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Review Seth's steps for rewriting and simplifying an expression.

Given [tex]\[8 x^6 \sqrt{200 x^{13}} \div 2 x^5 \sqrt{32 x^7}\][/tex]

Step 1: [tex]\[8 x^6 \sqrt{4 \cdot 25 \cdot 2 \cdot \left(x^6\right)^2 \cdot x} \div 2 x^5 \sqrt{16 \cdot 2 \cdot \left(x^3\right)^2 \cdot x}\][/tex]

Step 2: [tex]\[8 \cdot 2 \cdot 5 \cdot x^6 \cdot x^6 \sqrt{2 x} \div 2 \cdot 16 \cdot x^5 \cdot x^3 \sqrt{2 x}\][/tex]

Step 3: [tex]\[80 x^{12} \sqrt{2 x} \div 32 x^8 \sqrt{2 x}\][/tex]

Step 4: [tex]\[\frac{80 x^{12} \sqrt{2 x}}{32 x^8 \sqrt{2 x}}\][/tex]

Step 5: [tex]\[\frac{5}{2} x^4\][/tex]

Seth's first mistake was made in [tex]$\square$[/tex], where he

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Answer :

Sure, let's go through the steps in detail to find out where Seth made his mistake.

We start with the given expression:
[tex]\[ 8x^6 \sqrt{200x^{13}} \div 2x^5 \sqrt{32x^7} \][/tex]

### Step 1
Seth rewrites the radicals:
[tex]\[ 8x^6 \sqrt{4 \cdot 25 \cdot 2 \cdot (x^6)^2 \cdot x} \div 2x^5 \sqrt{16 \cdot 2 \cdot (x^3)^2 \cdot x} \][/tex]

This step is correct because:
[tex]\[ \sqrt{200x^{13}} = \sqrt{4 \cdot 25 \cdot 2 \cdot x^{12} \cdot x} \][/tex]
[tex]\[ \sqrt{32x^7} = \sqrt{16 \cdot 2 \cdot x^6 \cdot x} \][/tex]

### Step 2
Next, Seth simplifies the radicals:
[tex]\[ 8 \cdot 2 \cdot 5 \cdot x^6 \cdot x^6 \sqrt{2x} \div 2 \cdot 4 \cdot 2 \cdot x^5 \cdot x^3 \sqrt{2x} \][/tex]

This step is correct as well. The radicals were reduced correctly and he simplified the constants:
[tex]\[ \sqrt{200x^{13}} = \sqrt{4 \cdot 25 \cdot 2 \cdot x^{12} \cdot x} = 2 \cdot 5 \cdot x^6 \sqrt{2x} \][/tex]
[tex]\[ \sqrt{32x^7} = \sqrt{16 \cdot 2 \cdot x^6 \cdot x} = 4 \cdot x^3 \sqrt{2x} \][/tex]

### Step 3
Then, Seth proceeds with the division:
[tex]\[ 80x^{12} \sqrt{2x} \div 32x^8 \sqrt{2x} \][/tex]

However, there is a mistake. The right expression should be:
[tex]\[ 8 \times 2 \times 5 = 80 \text{ (numerator) }\][/tex]
And
[tex]\[ 2 \times 4 = 8 \text{ (denominator) }\][/tex]

### Correct Step 3 should be:
[tex]\[ 80 x^{12} \sqrt{2x} \div 8 x^8 \sqrt{2x} \][/tex]

### Step 4
Seth continues:
[tex]\[ \frac{80 x^{12} \sqrt{2x}}{32 x^8 \sqrt{2x}} \][/tex]

Since there was an error in the previous step with the 32, it should be:
[tex]\[ \frac{80 x^{12} \sqrt{2x}}{8 x^8 \sqrt{2x}} \][/tex]

The [tex]\(\sqrt{2x}\)[/tex] terms cancel each other out:
[tex]\[ \frac{80 x^{12}}{8 x^8} = 10 x^4 \][/tex]

### Step 5
[tex]\[ \frac{80 x^{12}}{32 x^8} = \frac{5}{2}x^4 \][/tex]

Therefore, Seth made his mistake in Step 3, where he incorrectly simplified the division of the constants.

So, the correct step where Seth made the mistake was in Step 3.

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