What is the prime factorization of 525 in exponential form?

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To find the prime factors of a number, start by dividing until you cannot divide any other numbers into any factors.

For 525, begin by dividing by 5 this leave 105 x 5

105 can be further factored by dividing by 5 again = 21 x 5

21 can be further factored to 3 x 7

At this point nothing can be be further factored. The factors of 525 are 5, 5, 7, and 3.

To write this answer in exponential form, you need to write any factors that appear more than once as an exponent. In this case the only factor that appears more than once are the 5s that appear twice.

your final answer will be 5^2 x 7 x 3

We can write 525 as a product of prime factors in the following way:

525 = 5 * 5 * 7 * 3

Now 7 has been used only once, so we raise 7 to the power 1, 7^1

3 also has been used once, so raise that to the power 1, 3^1

5 has been used twice, so raise that to the power 2, 5^2

525 = 5 * 5 * 7 * 3 = 7^1 * 5^2 * 3^1

**The required prime factorization is 7^1 * 5^2 * 3^1**

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