Question

What is the correct prime factorization (expansion) for 162−−−√
?(1 point)
Responses

2×2×2×3×3
2 times 2 times 2 times 3 times 3

2×3×3×3×3
2 times 3 times 3 times 3 times 3

2×2×3×3×3
2 times 2 times 3 times 3 times 3

2×2×2×2×3

Answer 1

To find the prime factorization of \( \sqrt{162} \), we first need to determine the prime factorization of 162 itself.

1. Start by dividing 162 by 2 (the smallest prime number):
\[
162 \div 2 = 81
\]
So, \( 162 = 2 \times 81 \).

2. 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:
\[
2 \text{ times } 3 \text{ times } 3 \text{ times } 3 \text{ times } 3
\]