An example of decomposing a number into prime factors. How to decompose a number into a product of prime factors

Have you come across such a term as "prime numbers" or " prime factors' but don't know what it is? Also, prime numbers are very popular in the film industry, so it is not uncommon to see them in films and TV shows. Let's figure out what prime numbers are in this article!

prime numbers is a positive integer (natural) number that can only be divided by one and itself. Numbers that have more than two natural divisors are composite.

• Example 1: The prime number 7 can only be divided by 1 and 7.
• Example 2: composite number 6 can be divided into 1, 2, 3, 6.

Prime numbers up to 100: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97

Prime numbers are a very popular topic in mathematics, and a huge number of problems, theorems, etc. are associated with it.

Prime Factors are factors (elements of the product), which are prime numbers. There are several school assignments associated with prime factors that can cause problems even for the older generation.

Factorize the numbers...

Quite a popular problem in mathematics. The most common examples:

Expand non-prime factors of the number 27, 54, 56, 65, 99, 162, 625, 1000. First of all, it should be said that the most common mistake in solving this problem is that the number of factors is not indicated, there are not necessarily 2 of them! If you made this mistake, you can try to solve the task yourself.

Answers:

• 27 = 3 x 3 x 3
• 54 = 2 x 3 x 3 x 3
• 56 = 2 x 2 x 2 x 7
• 65 = 5 x 13
• 99 = 3 x 3 x 11
• 162 = 2 x 3 x 3 x 3 x 3
• 625 = 5 x 5 x 5 x 5
• 1000 = 2 x 2 x 2 x 5 x 5 x 5

Any natural number can be decomposed into a product of prime factors. If you don't like to deal with big numbers, such as 5733, learn how to factor them into prime factors (in this case, it's 3 x 3 x 7 x 7 x 13). A similar task is often encountered in cryptography, which deals with information security problems. If you're not ready to create your own secure email system, learn how to factor numbers into prime factors first.

Steps

Part 1

Finding Prime Factors

1. Start with the original number. Choose a composite number greater than 3. It makes no sense to take a prime number, since it is only divisible by itself and one.

   • Example: decompose into a product prime numbers number 24.

2. Let's decompose given number to the product of two factors. Find two smaller numbers whose product is equal to the original number. You can use any multipliers, but it's easier to take prime numbers. One of good ways consists in trying to divide the original number first by 2, then by 3, then by 5 and check which of these prime numbers it is divisible by without a remainder.

   • Example: if you don't know the factors for the number 24, try dividing it by small prime numbers. So you will find that the given number is divisible by 2: 24 = 2 x 12. This is a good start.
   • Since 2 is a prime number, it is good to use it when decomposing even numbers.

3. Start building a multiplier tree. This simple procedure will help you factorize a number into prime factors. To get started, draw two "branches" down from the original number. At the end of each branch, write the found multipliers.

   • Example:

4. Factor the following row of numbers. Take a look at the two new numbers (second row of the multiplier tree). Are they both prime numbers? If one of them is not prime, also factor it into two factors. Draw two more branches and write two new multipliers in the third line of the tree.

   • Example: 12 is not a prime number, so it must be factored. We use the decomposition 12 = 2 x 6 and write it in the third line of the tree:
   • 2x6

5. Keep moving down the tree. If one of the new factors turns out to be a prime number, draw one "branch" from it and write the same number at its end. Prime numbers are not decomposed into smaller factors, so just transfer them to the level below.

   • Example: 2 is a prime number. Just move the 2 from the second to the third line:
   • 2 2 6

6. Keep factoring numbers until you are left with only prime numbers. Check each new line of the tree. If at least one of the new factors is not a prime number, factor it and write a new line. In the end, you will be left with only prime numbers.

   • Example: 6 is not a prime number, so it should also be factored. At the same time, 2 is a prime number, and we carry two 2s to the next level:
   • 2 2 6
   • / / /\
   • 2 2 2 3

7. Write the last line as a product of prime factors. In the end, you will be left with only prime numbers. When this happens, the prime factorization is complete. The last line is a set of prime numbers whose product gives the original number.

   • Check your answer: multiply the numbers in the last line. The result should be the original number.
   • Example: The last row of the factor tree contains the numbers 2 and 3. Both of these numbers are prime, so the expansion is complete. Thus, the decomposition of the number 24 into prime factors has the following form: 24 = 2 x 2 x 2 x 3.
   • The order of the multipliers does not matter. The expansion can also be written as 2 x 3 x 2 x 2.

8. Simplify your answer with power notation if you wish. If you are familiar with raising numbers to a power, you can write your answer in a simpler way. Remember that the base is written below, and the superscript number shows how many times this base should be multiplied by itself.

   • Example: how many times does the number 2 appear in the found expansion of 2 x 2 x 2 x 3? Three times, so the expression 2 x 2 x 2 can be written as 2 3 . In simplified notation, we get 23x3.

   Part 2

   Using Prime Factorization

   1. Find the greatest common divisor of two numbers. The greatest common divisor (GCD) of two numbers is the maximum number by which both numbers are divisible without a remainder. The following example shows how to use prime factorization to find the greatest common divisor of 30 and 36.

      • Let's decompose both numbers into prime factors. For the number 30, the expansion is 2 x 3 x 5. The number 36 is decomposed into prime factors as follows: 2 x 2 x 3 x 3.
      • Find a number that occurs in both expansions. We cross out this number in both lists and write it on a new line. For example, 2 occurs in two expansions, so we write 2 in a new line. After that, we are left with 30 = 2 x 3 x 5 and 36 = 2 x 2 x 3 x 3.
      • Repeat this action until there are no common factors left in the expansions. Both lists also include the number 3, so on a new line we can write 2 And 3 . After that, compare the expansions again: 30 = 2 x 3 x 5 and 36 = 2 x 2 x 3 x 3. As you can see, there are no common factors left in them.
      • To find the greatest common divisor, you must find the product of all common factors. In our example, these are 2 and 3, so gcd is 2 x 3 = 6 . This is the largest number by which the numbers 30 and 36 are divisible without a remainder.

   2. GCD can be used to simplify fractions. If you suspect that a fraction can be reduced, use the greatest common divisor. Use the above procedure to find the GCD of the numerator and denominator. Then divide the numerator and denominator of the fraction by that number. As a result, you will get the same fraction in a simpler form.

      • For example, let's simplify the fraction 30/36. As we stated above, for 30 and 36 GCD is 6, so we divide the numerator and denominator by 6:
      • 30 ÷ 6 = 5
      • 36 ÷ 6 = 6
      • 30 / 36 = 5 / 6

   3. Find the least common multiple of two numbers. The least common multiple (LCM) of two numbers is the smallest number that is evenly divisible by both given numbers. For example, the LCM of 2 and 3 is 6 because it is the smallest number that is divisible by 2 and 3. The following is an example of finding the LCM using prime factorization:

      • We start with two factorizations into prime factors. For example, for the number 126, the expansion can be written as 2 x 3 x 3 x 7. The number 84 is decomposed into prime factors in the form 2 x 2 x 3 x 7.
      • Let's compare how many times each factor occurs in expansions. Choose the list where the multiplier occurs the maximum number of times, and circle this place. For example, the number 2 occurs once in the expansion for 126 and twice in the list for 84, so circle 2x2 in the second multiplier list.
      • Repeat this action for each multiplier. For example, 3 occurs more often in the first expansion, so circle it 3x3. The number 7 appears once in both lists, so circle 7 (it doesn't matter in which list, if the given factor occurs in both lists the same number of times).
      • To find the LCM, multiply all circled numbers. In our example, the least common multiple of 126 and 84 is 2 x 2 x 3 x 3 x 7 = 252. This is the smallest number that is divisible by 126 and 84 without a remainder.

   4. Use LCM to add fractions. When adding two fractions, you need to bring them to a common denominator. To do this, find the LCM of two denominators. Then multiply the numerator and denominator of each fraction by such a number that the denominators of the fractions become equal to the LCM. After that, you can add fractions.

      • For example, you need to find the sum of 1 / 6 + 4 / 21 .
      • Using the above method, you can find the LCM for 6 and 21. It is equal to 42.
      • Let's transform the fraction 1/6 so that its denominator is 42. To do this, divide 42 by 6: 42 ÷ 6 = 7. Now multiply the numerator and denominator of the fraction by 7: 1/6 x 7/7 = 7/42.
      • To bring the second fraction to the denominator 42, divide 42 by 21: 42 ÷ 21 = 2. Multiply the numerator and denominator of the fraction by 2: 4 / 21 x 2 / 2 = 8 / 42.
      • After the fractions have been reduced to the same denominator, they can be easily added: 7/42 + 8/42 = 15/42.

This is one of the most elementary ways to simplify an expression. To apply this method, let's remember the distributive law of multiplication with respect to addition (do not be afraid of these words, you definitely know this law, you just might have forgotten its name).

The law says: in order to multiply the sum of two numbers by a third number, you need to multiply each term by this number and add the results, in other words,.

You can also do the reverse operation, and it is this reverse operation that interests us. As can be seen from the sample, the common factor a, can be taken out of the bracket.

A similar operation can be done both with variables, such as and, for example, and with numbers: .

Yes, this is too elementary an example, just like the example given earlier, with the expansion of a number, because everyone knows what numbers are, and are divisible by, but what if you got a more complicated expression:

How to find out what, for example, a number is divided into, no, with a calculator, anyone can, but without it it’s weak? And for this there are signs of divisibility, these signs are really worth knowing, they will help you quickly understand whether it is possible to take the common factor out of the bracket.

Signs of divisibility

It is not so difficult to remember them, most likely, most of them were already familiar to you, and something will be a new useful discovery, more details in the table:

Note: The table lacks a sign of divisibility by 4. If the last two digits are divisible by 4, then the whole number is divisible by 4.

Well, how do you like the sign? I advise you to remember it!

Well, let's get back to the expression, maybe take it out of the bracket and that's enough from it? No, it is customary for mathematicians to simplify, so to the fullest, take out EVERYTHING that is taken out!

And so, everything is clear with the player, but what about the numerical part of the expression? Both numbers are odd, so you can't divide by

You can use the sign of divisibility by, the sum of the digits, and, of which the number consists, is equal, and is divisible by, which means it is divisible by.

Knowing this, you can safely divide into a column, as a result of dividing by we get (signs of divisibility came in handy!). Thus, we can take the number out of the bracket, just like y, and as a result we have:

To make sure that everything is decomposed correctly, you can check the expansion by multiplication!

Also, the common factor can be taken out in power expressions. Here, for example, do you see the common factor?

All members of this expression have x's - we take out, all are divided by - we take out again, we look at what happened: .

2. Abbreviated multiplication formulas

Abbreviated multiplication formulas have already been mentioned in theory, if you can hardly remember what it is, then you should refresh them in your memory.

Well, if you consider yourself very smart and you are too lazy to read such a cloud of information, then just read on, look at the formulas and immediately take on the examples.

The essence of this decomposition is to notice some definite formula in the expression before you, apply it and thus obtain the product of something and something, that's all the decomposition. Following are the formulas:

Now try factoring the following expressions using the above formulas:

And here is what should have happened:

As you have noticed, these formulas are a very effective way of factoring, it is not always suitable, but it can be very useful!

3. Grouping or grouping method

Here's another example for you:

Well, what are you going to do with it? It seems to be divisible by and into something, and something into and into

But you can’t divide everything together into one thing, well there is no common factor, how not to look for what, and leave it without factoring?

Here you need to show ingenuity, and the name of this ingenuity is a grouping!

It is used just when not all members have common divisors. For grouping you need find groups of terms that have common divisors and rearrange them so that the same multiplier can be obtained from each group.

Of course, it is not necessary to rearrange in places, but this gives visibility, for clarity, you can take individual parts of the expression in brackets, it is not forbidden to put them as much as you like, the main thing is not to confuse the signs.

All this is not very clear? Let me explain with an example:

In a polynomial - put a member - after the member - we get

we group the first two terms together in a separate bracket and group the third and fourth terms in the same way, leaving the minus sign out of the bracket, we get:

And now we look separately at each of the two "heaps" into which we have broken the expression with brackets.

The trick is to break it into such piles from which it will be possible to take out the largest possible factor, or, as in this example, try to group the members so that after taking the factors out of the brackets from the piles, we have the same expressions inside the brackets.

From both brackets we take out the common factors of the members, from the first bracket, and from the second bracket, we get:

But it's not decomposition!

Pdonkey decomposition should remain only multiplication, but for now we have a polynomial simply divided into two parts ...

BUT! This polynomial has a common factor. This

outside the bracket and we get the final product

Bingo! As you can see, there is already a product and outside the brackets there is neither addition nor subtraction, the decomposition is completed, because we have nothing more to take out of the brackets.

It may seem like a miracle that after taking the factors out of the brackets, we still have the same expressions in the brackets, which, again, we took out of the brackets.

And this is not a miracle at all, the fact is that the examples in textbooks and in the exam are specially made in such a way that most of the expressions in tasks for simplification or factorization with the right approach to them, they are easily simplified and abruptly collapse like an umbrella when you press a button, so look for that very button in each expression.

Something I digress, what do we have there with simplification? The intricate polynomial took on a simpler form: .

Agree, not as bulky as it used to be?

4. Selection of a full square.

Sometimes, in order to apply the formulas for abbreviated multiplication (repeat the topic), it is necessary to transform the existing polynomialby presenting one of its terms as the sum or difference of two terms.

In which case you have to do this, you will learn from the example:

A polynomial in this form cannot be decomposed using abbreviated multiplication formulas, so it must be converted. Perhaps at first it will not be obvious to you which term to divide into which, but over time you will learn to immediately see the formulas for abbreviated multiplication, even if they are not present in their entirety, and you will quickly determine what is missing here to the full formula, but for now - learn , a student, more precisely a schoolboy.

For the full formula of the square of the difference, here you need instead. Let's represent the third term as a difference, we get: We can apply the difference square formula to the expression in brackets (not to be confused with the difference of squares!!!), we have: , to this expression, we can apply the formula for the difference of squares (not to be confused with the squared difference!!!), imagining how, we get: .

An expression not always factored into factors looks simpler and smaller than it was before decomposition, but in this form it becomes more mobile, in the sense that you can not worry about changing signs and other mathematical nonsense. Well, for you to decide on your own, the following expressions need to be factored.

Examples:

Answers:​

5. Factorization of a square trinomial

For the factorization of a square trinomial, see below in the decomposition examples.

Examples of 5 Methods for Factoring a Polynomial

1. Taking the common factor out of brackets. Examples.

Do you remember what the distributive law is? This is such a rule:

Example:

Factorize a polynomial.

Solution:

Another example:

Multiply.

Solution:

If the whole term is taken out of brackets, one remains in brackets instead of it!

2. Formulas for abbreviated multiplication. Examples.

The most commonly used formulas are the difference of squares, the difference of cubes and the sum of cubes. Remember these formulas? If not, urgently repeat the topic!

Example:

Factor the expression.

Solution:

In this expression, it is easy to find out the difference of cubes:

Example:

Solution:

3. Grouping method. Examples

Sometimes it is possible to interchange the terms in such a way that one and the same factor can be extracted from each pair of neighboring terms. This common factor can be taken out of the bracket and the original polynomial will turn into a product.

Example:

Factor out the polynomial.

Solution:

We group the terms as follows:
.

In the first group, we take the common factor out of brackets, and in the second - :
.

Now the common factor can also be taken out of brackets:
.

4. The method of selection of a full square. Examples.

If the polynomial can be represented as the difference of the squares of two expressions, all that remains is to apply the abbreviated multiplication formula (difference of squares).

Example:

Factor out the polynomial.

Solution:Example:

\begin(array)(*(35)(l))
((x)^(2))+6(x)-7=\underbrace(((x)^(2))+2\cdot 3\cdot x+9)_(square\ sums\ ((\left (x+3 \right))^(2)))-9-7=((\left(x+3 \right))^(2))-16= \\
=\left(x+3+4 \right)\left(x+3-4 \right)=\left(x+7 \right)\left(x-1 \right) \\
\end(array)

Factor out the polynomial.

Solution:

\begin(array)(*(35)(l))
((x)^(4))-4((x)^(2))-1=\underbrace(((x)^(4))-2\cdot 2\cdot ((x)^(2) )+4)_(square\ differences((\left(((x)^(2))-2 \right))^(2)))-4-1=((\left(((x)^ (2))-2 \right))^(2))-5= \\
=\left(((x)^(2))-2+\sqrt(5) \right)\left(((x)^(2))-2-\sqrt(5) \right) \\
\end(array)

5. Factorization of a square trinomial. Example.

A square trinomial is a polynomial of the form, where is an unknown, are some numbers, moreover.

Variable values ​​that turn the square trinomial to zero are called roots of the trinomial. Therefore, the roots of a trinomial are the roots of a quadratic equation.

Theorem.

Example:

Let's factorize the square trinomial: .

First, we solve the quadratic equation: Now we can write the factorization of this square trinomial into factors:

Now your opinion...

We have described in detail how and why to factorize a polynomial.

We gave a lot of examples of how to do it in practice, pointed out the pitfalls, gave solutions ...

What do you say?

How do you like this article? Do you use these tricks? Do you understand their essence?

Write in the comments and... get ready for the exam!

So far, it's the most important thing in your life.

Let's decompose the number 120 into prime factors

120 = 2 ∙ 2 ∙ 2 ∙ 3 ∙ 5

Solution
Let's expand the number 120

120: 2 = 60
60: 2 = 30 - is divisible by a prime number 2
30: 2 = 15 - is divisible by a prime number 2
15: 3 = 5
We complete the division, since 5 is a prime number

Answer: 120 = 2 ∙ 2 ∙ 2 ∙ 3 ​​∙ 5

Let's decompose the number 246 into prime factors

246 = 2 ∙ 3 ∙ 41

Solution
Let's expand the number 246 into prime factors and highlight them in green. We begin to select a divisor from prime numbers, starting with the smallest prime number 2, until the quotient is a prime number

246: 2 = 123 - is divisible by a prime number 2
123: 3 = 41 is divisible by the prime number 3.
We complete the division, since 41 are prime numbers

Answer: 246 = 2 ∙ 3 ​​∙ 41

Let's decompose the number 1463 into prime factors

1463 = 7 ∙ 11 ∙ 19

Solution
Let's expand the number 1463 into prime factors and highlight them in green. We begin to select a divisor from prime numbers, starting with the smallest prime number 2, until the quotient is a prime number

1463: 7 = 209 - is divisible by the prime number 7
209: 11 = 19
We complete the division, since 19 is a prime number

Answer: 1463 = 7 ∙ 11 ∙ 19

Let's decompose the number 1268 into prime factors

1268 = 2 ∙ 2 ∙ 317

Solution
Let's expand the number 1268 into prime factors and highlight them in green. We begin to select a divisor from prime numbers, starting with the smallest prime number 2, until the quotient is a prime number

1268: 2 = 634 - is divisible by a prime number 2
634: 2 = 317 is divisible by the prime number 2.
We complete the division, since 317 is a prime number

Answer: 1268 = 2 2 317

Let's decompose the number 442464 into prime factors

442464

Solution
Let's expand the number 442464 into prime factors and highlight them in green. We begin to select a divisor from prime numbers, starting with the smallest prime number 2, until the quotient is a prime number

442464: 2 = 221232 - is divisible by a prime number 2
221232: 2 = 110616 - is divisible by a prime number 2
110616: 2 = 55308 - is divisible by a prime number 2
55308: 2 = 27654 - is divisible by a prime number 2
27654: 2 = 13827 - is divisible by a prime number 2
13827: 3 = 4609 - is divisible by a prime number 3
4609: 11 = 419 is divisible by the prime number 11.
We complete the division, since 419 is a prime number

Answer: 442464 = 2 ∙ 2 ∙ 2 ∙ 2 ∙ 2 ∙ 3 ∙ 11 ∙ 419

Lesson in 6th grade on the topic

"Decomposition into Prime Factors"

Lesson Objectives:

Educational:

To form an idea of ​​\u200b\u200bdecomposing numbers into prime factors, the ability to use the appropriate algorithm in practice.

To form the skills and abilities of using the signs of divisibility when decomposing numbers into prime factors.

Developing:

Develop computational skills, the ability to generalize, analyze, identify patterns, compare.

Educational:

To cultivate attention, a culture of mathematical thinking, a serious attitude to educational work.

Lesson content:

1. Oral account.

2. Repetition of the material covered.

3. Explanation of new material.

4. Fixing the material.

5. Reflection.

6. Summing up the lesson.

During the classes

Motivation (self-determination) for learning activities.

Introduction:

Hello guys. The topic of our lesson is “Decomposition of numbers into prime factors”. Some of you are already familiar with it. And in order to better set the goal of the lesson, we will work with you a little orally.

Take action (verbal) .

Calculate:

1. 15 x (325 -325) + 236x1 - 30: 1 206

2. 207 - (0 x4376 -0:585) + 315: 315 208

3. (60 - 0:60) + (150:1 -48x0) 210

4. (707:707 +211х1):1 -0:123 212

Repetition of the studied material

Continue the resulting series for 3 numbers

(206; 208;210; 212;214;216;218)

Choose from them divisible numbers

on: 2 (206; 208; 210; 212; 214; 216; 218)

by 3: (210;216)

by 9: (216)

by 5: (210)

by 4: (208; 212; 216)

Formulate signs of divisibility

Questions: 1. What numbers are called prime?

2. What numbers are called composite?

3. What is the number 1?

4. Name all the prime numbers of the first two tens.

5. How many prime numbers are there?

6. Is the number 32 prime?

7. Is the number 73 prime?

Explanation of new material.

Let's solve a very interesting problem.

Lived, there were troubles and a grandmother. They had a Ryaba chicken. A hen lays every seventh golden egg, and every third silver one. Could it be?

(Answer: no, because 21 testicles can be gold and silver) Why?

What should we learn in today's lesson? (Decompose any numbers into prime factors)

Why do you think we need this? (to solve more complex examples and also reduce fractions)

Today, the topic of our lesson will help us better understand and solve such problems.

Solve the problem: It is necessary to allocate a rectangular plot of land with an area of ​​​​18 square meters. m., What can be the dimensions of this area, if they must be expressed in natural numbers?

Solution: 1. 18=1 x 18 = 2 x3 x3

2. 18= 2 x 9 = 2x3x3

3. 18=3 x 6 = 3 x2x 3

Work in pairs.

What have we done? (Represented as a product or factored). Is it possible to continue decomposition? But as? What did you get?

Question: what can be said about these multipliers?

All factors are prime numbers.

Open the textbook What needs to be done? Who can explain to me how it's done? (Discussion in pairs)

Using the analyzed example, we decompose the number 84 into prime factors (decomposition algorithm):

84 2 756 2 - the teacher shows on the blackboard.

42 2 378 2

21 3 189 3 84 = 2x2∙3∙7 = 2 2∙3∙7

7 7 63 3

1 21 3 756= 2x2x3x3x3x3

Factor the number 756 into prime factors. Compare with my solution. What did you notice?

On page 194, find the answer to the following question?

Any number is decomposed into a product of prime factors

the only way.

Consolidation of the studied material .

1. Factorize the numbers: 20; 188; 254.

let's check slide 12

20 2 188 2 254 2

10 2 94 2 127 127

5 5 47 47 1 1

1 1 1

№ 1. 20 = 2 2 ∙5; 188 = 2²∙47; 254 = 2∙127.

Cards are offered to everyone. Students decide and check with the original, which is on the teacher's desk. If you did it right, put yourself a plus sign in the pivot table. (Solve by 3)

Card number 2. Factorize the numbers: 30; 136; 438.

Card number 3. Factorize the numbers: 40; 125; 326.

Card number 4. Factorize the numbers: 50; 78; 285.

Card number 5. Factorize the numbers: 60; 654; 99.

Card number 6. Factorize the numbers: 70; 65; 136.

After the work is done, we will check.

№ 2. 30 = 2∙3∙5; 136 = 2 3 ∙17; 438 =2∙3∙73.

№3. 40 = 2 3 ∙5; 125 = 5 3 ; 326 = 2 ∙163

№4. 50 = 2∙5²; 78 = 2∙3∙13; 285 = 3∙5∙9.

№ 5. 60 = 2²∙3∙5; 654 = 2∙3∙109; 99 = 3²∙11

№ 6. 70 = 2∙5∙7; 65 = 5∙13; 136 = 2 3 ∙17.

Outcome.

What does it mean to factor a number into prime factors?

(To factor a natural number into prime factors is to represent a number as a product of prime numbers.)

2) Is the decomposition natural number for prime factors?

(Whatever way the decomposition of a natural number into prime factors is performed, we get its unique factorization, the order of the factors is not taken into account.)

Homework.

Factor any 4 numbers into prime factors.