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What is the product of 8, square root of, 14, end square root8
14
​
and 5, square root of, 12, end square root5
12
​
in simplest radical form?
Answer
Attempt 1 out of 5

Answer 1

To find the product of \( 8\sqrt{14} \) and \( 5\sqrt{12} \), we can follow these steps:

1. **Multiply the coefficients (whole numbers)**:
\[
8 \times 5 = 40
\]

2. **Multiply the radical parts**:
\[
\sqrt{14} \times \sqrt{12} = \sqrt{14 \times 12} = \sqrt{168}
\]

3. **Combine the results from steps 1 and 2**:
\[
40\sqrt{168}
\]

4. **Simplify \( \sqrt{168} \)**:
First, we can factor \( 168 \):
\[
168 = 4 \times 42 = 4 \times 6 \times 7
\]
Thus:
\[
\sqrt{168} = \sqrt{4 \times 42} = \sqrt{4} \times \sqrt{42} = 2\sqrt{42}
\]

5. **Replace \( \sqrt{168} \) in the product**:
\[
40\sqrt{168} = 40 \times 2\sqrt{42} = 80\sqrt{42}
\]

Therefore, the product of \( 8\sqrt{14} \) and \( 5\sqrt{12} \) in simplest radical form is:
\[
\boxed{80\sqrt{42}}
\]