What is a simple multiplier. Prime and Composite Numbers

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

One of the simplest ways to factor a number into prime factors is to check whether given number 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 from right side tables. 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 does it mean to factorize? This means finding numbers whose product is equal to the original number.

To understand what it means to factorize, consider an example.

An example of factoring a number

Factor the number 8.

The number 8 can be represented as a product of 2 by 4:

Representing 8 as a product of 2 * 4 and hence the factorization.

Note that this is not the only factorization of 8.

After all, 4 is factored as follows:

From here 8 can be represented:

8 = 2 * 2 * 2 = 2 3

Let's check our answer. Let's find what the factorization is equal to:

That is, we received the original number, the answer is correct.

Factorize the number 24

How to factorize the number 24?

A number is called prime if it is only divisible by 1 and itself.

The number 8 can be represented as a product of 3 by 8:

Here the number 24 is factored. But the task says "to factorize the number 24", i.e. we need prime factors. And in our expansion, 3 is a prime factor, and 8 is not a prime factor.

In this article you will find all the necessary information that answers the question, how to factorize a number. First given general idea on the decomposition of a number into prime factors, examples of expansions are given. Further shown canonical form decomposition 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, allowing you to quickly decompose small integers into prime factors using divisibility signs and a multiplication table.

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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 the product of some numbers takes place, and the clarifying word “prime” 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 the 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 prime numbers 2 , 3 and 5 , it is equal to 30 , so the factorization 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 . Separately, we emphasize 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 decomposition into prime factors, since the number 15 is composite.

The following question arises: “And 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. Given this fact and , it can be argued that the product of several prime factors is an integer positive number exceeding unity. Therefore, factorization takes place only for positive integers that are greater than 1.

But do all integers greater than one factor into prime factors?

It is clear that there is no way to decompose simple integers into prime factors. This is because prime numbers have only two positive divisors, one and itself, so they cannot be represented as a product of two or more prime numbers. If an integer z could be represented as a 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, it is believed that any prime number is itself its decomposition.

What about composite numbers? Do they unfold composite numbers into prime factors, and are all composite numbers subject to such a decomposition? An affirmative answer to a number of these questions is given by the fundamental theorem of arithmetic. The fundamental theorem of arithmetic states that any integer a that is greater than 1 can be decomposed into a product of prime factors p 1 , p 2 , ..., p n , while the expansion has the form a=p 1 p 2 ... p n , and this the decomposition is unique, if we do not take into account the order of the factors

Canonical decomposition 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 the prime factor p 1 occur s 1 times in the decomposition of the number a, 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 writing 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 form is 609 840=2 4 3 2 5 7 11 2.

The canonical decomposition 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 decomposing a number into prime factors

To successfully cope with the task of decomposing a number into prime factors, you need to be very good at the information in the article simple and composite numbers.

The essence of the process of expansion of a positive integer and greater than one number a is clear from the proof of the main theorem of arithmetic. The meaning is to sequentially find the smallest prime divisors p 1 , p 2 , …,p n numbers a, a 1 , a 2 , …, a n-1 , which allows you to get 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 a n =1 is obtained, then the equality a=p 1 ·p 2 ·…·p n will give us the required decomposition of the number a into prime factors. Here it should also be noted that p 1 ≤p 2 ≤p 3 ≤…≤p n.

It remains to deal with finding the smallest prime divisors at each step, and we will have an algorithm for decomposing a number into prime factors. The prime number table will help us find prime divisors. Let's show how to use it to get 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 divisible is its smallest prime divisor. If the number z is prime, then its smallest prime divisor will be the number z itself. It should also be recalled here that if z is not a prime number, then its smallest prime divisor does not exceed the number , where - 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).

For example, let's show how to find the smallest prime divisor of the number 87. We take the number 2. Divide 87 by 2, we get 87:2=43 (rest. 1) (if necessary, see the 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 table of prime numbers, this is the number 3 . We divide 87 by 3, we get 87:3=29. So 87 is evenly divisible by 3, so 3 is the smallest prime divisor of 87.

Note that in the general case, in order to factorize the number a, we need a table of prime numbers up to a number no 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, we will 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 algorithm for factoring a number into prime factors. The algorithm for expanding 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 is itself its decomposition into prime factors. If a 1 is equal to 1, then we have a=p 1 ·a 1 and go to the next step.
We find the smallest prime divisor p 2 of the number a 1 , for this we sequentially sort through the numbers from the table of prime numbers, starting with p 1 , after which we calculate a 2 =a 1:p 2 . If a 2 =1, then the desired decomposition of the number a into prime factors has the form a=p 1 ·p 2 . If a 2 is 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 desired decomposition of the number a into prime factors has the form a=p 1 ·p 2 ·p 3 . If a 3 is 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 the number a n-1 by sorting through the primes, 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 .

All the results obtained at each step of the algorithm for decomposing a number into prime factors are presented for clarity in the form of the following table, in which the numbers a, a 1, a 2, ..., a n are written sequentially to the left of the vertical bar, and to the right of the bar - the corresponding smallest prime divisors p 1 , p 2 , …, p n .

It remains only to consider a few examples of applying the obtained algorithm to decomposing numbers into prime factors.

Prime factorization examples

Now we will analyze in detail prime factorization examples. When decomposing, we will apply the algorithm from the previous paragraph. Let's start with simple cases, and gradually we will complicate them in order to face all the possible nuances that arise when decomposing numbers into prime factors.

Example.

Factor the number 78 into prime factors.

Solution.

We start searching for the first smallest prime divisor p 1 of the number a=78 . To do this, we begin to sequentially sort through the prime numbers from the table of prime numbers. We take the number 2 and divide by it 78, we get 78:2=39. The number 78 was divided by 2 without a remainder, so p 1 \u003d 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 go 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 enumeration of numbers from the table of primes, starting with p 1 =2 . Divide 39 by 2, we get 39:2=19 (remaining 1). Since 39 is not evenly divisible by 2, 2 is not its divisor. Then we take the next number from the table of prime numbers (the number 3) and divide by it 39, we get 39:3=13. Therefore, p 2 \u003d 3 is the smallest prime divisor of the number 39, while a 2 \u003d a 1: p 2 \u003d 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 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 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 (res. 6) and 13:11=1 (res. 2) . The next prime number is 13, and 13 is divisible by it without a remainder, therefore, the smallest prime divisor p 3 of the number 13 is the number 13 itself, and a 3 =a 2:p 3 =13:13=1. Since a 3 =1 , then this step of the algorithm is the last one, and the desired 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 factoring 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 divisor 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 .

Further 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 is 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 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 one, and the desired decomposition has the form 83 006=2·7·7·7·11·11 .

The result obtained can be written as a 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 has no prime divisor that does not exceed ( 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. We give examples for clarification.

For example, we need to decompose the number 10 into prime factors. We know from the multiplication table 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, we decompose the number 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 twice two is four, then we get the desired decomposition into prime factors 48=2 3 2 2 2 . Let's write this decomposition in the canonical form: 48=2 4 ·3 .

But when decomposing the number 3400 into prime factors, you can use the signs of divisibility. The signs of divisibility by 10, 100 allow us to assert that 3400 is divisible by 100, while 3400=34 100, and 100 is divisible by 10, while 100=10 10, therefore, 3400=34 10 10. And on the basis of the sign of divisibility 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 expansion are simple, so this expansion is the required one. It remains only to rearrange the factors so that they go in ascending order: 3 400=2 2 2 5 5 17 . We 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 represent the number 75 as a product of prime factors. The sign of divisibility by 5 allows us to assert that 75 is divisible by 5, while 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 desired decomposition of the number 75 into prime factors.

Bibliography.

Vilenkin N.Ya. etc. 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: Textbook for students of fiz.-mat. specialties of pedagogical institutes.

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 thing, 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 in front of us, 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. It's 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 original 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 original 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 also 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 correct 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.

If you like this site...

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

You can practice solving examples and find out your level. Testing with instant verification. Learning - with interest!)

you can get acquainted with functions and derivatives.