# What are the prime factors of 36?

#### Understand the Problem

The question is asking for the prime factors of the number 36. To solve this, we need to identify which prime numbers multiply together to give the product of 36.

The prime factors of 36 are $2^2 \times 3^2$.

The prime factors of 36 are $2^2 \times 3^2$.

#### Steps to Solve

1. Identify the smallest prime number
Start with the smallest prime number, which is 2. Check if 2 divides 36 evenly. Since $36 \div 2 = 18$, 2 is a prime factor.

2. Divide and repeat
Now take the quotient from the first division, which is 18, and check again with 2. Since $18 \div 2 = 9$, 2 is still a prime factor.

3. Move to the next prime number
Now take 9 and check for the next smallest prime number, which is 3. Since $9 \div 3 = 3$, 3 is a prime factor.

4. Continue with the quotient
Take the quotient from the previous step, which is 3. Check again with 3, as $3 \div 3 = 1$.

5. Compile the prime factors
Now that we have reached 1, we can compile our prime factors. The prime factors of 36 are 2 and 3, and we counted them as follows:

• 2 appears twice (from our original divisions).
• 3 appears twice (from our divisions).

So, we can express the prime factorization as:
$$36 = 2^2 \times 3^2$$

The prime factors of 36 are $2^2 \times 3^2$.