Lesson Video: Prime Factorization with Exponents | Nagwa Lesson Video: Prime Factorization with Exponents | Nagwa

Lesson Video: Prime Factorization with Exponents Mathematics • 6th Grade

In this video, we will learn how to use different strategies to find prime factorization using exponents.

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Video Transcript

In this video, we’ll explore how to use different strategies to find prime factorization using exponents. This is sometimes called writing a number as a product of its prime factors and can be hugely useful in helping us to calculate the greatest common factor or least common multiple of two or more numbers.

Let’s begin by recalling what we mean by a factor. A factor is a number that divides into another number without leaving a remainder. When listing factors of a number, we often list them in pairs for ease. For example, the product of one and 12 is 12. So a factor pair of 12 is one and 12. We also know that two times six is 12 and three times four is 12. So all the factors of 12 are one, 12, two, six, three, and four. And what about a prime number? A prime number is a number that has exactly two factors, one and itself. The first few prime numbers that you should be able to recognize are two, three, five, seven, 11, 13, 17, and 19. Now, a common mistake is to think that nine is a prime number. But its factors are one, three, and nine. It has three factors, not two.

Similarly, one is not a prime number, since it only has one factor, itself. When we write a number as a product of its prime factors, that’s called prime factorization, we look to find all the prime numbers that multiply together to make the original number. There are a couple of methods. We can draw a prime factor tree or use the division method. We’ll begin by taking a look at the factor tree method.

Which of the following is the prime factorization of 18? Is it (A) 18, (B) two times three times three, (C) two, (D) two times nine, or (E) three?

When we find the prime factorization of 18, we’re looking for all the prime numbers which multiply together to make 18. And we know the first few prime numbers to be two, three, five, seven, 11, and 13. It can be nice to begin by testing whether our smallest prime number, whether two, is a factor of the original number. Well, 18 is an even number. So we know two is definitely a factor. In fact, 18 divided by two is nine. And our factor tree begins. Its first two branches are two and nine. Now, since two is a prime number, we circle it. And this indicates we stop at this branch here.

Nine, however, is not a prime number. So we repeat this process for nine. Nine is not even, so it’s not divisible by two. And instead, we go to the next prime on our list. Three is a factor of nine. In fact, three squared is nine. So nine divided by three is three. This means the next two branches on our tree are three and three. Three is prime, so we circle both of these. And we see that we can go no further. We aren’t done, though. We know that these three prime numbers multiply together to make the original. So in other words, 18 is the product of them. It’s two times three times three, which is option (B).

Notice that we can also write this as two times three squared. Had the question asked us to write our original number in exponent form, then this is what we would have needed to have done. Be careful, though, a common mistake is just to think that we can give our answers in a list, that’s two, three, three, or as a sum, two plus three plus three. Well, both of these are incorrect answers. We must give the solution as a product.

In our next example, we’ll consider an alternative method. It’s called the division method.

Write 60 as a product of its prime factors, giving the answer in exponent form.

Writing a number as a product of its prime factors is sometimes called prime factorization. Product means time. So we want all the prime numbers which multiply together to make here the original number that’s 60. We’re going to use the division method. Now, the first step in this method is to find the smallest prime number that’s also a factor of the original number, so a factor of 60. Our first few primes are two, three, five, seven, and 11. Now, since 60 is an even number, we know that it must be divisible by two. It has a factor of two. And so we divide 60 by this number. 60 divided by two is 30. That’s equivalent to saying 60 is equal to two times 30.

Our next step is really similar. This time, though, we divide the quotient by the smallest prime number that’s a factor of it. Once again, 30 is an even number, so we can divide it by two. And when we do, we find that 30 divided by two is 15. This is equivalent to saying that 30 is equal to two times 15. So we replace 30 with two times 15. And we see that we’ve now written 60 as two times two times 15. We continue to repeat this process until our quotient, the number we get after dividing, is a prime number.

15 is not divisible by two, so we move to the next smallest prime number. That’s three. And when we divide 15 by three, we get five. Now, that is a prime number. So we finished. We replace 15 with its prime factors, with three and five. And we find that 60 is two times two times three times five.

We’re not quite done, though. The question asks us to write this in exponent form. Another way of writing two times two is writing two squared. Remember, when we square a number, we times it by itself. And so in exponent form, 60 is two squared times three times five.

In our next example, we’ll look at the prime factorization of a much larger number, this time going back to using a factor tree.

Write the prime factorization of 392 in exponent form.

The prime factorization of 392 is the product of all the prime numbers, which multiply together to make 392. And it’s always sensible to list out the first few prime numbers. They are two, three, five, seven, 11, and 13. We’re going to draw a prime factor tree to work out the prime factorization of 392. We begin by finding a factor of 392. A really easy one is to check whether the number is even. If it’s even, then it has a factor of two. 392 is even, which means we can divide it by two. And when we do, we get a result of 196. This means that, on our factor tree, the first two branches are two and 196. Remember, though, we said two is a prime number. That means we circle the number two. And it indicates to us that on this branch at least, we stop here.

Next, we look for a factor of 196. Well, 196 is also even. So it must have a factor of two. In fact, when we divide 196 by two, we get 98. And that means the next two branches in our tree are two and 98. We pop a circle around this two to show that it’s the end of this branch, and we carry on with the 98. 98 can also be divided by two to give us 49. And so our next two branches are two and 49. Circling the two, we look for a factor of 49. We know 49 in fact is divisible by seven. And since 49 divided by seven is seven, our next two branches are both seven. And since seven is a prime number, we circle both of these. And we’re done with our factor tree.

We now know that 392 is the product of all these prime numbers, everything in the circle. It’s two times two times two times seven times seven. But the question told us to give our answer in exponent form. Two times two times two is two cubed, and seven times seven is seven squared. So 392 can be written as two cubed times seven squared.

Now, it’s important to realize that had we chosen a different starting factor of 392, we would’ve still achieved the same answer. For example, imagine we’d first spotted that 392 is the same as 49 times eight. Neither of these are prime numbers, so we’re going to perform the usual process on both branches. We write 49 as seven times seven, which means that these two branches both have a seven. Seven is a prime, so we circle them and finish here. We might then spot that eight is the same as four times two. We circle the two, but four is not prime, so we continue. Four can be written as two times two. So our next two branches are both two, and we circle them because they’re prime and finish here. Once again, we have two times two times two times seven times seven.

Let’s now have a look at a prime factorization of a larger number using the division method.

Find the prime factorization of 792.

We’re going to use the division method to find the prime factorization of this number. When we perform this process, we want to find all the prime numbers that multiply together to make 792. And so we list the first few prime numbers. And then the first step is to find the smallest prime number that’s also a factor of the original number. 792 is even, so we know it has a factor of two. Let’s divide then 792 by two. 792 divided by two is 396. Now, we can equivalently say that 792 is equal to two times 396.

Step two is to repeat this process with the quotient, the number we got when we performed the division. So we need to find the smallest prime factor of 396. Well, once again, it’s even, so that’s two. We’re, therefore, going to calculate 396 divided by two. That’s 198. This means 396 is equal to two times 198. And it means, in turn, we can rewrite 792 as two times two times 198.

We continue to divide our quotients by their smallest prime factor. And we only stop when our quotient is also prime. So we need to divide 198 by its smallest prime factor, which is once again two. 198 divided by two is 99. So equivalently, we can say that 198 must be equal to two times 99. We now need to divide 99 by its smallest prime factor. But 99 is not even. So it’s not divisible by two. We do, however, know that it’s divisible by three. And of course, we might recall that we can check for divisibility by three by adding the digits of the number together. If their sum is divisible by three, then the original number is also divisible by three. In this case, nine plus nine is 18. Now, 18 is divisible by three. So 99 must be divisible by three.

And in fact, when we divide 99 by three, we get 33. So we rewrite 99 is three times 33. And 792 is now two times two times two times three times 33. The smallest prime factor of 33, the smallest factor that’s also a prime number, is three. And when we divide 33 by three, we get 11. Now, 11 itself is also a prime number. It’s in our list here, so we finish dividing. We write 33 is three times 11, and this means 792 is two times two times two times three times three times 11. We can go even further and write this in exponent form. Two times two times two is two cubed and three times three is three squared. We can use a dot instead of a multiplication symbol, and we see that 792 is two cubed times three squared times 11.

In this video, we’ve learned that writing a number as a product of its prime factors, also called prime factorization, is finding all the prime numbers that multiply together to make the original number. We also learned that we absolutely must write these as a product, not a list or a sum. So 30 is two times three times five. We also saw that we have a couple of methods that we can use, and these are called the prime factor tree method or the division method.

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