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

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

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

2×3×3×3×3

1 answer

To find the prime factorization of \(\sqrt{162}\), we first factor 162 into its prime factors.

  1. Start with 162:

    • \(162\) is even, so divide by \(2\):
      \(162 \div 2 = 81\)
  2. Next, factor \(81\):

    • \(81 = 9 \times 9 = 3 \times 3 \times 3 \times 3 = 3^4\)

Putting everything together: \[ 162 = 2^1 \times 3^4 \]

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

However, since we are asked for the prime factorization in expanded form, let's express \(\sqrt{162}\) in terms of its factors.

The prime factorization of \(162\) is \(2 \times 3^4\), so the complete prime factorization of \(162\) is: \[ 2 \times 3 \times 3 \times 3 \times 3 \]

However, since we are looking for the prime factorization of \(\sqrt{162}\) in terms of integers: \[ \sqrt{162} = 3 \times 3 \times 3 = 9 \sqrt{2} \]

Since our focus was on the prime factorization of the number as \(3\) to some power (but for the basic calculation of \(\sqrt{162}\))

Thus, the closest choice reflecting the correct prime factorization should include only whole integers without the \(\sqrt{\text{2}}\).

Looking at the original options:

  • \(2 \times 3 \times 3 \times 3\): This does not appear accurately in the provided options; thus, the best candidate might be \(2\times 2\times 3\times 3\times 3\), given that noticing \(\sqrt{2}\) separated would occur if kept with.

Further simplifications maintain understanding that one should primarily focus on what is squared effectively under approximations like such, for clarity backloop references.

So: The best option based on prime factors of 162 in integer form would be to represent them as high roots while qualifying those factors as: Omitting extraneous markup digits associated with rounding digits signifies factors in prime involved expansions.

Thus, end choice and validity mode remains the first one. Answer: \(2 \times 2 \times 2 \times 3 \times 3\) is the correct representation but to clarify on rooted forms differs in visuals.

Always clarify from simplicity focused on even spread mechanics, deriving specific inner types is always a plus for theorem references.