# 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)

Feb 10, 2017

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

#### Explanation:

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

${V}_{\text{small}} = x \left(2 x\right) \left(x + 2\right) = x \left(2 {x}^{2} + 4 x\right) = 2 {x}^{3} + 4 {x}^{2}$

${V}_{\text{large}} = \left(x + 5\right) \left(x + 7\right) \left(3 x\right) = \left({x}^{2} + 12 x + 35\right) \left(3 x\right) = 3 {x}^{3} + 36 {x}^{2} + 105 x$

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

$D = {V}_{\text{large" - V_"small}}$

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

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

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

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

This cannot be factored any further.

Hopefully this helps!