In this article you will find all necessary information answering the question how to factor a number into prime factors... First given general idea on the decomposition of a number into prime factors, examples of decompositions are given. The following shows the canonical form of the factorization of a number into prime factors. After that, an algorithm for decomposing arbitrary numbers into prime factors is given and examples of decomposing numbers using this algorithm are given. Also considered alternative ways that allow you to quickly decompose small integers into prime factors using divisibility criteria and multiplication tables.
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What does it mean to factor a number into prime factors?
First, let's figure out what prime factors are.
It is clear that since the word “factors” is present in this phrase, then there is a product of some numbers, and the qualifying word “simple” means that each factor is a prime number. For example, in a product of the form 2 · 7 · 7 · 23 there are four prime factors: 2, 7, 7, and 23.
What does it mean to factor a number into prime factors?
It means that given number must be represented as a product of prime factors, and the value of this product must be equal to the original number. As an example, consider the product of three primes 2, 3 and 5, it is equal to 30, so the factorization of 30 into prime factors is 2 · 3 · 5. Usually, the decomposition of a number into prime factors is written as an equality, in our example it will be like this: 30 = 2 · 3 · 5. We emphasize separately that prime factors in the expansion can be repeated. This is clearly illustrated by the following example: 144 = 2 · 2 · 2 · 2 · 3 · 3. But the representation of the form 45 = 3 · 15 is not a prime factorization, since the number 15 is composite.
The following question arises: "What numbers in general can be decomposed into prime factors"?
In search of an answer to it, we present the following reasoning. Prime numbers are, by definition, among those larger than ones. Considering this fact and, it can be argued that the product of several prime factors is a positive integer greater than one. Therefore, prime factorization only takes place for positive integers greater than 1.
But do all integers greater than one factor out into prime factors?
It is clear that there is no way to decompose prime integers into prime factors. This is because prime numbers have only two positive divisors - one and themselves, so they cannot be represented as a product of two or more prime numbers. If the integer z could be represented as a product of prime numbers a and b, then the notion of divisibility would allow us to conclude that z is divisible by both a and b, which is impossible due to the simplicity of the number z. However, it is believed that any prime number itself is its expansion.
What about composite numbers? Do composite numbers decompose into prime factors, and are all composite numbers subject to such a decomposition? A number of these questions are answered in the affirmative by the main theorem of arithmetic. The main theorem of arithmetic states that any integer a that is greater than 1 can be decomposed into the product of prime factors p 1, p 2, ..., pn, and the decomposition is of the form a = p 1 p 2 ... the decomposition is unique, if the order of the factors is not taken into account
Canonical prime factorization
In the expansion of a number, prime factors can be repeated. Duplicate prime factors can be written more compactly using. Suppose that in the expansion of a number a prime factor p 1 occurs s 1 times, a prime factor p 2 - s 2 times, and so on, p n - s n times. Then the prime factorization of the number a can be written as a = p 1 s 1 p 2 s 2… p n s n... This form of recording is the so-called canonical prime factorization.
Let's give an example of the canonical factorization of a number into prime factors. Let us know the decomposition 609 840 = 2 2 2 2 3 3 5 7 11 11, its canonical notation is 609 840 = 2 4 3 2 5 7 11 2.
The canonical factorization of a number into prime factors allows you to find all the divisors of a number and the number of divisors of a number.
Algorithm for factoring a number into prime factors
To successfully cope with the problem of factoring a number into prime factors, you need to be very familiar with the information in the article on prime and composite numbers.
The essence of the process of decomposition of an integer positive and greater than one number a is clear from the proof of the main theorem of arithmetic. The idea is to sequentially find the smallest prime divisors p 1, p 2, ..., pn of numbers a, a 1, a 2, ..., a n-1, which allows us to obtain a series of equalities a = p 1 a 1, where a 1 = a: p 1, a = p 1 a 1 = p 1 p 2 a 2, where a 2 = a 1: p 2,…, a = p 1 p 2… = a n-1: pn. When we get a n = 1, then the equality a = p 1 · p 2 ·… · p n will give us the required decomposition of the number a into prime factors. It should be noted here that p 1 ≤p 2 ≤p 3 ≤… ≤p n.
It remains to figure out how to find the smallest prime factors at each step, and we will have an algorithm for factoring the number into prime factors. The table of prime numbers will help us find prime factors. Let us show how to use it to obtain the smallest prime divisor of the number z.
Sequentially we take primes from the table of primes (2, 3, 5, 7, 11, and so on) and divide the given number z by them. The first prime number z is divided by one integer will be its smallest prime divisor. If the number z is prime, then its smallest prime divisor will be the number z itself. It should be recalled here that if z is not prime number, then its smallest prime divisor does not exceed the number, where is from z. Thus, if among the prime numbers not exceeding, there was not a single divisor of the number z, then we can conclude that z is a prime number (for more details, see the theory section under the heading this number is prime or composite).
As an example, we'll show you how to find the smallest prime divisor of 87. We take the number 2. Divide 87 by 2, we get 87: 2 = 43 (rest. 1) (if necessary, see the article). That is, dividing 87 by 2 results in a remainder of 1, so 2 is not a divisor of 87. We take the next prime number from the table of primes, which is 3. We divide 87 by 3, we get 87: 3 = 29. Thus, 87 is evenly divisible by 3, so 3 is the smallest prime divisor of 87.
Note that in the general case, to factor a number a into prime factors, we need a table of primes up to a number not less than. We will have to refer to this table at every step, so you need to have it at hand. For example, to factor 95 into prime factors, a table of primes up to 10 will suffice (since 10 is greater than). And to decompose the number 846 653, you will already need a table of primes up to 1,000 (since 1,000 is more than).
We now have sufficient information to write prime factorization algorithm... The decomposition algorithm for the number a is as follows:
- Sequentially going through the numbers from the table of primes, we find the smallest prime divisor p 1 of the number a, after which we calculate a 1 = a: p 1. If a 1 = 1, then the number a is prime, and it is itself its prime factorization. If a 1 is not equal to 1, then we have a = p 1 · a 1 and go to the next step.
- Find the smallest prime divisor p 2 of a 1, for this we sequentially iterate over the numbers from the table of primes, starting with p 1, and then calculate a 2 = a 1: p 2. If a 2 = 1, then the required factorization of the number a into prime factors has the form a = p 1 · p 2. If a 2 is not equal to 1, then we have a = p 1 · p 2 · a 2 and go to the next step.
- Going through the numbers from the table of primes, starting with p 2, we find the smallest prime divisor p 3 of the number a 2, after which we calculate a 3 = a 2: p 3. If a 3 = 1, then the required factorization of the number a into prime factors has the form a = p 1 · p 2 · p 3. If a 3 is not equal to 1, then we have a = p 1 · p 2 · p 3 · a 3 and go to the next step.
- Find the smallest prime divisor p n of a n-1 by going through prime numbers, starting with p n-1, and also a n = a n-1: p n, and a n is equal to 1. This step is the last step of the algorithm, here we get the required decomposition of the number a into prime factors: a = p 1 · p 2 ·… · p n.
For clarity, all the results obtained at each step of the algorithm for decomposing a number into prime factors are presented in the form of the following table, in which, to the left of the vertical line, the numbers a, a 1, a 2, ..., an are written sequentially in a column, and to the right of the line - the corresponding least prime divisors p 1, p 2,…, pn.
It remains only to consider a few examples of the application of the obtained algorithm for the decomposition of numbers into prime factors.
Prime Factoring Examples
Now we will analyze in detail examples of factoring numbers into prime factors... In the decomposition, we will apply the algorithm from the previous paragraph. Let's start with simple cases, and gradually complicate them in order to face all the possible nuances that arise when factoring numbers into prime factors.
Example.
Divide 78 into prime factors.
Solution.
We start looking for the first smallest prime divisor p 1 of the number a = 78. To do this, we begin to sequentially iterate over the prime numbers from the table of prime numbers. We take the number 2 and divide 78 by it, we get 78: 2 = 39. The number 78 was divided by 2 without a remainder, so p 1 = 2 is the first found prime divisor of 78. In this case, a 1 = a: p 1 = 78: 2 = 39. So we come to the equality a = p 1 · a 1 having the form 78 = 2 · 39. Obviously, a 1 = 39 is different from 1, so we pass to the second step of the algorithm.
Now we are looking for the smallest prime divisor p 2 of the number a 1 = 39. We start iterating over the numbers from the table of primes, starting with p 1 = 2. Divide 39 by 2, we get 39: 2 = 19 (rest. 1). Since 39 is not divisible by 2, 2 is not a divisor of it. Then we take next number from the table of prime numbers (number 3) and divide by 39, we get 39: 3 = 13. Therefore, p 2 = 3 is the smallest prime divisor of 39, while a 2 = a 1: p 2 = 39: 3 = 13. We have the equality a = p 1 · p 2 · a 2 in the form 78 = 2 · 3 · 13. Since a 2 = 13 is different from 1, then go to the next step of the algorithm.
Here we need to find the smallest prime divisor of the number a 2 = 13. In search of the smallest prime divisor p 3 of 13, we will sequentially iterate over the numbers from the table of primes, starting with p 2 = 3. The number 13 is not divisible by 3, since 13: 3 = 4 (rest. 1), also 13 is not divisible by 5, 7 and 11, since 13: 5 = 2 (rest. 3), 13: 7 = 1 (rest. 6) and 13:11 = 1 (rest. 2). The next prime number is 13, and 13 is divisible by it without a remainder, therefore, the smallest prime divisor p 3 of 13 is the number 13 itself, and a 3 = a 2: p 3 = 13: 13 = 1. Since a 3 = 1, this step of the algorithm is the last, and the required factorization of 78 into prime factors has the form 78 = 2 · 3 · 13 (a = p 1 · p 2 · p 3).
Answer:
78 = 2 3 13.
Example.
Present the number 83,006 as a product of prime factors.
Solution.
At the first step of the algorithm for decomposing a number into prime factors, we find p 1 = 2 and a 1 = a: p 1 = 83 006: 2 = 41 503, whence 83 006 = 2 · 41 503.
At the second step, we find out that 2, 3 and 5 are not prime divisors of the number a 1 = 41 503, and the number 7 is, since 41 503: 7 = 5 929. We have p 2 = 7, a 2 = a 1: p 2 = 41 503: 7 = 5 929. Thus, 83 006 = 2 7 5 929.
The smallest prime factor of a 2 = 5 929 is 7, since 5 929: 7 = 847. Thus, p 3 = 7, a 3 = a 2: p 3 = 5 929: 7 = 847, whence 83 006 = 2 7 7 847.
Then we find that the smallest prime divisor p 4 of the number a 3 = 847 is 7. Then a 4 = a 3: p 4 = 847: 7 = 121, therefore 83 006 = 2 7 7 7 7 121.
Now we find the smallest prime divisor of the number a 4 = 121, it is the number p 5 = 11 (since 121 is divisible by 11 and not divisible by 7). Then a 5 = a 4: p 5 = 121: 11 = 11, and 83 006 = 2 · 7 · 7 · 7 · 11 · 11.
Finally, the smallest prime factor of a 5 = 11 is p 6 = 11. Then a 6 = a 5: p 6 = 11: 11 = 1. Since a 6 = 1, then this step of the algorithm for decomposing a number into prime factors is the last, and the required decomposition has the form 83 006 = 2 · 7 · 7 · 7 · 11 · 11.
The result obtained can be written as the canonical factorization of a number into prime factors 83 006 = 2 · 7 3 · 11 2.
Answer:
83 006 = 2 7 7 7 11 11 = 2 7 3 11 2 991 is a prime number. Indeed, it does not have a single prime divisor not exceeding (can be roughly estimated as, since it is obvious that 991<40 2
), то есть, наименьшим делителем числа 991
является оно само. Тогда p 3 =991
и a 3 =a 2:p 3 =991:991=1
. Следовательно, искомое разложение числа 897 924 289
на простые множители имеет вид 897 924 289=937·967·991
.
Answer:
897 924 289 = 937 967 991.
Using divisibility criteria for prime factorization
In simple cases, you can decompose a number into prime factors without using the decomposition algorithm from the first paragraph of this article. If the numbers are not large, then for their decomposition into prime factors it is often enough to know the divisibility criteria. Here are some examples for clarification.
For example, we need to factor 10 into prime factors. From the multiplication table, we know that 2 · 5 = 10, and the numbers 2 and 5 are obviously prime, so the prime factorization of 10 is 10 = 2 · 5.
Another example. Using the multiplication table, factor 48 into prime factors. We know that six eight is forty-eight, that is, 48 = 6 · 8. However, neither 6 nor 8 are prime numbers. But we know that twice three is six, and twice four is eight, that is, 6 = 2 · 3 and 8 = 2 · 4. Then 48 = 6 8 = 2 3 2 4. It remains to remember that two times two is four, then we get the required decomposition into prime factors 48 = 2 · 3 · 2 · 2 · 2. We write this decomposition in the canonical form: 48 = 2 4 · 3.
But when decomposing the number 3 400 into prime factors, you can use the divisibility criteria. Divisibility by 10, 100 allows us to assert that 3400 is divisible by 100, while 3400 = 34100, and 100 is divisible by 10, while 100 = 1010, therefore, 3400 = 341010. And on the basis of the divisibility criterion by 2, it can be argued that each of the factors 34, 10 and 10 is divisible by 2, we get 3 400 = 34 10 10 = 2 17 2 5 2 5... All factors in the resulting decomposition are prime, so this decomposition is the desired one. It remains only to rearrange the factors so that they go in ascending order: 3400 = 2 · 2 · 2 · 5 · 5 · 17. We also write the canonical factorization of this number into prime factors: 3 400 = 2 3 · 5 2 · 17.
When decomposing a given number into prime factors, you can use in turn both the divisibility criteria and the multiplication table. Let's represent the number 75 as a product of prime factors. Divisibility by 5 allows us to assert that 75 is divisible by 5, and we get that 75 = 5 15. And from the multiplication table we know that 15 = 3 · 5, therefore, 75 = 5 · 3 · 5. This is the required prime factorization of 75.
Bibliography.
- Vilenkin N. Ya. and other Mathematics. Grade 6: textbook for educational institutions.
- Vinogradov I.M. Fundamentals of number theory.
- Mikhelovich Sh.Kh. Number theory.
- Kulikov L.Ya. and others. Collection of problems in algebra and number theory: a textbook for students of physics and mathematics. specialties of pedagogical institutes.
Online calculator.
Selection of a square of a binomial and factorization of a square trinomial.
This math program extracts the square binomial from the square trinomial, i.e. does a conversion like: \ (ax ^ 2 + bx + c \ rightarrow a (x + p) ^ 2 + q \) and factors a square trinomial: \ (ax ^ 2 + bx + c \ rightarrow a (x + n) (x + m) \)
Those. problems are reduced to finding the numbers \ (p, q \) and \ (n, m \)
The program not only gives the answer to the problem, but also displays the solution process.
This program can be useful for senior students of secondary schools in preparation for tests and exams, when checking knowledge before the exam, for parents to control the solution of many problems in mathematics and algebra. Or maybe it's too expensive for you to hire a tutor or buy new textbooks? Or do you just want to get your math or algebra homework done as quickly as possible? In this case, you can also use our programs with a detailed solution.
In this way, you can conduct your own teaching and / or the teaching of your younger siblings, while the level of education in the field of the problems being solved increases.
If you are not familiar with the rules for entering a square trinomial, we recommend that you familiarize yourself with them.
Rules for entering a square polynomial
Any Latin letter can be used as a variable.
For example: \ (x, y, z, a, b, c, o, p, q \) etc.
Numbers can be entered as whole or fractional numbers.
Moreover, fractional numbers can be entered not only in the form of a decimal, but also in the form of an ordinary fraction.
Rules for entering decimal fractions.
In decimal fractions, the fractional part from the whole can be separated by either a point or a comma.
For example, you can enter decimal fractions like this: 2.5x - 3.5x ^ 2
Rules for entering ordinary fractions.
Only an integer can be used as the numerator, denominator and whole part of a fraction.
The denominator cannot be negative.
When entering a numeric fraction, the numerator is separated from the denominator by a division sign: /
The whole part is separated from the fraction by an ampersand: &
Input: 3 & 1/3 - 5 & 6 / 5x + 1 / 7x ^ 2
Result: \ (3 \ frac (1) (3) - 5 \ frac (6) (5) x + \ frac (1) (7) x ^ 2 \)
When entering an expression brackets can be used... In this case, when solving, the entered expression is first simplified.
For example: 1/2 (x-1) (x + 1) - (5x-10 & 1/2)
Detailed solution example
Selection of a square of a binomial.$$ ax ^ 2 + bx + c \ rightarrow a (x + p) ^ 2 + q $$ $$ 2x ^ 2 + 2x-4 = $$ $$ 2x ^ 2 + 2 \ cdot 2 \ cdot \ left ( \ frac (1) (2) \ right) \ cdot x + 2 \ cdot \ left (\ frac (1) (2) \ right) ^ 2- \ frac (9) (2) = $$ $$ 2 \ left (x ^ 2 + 2 \ cdot \ left (\ frac (1) (2) \ right) \ cdot x + \ left (\ frac (1) (2) \ right) ^ 2 \ right) - \ frac (9 ) (2) = $$ $$ 2 \ left (x + \ frac (1) (2) \ right) ^ 2- \ frac (9) (2) $$ Answer:$$ 2x ^ 2 + 2x-4 = 2 \ left (x + \ frac (1) (2) \ right) ^ 2- \ frac (9) (2) $$ Factorization.$$ ax ^ 2 + bx + c \ rightarrow a (x + n) (x + m) $$ $$ 2x ^ 2 + 2x-4 = $$
$$ 2 \ left (x ^ 2 + x-2 \ right) = $$
$$ 2 \ left (x ^ 2 + 2x-1x-1 \ cdot 2 \ right) = $$ $$ 2 \ left (x \ left (x +2 \ right) -1 \ left (x +2 \ right ) \ right) = $$ $$ 2 \ left (x -1 \ right) \ left (x +2 \ right) $$ Answer:$$ 2x ^ 2 + 2x-4 = 2 \ left (x -1 \ right) \ left (x +2 \ right) $$
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A bit of theory.
Extraction of a square binomial from a square trinomial
If the square trinomial ax 2 + bx + c is represented in the form a (x + p) 2 + q, where p and q are real numbers, then they say that from of a square trinomial, the square of a binomial.
Select from the trinomial 2x 2 + 12x + 14 the square of the binomial.
\ (2x ^ 2 + 12x + 14 = 2 (x ^ 2 + 6x + 7) \)
To do this, we represent 6x as a product of 2 * 3 * x, and then add and subtract 3 2. We get:
$$ 2 (x ^ 2 + 2 \ cdot 3 \ cdot x + 3 ^ 2-3 ^ 2 + 7) = 2 ((x + 3) ^ 2-3 ^ 2 + 7) = $$ $$ = 2 ((x + 3) ^ 2-2) = 2 (x + 3) ^ 2-4 $$
That. we singled out the square binomial from the square trinomial, and show that:
$$ 2x ^ 2 + 12x + 14 = 2 (x + 3) ^ 2-4 $$
Factoring a square trinomial
If the square trinomial ax 2 + bx + c is represented in the form a (x + n) (x + m), where n and m are real numbers, then the operation is said to have been performed square trinomial factorization.
Let's show with an example how this transformation is done.
Factor the square trinomial 2x 2 + 4x-6.
Let's take out the coefficient a from brackets, i.e. 2:
\ (2x ^ 2 + 4x-6 = 2 (x ^ 2 + 2x-3) \)
Let's transform the expression in parentheses.
To do this, we represent 2x as the difference 3x-1x, and -3 as -1 * 3. We get:
$$ = 2 (x ^ 2 + 3 \ cdot x -1 \ cdot x -1 \ cdot 3) = 2 (x (x + 3) -1 \ cdot (x + 3)) = $$
$$ = 2 (x-1) (x + 3) $$
That. we factorized the square trinomial, and show that:
$$ 2x ^ 2 + 4x-6 = 2 (x-1) (x + 3) $$
Note that factorization of a quadratic trinomial is possible only when the quadratic equation corresponding to this trinomial has roots.
Those. in our case, factoring the trinomial 2x 2 + 4x-6 is possible if the quadratic equation 2x 2 + 4x-6 = 0 has roots. In the process of factoring, we found that the equation 2x 2 + 4x-6 = 0 has two roots 1 and -3, because for these values, the equation 2 (x-1) (x + 3) = 0 becomes a true equality.
Factoring a polynomial. Part 2
In this article, we will continue to talk about how factor out a polynomial. We have already said that factorization is a versatile trick for solving complex equations and inequalities. The first thought that should come to mind when solving equations and inequalities in which there is zero on the right side is to try to factor the left side.
Let's list the main methods of factoring a polynomial:
- parenthesis of a common factor
- using abbreviated multiplication formulas
- by the factorization formula of the square trinomial
- grouping method
- division of a polynomial by a binomial
- method of undefined coefficients.
We have already covered in detail. In this article, we will focus on the fourth method, method of grouping.
If the number of terms in the polynomial exceeds three, then we try to apply grouping method... It is as follows:
1.We group the terms in a certain way so that then each group can be factorized in some way. The criterion that the terms are grouped correctly is the presence of the same factors in each group.
2. Factor out the same factors.
Since this method is used most often, we will analyze it with examples.
Example 1. 
Solution. 1. Let's combine the terms into groups:
2. Take out the common factor from each group:
3. Let's take out the factor common for both groups:
Example 2. Factor expression: 
1. Let us group the last three terms and factor them according to the formula of the square of the difference:
2. Let us factorize the resulting expression by the formula for the difference of squares:
Example 3. Solve the equation:
There are four terms on the left side of the equation. Let's try to factor the left side using grouping.
1. To make the structure of the left-hand side of the equation clearer, we introduce a change of variable:,
We get an equation of this form:
2. Factor the left side using grouping:
Attention! In order not to be mistaken with the signs, I recommend combining the terms into groups "as is", that is, without changing the signs of the coefficients, and the next action, if necessary, to place the "minus" outside the parenthesis.
3. So, we got the equation:
4. Let's go back to the original variable:
Let's divide both parts into. We get:. From here
Answer: 0
Example 4. Solve the equation:
To make the structure of the equation more "transparent", we introduce a variable change:
We get the equation:
Factor the left side of the equation. To do this, we group the first and second terms and put them outside the parenthesis:
put out of the bracket:
Let's go back to the equation:
Hence or,
Let's go back to the original variable:
Let's look at specific examples of how to factor a polynomial.
The decomposition of polynomials will be carried out in accordance with.
Factor polynomials:
Check if there is a common factor. is, it is equal to 7cd. Let's take it out of the brackets:
The expression in parentheses consists of two terms. There is no common factor anymore, the expression is not a formula for the sum of cubes, which means that the decomposition is complete.
Check if there is a common factor. No. The polynomial consists of three terms, so we check if there is a perfect square formula. Two terms are squares of expressions: 25x² = (5x) ², 9y² = (3y) ², the third term is equal to twice the product of these expressions: 2 ∙ 5x ∙ 3y = 30xy. Hence, this polynomial is a perfect square. Since the doubled product is with a minus sign, then this is -:
We check if it is possible to take the common factor out of the parentheses. There is a common factor, it is equal to a. Let's take it out of the brackets:
There are two terms in brackets. Check if there is a formula for the difference of squares or the difference between cubes. a² - square a, 1 = 1². This means that the expression in brackets can be written using the formula for the difference of squares:
There is a common factor, it is 5. We take it out of the parentheses:
in brackets - three terms. Check if the expression is not a perfect square. Two terms are squares: 16 = 4² and a² is the square of a, the third term is equal to twice the product of 4 and a: 2 ∙ 4 ∙ a = 8a. Hence, it is a complete square. Since all terms are with a "+" sign, the expression in parentheses is the full square of the sum:
We take out the common factor -2x outside the brackets:
In brackets is the sum of two terms. Check if the given expression is a sum of cubes. 64 = 4³, x³- cube x. Hence, the binomial can be expanded by the formula:
There is a common factor. But, since the polynomial consists of 4 terms, we will first, and only then, factor out the common factor. Let's group the first term with the fourth, in the second - with the third:
From the first parentheses we take out the common factor 4a, from the second - 8b:
There is no common factor yet. To get it, from the second brackets we take out “-“ outside the brackets, while each character in the brackets will change to the opposite:
Now we take out the common factor (1-3a) outside the brackets:
In the second parentheses there is a common factor of 4 (this is the same factor that we did not put outside the parentheses at the beginning of the example):
Since the polynomial consists of four terms, we perform the grouping. Let's group the first term with the second, the third - with the fourth:
In the first brackets there is no common factor, but there is a formula for the difference of squares, in the second brackets there is a common factor -5:
A common factor (4m-3n) has appeared. We take it out of the parentheses.
In general, this task involves a creative approach, since there is no universal method for solving it. But still, let's try to give a few tips.
In the overwhelming majority of cases, the factorization of a polynomial is based on a corollary from Bezout's theorem, that is, a root is found or selected and the degree of the polynomial is reduced by one by dividing by. A root is sought for the resulting polynomial, and the process is repeated until it is completely decomposed.
If the root cannot be found, then specific decomposition methods are used: from grouping to the introduction of additional mutually exclusive terms.
The further presentation is based on the skills of solving equations of higher degrees with integer coefficients.
Factor out the common factor.
Let's start with the simplest case when the free term is equal to zero, that is, the polynomial has the form.
Obviously, the root of such a polynomial is, that is, the polynomial can be represented in the form.
This method is nothing more than factoring out the common factor.
Example.
Factor a third-degree polynomial.
Solution.
Obviously, it is a root of a polynomial, that is NS can be taken outside the brackets:
Find the roots of the square trinomial 
Thus, 
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Factoring a polynomial with rational roots.
First, consider a method for decomposing a polynomial with integer coefficients of the form, the coefficient at the highest power is equal to one.
In this case, if the polynomial has integer roots, then they are the divisors of the free term.
Example.
Solution.
Let's check if there are whole roots. To do this, we write out the divisors of the number -18
:. That is, if the polynomial has integer roots, then they are among the written out numbers. Let's check these numbers one by one according to Horner's scheme. Its convenience also lies in the fact that, as a result, we obtain the coefficients of the expansion of the polynomial: 
That is, x = 2 and x = -3 are the roots of the original polynomial and it can be represented as a product:
It remains to expand the square trinomial.
The discriminant of this trinomial is negative, therefore, it has no real roots.
Answer:
Comment:
instead of Horner's scheme, one could use the selection of the root and the subsequent division of the polynomial by the polynomial.
Now consider the decomposition of a polynomial with integer coefficients of the form, and the coefficient at the highest degree is not equal to one.
In this case, the polynomial can have fractionally rational roots.
Example.
Factor expression.
Solution.
By performing variable replacement y = 2x, we pass to a polynomial with a coefficient equal to one at the highest degree. To do this, first we multiply the expression by 4
.
If the resulting function has integer roots, then they are among the divisors of the free term. Let's write them down:
Let us calculate successively the values of the function g (y) at these points until zero is obtained. 