Any composite number can be factorized into prime factors. There can be several methods of decomposition. Either method produces the same result.
How to factor a number into prime factors most in a convenient way? Let's look at how best to do this using specific examples.
Examples. 1) Factor the number 1400 into prime factors.
1400 is divisible by 2. 2 is a prime number; there is 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 is 35 - we divide it by 5 again. Total is 7. It can only be divided by 7. We get 1, division over.
The same number can be factorized differently:
1400 is conveniently divided by 10. 10 is not prime number, so it needs to be factorized into simple factors: 10=2∙5. The result is 140. We divide it again by 10=2∙5. We get 14. If 14 is divided by 14, then it should also be decomposed into a 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 into prime factors. We divide by what is more convenient, for example, by 10. You just need to remember to decompose the compound divisors into simple factors.
2) Factor the number 1620 into prime factors.
The most convenient way to divide the number 1620 is 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 by 9 it is more convenient. Since 9 is not a prime number, we expand it as 9=3∙3. We get 9. We also divide it by 9 and expand it into the product of prime factors.
Let's factor 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
- divisible by the prime number 2
30:
2
= 15
- divisible by the 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 factor the number 246 into prime factors
246 = 2 ∙ 3 ∙ 41
Solution
Let's break down 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 turns out to be a prime number
246:
2
= 123
- divisible by the prime number 2
123:
3
=
41
- divisible by the prime number 3.
We complete the division since 41 is a prime number
Answer: 246 = 2 ∙ 3 ∙ 41
Let's factor 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 turns out to be a prime number
1463:
7
= 209
- 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 factor 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 turns out to be a prime number
1268:
2
= 634
- divisible by the prime number 2
634:
2
=
317
- divisible by the prime number 2.
We complete the division since 317 is a prime number
Answer: 1268 = 2 ∙ 2 ∙ 317
Let's factor 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 turns out to be a prime number
442464:
2
= 221232
- divisible by the prime number 2
221232:
2
= 110616
- divisible by the prime number 2
110616:
2
= 55308
- divisible by the prime number 2
55308:
2
= 27654
- divisible by the prime number 2
27654:
2
= 13827
- divisible by the prime number 2
13827:
3
= 4609
- divisible by the prime number 3
4609:
11
=
419
- 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
Any composite number can be represented as a product of its prime divisors:
28 = 2 2 7
The right-hand sides of the resulting equalities are called prime factorization numbers 15 and 28.
To factor a given composite number into prime factors means to represent this number as a product of its prime factors.
The decomposition of a given number into prime factors is performed as follows:
- First you need to select the smallest prime number from the table of prime numbers that divides the given composite number without a remainder, and perform the division.
- Next, you need to again select the smallest prime number by which the already obtained quotient will be divided without a remainder.
- The second action is repeated until one is obtained in the quotient.
As an example, let's factorize the number 940 into prime factors. Find the smallest prime number that divides 940. This number is 2:
Now we select the smallest prime number that is divisible by 470. This number is again 2:
The smallest prime number that is divisible by 235 is 5:
The number 47 is prime, which means that the smallest prime number that can be divided by 47 is the number itself:
Thus, we get the number 940, factored into prime factors:
940 = 2 470 = 2 2 235 = 2 2 5 47
If the decomposition of a number into prime factors resulted in several identical factors, then for brevity, they can be written in the form of a power:
940 = 2 2 5 47
It is most convenient to write decomposition into prime factors as follows: first we write down this composite number and draw a vertical line to the right of it:
To the right of the line we write the smallest prime divisor by which the given composite number is divided:
We perform the division and write the resulting quotient under the dividend:
We act with the quotient in the same way as with the given composite number, i.e., we select the smallest prime number by which it is divisible without a remainder and perform the division. And we repeat this until we get a unit in the quotient:
Please note that sometimes it can be quite difficult to factor a number into prime factors, since when factoring we may encounter a large number, which is difficult to immediately determine whether it is simple or compound. And if it is composite, then it is not always easy to find its smallest prime divisor.
Let’s try, for example, to factorize the number 5106 into prime factors:
Having reached the quotient 851, it is difficult to immediately determine its smallest divisor. We turn to the table of prime numbers. If there is a number in it that puts us in difficulty, then it is divisible only by itself and one. The number 851 is not in the table of prime numbers, which means it is composite. All that remains is to divide it by sequential search into prime numbers: 3, 7, 11, 13, ..., and so on until we find a suitable prime divisor. By brute force we find that 851 is divisible by the number 23.
In this article you will find all necessary information answering the question how to factor a number into prime factors. First, a general idea of the decomposition of a number into prime factors is given, and examples of decompositions are given. The following shows the canonical form of decomposing a number into prime factors. After this, an algorithm is given for decomposing arbitrary numbers into prime factors and examples of decomposing numbers using this algorithm are given. Alternative methods are also considered that allow you to quickly factor small integers into prime factors using divisibility tests and multiplication tables.
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What does it mean to factor a number into prime factors?
First, let's look at 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?
This means that this 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 prime numbers 2, 3 and 5, it is equal to 30, thus the decomposition of the number 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 a representation of the form 45=3·15 is not a decomposition into prime factors, since the number 15 is a composite number.
The following question arises: “What numbers can be decomposed into prime factors?”
In search of an answer to it, we present the following reasoning. Prime numbers, by definition, are among those greater than one. Considering this fact and , it can be argued that the product of several prime factors is a positive integer greater than one. Therefore, factorization into prime factors occurs only for positive integers that are greater than 1.
But can all integers greater than one be factored into prime factors?
It is clear that it is not possible to factor simple integers into prime factors. This is because prime numbers have only two positive factors - one and itself, so they cannot be represented as the product of two or more prime numbers. If the integer z could be represented as the product of prime numbers a and b, then the concept 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, they believe that any prime number is itself a decomposition.
What about composite numbers? Are composite numbers decomposed into prime factors, and are all composite numbers subject to such decomposition? The fundamental theorem of arithmetic gives an affirmative answer to a number of these questions. The basic 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, ..., p n, and the decomposition has the form a = p 1 · p 2 ·… · p n, and this the expansion is unique, if you do not take into account the order of the factors
Canonical factorization of a number into prime factors
In the expansion of a number, prime factors can be repeated. Repeating prime factors can be written more compactly using . Let in the decomposition of a number the prime factor p 1 occur s 1 times, the 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 factorization of a number into prime factors.
Let us give an example of the canonical decomposition 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 has the form 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 the number and the number of divisors of the number.
Algorithm for factoring a number into prime factors
To successfully cope with the task of decomposing a number into prime factors, you need to have a very good knowledge of the information in the article prime and composite numbers.
The essence of the process of decomposing a positive integer number a that exceeds one is clear from the proof of the fundamental theorem of arithmetic. The point is to sequentially find the smallest prime divisors p 1, p 2, ..., p n of the 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 ·…·p n ·a n , where a n =a n-1:p n . When it turns out a n =1, then the equality a=p 1 ·p 2 ·…·p n will give us the desired decomposition of the number a into prime factors. It should also 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 decomposing a number into prime factors. A 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.
We sequentially take prime numbers from the table of prime numbers (2, 3, 5, 7, 11, and so on) and divide the given number z by them. The first prime number by which z is evenly divided 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 a 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 (more about this is written in the theory section under the heading This number is prime or composite).
As an example, we will show how to find the smallest prime divisor of the number 87. Let's take the number 2. Divide 87 by 2, we get 87:2=43 (remaining 1) (if necessary, see article). That is, when dividing 87 by 2, the remainder is 1, so 2 is not a divisor of the number 87. We take the next prime number from the prime numbers table, this is the number 3. Divide 87 by 3, we get 87:3=29. Thus, 87 is divisible by 3, therefore, the number 3 is the smallest prime divisor of the number 87.
Note that in general case To factor the number a into prime factors, we need a table of prime numbers up to a number not less than . We will have to refer to this table at every step, so we need to have it at hand. For example, to factorize the number 95 into prime factors, we will only need a table of prime numbers up to 10 (since 10 is greater than ). And to decompose the number 846,653, you will already need a table of prime numbers up to 1,000 (since 1,000 is greater than ).
We now have enough information to write down algorithm for factoring a number into prime factors. The algorithm for decomposing the number a is as follows:
- Sequentially sorting through the numbers from the table of prime numbers, 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 itself is its decomposition into prime factors. If a 1 is not equal to 1, then we have a=p 1 ·a 1 and move on to the next step.
- We find the smallest prime divisor p 2 of the number a 1 , to do this we sequentially sort through the numbers from the table of prime numbers, starting with p 1 , and then calculate a 2 =a 1:p 2 . If a 2 =1, then the required decomposition 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 move on to the next step.
- Going through the numbers from the table of prime numbers, 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 decomposition 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 move on to the next step.
- We find the smallest prime divisor p n of the number a n-1 by sorting through the prime numbers, starting with p n-1, as well as 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 obtain 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 the numbers a, a 1, a 2, ..., a n are written sequentially in a column to the left of the vertical line, and to the right of the line - the corresponding smallest prime divisors p 1, p 2, ..., p n.
All that remains is to consider a few examples of the application of the resulting algorithm for decomposing numbers into prime factors.
Examples of prime factorization
Now we will look in detail examples of factoring numbers into prime factors. When decomposing, we will use the algorithm from the previous paragraph. Let's start with simple cases, and gradually complicate them in order to encounter all the possible nuances that arise when decomposing numbers into prime factors.
Example.
Factor the number 78 into its prime factors.
Solution.
We begin the search for the first smallest prime divisor p 1 of the number a=78. To do this, we begin to sequentially sort through 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 is divided by 2 without a remainder, so p 1 =2 is the first found prime divisor of the number 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 move on 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 begin enumerating numbers from the table of prime numbers, starting with p 1 =2. Divide 39 by 2, we get 39:2=19 (remaining 1). Since 39 is not evenly divisible by 2, then 2 is not its divisor. Then we take the next number from the table of prime numbers (number 3) and divide 39 by it, we get 39:3=13. Therefore, p 2 =3 is the smallest prime divisor of the number 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, we move on 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 the number 13, we will sequentially sort through the numbers from the table of prime numbers, 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 decomposition of the number 78 into prime factors has the form 78=2·3·13 (a=p 1 ·p 2 ·p 3 ).
Answer:
78=2·3·13.
Example.
Express 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, from which 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, but 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 divisor of the number a 2 =5 929 is the number 7, since 5 929:7 = 847. Thus, p 3 =7, a 3 =a 2:p 3 =5 929:7 = 847, from which 83 006 = 2·7·7·847.
Next we find that the smallest prime divisor p 4 of the number a 3 =847 is equal to 7. Then a 4 =a 3:p 4 =847:7=121, so 83 006=2·7·7·7·121.
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 divisor of the number a 5 =11 is the number p 6 =11. Then a 6 =a 5:p 6 =11:11=1. Since a 6 =1, this step of the algorithm for decomposing a number into prime factors is the last, and the desired decomposition has the form 83 006 = 2·7·7·7·11·11.
The result obtained can be written as the canonical decomposition of the 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 tests 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 to decompose them into prime factors it is often enough to know the signs of divisibility. Let's give examples for clarification.
For example, we need to factor the number 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 the number 10 looks like 10=2·5.
Another example. Using the multiplication table, we will factor the number 48 into prime factors. We know that six is eight - 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 desired decomposition into prime factors 48 = 2·3·2·2·2. Let's write this expansion in canonical form: 48=2 4 ·3.
But when factoring the number 3,400 into prime factors, you can use the divisibility criteria. The signs of divisibility by 10, 100 allow us to state that 3,400 is divisible by 100, with 3,400=34·100, and 100 is divisible by 10, with 100=10·10, therefore, 3,400=34·10·10. And based on the test of divisibility by 2, we can say 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 expansion are simple, so this expansion is the desired one. All that remains is to rearrange the factors so that they go in ascending order: 3 400 = 2·2·2·5·5·17. Let us also write down the canonical decomposition 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 signs of divisibility and the multiplication table. Let's imagine the number 75 as a product of prime factors. The test of divisibility by 5 allows us to state that 75 is divisible by 5, and we obtain that 75 = 5·15. And from the multiplication table we know that 15=3·5, therefore, 75=5·3·5. This is the required decomposition of the number 75 into prime factors.
Bibliography.
- Vilenkin N.Ya. and others. Mathematics. 6th grade: textbook for general education institutions.
- Vinogradov I.M. Fundamentals of number theory.
- Mikhelovich Sh.H. Number theory.
- Kulikov L.Ya. and others. Collection of problems in algebra and number theory: Textbook for students of physics and mathematics. specialties of pedagogical institutes.
This is one of the most basic ways to simplify an expression. To apply this method, let’s remember the distributive law of multiplication relative to addition (don’t 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 resulting 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 a very elementary example, just like the example given earlier, with the decomposition of a number, because everyone knows that numbers are divisible by, but what if you get a more complicated expression:
How do you find out what, for example, a number is divisible by? No, anyone can do it with a calculator, but without it it’s difficult? And for this there are signs of divisibility, these signs are really worth knowing, they will help you quickly understand whether the common factor can be taken out of the bracket.
Signs of divisibility
It’s not so difficult to remember them; most likely, most of them were already familiar to you, and some will be a new useful discovery, more details in the table:
Note: The table is missing the divisibility test by 4. If the last two digits are divisible by 4, then the entire number is divisible by 4.
Well, how do you like the sign? I advise you to remember it!
Well, let's return to the expression, maybe he can take it out of the bracket and that's enough? No, mathematicians tend to simplify, so to the fullest, endure EVERYTHING that is endured!
And so, everything is clear with the game, but what about the numerical part of the expression? Both numbers are odd, so you can't divide by
You can use the divisibility test: the sum of the digits, and, that make up the number is equal, and divisible by, means divisible by.
Knowing this, you can safely divide into a column, and as a result of dividing by we get (the signs of divisibility are useful!). Thus, we can take the number out of brackets, just like y, and as a result we have:
To make sure that everything has been expanded correctly, you can check the expansion by multiplying!
The common factor can also be expressed in power terms. Here, for example, do you see the common multiplier?
All members of this expression have xes - we take them out, they are all divided by - we take them out again, look at what happened: .
2. Abbreviated multiplication formulas
Abbreviated multiplication formulas have already been mentioned in theory; if you have difficulty remembering what they are, then you should refresh your memory.
Well, if you consider yourself very smart and 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 a certain formula in the expression in front of you, apply it and thus obtain the product of something and something, that’s all the decomposition. The following are the formulas:
Now try, factor the following expressions using the above formulas:
Here's 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:
So what are you going to do with it? It seems that something is divided into and into, and something into and into
But you can’t divide everything together into one thing, well there is no common factor here, no matter how you look, what should you leave it like that, without factoring it into factors?
Here you need to show ingenuity, and the name of this ingenuity is grouping!
It is used precisely when not all members have common divisors. For grouping you need find groups of terms that have common factors and rearrange them so that the same factor can be obtained from each group.
Of course, it is not necessary to rearrange them, but this gives clarity; for clarity, you can put 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.
Is all this not very clear? Let me explain with an example:
In a polynomial - we put the term - after the term - we get
we group the first two terms together in a separate bracket and also group the third and fourth terms, taking the minus sign out of the bracket, we get:
Now we look separately at each of the two “piles” into which we divided the expression with brackets.
The trick is to break it down into piles from which the largest factor can be taken out, or, as in this example, try to group the terms so that after removing the factors from the piles out of brackets, we still have the same expressions inside the brackets.
From both brackets we take out the common factors of the terms, from the first bracket, and from the second, we get:
But this is not decomposition!
Pdonkey decomposition should only remain multiplication, but for now our polynomial is simply divided into two parts...
BUT! This polynomial has a common factor. This
beyond the bracket and we get the final product
Bingo! As you can see, there is already a product here and outside the brackets there is no addition or subtraction, the decomposition is complete, because We have nothing more to take out of brackets.
It may seem like a miracle that after taking the factors out of brackets, we were left with identical expressions in brackets, which we again put out of brackets.
And this is not a miracle at all, the fact is that the examples in textbooks and in the Unified State Examination are specially made so that most expressions in tasks for simplification or factorization with the right approach to them, they are easily simplified and collapse sharply like an umbrella when you press a button, so look for that very button in every expression.
I got distracted, what are we doing with simplification? The intricate polynomial took on a simpler form: .
Agree, it’s not as bulky as it was?
4. Selecting a complete square.
Sometimes, to apply abbreviated multiplication formulas (repeat the topic), it is necessary to transform an existing polynomial, presenting one of its terms as the sum or difference of two terms.
In what case you have to do this, you will learn from the example:
A polynomial in this form cannot be expanded using abbreviated multiplication formulas, so it must be transformed. Perhaps at first it will not be obvious to you which term should be divided into which, but over time you will learn to immediately see the formulas for abbreviated multiplication, even if they are not entirely present, and you will quickly determine what is missing from the full formula, but for now - learn , a student, or rather a schoolboy.
For the complete formula for the squared difference, here you need instead. Let's imagine the third term as a difference, we get: To the expression in parentheses, you can apply the formula for the square of the difference (not to be confused with the difference of squares!!!), we have: , to this expression we can apply the difference of squares formula (not to be confused with the squared difference!!!), imagining how, we get: .
An expression factorized does not always look simpler and smaller than it was before the expansion, but in this form it becomes more flexible, in the sense that you don’t have to worry about changing signs and other mathematical nonsense. Well, for you to decide for yourself, the following expressions need to be factorized.
Examples:
Answers:
5. Factoring a quadratic trinomial
For the decomposition of a quadratic trinomial into factors, see further examples of decomposition.
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 the rule:
Example:
Factor the polynomial.
Solution:
Another example:
Factor it out.
Solution:
If the entire term is taken out of the brackets, a unit remains in the brackets instead!
2. Abbreviated multiplication formulas. Examples.
The formulas we most often use are difference of squares, difference of cubes and sum of cubes. Do you remember these formulas? If not, repeat the topic urgently!
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 you can swap terms so that the same factor can be extracted from each pair of adjacent terms. This common factor can be taken out of the bracket and the original polynomial will turn into a product.
Example:
Factor the polynomial.
Solution:
Let's 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. Method for selecting a complete 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 the polynomial.
Solution:Example:
\begin(array)(*(35)(l))
((x)^(2))+6(x)-7=\underbrace(((x)^(2))+2\cdot 3\cdot x+9)_(square\ sum\ ((\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 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. Factoring a quadratic trinomial. Example.
A square trinomial is a polynomial of the form, where - the unknown, - some numbers, and.
The values of the variable that make the quadratic trinomial vanish are called the roots of the trinomial. Therefore, the roots of a trinomial are the roots of a quadratic equation.
Theorem.
Example:
Let's factorize the quadratic trinomial: .
First, let's solve the quadratic equation: Now we can write the factorization of this quadratic trinomial:
Now your opinion...
We have described in detail how and why to factor a polynomial.
We gave a lot of examples of how to do this in practice, pointed out pitfalls, gave solutions...
What do you say?
What do you think of this article? Do you use these techniques? Do you understand their essence?
Write in the comments and... prepare for the exam!
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