What is Prime Factorization?
Prime factorization allows us to write any number as a product of prime factors.
For example, consider the number 360. Let us write this number as follows:
\[\begin{array}{l}360 = 2 \times 2 \times 2 \times 3 \times 3 \times 5\\\;\;\;\;\;\, = {2^3} \times {3^2} \times {5^1}\end{array}\]
Basically, we have expressed 360 as a product of Prime Numbers. Let us take some other (composite) numbers and express them as products of prime numbers:
\[\begin{array}{l}\;\,80 = 2 \times 2 \times 2 \times 2 \times 5\\\;\;\;\;\;\,\,\, = {2^4} \times {5^1}\\144 = 2 \times 2 \times 2 \times 2 \times 3 \times 3\\\;\;\;\;\;\,\,\, = {2^4} \times {3^2}\\600 = 2 \times 2 \times 2 \times 3 \times 3 \times 5\\\;\;\;\;\;\,\,\,= {2^3} \times {3^2} \times {5^1}\end{array}\]
Expressing a number this way (as a product of primes) is called the prime factorization of that number. Note that the prime factorization of a prime number is trivial, since the only nonunity divisor of any prime number is that number itself.
✍Note: if a number is not prime, it has a Prime Numbers as its factor.
Can every number be primefactorized in a unique way?
Two questions now arise:

Can every composite number be expressed as a product of primes, that is, can every composite number be prime factorized?

If we prime factorize a composite number, is that factorization unique?
The answer is YES for both the above questions. The uniqueness of prime factorization is an incredibly important result, thus earning the name of the fundamental theorem of arithmetic.
Solved Examples:
Example 1: What are the prime factors of 12?
Solution: The factors of 12 are 1, 2, 3, 4, 6, and 12. The prime factors are 2 and 3.
Example 2: If \(N = a \times b \times c\) such that \(a\), \(b\) and \(c\) are three different prime numbers, how many positive divisors does \(N\) have excluding 1 and itself.
Solution: Here, \(N = a \times b \times c\), we can conclude that \(a\), \(b\) and \(c\) are the factors of \(N\). Since \(a\), \(b\) and \(c\) are also prime numbers, we can't factor them to get any other number, so that gives us a total of 3 numbers.
But we know that if \(a\) and \(b\) are factors of \(N\), then \(a \times b\) is also a factor of \(N\). So a combination of two factors out of these three factors is also a divisor of \(N\). In other words, we have \(a \times b\), \(b \times c\), \(c \times a\) as factors of \(N\), which are another 3 in addition to the 3 above.
Note that \(a \times b \times c\) is also a combination that is a factor of \(N\), but it equals the number itself and is therefore omitted.
So we have a total of 6 divisors, excluding 1 and the number itself.
Challenge: Write all the factors of 120 and separate prime factors from them.
⚡Tip: Similar to example 1.