To find the prime factorization of \( \sqrt{162} \), we first need to determine the prime factorization of 162 itself.
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Start by dividing 162 by 2 (the smallest prime number): \[ 162 \div 2 = 81 \] So, \( 162 = 2 \times 81 \).
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Next, factor 81: \[ 81 = 3 \times 27 \] Continuing to factor 27: \[ 27 = 3 \times 9 \] And factor 9: \[ 9 = 3 \times 3 \] So, \( 81 = 3^4 \).
Putting this all together: \[ 162 = 2 \times 3^4 \]
Now, \( \sqrt{162} \) can be expressed as: \[ \sqrt{162} = \sqrt{2 \times 3^4} = \sqrt{2} \times \sqrt{3^4} = \sqrt{2} \times 3^2 = 3 \times 3 \times \sqrt{2} \]
However, since we're asked for the prime factorization, we focus only on the prime factors before the square root.
The prime factors of 162 are \( 2^1 \) and \( 3^4 \), which can be multiplied respectively as:
- \( 2 = 2 \)
- \( 3 \times 3 \times 3 \times 3 \)
In expanded form, this is: \[ 2 \times 3 \times 3 \times 3 \times 3 \]
Looking through the provided options, the correct prime factorization of \( \sqrt{162} \) is: \[ 2 \times 3 \times 3 \times 3 \times 3 \quad \text{(Option 2)} \] So the rephrased response is: \[ 2 \text{ times } 3 \text{ times } 3 \text{ times } 3 \text{ times } 3 \]