To find the prime factorization for \( \sqrt{162} \), we first need to factor 162.
162 can be factored into primes as follows:
-
Start by dividing by 2:
\( 162 \div 2 = 81 \)
So, \( 162 = 2 \times 81 \) -
Next, factor 81:
\( 81 = 3 \times 27 \)
\( 27 = 3 \times 9 \)
\( 9 = 3 \times 3 \)
So, \( 81 = 3 \times 3 \times 3 \times 3 = 3^4 \)
Putting it all together, we have:
\[ 162 = 2^1 \times 3^4 \]
Now, we take the square root of both sides:
\[ \sqrt{162} = \sqrt{2^1 \times 3^4} = \sqrt{2^1} \times \sqrt{3^4} = \sqrt{2} \times 3^2 = 3^2 \sqrt{2} \]
However, since the question asks for the prime factorization in expanded form directly from just the number 162, this means we are identifying the basic primes involved.
The prime factorization of 162 is:
- \( 2 \times 3 \times 3 \times 3 \times 3 \)
In terms of exponents, this is \( 2^1 \times 3^4 \).
In expanded form (as in the multiple-choice responses), this corresponds to:
2×3×3×3×3
So the correct answer is:
2 times 3 times 3 times 3 times 3.