To find the prime factorization of \(\sqrt{162}\), we can first factor 162 into its prime components:
- Start with 162 and divide by 2 (the smallest prime): \[ 162 \div 2 = 81 \quad \text{(two is a prime factor)} \]
- Now factor 81, which is \(3^4\): \[ 81 \div 3 = 27 \ 27 \div 3 = 9 \ 9 \div 3 = 3 \ 3 \div 3 = 1 \] So, \(81 = 3 \times 3 \times 3 \times 3 = 3^4\).
Now combining these results: \[ 162 = 2 \times 3^4 \]
Next, when taking the square root: \[ \sqrt{162} = \sqrt{2 \times 3^4} = \sqrt{2} \times \sqrt{3^4} \]
Since \(\sqrt{3^4} = 3^2 = 9\), we get: \[ \sqrt{162} = 9\sqrt{2} \]
Now we identify the prime factors: The prime factorization of \(162\) is \(2^1 \times 3^4\), which can also be written as \(2 \times 3 \times 3 \times 3 \times 3\) or \(2 \times 3^4\).
However, according to your choices, the correct expansion represented in the list would be: \[ 2 \times 3 \times 3 \times 3 \times 3 \] This can also be represented as: \[ 2 \text{ times } 3 \text{ times } 3 \text{ times } 3 \text{ times } 3 \]
Thus, the correct answer is: 2×3×3×3×3.