Question

What is the correct prime factorization (Expansion) for \sqrt(162) ?

A. 2 x 3 x 3 x 3 x 3
B. 2 x 2 x 2 x 3 x 3
C. 2 x 2 x 3 x 3 x 3
D. 2 x 2 x 2 x 2 x 3

Answer 1

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

1. Divide \(162\) by \(2\) (the smallest prime):
\[
162 \div 2 = 81
\]
So, \(2\) is a factor. We have \(2\) as a factor once.

2. Now factor \(81\):
\[
81 = 3 \times 27
\]
\[
27 = 3 \times 9
\]
\[
9 = 3 \times 3
\]
So, \(81 = 3^4\).

Combining these, we have:
\[
162 = 2^1 \times 3^4
\]

Now, we need to find \(\sqrt{162}\):
\[
\sqrt{162} = \sqrt{2^1 \times 3^4} = \sqrt{2^1} \times \sqrt{3^4}
\]
\[
= 2^{1/2} \times 3^{4/2} = 2^{1/2} \times 3^2 = \sqrt{2} \times 9 = 9\sqrt{2}
\]

However, if we are looking for the prime factorization of \(162\) in expanded form before taking the square root, it is:
\[
162 = 2^1 \times 3^4 = 2 \times 3 \times 3 \times 3 \times 3
\]

That matches with choice \(A\):
\[
2 \times 3 \times 3 \times 3 \times 3
\]

Thus, the correct prime factorization expansion for \(\sqrt{162}\) is:

**Answer: A. \(2 \times 3 \times 3 \times 3 \times 3\)**.