Question

How do you find the prime factorization of 72?

Answer 1

`72 = 2^3xx3^2`

Explanation:

Divide the number to be factored by the prime numbers in turn. Test that result is a natural number. If so, it is a prime factor. Continue until the factorisation is complete.

In this example:
72 is even so 2 is a factor; `72/2 = 36`
36 and `36/2 = 18` are also even so we have two more 2's as prime factors.

`18/2 = 9 = 3xx3` Hence we have two 3's as prime factors.

Therefore the prime factors of 72 are `2xx2xx2xx3xx3 =2^3xx3^2`

Answer 2

The prime factorization for `72` is `color(red)(2)xxcolor(red)(2)xxcolor(red)(2)xxcolor(red)(3)xxcolor(red)(3)`.

Explanation:

The prime factors are prime numbers that when multiplied result in the number that is being factored. Continually divide by prime numbers until you reach the last prime number.

Divide `72` by the prime number `color(red)(2)`.

`72/2=36`

`36xxcolor(red)2=72`

Divide `36` by the prime number `color(red)(2)`.

`36/2=18`

`18xxcolor(red)2=36`

Divide `18` by the prime number `color(red)(2)`.

`18/2=9`

`9xxcolor(red)2=18`

Divide `9` by the prime number `color(red)(3)`.

`9/3=3`

`color(red)(3)xxcolor(red)3=9`

The prime factorization for `72` is `color(red)(2)xxcolor(red)(2)xxcolor(red)(2)xxcolor(red)(3)xxcolor(red)(3)`.

Answer 3

`2^3xx3^2`

Explanation:

If you are really stuck you can always draw a factor tree and see what you can build from than.

From this you can see that the prime factors are:

`2^3xx3^2`