Every natural number, except one, has two or more divisors. For example, the number 7 is divisible without a remainder only by 1 and 7, that is, it has two divisors. And the number 8 has divisors 1, 2, 4, 8, that is, as many as 4 divisors at once.
What is the difference between prime and composite numbers?
Numbers that have more than two divisors are called composite numbers. Numbers that have only two divisors: one and the number itself are called prime numbers.
The number 1 has only one division, namely the number itself. One is neither a prime nor a composite number.
- For example, the number 7 is prime and the number 8 is composite.
First 10 prime numbers: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29. The number 2 is the only even prime number, all other prime numbers are odd.
The number 78 is composite, since in addition to 1 and itself, it is also divisible by 2. When divided by 2, we get 39. That is, 78 = 2*39. In such cases, they say that the number was factored into factors of 2 and 39.
Any composite number can be decomposed into two factors, each of which is greater than 1. This trick will not work with a prime number. So it goes.
Factoring a number into prime factors
As noted above, any composite number can be decomposed into two factors. Let's take, for example, the number 210. This number can be decomposed into two factors 21 and 10. But the numbers 21 and 10 are also composite, let's decompose them into two factors. We get 10 = 2*5, 21=3*7. And as a result, the number 210 was decomposed into 4 factors: 2,3,5,7. These numbers are already prime and cannot be expanded. That is, we factored the number 210 into prime factors.
When factoring composite numbers into prime factors, they are usually written in ascending order.
It should be remembered that any composite number can be decomposed into prime factors and in a unique way, up to permutation.
- Usually, when decomposing a number into prime factors, divisibility criteria are used.
Let's factor the number 378 into prime factors
We will write down the numbers, separating them with a vertical line. The number 378 is divisible by 2, since it ends in 8. When divided, we get the number 189. The sum of the digits of the number 189 is divisible by 3, which means the number 189 itself is divisible by 3. The result is 63.
The number 63 is also divisible by 3, according to divisibility. We get 21, the number 21 can again be divided by 3, we get 7. Seven is divided only by itself, we get one. This completes the division. To the right after the line are the prime factors into which the number 378 is decomposed.
378|2
189|3
63|3
21|3
This online calculator is designed to factorize a function.
For example, factorize: x 2 /3-3x+12. Let's write it as x^2/3-3*x+12. You can also use this service, where all calculations are saved in Word format.
For example, decompose into terms. Let's write it as (1-x^2)/(x^3+x) . To see the progress of the solution, click Show steps. If you need to get the result in Word format, use this service.
Note: the number "pi" (π) is written as pi; square root as sqrt , for example sqrt(3) , tangent tg is written tan . To view the answer, see Alternative.
- If a simple expression is given, for example, 8*d+12*c*d, then factoring the expression means representing the expression in the form of factors. To do this, you need to find common factors. Let's write this expression as: 4*d*(2+3*c) .
- Present the product in the form of two binomials: x 2 + 21yz + 7xz + 3xy. Here you already need to find several common factors: x(x+7z) + 3y(x + 7z). We take out (x+7z) and get: (x+7z)(x + 3y) .
see also Dividing polynomials with a corner(all division steps are shown in a column)
Useful when studying the rules of factorization will be abbreviated multiplication formulas, with the help of which it will be clear how to open brackets with a square:
- (a+b) 2 = (a+b)(a+b) = a 2 +2ab+b 2
- (a-b) 2 = (a-b)(a-b) = a 2 -2ab+b 2
- (a+b)(a-b) = a 2 - b 2
- a 3 +b 3 = (a+b)(a 2 -ab+b 2)
- a 3 -b 3 = (a-b)(a 2 +ab+b 2)
- (a+b) 3 = (a+b)(a+b) 2 = a 3 +3a 2 b + 3ab 2 +b 3
- (a-b) 3 = (a-b)(a-b) 2 = a 3 -3a 2 b + 3ab 2 -b 3
Factorization Methods
After learning a few tricks factorization The following classification of solutions can be made:- Using abbreviated multiplication formulas.
- Finding a common factor.
It all starts with geometric progression. At the first lecture on rows (see section 18.1. Basic definitions) we have proven that this function is the sum of the series 
, and the series converges to the function at 
. So,
.
Let us list several varieties of this series. Replacing X on - X , we get
when replacing X
on 
we get
etc.; The convergence region of all these series is the same: 
.
2.
.
All derivatives of this function at the point X
=0 are equal 
, so the series looks like
.
The area of convergence of this series is the entire numerical axis (example 6 of section 18.2.4.3. Radius of convergence, interval of convergence and region of convergence of a power series), That's why 
at 
. As a consequence, the remainder term of the Taylor formula 
. Therefore the series converges to
at any point X
.
3.
.
This series converges absolutely at 
, and its sum is really equal 
. The remainder term of the Taylor formula has the form 
, Where 
or 
- limited function, and 
(this is the general term of the previous expansion).
4.
.
This expansion can be obtained, like the previous ones, by sequentially calculating derivatives, but we will proceed differently. Let us differentiate the previous series term by term:
Convergence to a function on the entire axis follows from the theorem on term-by-term differentiation of a power series.
5. Prove independently that on the entire numerical axis, .
6.
.
The series for this function is called binomial series. Here we will calculate derivatives.
...The Maclaurin series has the form
We are looking for the interval of convergence: therefore, the interval of convergence is 
. We will not study the remainder term and the behavior of the series at the ends of the convergence interval; it turns out that when 
The series converges absolutely at both points 
, at 
the series conditionally converges at a point 
and diverges at a point 
, at 
diverges at both points.
7.
.
Here we will use the fact that
. Since , then, after term-by-term integration,
The area of convergence of this series is the half-interval 
, convergence to a function at interior points follows from the theorem on term-by-term integration of a power series, at the point X
=1 - from the continuity of both the function and the sum of the power series at all points, arbitrarily close to X
=1 left. Note that taking X
=1, we will find the sum of the series .
8.
Integrating the series term by term, we obtain an expansion for the function 
. Perform all the calculations yourself, write out the convergence region.
9.
Let us write down the expansion of the function 
according to the binomial series formula with 
: . Denominator 
represented as , double factorial 
means the product of all natural numbers of the same parity as 
, not exceeding 
. The expansion converges to the function at 
. Integrating it term by term from 0 to X
, we will receive . It turns out that this series converges to the function on the entire interval 
; at X
=1 we get another beautiful representation of the number 
:
.
18.2.6.2. Solving problems involving series expansion of functions. Most problems in which you need to expand an elementary function into a power series 
, is solved by using standard expansions. Fortunately, every basic elementary function has a property that allows you to do this. Let's look at a number of examples.
1. Expand the function 
by degrees 
.
Solution. . The series converges at 
.
2. Expand the function 
by degrees 
.
Solution. 
. Convergence area: 
.
3. Expand the function 
by degrees 
.
Solution. . The series converges at 
.
4. Expand the function 
by degrees 
.
Solution. . The series converges at 
.
5. Expand the function 
by degrees 
.
Solution. . Convergence region 
.
6. Expand the function 
by degrees 
.
Solution. The expansion into a series of simple rational fractions of the second type is obtained by term-by-term differentiation of the corresponding expansions of fractions of the first type. In this example. Further, by term-by-term differentiation, we can obtain expansions of the functions 
,
etc.
7. Expand the function 
by degrees 
.
Solution. If a rational fraction is not a simple fraction, it is first represented as a sum of simple fractions: 
, and then proceed as in example 5: where 
.
Naturally, this approach is not applicable, for example, to decompose the function 
by degrees X
. Here, if you need to get the first few terms of the Taylor series, the easiest way is to find the values at the point X
=0 required number of first derivatives.
What does factoring mean? How to do it? What can you learn from factoring a number into prime factors? The answers to these questions are illustrated with specific examples.
Definitions:
A number that has exactly two different divisors is called prime.
A number that has more than two divisors is called composite.
To factor 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 decomposition of a prime number, one of the factors is equal to one, and the other is equal to the number itself.
- It makes no sense to talk about factoring unity.
- A composite number can be factored into factors, each of which is different from 1.
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Let's factor the number 150. For example, 150 is 15 times 10. 15 is a composite number. It can be factored into prime factors of 5 and 3. 10 is a composite number. It can be factored into prime factors of 5 and 2. By writing their decompositions into prime factors instead of 15 and 10, we obtained the decomposition of the number 150. |
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The number 150 can be factorized in another way. For example, 150 is the product of the numbers 5 and 30. 5 is a prime number. 30 is a composite number. It can be thought of as the product of 10 and 3. 10 is a composite number. It can be factored into prime factors of 5 and 2. We obtained the factorization of 150 into prime factors in a different way. |
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Note that the first and second expansions are the same. They differ only in the order of the factors. It is customary to write factors in ascending order. |
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Every composite number can be factorized into prime factors in a unique way, up to the order of the factors. |
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When factoring large numbers into prime factors, use column notation:
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The smallest prime number that is divisible by 216 is 2. Divide 216 by 2. We get 108. |
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The resulting number 108 is divided by 2. Let's do the division. The result is 54. |
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According to the test of divisibility by 2, the number 54 is divisible by 2. After dividing, we get 27. |
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The number 27 ends with the odd digit 7. It Not divisible by 2. The next prime number is 3. Divide 27 by 3. We get 9. Least prime The number that 9 is divisible by is 3. Three is itself a prime number; it is divisible by itself and one. Let's divide 3 by ourselves. In the end we got 1. |
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- A number is divisible only by those prime numbers that are part of its decomposition.
- A number is divisible only into those composite numbers whose decomposition into prime factors is completely contained in it.
Let's look at examples:
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4900 is divisible by the prime numbers 2, 5 and 7 (they are included in the expansion of the number 4900), but is not divisible by, for example, 13. |
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11 550 75. This is so because the decomposition of the number 75 is completely contained in the decomposition of the number 11550. The result of division will be the product of factors 2, 7 and 11. 11550 is not divisible by 4 because there is an extra two in the expansion of four. |
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
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The decomposition of the number b is completely contained in the decomposition of the number a. |
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The result of dividing a by b is the product of the three numbers remaining in the expansion of a. So the answer is: 30. |
Bibliography
- Vilenkin N.Ya., Zhokhov V.I., Chesnokov A.S., Shvartsburd S.I. Mathematics 6. - M.: Mnemosyne, 2012.
- Merzlyak A.G., Polonsky V.V., Yakir M.S. Mathematics 6th grade. - Gymnasium. 2006.
- Depman I.Ya., Vilenkin N.Ya. Behind the pages of a mathematics textbook. - M.: Education, 1989.
- Rurukin A.N., Tchaikovsky I.V. Assignments for the mathematics course for grades 5-6. - M.: ZSh MEPhI, 2011.
- Rurukin A.N., Sochilov S.V., Tchaikovsky K.G. Mathematics 5-6. A manual for 6th grade students at the MEPhI correspondence school. - M.: ZSh MEPhI, 2011.
- Shevrin L.N., Gein A.G., Koryakov I.O., Volkov M.V. Mathematics: Textbook-interlocutor for 5-6 grades of secondary school. - M.: Education, Mathematics Teacher Library, 1989.
- Internet portal Matematika-na.ru ().
- Internet portal Math-portal.ru ().
Homework
- Vilenkin N.Ya., Zhokhov V.I., Chesnokov A.S., Shvartsburd S.I. Mathematics 6. - M.: Mnemosyne, 2012. No. 127, No. 129, No. 141.
- Other tasks: No. 133, No. 144.
