Factorization rule. Factoring Large Numbers
Very often, the numerator and denominator of a fraction are algebraic expressions that must first be decomposed into factors, and then, finding the same among them, divide both the numerator and the denominator into them, that is, reduce the fraction. A whole chapter of a textbook on algebra in the 7th grade is devoted to tasks to factorize a polynomial. Factoring can be done 3 ways, as well as a combination of these methods.
1. Application of abbreviated multiplication formulas
As is known to multiply a polynomial by a polynomial, you need to multiply each term of one polynomial by each term of the other polynomial and add the resulting products. There are at least 7 (seven) common cases of multiplication of polynomials that are included in the concept. For example,
Table 1. Factorization in the 1st way
2. Taking the common factor out of the bracket
This method is based on the application of the distributive law of multiplication. For example,
We divide each term of the original expression by the factor that we take out, and at the same time we get the expression in brackets (that is, the result of dividing what was by what we take out remains in brackets). First of all, you need correctly determine the multiplier, which must be bracketed.
The polynomial in brackets can also be a common factor:
When performing the “factorize” task, one must be especially careful with the signs when taking the common factor out of brackets. To change the sign of each term in a parenthesis (b - a), we take out the common factor -1 , while each term in the bracket is divided by -1: (b - a) = - (a - b) .
In the event that the expression in brackets is squared (or to any even power), then numbers inside brackets can be swapped completely free, since the minuses taken out of brackets will still turn into a plus when multiplied: (b - a) 2 = (a - b) 2, (b - a) 4 = (a - b) 4 and so on…
3. Grouping method
Sometimes not all terms in the expression have a common factor, but only some. Then you can try group terms in brackets so that some factor can be taken out of each. Grouping method is double bracketing of common factors.
4. Using several methods at once
Sometimes you need to apply not one, but several ways to factorize a polynomial into factors at once.
This is a synopsis on the topic. "Factorization". Choose next steps:
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Expanding polynomials to get a product sometimes seems confusing. But it is not so difficult if you understand the process step by step. The article details how to factorize square trinomial.
Many do not understand how to factorize a square trinomial, and why this is done. At first it may seem that this is a useless exercise. But in mathematics, nothing is done just like that. The transformation is necessary to simplify the expression and the convenience of calculation.
A polynomial having the form - ax² + bx + c, is called a square trinomial. The term "a" must be negative or positive. In practice, this expression is called a quadratic equation. Therefore, sometimes they say differently: how to expand a quadratic equation.
Interesting! A square polynomial is called because of its largest degree - a square. And a trinomial - because of the 3 component terms.
Some other kinds of polynomials:
- linear binomial (6x+8);
- cubic quadrilateral (x³+4x²-2x+9).
Factorization of a square trinomial
First, the expression is equal to zero, then you need to find the values of the roots x1 and x2. There may be no roots, there may be one or two roots. The presence of roots is determined by the discriminant. Its formula must be known by heart: D=b²-4ac.
If the result of D is negative, there are no roots. If positive, there are two roots. If the result is zero, the root is one. The roots are also calculated by the formula.
If the calculation of the discriminant results in zero, you can apply any of the formulas. In practice, the formula is simply abbreviated: -b / 2a.
Formulas for different values discriminant are different.
If D is positive:
If D is zero:
Online calculators
The Internet has online calculator. It can be used to factorize. Some resources provide the opportunity to see the solution step by step. Such services help to better understand the topic, but you need to try to understand well.
Useful video: Factoring a square trinomial
Examples
We invite you to view simple examples how to factorize a quadratic equation.
Example 1
Here it is clearly shown that the result will be two x, because D is positive. They need to be substituted into the formula. If the roots are negative, the sign in the formula is reversed.
We know the formula for factoring a square trinomial: a(x-x1)(x-x2). We put the values in brackets: (x+3)(x+2/3). There is no number before the term in the exponent. This means that there is a unit, it is lowered.
Example 2
This example clearly shows how to solve an equation that has one root.
Substitute the resulting value:
Example 3
Given: 5x²+3x+7
First, we calculate the discriminant, as in the previous cases.
D=9-4*5*7=9-140= -131.
The discriminant is negative, which means there are no roots.
After receiving the result, it is worth opening the brackets and checking the result. The original trinomial should appear.
Alternative solution
Some people have never been able to make friends with the discriminant. There is another way to factorize a square trinomial. For convenience, the method is shown in an example.
Given: x²+3x-10
We know that we should end up with 2 parentheses: (_)(_). When the expression looks like this: x² + bx + c, we put x at the beginning of each bracket: (x_) (x_). The remaining two numbers are the product that gives "c", i.e. -10 in this case. To find out what these numbers are, you can only use the selection method. Substituted numbers must match the remaining term.
For example, multiplying the following numbers gives -10:
- -1, 10;
- -10, 1;
- -5, 2;
- -2, 5.
- (x-1)(x+10) = x2+10x-x-10 = x2+9x-10. No.
- (x-10)(x+1) = x2+x-10x-10 = x2-9x-10. No.
- (x-5)(x+2) = x2+2x-5x-10 = x2-3x-10. No.
- (x-2)(x+5) = x2+5x-2x-10 = x2+3x-10. Fits.
So, the transformation of the expression x2+3x-10 looks like this: (x-2)(x+5).
Important! You should be careful not to confuse the signs.
Decomposition of a complex trinomial
If "a" is greater than one, difficulties begin. But everything is not as difficult as it seems.
In order to factorize, one must first see if it is possible to factor something out.
For example, given the expression: 3x²+9x-30. Here the number 3 is taken out of brackets:
3(x²+3x-10). The result is the already known trinomial. The answer looks like this: 3(x-2)(x+5)
How to decompose if the term that is squared is negative? IN this case the number -1 is taken out of the bracket. For example: -x²-10x-8. The expression will then look like this:
The scheme differs little from the previous one. There are only a few new things. Let's say the expression is given: 2x²+7x+3. The answer is also written in 2 brackets, which must be filled in (_) (_). X is written in the 2nd bracket, and what is left in the 1st. It looks like this: (2x_)(x_). Otherwise, the previous scheme is repeated.
The number 3 gives the numbers:
- -1, -3;
- -3, -1;
- 3, 1;
- 1, 3.
We solve equations by substituting the given numbers. The last option fits. So the transformation of the expression 2x²+7x+3 looks like this: (2x+1)(x+3).
Other cases
It is not always possible to transform an expression. In the second method, the solution of the equation is not required. But the possibility of converting terms into a product is checked only through the discriminant.
Worth the practice of deciding quadratic equations so that there are no difficulties when using formulas.
Useful video: factorization of a trinomial
Conclusion
You can use it in any way. But it is better to work both to automatism. Also, those who are going to connect their lives with mathematics need to learn how to solve quadratic equations well and decompose polynomials into factors. All the following mathematical topics are built on this.
What's happened 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 rule of thumb:
If we have a terrible fractional expression, you 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 ways decomposition. 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, in more general view:
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.
Practical use Let's take a look at examples. At first, the variant is simple, even primitive.) But on 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 exciting process It is customary to lay out everything 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, the grouping is difficult to name in an independent way factorizations. It's more of a way to get out complex example.) It is necessary 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! This (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)
All! 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 it is necessary to interchange the terms, to consider different variants grouping until a good one is found. 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 different, 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.
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.
It's great, right?) elegant solution possible if left side equations 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.
When solving equations and inequalities, it often becomes necessary to factor a polynomial whose degree is three or higher. In this article, we will look at the easiest way to do this.
As usual, let's turn to theory for help.
Bezout's theorem states that the remainder of dividing a polynomial by a binomial is .
But it is not the theorem itself that is important for us, but corollary from it:
If the number is the root of a polynomial, then the polynomial is divisible without remainder by the binomial.
We are faced with the task of somehow finding at least one root of the polynomial, then dividing the polynomial by , where is the root of the polynomial. As a result, we get a polynomial whose degree is one less than the degree of the original one. And then, if necessary, you can repeat the process.
This task is divided into two: how to find the root of a polynomial, and how to divide a polynomial into a binomial.
Let's take a closer look at these points.
1. How to find the root of a polynomial.
First, we check whether the numbers 1 and -1 are the roots of the polynomial.
The following facts will help us here:
If the sum of all the coefficients of a polynomial is zero, then the number is the root of the polynomial.
For example, in a polynomial the sum of the coefficients is equal to zero: . It is easy to check what is the root of a polynomial.
If the sum of the coefficients of a polynomial at even degrees is equal to the sum of the coefficients at odd degrees, then the number is a root of the polynomial. The free term is considered a coefficient at an even degree, since , a is an even number.
For example, in a polynomial the sum of coefficients at even degrees is : , and the sum of coefficients at odd degrees is : . It is easy to check what is the root of a polynomial.
If neither 1 nor -1 are roots of the polynomial, then we move on.
For a reduced degree polynomial (that is, a polynomial in which the leading coefficient is the coefficient at - equal to one) the Vieta formula is valid:
Where are the roots of the polynomial.
There are also Vieta formulas concerning the remaining coefficients of the polynomial, but it is this one that interests us.
From this Vieta formula it follows that if the roots of the polynomial are integer, then they are divisors of its free term, which is also an integer.
Based on this, we need to decompose the free term of the polynomial into factors, and sequentially, from smaller to larger, check which of the factors is the root of the polynomial.
Consider, for example, the polynomial
Free member divisors: ; ; ;
The sum of all the coefficients of the polynomial is equal, therefore, the number 1 is not the root of the polynomial.
The sum of the coefficients at even powers:
The sum of coefficients at odd powers:
Therefore, the number -1 is also not a root of the polynomial.
Let's check if the number 2 is the root of the polynomial: therefore, the number 2 is the root of the polynomial. Hence, according to Bezout's theorem, the polynomial is divisible without remainder by the binomial.
2. How to divide a polynomial into a binomial.
A polynomial can be divided into a binomial by a column.
We divide the polynomial into a binomial column:
There is another way to divide a polynomial into a binomial - Horner's scheme.
Watch this video to understand how to divide a polynomial by a binomial by a column, and using Horner's scheme.
I note that if, when dividing by a column, some degree of the unknown is absent in the original polynomial, we write 0 in its place - just as when compiling a table for the Horner scheme.
So, if we need to divide a polynomial into a binomial and as a result of division we get a polynomial, then we can find the coefficients of the polynomial using the Horner scheme:
We can also use Horner's scheme in order to check whether the given number is the root of the polynomial: if the number is the root of the polynomial, then the remainder of dividing the polynomial by is zero, that is, in the last column of the second row of the Horner scheme, we get 0.
Using Horner's scheme, we "kill two birds with one stone": at the same time we check whether the number is the root of a polynomial and divide this polynomial by a binomial.
Example. Solve the equation:
1. We write out the divisors of the free term, and we will look for the roots of the polynomial among the divisors of the free term.
Divisors of 24:
2. Check if the number 1 is the root of the polynomial.
The sum of the coefficients of a polynomial, therefore, the number 1 is the root of the polynomial.
3. Divide the original polynomial into a binomial using Horner's scheme.
A) Write the coefficients of the original polynomial in the first row of the table.
Since the containing member is missing, in the column of the table in which the coefficient of at should be written, we write 0. On the left, we write the found root: the number 1.
B) Fill in the first row of the table.
In the last column, as expected, we got zero, we divided the original polynomial into a binomial without a remainder. The coefficients of the polynomial resulting from the division are shown in blue in the second row of the table:
It is easy to check that the numbers 1 and -1 are not roots of the polynomial
C) Let's continue the table. Let's check if the number 2 is the root of the polynomial:
So the degree of the polynomial, which is obtained as a result of division by one, is less than the degree of the original polynomial, therefore the number of coefficients and the number of columns are less by one.
In the last column, we got -40 - a number, not zero, therefore, the polynomial is divisible by a binomial with a remainder, and the number 2 is not a root of the polynomial.
C) Let's check if the number -2 is the root of the polynomial. Since the previous attempt was unsuccessful, so that there is no confusion with the coefficients, I will erase the line corresponding to this attempt:
Great! In the remainder, we got zero, therefore, the polynomial was divided into a binomial without a remainder, therefore, the number -2 is the root of the polynomial. The coefficients of the polynomial, which is obtained by dividing the polynomial by the binomial, are shown in green in the table.
As a result of division, we got a square trinomial 
, whose roots are easily found by Vieta's theorem:
So, the roots of the original equation:
{
}
Answer: ( 
}
In this article you will find all the necessary information that answers the question, how to factorize a number. First, a general idea of \u200b\u200bthe decomposition of a number into prime factors is given, examples of expansions are given. The canonical form of factoring a number into prime factors is shown next. 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 the 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 clearly illustrates next 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. Considering this fact and , it can be argued that the product of several prime factors is a positive integer greater than one. 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 composite numbers decompose 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 the 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 prime number, then its least 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. We divide 87 by 2, we get 87:2=43 (remaining 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 possible nuances arising from the decomposition of 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 desired 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 required 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.
