What are the ways of decomposition of multipliers. Complex cases of decomposition of polynomials on multipliers
This is one of the most elementary ways to simplify the expression. To apply this method, let us recall the distribution law of multiplication relative to addition (do not be afraid of these words, you definitely know this law, I just could forget his name).
The law says: In order to multiply the amount of two numbers to the third number, you need to multiply each alignment to this number and the results obtained are folded, in other words.
You can also do the opposite operation, this is exactly this reverse operation of us and interests us. As can be seen from the sample, the general factor A, can be taken out of the bracket.
Such an operation can be done both with variables, such as, for example, and with numbers :.
Yes, this is too elementary example, as well as the previous example, with the decomposition of the number, because everyone knows that numbers, and are divided into, and what if you got an expression more complicated:
How to find out what, for example, is divided by the number, does it, with a calculator, can anyone be able, and without it weak? And for this there are signs of divisibility, these signs are really worth knowing, they will help you quickly understand whether the general multiplier should be taken out of the bracket.
Signs of divisibility
They are not so difficult to remember, most likely most of them were familiar with that, and something will be a new useful discovery, more in the table:
Note: The table lacks a sign of divisibility by 4. If the two last figures are divided into 4, then the entire number is divided into 4.
How do you like the sign? I advise her to remember!
Well, let's go back to the expression, maybe it may be enough for a bracket and enough with it? No, mathematicians are customary to simplify, so in full, having all what is taken out!
And so, with Igrek, everything is clear, and what with a numerical part of the expression? Both numbers are odd, so it will not be possible to divide,
You can use a sign of divisibility on, the amount of numbers, and, of which the number is equal, and is divided into, it means it is divided by.
Knowing it, you can safely divide into the column, as a result of division on receiving (signs of divisibility were useful!). Thus, the number we can take out the bracket, as well as y and as a result we have:
To make sure that you laid everything right, you can check the decomposition, multiplying!
Also, the general multiplier can be taken out in power expressions. Here, for example, see a general multiplier?
All members of this expression have Xers - we endure, everyone is divided into - we take again, we look at what happened :.
2. Formulas of abbreviated multiplication
The formulas of the abbreviated multiplication have already been mentioned in theory if you hardly remember what it is, then you should refresh them in memory.
Well, if you consider yourself very clever and too lazy to read such a cloud of information, just read further, look at the formula and immediately try for examples.
The essence of this decomposition is to notice a certain formula in the existing expression existing in front of you, to apply it and get it, so the product of something and something, that's all decomposition. The following are formulas:
And now try, spread the following expressions on multipliers using the above formulas:
But what should happen:
As you managed to notice, these formulas are a very effective way of decomposition of multipliers, it is not always suitable, but can be very useful!
3. Grouping or grouping method
And here is another faithful:
so what will you do with it? It seems to be divided into something and something on and on
But all together do not divide one thing, well there is no general factorHow do not look for, so leave, not laying out for multipliers?
Here it is necessary to show a mixture, and the name of this smelling is a grouping!
It is applied just when there are no common divisors from all members. For grouping it is necessary find groups of terms having common dividers And to rearrange them so that one and the same multiplier can be obtained from each group.
It is not necessary to rearrange in some places, but it gives clarity, for clarity it is possible to take some parts of the expression in the brackets, it is not forbidden to install as much as you like, the main thing is not intimidated.
Isn't it clear all this? I will explain on the example:
In the polynomial - we put a member - after a member - we get
we group the first two members together in a separate bracket and also group the third and fourth members, I will receive a "minus" sign for the bracket, we get:
And now we look separately on each of the two "pile", for which we broke the expression with brackets.
The trick is to break on such bugs from which it will be possible to carry out the maximum multiplier, or, as in this example, try to group members so that after making the facility of multipliers for a bracket, we remained the same expressions.
From both brackets, we carry out general multipliers of members from the first bracket, and from the second, we get:
But this is not a decomposition!
Pdonkey decomposition should remain only multiplicationAs long as we have a polynomial just divided into two parts ...
BUT! This polynomial has a general multiplier. it
for bracket and get a final work
Bingo! As you can see, there is already a piece and out of brackets, neither addition, nor subtraction, decomposition is completed, because We have nothing more for the brackets.
It may seem miracle that after making multipliers for brackets, we left the same expressions in brackets, which again we carried out behind the bracket.
And it is not a miracle at all, the fact is that examples in textbooks and the exam in the EE are specifically made so that most expressions in the tasks for simplification or factorization With the right approach, it is easily simplified and sharply collapsed as an umbrella when you press the button, here and look for the same button in each expression.
Something I distracted, what about us with a simplification? The intricate polynomial took a simpler form :.
Agree, not so cumbersome, how was it?
4. Isolation of a full square.
Sometimes it is necessary to transform the existing polynomial to use the formulas of abbreviated multiplication, representing one of its terms in the form of a sum or the difference of two members.
In which case, you have to do it, learn from the example:
The polynomial in this form cannot be decomposed using the formulas of abbreviated multiplication, so it must be converted. Perhaps, at first it will not be obvious to what a member to smash, but over time you will learn to immediately see the formulas of abbreviated multiplication, even if they are not present not entirely, and will be quite quickly determined, which is not enough to fully in full formula, but for now - learn , student, or rather a schoolboy.
For the full formula of the square of the difference here instead. Imagine the third member as a difference, we get: to the expression in brackets you can apply the formula of the square of the difference (not to be confused with the difference of squares !!!)We:, to this expression, you can apply the formula of the square difference (not to be confused with the square of the difference !!!), I submit, how, we get :.
The expression is not always unfolded on the factors looks easier and less than it was before decomposition, but in this form it becomes more movable, in the sense that you can not steam about the change of signs and other mathematical nonsense. Well, here for an independent decision, the following expressions need to be decomposed on multipliers.
Examples:
Answers:
5. Decomposition of the square three decar on multipliers
On the decomposition of a square three decomposition on the factors see further in the examples of decomposition.
Examples of 5 methods of decomposition of polynomial to multipliers
1. Removing a common factor for brackets. Examples.
Do you remember what a distribution law is? This is a rule:
Example:
Dispatch polynomials to multipliers.
Decision:
Another example:
Spread on multipliers.
Decision:
If the term is fully ended behind the brackets, the unit remains in brackets instead!
2. Formulas of abbreviated multiplication. Examples.
Most often, we use the formulas difference of squares, the difference of cubes and the amount of cubes. Do you remember these formulas? If not, urgently repeat the topic!
Example:
Explore the expression on multipliers.
Decision:
In this expression, it is easy to know the difference of cubes:
Example:
Decision:
3. Grouping method. Examples
Sometimes it can be changed in places in such a way that the same multiplier can be allocated from each pair of neighboring terms. This common factor can be reached by bracket and the initial polynomial will turn into a work.
Example:
Spread the multi-multistropers.
Decision:
Grouting the components as follows:
.
In the first group, I will bring a general multiplier for the bracket, and in the second -:
.
Now the general factory can also be submitted for braces:
.
4. Method of high-square isolation. Examples.
If the polynomial can be represented in the form of the square of the squares of two expressions, it will remain only to apply the formula of the abbreviated multiplication (the difference of squares).
Example:
Spread the multi-multistropers.
Decision:Example:
\\ Begin (Array) (* (35) (L))
((x) ^ (2)) + 6 (x) -7 \u003d \\ underbrace (((x) ^ (2)) + 2 \\ Cdot 3 \\ Cdot x + 9) _ (square \\ 1 \\ ((\\ left (X + 3 \\ RIGHT)) ^ (2))) - 9-7 \u003d ((\\ left (x + 3 \\ right)) ^ (2)) - 16 \u003d \\\\
\u003d \\ left (x + 3 + 4 \\ right) \\ left (x + 3-4 \\ right) \u003d \\ left (x + 7 \\ right) \\ left (x-1 \\ right) \\\\
\\ END (Array)
Spread the multi-multistropers.
Decision:
\\ Begin (Array) (* (35) (L))
((x) ^ (4)) - 4 ((x) ^ (2)) - 1 \u003d \\ underbrace (((x) ^ (4)) - 2 \\ Cdot 2 \\ Cdot ((x) ^ (2) ) +4) _ (square \\ difference ((\\ left (((x) ^ (2)) - 2 \\ right)) ^ (2))) - 4-1 \u003d ((\\ left (((x) ^ (2)) - 2 \\ RIGHT)) ^ (2)) - 5 \u003d \\\\
\u003d \\ left (((x) ^ (2)) - 2+ \\ SQRT (5) \\ RIGHT) \\ left (((x) ^ (2)) - 2- \\ sqrt (5) \\ Right) \\\\
\\ END (Array)
5. Decomposition of the square three decar on multipliers. Example.
Square three-melen is a polynomial view where the unknown is some numbers, and.
The values \u200b\u200bof the variable that turn the square three-shred to zero are called the roots of three-shoes. Consequently, the roots of three-shots are the roots of the square equation.
Theorem.
Example:
Spread on multipliers Square threesties :.
First, we solve a square equation: Now you can record the decomposition of this square three decompositions on factors:
Now your opinion ...
We painted in detail how and to lay the polynomial to multipliers.
We led a lot of examples how to do it in practice, pointed to the pitfalls, gave solutions ...
What do you say?
How do you like this article? Do you use these techniques? Do you understand their essence?
Write in comments and ... Prepare for the exam!
So far, he is the most important in your life.
What to do if in the process of solving the problem from the exam or on the entrance exam in mathematics you received a polynomial, which is not possible to decompose the multipliers with the standard methods that you learned to school? In this article, a mathematics tutor will tell about one effective way, the study of which is beyond the school program, but with the help of which the polynomial is not much difficult to decompose the polynomial. Take this article to the end and look at the applied video tutorial. Knowledge that you will receive will help you on the exam.
Decomposition of a polynomial to division methods
In the event that you received a polynomial more than the second degree and could guess the value of the variable in which this polynomial becomes zero (for example, this value is equal to), know! This polynomial can be divided without a residue.
For example, it is easy to see that the fourth degree polynomial appeals to zero. It means that it can be divided into without a residue, having obtained a polynomial of the third degree (less per unit). That is, imagine:
where A., B., C. and D. - Some numbers. Recall brackets:
Since the coefficients with the same degrees should be the same, we get:
So, got:
Go ahead. It is enough to sort out a few small integers, which see that the polynomial of the third degree is again divided by. This obtains a second degree polynomial (less per unit). Then go to a new entry:
where E., F. and G. - Some numbers. We reveal brackets and come to the following expression:
Again from the condition of equality of coefficients with the same degrees we get:
Then we get:
That is, the initial polynomial can be decomposed on the factors as follows:
In principle, if desired, using the formula, the difference of squares, the result can also be submitted as follows:
This is such a simple and effective way to decompose polynomials on multipliers. Remember it, he can come in handy on the exam or Olympiad in mathematics. Check if you have learned to use this method. Try to solve the following task yourself.
Spread the polynomial to multipliers:
Write your answers in the comments.
Material prepared, Sergey Valerievich
- 1. Treatment of a common factor for brackets and a grouping method. In some cases, it is advisable to replace some members for the amount (difference) of such terms or introduce mutually destroying members.
- 2. The use of formulas of abbreviated multiplication.Sometimes you have to endure multipliers for brackets, group members, allocate full square and only then the amount of cubes, the difference of squares or the difference of cubes to represent in the form of a work.
- 3. Using the theorem of the mow and the method of uncertain coefficients.
Example . Dispatch on multipliers:
P 3 (x) \u003d x 3 + 4x 2 + 5x + 2;
Since p 3 (-1) \u003d 0, then the polynomial P 3 (x) is divided by x + 1. The method of indefinite coefficients will find private from the division of the polynomial
P 3 (x) \u003d x 3 + 4x 2 + 5x + 2 per bounce x + 1.
Let the private eating a polynomial x 2 +. Since x 3 + 4x 2 + 5x + 2 \u003d (x + 1) · (x 2 +) \u003d
X 3 + (+ 1) · x 2 + () · x +, we get the system:
From where. Consequently, P 3 (x) \u003d (x + 1) · (x 2 + 3x + 2).
Since x 2 + 3x + 2 \u003d x 2 + x + 2x + 2 \u003d x · (x + 1) + 2 · (x + 1) \u003d (x + 1) · (x + 2), then p 3 (x ) \u003d (x + 1) 2 · (x + 2).
4. The use of the theorem of the mud and division of the "Stage".
Example . Decompose on multipliers
P 4 (x) \u003d 5 · x 4 + 9 · x 3 -2 · x 2 -4 · x -8.
Decision . Since P 4 (1) \u003d 5 + 9-2-4-8 \u003d 0, then P 4 (x) is divided into (x-1). Division "column" we will find private
Hence,
P 4 (x) \u003d (x-) · (5 · x 3 + 14x 2 + 12x + 8) \u003d
\u003d (x - 1) · p 3 (x).
Since P 3 (-2) \u003d -40 + 56-24 + 8 \u003d 0, then the polynomial P 3 (x) \u003d 5 · x 3 + 14x 2 + 12x + 8 is divided by x + 2.
Find a private division of the "Stage":
Hence,
P 3 (x) \u003d (x + 2) · (5 · x 2 + 4x + 4).
Since the discriminant of square three decreases 5 · x 2 + 4x + 4 is d \u003d -24<0, то этот
square threefold on linear multipliers does not decompose.
So, p 4 (x) \u003d (x - 1) · (x + 2) · (5 · x 2 + 4x + 4)
5. Using the mouture theorem and the Gorner scheme. Private methods obtained by these methods can be dispersed on multipliers to any other or in the same way.
Example . Dispatch on multipliers:
P 3 (x) \u003d 2 · x 3 -5 · x 2 -196 · x + 99;
Decision .
If this polynomial has rational roots, then they can only be among numbers 1/2, 1, 3/2, 3, 9/2, 11/2, 9, 33, 99, 11.
To find the root of this polynomial, we will use the following statement:
If at the ends of some segment the value of the polynomial has different signs, then on the interval (a; b) There is at least one root of this polynomial.
For this polynomial P 3 (0) \u003d 99, P 3 (1) \u003d - 100. Therefore, there is at least one root of this polynomial on the interval (0; 1). Therefore, among those written above 24 numbers, it is advisable first to check those numbers that belong to the interval
(0; 1). Of these numbers, only the number belongs to this interval.
The value of P 3 (x) with x \u003d 1/2 can be found not only by direct substitution, but also in other ways, for example, according to the mountain scheme, since P () is equal to the residue from the division of the polynomial P (x) to X-. Moreover, in many examples, this method is preferable, since both the coefficients are also located at the same time.
According to the mountain scheme for this example, we obtain:
Since p 3 (1/2) \u003d 0, x \u003d 1/2 is the root of the polynomial P 3 (x), and the polynomial P 3 (x) is divided into X-1/2, i.e. 2 · x 3 -5 · x 2 -196 · x + 99 \u003d (x-1/2) · (2 \u200b\u200b· x 2 -4 · x-198).
Since 2 · x 2 -4 · x-198 \u003d 2 · (x 2 -2 · x + 1-100) \u003d 2 · ((x-1) 2 -10 2) \u003d 2 · (x + 9) · ( x-11) then
P 3 (x) \u003d 2 · x 3 -5 · x 2 -196 · x + 99 \u003d 2 · (x-1/2) · (x + 9) · (x-11).
The concept of ring rings
Let be TOand L. Commutative rings
Definition 1. : Ring TO called simple rings expansion K. Using elements x. and write:
L \u003d k [x]if conditions are satisfied:
page rings
Basic set K [x]denote somubs L, K [x].
Definition 2. : Simple expansion L \u003d k [x] rings K. via x. - Simple Transcendent Ring Extension K. via x.if conditions are satisfied:
page rings
If, then
Definition 3. : Element x. called transcendent over the ring K.if condition is satisfied: if, then
Sentence. Let be K [x] Simple transcendental expansion. If and, where, then
Evidence . By condition, the second expression will be subtracted from the first expression, we get: since the element x. Transcendentien Nad K., then from (3) we get:.
Output. Any element of a simple transcendental expansion of unequal zero, commutative ring K. Using an item x. Allows the only representation in the form of a linear combination of integer non-negative degrees of the element x.
Definition: Ring of polynomial from an unknown x. over, unequal zero, ring K. It is called a simple transcendental expansion of a non-zero commutative ring K. Using an item x..
Theorem . For any no zero commutative ring K, There is its simple transcendental extension using an element. x, k [x]
Operations on polynomials
Let K [x] be a ring of polynomials not zero commutative ring K.
Definition 1: The polynomials f and g belonging to k [x] are called equal and write F \u003d G if all the plants of the polynomial F and G are equal to each other, standing at some degrees of the unknown x.
Corollary . In the record of the polynomial, the order of the alignment is not significant. Attributing and excluding from the recording of the polynomial, the component with a zero coefficient will not change the polynomial.
Definition 2. The amount of polynomials F and G is called a polynomial F + G, determined by equality:
Definition 3. : - The product of polynomials is denoted, which is determined by rule:
The degree of polynomials
Let the commutative ring. K [x] Ring of polynomials over the field K. : ,
Definition : Let - any polynomial. If, the whole non-negative number n is the degree of polynomials f.. At the same time they write n \u003d deg f..
Numbers are the coefficients of the polynomial where - the senior coefficient.
If a, f. - normalized. The degree of zero polynomial is uncertain.
Properties of the degree of polynomial
K. - area of \u200b\u200bintegrity
Evidence :
So like. TO - The area of \u200b\u200bintegrity.
Corollary 1. : k [x] over the field TO (Integrity) in turn is an integrity area. For any integrity area there is a scope of particular.
Corollary 2. : For any k [x] over the integrity area TO There is a private field.
Division on bounce and roots of polynomial.
Let the element be called a polynomial value f. From the argument.
Theorem Bezu : For any polynomial and element, there is an item :.
Evidence : Let - any polynomial
Corollary : The residue from the division of the polynomial is equal.
Definition : The element is called the root of the polynomial f., if a.
Theorem : Let the element be the root f. then and only when divisses f.
Evidence:
Need. Let, from the theorem, it follows from the theorem that, from the properties of divisibility it follows that
Sufficiency. Let that. Ch.T.D.
The maximum number of roots of the polynomial over the integrity area.
Theorem : Let K be the area of \u200b\u200bintegrity. Number of roots of polynomial f. In the area of \u200b\u200bintegrity k. No more degree n. polynomial f..
Evidence :
Induction by the degree of polynomial. Let the polynomial f. It has zero roots, and their number does not exceed.
Let the theorem is proved for any.
We show that from paragraph 2, the truth of the approval of the theorem for polynomials is followed.
Let two cases be possible:
- A) polynomial f. It does not have roots, therefore, the statement of the theorem is true.
- B) polynomial f. has at least the root, on the theorem without the mouture, since k. - area of \u200b\u200bintegrity, then by property 3 (degree of polynomial), it follows that
As, k - The area of \u200b\u200bintegrity.
Thus, all the roots of the polynomial is the root of the polynomial g. Since, in induction assumption, the number of all the roots of the polynomial g. not more n., hence, f. has no more ( n +.1) root.
Corollary : Let be k. - area of \u200b\u200bintegrity, if the number of roots of the polynomial f. More numbers n,where, that f. - zero polynomial.
Algebraic and functional equality of polynomials
Let, be some kind of polynomial, it defines some function
in general, any polynomial can define one function.
Theorem : Let be k.- The area of \u200b\u200bintegrity is thus for the equality of polynomials and equality (identical equality ()) defined and.
Evidence :
Need. Let both - the area of \u200b\u200bintegrity ,.
Let, that is
Sufficiency. Let's pretend that. Consider because k. Integrity area, then polynomial h. has the number of roots, from the investigation it follows that h. Zero polynomial. So, bt.t.
Discussion theorem with the residue
Definition : Euclidean Ring K. called such an area of \u200b\u200bintegrity k,that the function determines the function h,the adjacent integer non-negative values \u200b\u200band satisfies the condition
In the process of finding elements for these elements is called a division with the residue, is an incomplete private, - the balance of division.
Let - the ring of polynomials over the field.
Theorem (on division with the residue) : Let - the ring of polynomials over the field and the single pair of polynomials is a polynomial, such that the condition is satisfied or. or
Evidence : The existence of a polynomial. Let, that is. The theorem is valid, obviously, if zero or, since or. We prove the theorem when. The proof by induction of the degree of polynomials, suppose that the theorem is proved (except for uniqueness), for polynomial. We show that in this case the approval of the theorem is made for. Indeed, let - the eldest coefficient of the polynomial, therefore, the polynomial will have the same senior coefficient and the degree of the degree that has a polynomial, therefore the polynomial will have or is a zero polynomial. If, then, therefore, when we get. If, in inductive assumption, therefore, that is, when we get or. The existence of the polynomial is proved.
We show that such a pair of polynomials is the only one.
Let there be either, subtract :. Two cases are possible or.
On the other hand. By condition or, or.
If a. The contradiction is obtained, so. The uniqueness is proved.
Corollary 1. : Ring of polynomials over the field, is the Euclidean space.
Corollary 2. : Ring of polynomials over, is the ring of the main ideals (any ideal has a single generator)
Any Euclidean Ring Factorically: Ring of polynomial over, is called a factorial ring.
Algorithm Euclida. Node of two polynomials
Let the ring of polynomials above.
Definition 1. : Albeit, if there is a polynomial, the residue from the division is zero, the called the polynomial divider and is indicated: ().
Definition 2. : The largest general divider of polynomials is called polynomial:
and (- a common divider and).
(on any common divider and).
The largest general divider of polynomials and is denoted by the Node (;). Common divisters of any polynomials include all polynomials of zero degree from, that is, no zero field. It may turn out that two data of the polynomial and do not have common divisors that are not zero polynomials.
Definition : If polynomials and do not have common divisors of non-numerous zero degree, then they are called mutually simple.
Lemma : If polynomials are from above the field, there is a place, the largest common divisor of polynomials and node are associated. ~
Recording ( a ~ B.) means that (s) by definition.
Evidence : Let I.
and, hence it follows that we will teach that - the general divider of the polynomial and.
general divider and get
Algorithm Euclida
The decomposition of polynomials on multipliers is a identical transformation, as a result of which the polynomial is transformed into a product of several factors - polynomials or single-wing.
There are several ways to decompose polynomials on multipliers.
Method 1. Displacing a common factor for bracket.
This transformation is based on the distribution law of multiplication: AC + BC \u003d C (A + B). The essence of the conversion is to allocate in the two components under consideration the general factor and "out" it for brackets.
We will decompose the polynomials of the polynomial 28x 3 - 35x 4.
Decision.
1. Find elements 28x 3 and 35x 4 common divisor. For 28 and 35 it will be 7; For x 3 and x 4 - x 3. In other words, our total multiplier of 7x 3.
2. Each of the elements represent the work of multipliers, one of which
7x 3: 28x 3 - 35x 4 \u003d 7x 3 ∙ 4 - 7x 3 ∙ 5x.
3. We take out a general multiplier for brackets
7x 3: 28x 3 - 35x 4 \u003d 7x 3 ∙ 4 - 7x 3 ∙ 5x \u003d 7x 3 (4 - 5x).
Methodification 2. The use of formulas of abbreviated multiplication. "Mastery" by the possession of this method is to notice one of the formulas of abbreviated multiplication.
Spread on multipliers of polynomials x 6 - 1.
Decision.
1. To this expression, we can apply the formula for the difference in squares. To do this, imagine x 6 as (x 3) 2, and 1 as 1 2, i.e. 1. The expression will take the form:
(x 3) 2 - 1 \u003d (x 3 + 1) ∙ (x 3 - 1).
2. To the resulting expression, we can apply the formula of the amount and difference of cubes:
(x 3 + 1) ∙ (x 3 - 1) \u003d (x + 1) ∙ (x 2 - x + 1) ∙ (x - 1) ∙ (x 2 + x + 1).
So,
x 6 - 1 \u003d (x 3) 2 - 1 \u003d (x 3 + 1) ∙ (x 3 - 1) \u003d (x + 1) ∙ (x 2 - x + 1) ∙ (x - 1) ∙ (x 2 + x + 1).
Method 3. Grouping. The method of grouping is to combine the components of the polynomial in such a way that they are easy to perform actions (addition, subtraction, total multiplier).
We will decompose the polynomials of x 3 - 3x 2 + 5x - 15 on multipliers.
Decision.
1. Grouting the components in this way: 1st with the 2nd, and 3rd with the 4th
(x 3 - 3x 2) + (5x - 15).
2. In the resulting expression, we will carry out general multipliers for brackets: X 2 in the first case and 5 - in the second.
(x 3 - 3x 2) + (5x - 15) \u003d x 2 (x - 3) + 5 (x - 3).
3. We take out the general factor X - 3 for brackets and get:
x 2 (x - 3) + 5 (x - 3) \u003d (x - 3) (x 2 + 5).
So,
x 3 - 3 x 2 + 5x - 15 \u003d (x 3 - 3x 2) + (5x - 15) \u003d x 2 (x - 3) + 5 (x - 3) \u003d (x - 3) ∙ (x 2 + 5 ).
Fasten the material.
Dispatch polynomial A 2 - 7AB + 12B 2 on multipliers.
Decision.
1. Imagine 7ab 7AB as the sum of 3ab + 4ab. The expression will take the form:
a 2 - (3ab + 4ab) + 12b 2. 
We will reveal the brackets and get:
a 2 - 3AB - 4AB + 12B 2.
2. Grouting the components of the polynomial in this way: the 1st with the 2nd and 3rd with the 4th. We get:
(A 2 - 3AB) - (4ab - 12B 2).
3. I will bring general multipliers for brackets:
(A 2 - 3AB) - (4ab - 12b 2) \u003d A (A - 3B) - 4B (A - 3B).
4. I will bring a general multiplier for brackets (A - 3B):
a (A - 3B) - 4B (A - 3B) \u003d (A - 3 B) ∙ (A - 4B).
So,
a 2 - 7AB + 12B 2 \u003d
\u003d A 2 - (3ab + 4ab) + 12B 2 \u003d
\u003d a 2 - 3ab - 4ab + 12b 2 \u003d
\u003d (A 2 - 3AB) - (4ab - 12B 2) \u003d
\u003d A (A - 3B) - 4B (A - 3B) \u003d
\u003d (A - 3 B) ∙ (A - 4B).
the site, with full or partial copying of the material reference to the original source is required.
Any algebraic polynomial degree n can be represented as a product of n-linear factor of the species and a constant number, which is the coefficients of the polynomial at the senior stage x, i.e.
where 
- are the roots of the polynomial.
The root of the polynomial call the number (real or complex), which turns the polynomial to zero. The roots of the polynomial can be both valid roots and complex-conjugate roots, then the polynomial can be presented in the following form:
Consider the methods of decomposition of polynomials of the degree "N" into the work of multipliers of the first and second degree.
Method number 1.Method of uncertain coefficients.
The coefficients of such a converted expression are determined by the method of uncertain coefficients. The essence of the method is reduced to the fact that there is a pre-known form of multipliers to which this polynomial decomposes. When using the method of uncertain coefficients, the following statements are valid:
P.1. Two polynomials are identically equal in the event that their coefficients are equal with the same degrees x.
P. Any polynomial of the third degree decomposes into the product of linear and square multipliers.
P.3. Any polynomial of the fourth degree decomposes into the work of two polynomials of the second degree.
Example 1.1. It is necessary to decompose a cubic expression on multipliers:
P.1. In accordance with adopted statements for a cubic expression, identical equality is fair:
P. The right part of the expression can be presented in the form of the components as follows:
P.3. We compile a system of equations from the condition of equality of coefficients at the corresponding degrees of the cubic expression.
This system of equations can be solved by selecting coefficients (if there is a simple academic problem) or methods for solving nonlinear systems of equations. Resolving this system of equations, we obtain that uncertain coefficients are determined as follows:
Thus, the initial expression is declined to multipliers in the following form:
This method can be used both with analytical calculations and with computer programming to automate the root search process of the equation.
Method number 2.Vieta formulas
Vieta formulas are formulas that bind the coefficients of algebraic equations of degree n and its roots. These formulas were implicitly presented in the works of French mathematics Francois Vieta (1540 - 1603). Due to the fact that Viet considered only positive real roots, so he had no opportunity to write these formulas in general explicit form.
For any algebraic polynomial degree n, which has N-valid roots,
fair the following relationships that bind the roots of the polynomial with its coefficients:
Vieta's formulas are conveniently used to verify the correctness of the roots of the polynomial, as well as to compile a polynomial on the specified roots.
Example 2.1. Consider how the roots of the polynomial are connected with its coefficients on the example of a cubic equation
In accordance with the formulas of the Vieta, the relationship of the roots of the polynomial with its coefficients is the following form:
Similar relations can be made for any polynomial degree n.
Method number 3. Decomposition of the square equation for factors with rational roots
From the last formula of Vieta it follows that the roots of the polynomial are divisors of its free member and the older coefficient. In this regard, if in the condition of the problem sets a polynomial degree n with whole coefficients
this polynomial has a rational root (inconspicuous fraction), where P is a free member divider, and the Q is a dealer of the older coefficient. In this case, the polynomial of the degree n can be represented in the form (theorem of the moutitude):
The polynomial, the degree of which is 1 less than the degree of initial polynomial, is determined by the division of a polynomial of degree n bounce, for example, using a mountain scheme or the easiest way to be a "column".
Example 3.1. It is necessary to decompose the polynomial to multipliers
P.1. Due to the fact that the coefficient with the senior terms is equal to one, the rational roots of this polynomial are divisors of a free member of the expression, i.e. can be integers 
. We substitute each of the presented numbers into the initial expression, we find that the root of the polynomial represented is equal.
Perform the division of the original polynomial to bounce:
We use the Gorner scheme
The source polynomial coefficients are displayed in the top line, and the first cell of the top line remains empty.
In the first cell of the second line, the root found is recorded (in the example under consideration the number "2") is recorded, and the following values \u200b\u200bin the cells are calculated in a certain way and they are the coefficients of the polynomial, which will result in the division of the polynomial on the bouncer. Unknown coefficients are defined as follows:
In the second cell, the second line is transferred from the corresponding cell of the first line (in the example of the example, the number "1" is recorded).
The third line of the second line records the value of the first cell on the second cell of the second line plus the value from the third cell of the first line (in the example of the example 2 ∙ 1 -5 \u003d -3).
On the fourth cell of the second line, the value of the first cell is written to the third cell of the second line plus the value from the fourth cell of the first line (in the example 2 ∙ (-3) +7 \u003d 1).
Thus, the initial polynomial is declined to multipliers:
Method number 4.Using the formulas of abbreviated multiplication
The formulas of abbreviated multiplication are used to simplify calculations, as well as the decomposition of polynomials on multipliers. Reduced multiplication formulas make it possible to simplify the solution of individual tasks.
Formulas used to decompose multipliers
