how to factorization 2310

To be "technically" correct, I suppose it should be included!!

Not if we are talking prime factors only. The number 1 is no longer considered to be a prime number (hasn't been since the end of the 19th century) - that's why the factorize keyword doesn't list it!

If we are talking about all factors of 2310, then 1 is included, as is 2310 itself, and all the other numbers (combinations of the prime factors) that divide 2310 exactly, such as 6, 10, 14, 22, 15, ...etc.

Use the keyword factorize here:

$${factorize}{\left({\mathtt{2\,310}}\right)} \Rightarrow \left\{ \begin{array}{l}{\mathtt{2}}\\ {\mathtt{3}}\\ {\mathtt{5}}\\ {\mathtt{7}}\\ {\mathtt{11}}\\ \end{array} \right\}$$

This gives you the prime factors.

Let me add a little to Alan's answer.....Specifically....how do we arrive at those numbers??

Note we can write 2310 as

231 * 10  =

231 * 5 * 2       And there's a number theory property that says that if we can add the digits of a number, and that sum is divisible by three, then the number is divisible by three. So 231 = 2 + 3 + 1 = 6, and that's divisible by three. So, 231 is divisible by three. So we have

77 * 3 * 5 * 2 =

11 * 7 *3 * 5 * 2

Some answerers on here would also include "1" as a factor, but since "1" is a factor of any number, I usually don't include it.  But, if you want to throw it into the mix, be my guest. To be "technically" correct, I suppose it should be included!!