3. REASON QUANTITATIVELY A rectangular box has dimensions of $$ x $$ inches, $$ (x+5) $$ inches, and $$ (x-2) $$ inches. If the volume of the box is $$ 30 x $$ cubic inches, explain how to find the dimensions of the box. Show your work. $$ (x+5)^{2}+(x-2)^{2}=30 x $$

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To determine the dimensions of the rectangular box, we'll use the information provided:

- **Dimensions of the box:** $$ x $$ inches, $$ (x + 5) $$ inches, and $$ (x - 2) $$ inches.
- **Volume of the box:** $$ 30x $$ cubic inches.

### Step 1: Set Up the Volume Equation

The volume $$ V $$ of a rectangular box is calculated by multiplying its length, width, and height. Given the dimensions, the volume can be expressed as:

$$
V = x \times (x + 5) \times (x - 2)
$$

According to the problem, the volume is also equal to $$ 30x $$ cubic inches. Therefore, we can write the equation:

$$
x \times (x + 5) \times (x - 2) = 30x
$$

### Step 2: Simplify the Equation

First, observe that both sides of the equation have a factor of $$ x $$. Assuming $$ x \neq 0 $$ (since a dimension of length 0 doesn't make physical sense), we can divide both sides by $$ x $$:

$$
(x + 5)(x - 2) = 30
$$

### Step 3: Expand the Left Side

Now, expand the left side of the equation using the distributive property (also known as the FOIL method for binomials):

$$
(x + 5)(x - 2) = x \cdot x + x \cdot (-2) + 5 \cdot x + 5 \cdot (-2)
$$
$$
= x^2 - 2x + 5x - 10
$$
$$
= x^2 + 3x - 10
$$

So, the equation becomes:

$$
x^2 + 3x - 10 = 30
$$

### Step 4: Form a Quadratic Equation

Subtract $$ 30 $$ from both sides to set the equation to zero:

$$
x^2 + 3x - 10 - 30 = 0
$$
$$
x^2 + 3x - 40 = 0
$$

### Step 5: Solve the Quadratic Equation

We can solve the quadratic equation $$ x^2 + 3x - 40 = 0 $$ using the quadratic formula:

$$
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
$$

For our equation, $$ a = 1 $$, $$ b = 3 $$, and $$ c = -40 $$. Plugging these values in:

$$
x = \frac{-3 \pm \sqrt{3^2 - 4 \times 1 \times (-40)}}{2 \times 1}
$$
$$
x = \frac{-3 \pm \sqrt{9 + 160}}{2}
$$
$$
x = \frac{-3 \pm \sqrt{169}}{2}
$$
$$
x = \frac{-3 \pm 13}{2}
$$

This gives us two potential solutions:

1. **Positive Solution:**
   $$
   x = \frac{-3 + 13}{2} = \frac{10}{2} = 5
   $$
2. **Negative Solution:**
   $$
   x = \frac{-3 - 13}{2} = \frac{-16}{2} = -8
   $$

Since a physical dimension cannot be negative, we discard $$ x = -8 $$ and accept $$ x = 5 $$ inches.

### Step 6: Find the Other Dimensions

Now that we've found $$ x = 5 $$ inches, we can find the other dimensions:

1. **Second Dimension:**
   $$
   x + 5 = 5 + 5 = 10 \text{ inches}
   $$
2. **Third Dimension:**
   $$
   x - 2 = 5 - 2 = 3 \text{ inches}
   $$

### Step 7: Verify the Volume

To ensure our solution is correct, let's verify the volume with the found dimensions:

$$
\text{Volume} = x \times (x + 5) \times (x - 2) = 5 \times 10 \times 3 = 150 \text{ cubic inches}
$$

According to the problem, the volume should be $$ 30x = 30 \times 5 = 150 $$ cubic inches. Both calculations agree, confirming that our solution is correct.

### **Final Dimensions of the Box:**

- **Length:** $$ 5 $$ inches
- **Width:** $$ 10 $$ inches
- **Height:** $$ 3 $$ inches
To find the dimensions of the box, set up the volume equation $$ x \times (x + 5) \times (x - 2) = 30x $$. Simplify to $$ x^2 + 3x - 40 = 0 $$, solve for $$ x = 5 $$ inches. The other dimensions are $$ 10 $$ inches and $$ 3 $$ inches.

3. REASON QUANTITATIVELY A rectangular box has dimensions of \( x \) inches, \( (x+5) \) inches, and \( (x-2) \) inches. If the volume of the box is \( 30 x \) cubic inches, explain how to find the dimensions of the box. Show your work. \[ (x+5)^{2}+(x-2)^{2}=30 x \]

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