2 x 3^5 x 5 are all the prime factors of 2430.
You can discover this yourself by knowing what clues to look for..
If your number is even, i.e., ends in 0, or 2, or 4, 6, or 8, then 2 is a factor.
So, divide it by 2 to get 1215. The result can not be further divided by 2. So, 2 is only a factor one time. If other factors exist, they produce 1215.
Thus, we will work the result, 1215, to see what other factors may be in it.
If a number ends in 0 or 5 then 5 is a factor. Clearly, 1215 does end in 5.
So, divide the result 1215 by 5 to get 243.
Thus far, the factors identified are 2, and 5 and 243. But what easy test can we do on 243 to check for any other factors?
If the sum of its digits, i.e., 2 + 4 + 3 which is 9 is divisible by 9 or 3, then they are a factor. 9 is not a prime, but it shows that 3 is a factor more than once.
Divide 243 by 3, and you see it goes evenly 81 times.
Continue working this new result. Notice its sum of digits 8 + 1 is 9. So, keep dividing each result by 3.as many times as it will go evenly.
All in all, 3 goes this many times into 243 and each result: 3 x 3 x 3 x 3 x 3.
This is 3 to the 5th power, written as 3^5.