To find the prime factorization of \(\sqrt{162}\), we first need to factor 162 into its prime factors.
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Divide 162 by 2: \[ 162 \div 2 = 81 \] So, \(162 = 2 \times 81\).
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Now, factor 81: \[ 81 = 3 \times 27 \] And, \(27 = 3 \times 9\) and \(9 = 3 \times 3\). Therefore, \(81 = 3 \times 3 \times 3 \times 3 = 3^4\).
So, the complete prime factorization of 162 is: \[ 162 = 2 \times 3^4 \]
Now, we need to find the square root: \[ \sqrt{162} = \sqrt{2 \times 3^4} = \sqrt{2} \times \sqrt{3^4} = \sqrt{2} \times 3^2 = 3\sqrt{2} \times 9 \]
However, the question refers to the prime factorization, which we have found to be \(2 \times 3^4\).
If we expand \(3^4\): \[ 3^4 = 3 \times 3 \times 3 \times 3 \]
Now combining it with \(2\): \[ 162 = 2 \times 3 \times 3 \times 3 \times 3 \]
This can also be expressed as: \[ 2 \times 3 \times 3 \times 3 \times 3 \]
So looking through the given responses:
- The response 2×3×3×3×3 corresponds to the prime factorization of \(162\).
Hence, the correct response is: 2 times 3 times 3 times 3 times 3.