3} + 36 x^{2} + 105 x$ $V_"large" = (x + 5)(x + 7)(3x) = (x^2 + 12x + 35)(3x) = 3x^3 + 36x^2 + 105x$

Now, subtract the largest volume from the small volume to get the difference in volume. Call the difference $D$ $D$ .

$D = V_{large} - V_{small}$ $D = V_"large" - V_"small"$

$D = 3 x^{3} + 36 x^{2} + 105 x - (2 x^{3} + 4 x^{2})$ $D = 3x^3 + 36x^2 + 105x - (2x^3 + 4x^2)$

$D = 3 x^{3} + 36 x^{2} + 105 x - 2 x^{3} - 4 x^{2}$ $D = 3x^3 + 36x^2 + 105x - 2x^3 - 4x^2$

$D = x^{3} + 32 x^{2} + 105 x$ $D = x^3 + 32x^2 + 105x$

$D = x (x^{2} + 32 x + 105)$ $D = x(x^2 + 32x + 105)$

This cannot be factored any further.

Hopefully this helps!

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