Question

what numbers i can use to make in exponents a prime factorization of 198

Answer 1

To perform a prime factorization of the number \(198\), we can break it down into its prime factors.

1. Start by dividing \(198\) by the smallest prime number, which is \(2\):
\[
198 \div 2 = 99
\]
Thus, \(198 = 2 \times 99\).

2. Next, we factor \(99\). The smallest prime that divides \(99\) is \(3\):
\[
99 \div 3 = 33
\]
Therefore, \(99 = 3 \times 33\) and we have \(198 = 2 \times 3 \times 33\).

3. Now, we continue factoring \(33\). The smallest prime dividing \(33\) is \(3\) again:
\[
33 \div 3 = 11
\]
So \(33 = 3 \times 11\) and we have \(198 = 2 \times 3 \times 3 \times 11\), or \(198 = 2 \times 3^2 \times 11\).

4. Now, \(11\) is a prime number and cannot be factored further.

Therefore, the prime factorization of \(198\) is:
\[
198 = 2^1 \times 3^2 \times 11^1
\]

In summary, the prime factors of \(198\) are \(2\), \(3\), and \(11\) in exponential form \(2^1, 3^2, 11^1\).