Decomposition of a number into prime factors. Decomposition of numbers into prime factors, methods and examples of decomposition

What does it mean to factorize? How to do it? What can be learned from decomposing a number into prime factors? The answers to these questions are illustrated with specific examples.

Definitions:

A prime number is a number that has exactly two distinct divisors.

A composite number is a number that has more than two divisors.

To factorize a natural number means to represent it as a product of natural numbers.

To factor a natural number into prime factors means to represent it as a product of prime numbers.

Notes:

• In the expansion of a prime number, one of the factors is equal to one, and the other is equal to this number itself.
• It makes no sense to talk about the decomposition of unity into factors.
• A composite number can be decomposed into factors, each of which is different from 1.

When decomposing large numbers into prime factors, a column entry is used:

	The smallest prime number that 216 is divisible by is 2. Divide 216 by 2. We get 108.
	The resulting number 108 is divisible by 2. Let's do the division. We get 54 as a result.
	According to the test of divisibility by 2, the number 54 is divisible by 2. After dividing, we get 27.
	The number 27 ends with an odd number 7. It Not divisible by 2. The next prime number is 3. Divide 27 by 3. We get 9. The smallest prime The number that 9 is divisible by is 3. Three is itself a prime number, divisible by itself and by one. Let's divide 3 by ourselves. As a result, we got 1.

• A number is divisible only by those prime numbers that are part of its decomposition.
• A number is divisible only by those composite numbers, the decomposition of which into prime factors is completely contained in it.

Consider examples:

Find the quotient of dividing the number a by the number b, if these numbers are decomposed into prime factors as follows a=2∙2∙2∙3∙3∙3∙5∙5∙19; b=2∙2∙3∙3∙5∙19

Bibliography

1. Vilenkin N.Ya., Zhokhov V.I., Chesnokov A.S., Shvartsburd S.I. Mathematics 6. - M.: Mnemosyne, 2012.
2. Merzlyak A.G., Polonsky V.V., Yakir M.S. Mathematics 6th grade. - Gymnasium. 2006.
3. Depman I.Ya., Vilenkin N.Ya. Behind the pages of a mathematics textbook. - M.: Enlightenment, 1989.
4. Rurukin A.N., Tchaikovsky I.V. Tasks for the course of mathematics grade 5-6. - M.: ZSh MEPhI, 2011.
5. Rurukin A.N., Sochilov S.V., Tchaikovsky K.G. Mathematics 5-6. A manual for students of the 6th grade of the MEPhI correspondence school. - M.: ZSh MEPhI, 2011.
6. Shevrin L.N., Gein A.G., Koryakov I.O., Volkov M.V. Mathematics: Textbook-interlocutor for 5-6 grades of high school. - M .: Education, Mathematics Teacher Library, 1989.

1. Internet portal Matematika-na.ru ().
2. Internet portal Math-portal.ru ().

Homework

1. Vilenkin N.Ya., Zhokhov V.I., Chesnokov A.S., Shvartsburd S.I. Mathematics 6. - M.: Mnemozina, 2012. No. 127, No. 129, No. 141.
2. Other tasks: No. 133, No. 144.

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Factoring a Number into Prime Factors - Theory, Algorithm, Examples and Solutions

One of the simplest ways to factorize a number is to check if the given number is divisible by 2, 3, 5 ,... etc., i.e. check if a number is divisible by a series of prime numbers. If number n is not divisible by any prime number up to , then this number is prime, because if the number is composite, then it has at least two factors, and both cannot be greater than .

Let's imagine the number decomposition algorithm n to prime factors. Prepare a table of prime numbers in advance s=. Denote a series of prime numbers through p 1 , p 2 , p 3 , ...

Algorithm for decomposing a number into prime divisors:

Example 1. Decompose the number 153 into prime factors.

Solution. It is enough for us to have a table of prime numbers up to , i.e. 2, 3, 5, 7, 11.

Divide 153 by 2. 153 is not divisible by 2 without a remainder. Next, we divide 153 by the next element of the prime number table, i.e. by 3. 153:3=51. Fill in the table:

Next, we check if the number 17 is divisible by 3. The number 17 is not divisible by 3. It is not divisible by the numbers 5, 7, 11 either. The next divisor is greater . Therefore 17 is a prime number that is only divisible by itself: 17:17=1. The procedure has been stopped. fill in the table:

We select those divisors on which the numbers 153, 51, 17 were divided without a remainder, i.e. all numbers on the right side of the table. These are divisors 3, 3, 17. Now the number 153 can be represented as a product of prime numbers: 153=3 3 17.

Example 2. Decompose the number 137 into prime factors.

Solution. Calculate . So we need to check the divisibility of the number 137 by prime numbers up to 11: 2,3,5,7,11. Alternately dividing the number 137 by these numbers, we find out that the number 137 is not divisible by any of the numbers 2,3,5,7,11. Therefore 137 is a prime number.

What factorization? It's a way of turning an awkward and complicated example into a simple and cute one.) Very powerful trick! It occurs at every step both in elementary mathematics and in higher mathematics.

Such transformations in mathematical language are called identical transformations of expressions. Who is not in the subject - take a walk on the link. There is very little, simple and useful.) The meaning of any identical transformation is to write the expression in a different form while preserving its essence.

Meaning factorizations extremely simple and understandable. Right from the title itself. You can forget (or not know) what a multiplier is, but can you figure out that this word comes from the word "multiply"?) Factoring means: represent an expression as a multiplication of something by something. Forgive me mathematics and the Russian language ...) And that's it.

For example, you need to decompose the number 12. You can safely write:

So we presented the number 12 as a multiplication of 3 by 4. Please note that the numbers on the right (3 and 4) are completely different than on the left (1 and 2). But we are well aware that 12 and 3 4 same. The essence of the number 12 from the transformation hasn't changed.

Is it possible to decompose 12 in another way? Easily!

12=3 4=2 6=3 2 2=0.5 24=........

The decomposition options are endless.

Decomposing numbers into factors is a useful thing. It helps a lot, for example, when dealing with roots. But the factorization of algebraic expressions is not something that is useful, it is - necessary! Just for example:

Simplify:

Those who do not know how to factorize the expression, rest on the sidelines. Who knows how - simplifies and gets:

The effect is amazing, right?) By the way, the solution is quite simple. You will see for yourself below. Or, for example, such a task:

Solve the equation:

x 5 - x 4 = 0

Decided in the mind, by the way. With the help of factorization. Below we will solve this example. Answer: x 1 = 0; x2 = 1.

Or, the same, but for the older ones):

Solve the equation:

In these examples, I have shown main purpose factorizations: simplification of fractional expressions and solution of some types of equations. I recommend to remember the rule of thumb:

If we have a terrible fractional expression, we can try to factorize the numerator and denominator. Very often, the fraction is reduced and simplified.

If we have an equation in front of us, where on the right is zero, and on the left - don’t understand what, you can try to factorize the left side. Sometimes it helps.)

Basic methods of factorization.

Here are the most popular ways:

4. Decomposition of a square trinomial.

These methods must be remembered. Exactly in that order. Complex examples are checked for all possible decomposition methods. And it’s better to check in order, so as not to get confused ... Let’s start in order.)

1. Taking the common factor out of brackets.

Simple and reliable way. It doesn't get bad from him! It happens either well or not at all.) Therefore, he is the first. We understand.

Everyone knows (I believe!) the rule:

a(b+c) = ab+ac

Or, more generally:

a(b+c+d+.....) = ab+ac+ad+....

All equalities work both from left to right, and vice versa, from right to left. You can write:

ab+ac = a(b+c)

ab+ac+ad+.... = a(b+c+d+.....)

That's the whole point of putting the common factor out of brackets.

On the left side a - common factor for all terms. Multiplied by everything.) Right is the most a is already outside the brackets.

We will consider the practical application of the method with examples. At first, the variant is simple, even primitive.) But in this variant I will mark (in green) very important points for any factorization.

Multiply:

ah+9x

Which general is the multiplier in both terms? X, of course! We will take it out of brackets. We do so. We immediately write x outside the brackets:

ax+9x=x(

And in brackets we write the result of division each term on this very x. In order:

That's all. Of course, it is not necessary to paint in such detail, This is done in the mind. But to understand what's what, it is desirable). We fix in memory:

We write the common factor outside the brackets. In parentheses, we write the results of dividing all the terms by this very common factor. In order.

Here we have expanded the expression ah+9x for multipliers. Turned it into multiplying x by (a + 9). I note that in the original expression there was also a multiplication, even two: a x and 9 x. But it has not been factorized! Because in addition to multiplication, this expression also contained addition, the "+" sign! And in the expression x(a+9) nothing but multiplication!

How so!? - I hear the indignant voice of the people - And in brackets!?)

Yes, there is addition inside the brackets. But the trick is that while the brackets are not opened, we consider them like one letter. And we do all the actions with brackets in their entirety, like one letter. In this sense, in the expression x(a+9) nothing but multiplication. This is the whole point of factorization.

By the way, is there any way to check if we did everything right? Easy! It is enough to multiply back what was taken out (x) by brackets and see if it worked out initial expression? If it worked out, everything is tip-top!)

x(a+9)=ax+9x

Happened.)

There is no problem in this primitive example. But if there are several terms, and even with different signs ... In short, every third student messes up). Therefore:

If necessary, check the factorization by inverse multiplication.

Multiply:

3ax+9x

We are looking for a common factor. Well, everything is clear with X, it can be endured. Is there any more general factor? Yes! This is a trio. You can also write the expression like this:

3x+3 3x

Here it is immediately clear that the common factor will be 3x. Here we take it out:

3ax+3 3x=3x(a+3)

Spread out.

And what happens if you take only x? Nothing special:

3ax+9x=x(3a+9)

This will also be a factorization. But in this fascinating process, it is customary to lay everything out until it stops, while there is an opportunity. Here in brackets there is an opportunity to take out a triple. Get:

3ax+9x=x(3a+9)=3x(a+3)

The same thing, only with one extra action.) Remember:

When taking the common factor out of brackets, we try to take out maximum common multiplier.

Let's continue the fun?

Factoring the expression:

3ax+9x-8a-24

What will we take out? Three, X? No-ee... You can't. I remind you that you can only take general multiplier that is in all terms of the expression. That's why he general. There is no such multiplier here ... What, you can not lay out!? Well, yes, we were delighted, how ... Meet:

2. Grouping.

Actually, grouping can hardly be called an independent way of factorization. This is rather a way to get out of a complex example.) You need to group the terms so that everything works out. This can only be shown with an example. So we have an expression:

3ax+9x-8a-24

It can be seen that there are some common letters and numbers. But... General there is no multiplier to be in all terms. Do not lose heart and we break the expression into pieces. We group. So that in each piece there was a common factor, there was something to take out. How do we break? Yes, just parentheses.

Let me remind you that brackets can be placed anywhere and any way. If only the essence of the example didn't change. For example, you can do this:

3ax+9x-8a-24=(3ax + 9x) - (8a + 24)

Please pay attention to the second brackets! They are preceded by a minus sign, and 8a and 24 become positive! If, for verification, we open the brackets back, the signs will change, and we get initial expression. Those. the essence of the expression from brackets has not changed.

But if you just put in parentheses, not taking into account the sign change, for example, like this:

3ax+9x-8a-24=(3ax + 9x) -(8a-24 )

it will be a mistake. Right - already other expression. Expand the brackets and everything will become clear. You can decide no further, yes ...)

But back to factorization. Look at the first brackets (3ax + 9x) and think, is it possible to endure something? Well, we solved this example above, we can take it out 3x:

(3ax+9x)=3x(a+3)

We study the second brackets, there you can take out the eight:

(8a+24)=8(a+3)

Our entire expression will be:

(3ax + 9x) - (8a + 24) \u003d 3x (a + 3) -8 (a + 3)

Multiplied? No. The decomposition should result in only multiplication, and we have a minus sign spoils everything. But... Both terms have a common factor! it (a+3). It was not in vain that I said that the brackets as a whole are, as it were, one letter. So these brackets can be taken out of the brackets. Yes, that's exactly what it sounds like.)

We do as described above. Write the common factor (a+3), in the second brackets we write the results of dividing the terms by (a+3):

3x(a+3)-8(a+3)=(a+3)(3x-8)

Everything! On the right, there is nothing but multiplication! So the factorization is completed successfully!) Here it is:

3ax + 9x-8a-24 \u003d (a + 3) (3x-8)

Let's recap the essence of the group.

If the expression does not general multiplier for all terms, we split the expression with brackets so that inside the brackets the common factor was. Let's take it out and see what happens. If we are lucky, and exactly the same expressions remain in the brackets, we take these brackets out of the brackets.

I will add that grouping is a creative process). It doesn't always work the first time. It's OK. Sometimes you have to swap terms, consider different grouping options until you find a good one. The main thing here is not to lose heart!)

Examples.

Now, having enriched with knowledge, you can solve tricky examples.) At the beginning of the lesson, there were three of these ...

Simplify:

In fact, we have already solved this example. Imperceptibly to myself.) I remind you: if we are given a terrible fraction, we try to decompose the numerator and denominator into factors. Other simplification options simply no.

Well, the denominator is not decomposed here, but the numerator... We have already decomposed the numerator in the course of the lesson! Like this:

3ax + 9x-8a-24 \u003d (a + 3) (3x-8)

We write the result of expansion into the numerator of the fraction:

According to the rule of reduction of fractions (the main property of a fraction), we can divide (simultaneously!) The numerator and denominator by the same number, or expression. Fraction from this does not change. So we divide the numerator and denominator by the expression (3x-8). And here and there we get units. Final simplification result:

I emphasize in particular: reduction of a fraction is possible if and only if in the numerator and denominator, in addition to multiplying expressions there is nothing. That is why the transformation of the sum (difference) into multiplication so important to simplify. Of course, if the expressions various, then nothing will be reduced. Byvet. But the factorization gives a chance. This chance without decomposition - simply does not exist.

Equation example:

Solve the equation:

x 5 - x 4 = 0

Taking out the common factor x 4 for brackets. We get:

x 4 (x-1)=0

We assume that the product of the factors is equal to zero then and only then when any of them is equal to zero. If in doubt, find me a couple of non-zero numbers that, when multiplied, will give zero.) So we write, first the first factor:

With this equality, the second factor does not bother us. Anyone can be, anyway, in the end, zero will turn out. What is the number to the fourth power of zero? Only zero! And nothing else ... Therefore:

We figured out the first factor, we found one root. Let's deal with the second factor. Now we don't care about the first multiplier.):

Here we found a solution: x 1 = 0; x2 = 1. Any of these roots fit our equation.

A very important note. Note that we have solved the equation bit by bit! Each factor was set to zero. regardless of other factors. By the way, if in such an equation there are not two factors, as we have, but three, five, as many as you like, we will decide similar. Piece by piece. For example:

(x-1)(x+5)(x-3)(x+2)=0

The one who opens the brackets, multiplies everything, will forever hang on this equation.) The correct student will immediately see that there is nothing on the left except multiplication, on the right - zero. And he will begin (in his mind!) To equate to zero all the brackets in order. And he will get (in 10 seconds!) the right solution: x 1 = 1; x 2 \u003d -5; x 3 \u003d 3; x4 = -2.

Great, right?) Such an elegant solution is possible if the left side of the equation split into multiples. Is the hint clear?)

Well, the last example, for the older ones):

Solve the equation:

It is somewhat similar to the previous one, don't you think?) Of course. It's time to remember that in seventh grade algebra, sines, logarithms, and anything else can be hidden under letters! Factoring works in all mathematics.

Taking out the common factor lg4x for brackets. We get:

lg 4x=0

This is one root. Let's deal with the second factor.

Here is the final answer: x 1 = 1; x2 = 10.

I hope you've realized the power of factoring in simplifying fractions and solving equations.)

In this lesson, we got acquainted with the removal of the common factor and grouping. It remains to deal with the formulas for abbreviated multiplication and the square trinomial.

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By the way, I have a couple more interesting sites for you.)

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you can get acquainted with functions and derivatives.

Any composite number can be decomposed into prime factors. There are several ways of decomposition. Either method produces the same result.

How to decompose a number into prime factors in the most convenient way? Let's consider how to do it better, using specific examples.

Examples. 1) Decompose the number 1400 into prime factors.

1400 is divisible by 2. 2 is a prime number, no need to factor it. We get 700. Divide it by 2. We get 350. We also divide 350 by 2. The resulting number 175 can be divided by 5. The result - z5 - again divide by 5. Total - 7. It can only be divided by 7. We got 1, dividing finished.

The same number can be decomposed into prime factors differently:

1400 is conveniently divided by 10. 10 is not a prime number, so it must be factored into prime factors: 10=2∙5. The result is 140. We again divide it by 10=2∙5. We get 14. If 14 is divided by 14, then it should also be decomposed into the product of prime factors: 14=2∙7.

Thus, we again came to the same decomposition as in the first case, but faster.

Conclusion: when decomposing a number, it is not necessary to divide it only by prime divisors. We divide by what is more convenient, for example, by 10. We only need to remember to decompose the composite divisors into simple factors.

2) Decompose the number 1620 into prime factors.

The number 1620 is most conveniently divided by 10. Since 10 is not a prime number, we represent it as a product of prime factors: 10=2∙5. We got 162. It is convenient to divide it by 2. The result is 81. The number 81 can be divided by 3, but 9 is more convenient. Since 9 is not a prime number, we decompose it as 9=3∙3. We got 9. We also divide it by 9 and decompose it into the product of prime factors.

Very often, the numerator and denominator of a fraction are algebraic expressions that must first be decomposed into factors, and then, having found the same among them, divide both the numerator and the denominator into them, that is, reduce the fraction. A whole chapter of a textbook on algebra in the 7th grade is devoted to tasks to factorize a polynomial. Factoring can be done 3 ways, as well as a combination of these methods.

1. Application of abbreviated multiplication formulas

As is known to multiply a polynomial by a polynomial, you need to multiply each term of one polynomial by each term of the other polynomial and add the resulting products. There are at least 7 (seven) common cases of multiplication of polynomials that are included in the concept. For example,

Table 1. Factorization in the 1st way

2. Taking the common factor out of the bracket

This method is based on the application of the distributive law of multiplication. For example,

We divide each term of the original expression by the factor that we take out, and at the same time we get the expression in brackets (that is, the result of dividing what was by what we take out remains in brackets). First of all, you need correctly determine the multiplier, which must be bracketed.

A common factor can also be a polynomial in brackets:

When performing the “factorize” task, one must be especially careful with the signs when taking the common factor out of brackets. To change the sign of each term in a parenthesis (b - a), we take out the common factor -1 , while each term in the bracket is divided by -1: (b - a) = - (a - b) .

In the event that the expression in brackets is squared (or to any even power), then numbers inside brackets can be swapped completely free, since the minuses taken out of brackets will still turn into a plus when multiplied: (b - a) 2 = (a - b) 2, (b - a) 4 = (a - b) 4 and so on…

3. Grouping method

Sometimes not all terms in the expression have a common factor, but only some. Then you can try group terms in brackets so that some factor can be taken out of each. Grouping method is double bracketing of common factors.

4. Using several methods at once

Sometimes you need to apply not one, but several ways to factorize a polynomial into factors at once.

This is a synopsis on the topic. "Factorization". Choose next steps:

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