Decomposition of a square polynomial to multipliers. "Decomposition of a square triple to multipliers
At this lesson, we will learn to lay square on linear multipliers with you. To do this, you need to remember the theorem of the Vieta and the opposite. This skill will help us quickly and conveniently lay square trothes on linear multipliers, and simplify the reduction of fractions consisting of expressions.
So let's return to the square equation, where.
What is in our left side is called square triple.
Fair theorem: If - the roots of square triple, then the identity is true
Where is the senior coefficient, the roots of the equation.
So, we have a square equation - a square triple, where the roots of the square equation are also called the roots of square triple. Therefore, if we have square trothes roots, then this triple declines to linear multipliers.
Evidence:
Proof of this fact is performed using the Vieta Theorem, considered by us in previous lessons.
Let's remember what the Vieta Theorem says:
If - the roots of square triple, which, then.
From this theorem, the following statement implies that.
We see that, by the Vieta Theorem,, i.e., substituting these values \u200b\u200bin the formula above, we get the following expression
q.E.D.
Recall that we have proved the theorem that if the roots of the square triple, then the decomposition is fair.
Now let's remember an example of a square equation to which we picked up the roots using the Vieta theorem. From this fact, we can get the following equality due to the proven theorem:
Now let's check the correctness of this fact with simple disclosures of the brackets:
We see that we laid out true for multipliers, and any triple, if it has a root, can be decomposed on this theorem on linear factors by the formula
However, let's check, for any equation, such a discontinuity is possible:
Take, for example, the equation. To begin with, check the discriminant sign
And we remember that to perform the learned theorem D should be greater than 0, therefore, in this case, the expansion of multipliers according to the theorem studied is impossible.
Therefore, we formulate a new theorem: if the square triple does not have roots, it is impossible to decompose on linear multipliers.
So, we looked at the theorem of the Vieta, the possibility of decomposing a square triple to linear multipliers, and now decide several tasks.
Task number 1.
In this group, we will solve the task inverse to the set. We had an equation, and we found his roots, laying out for multipliers. Here we will act on the contrary. Suppose we have a square equation roots
The inverse task is: make a square equation to be rooted.
To solve this problem, there are 2 methods.
Because - the roots of the equation, then 
- This is a square equation, the roots of which are the specified numbers. Now reveal the brackets and check:
It was the first way we created a square equation with a given root, in which there are no other roots, since any square equation has no more than two roots.
This method involves the use of the reverse theorem of the Vieta.
If the roots of the equation, then they satisfy the condition that.
For a given square equation 
, i.e. in this case, and.
Thus, we created a square equation that has a given root.
Task number 2.
It is necessary to reduce the fraction.
We have triple in the numerator and triple in the denominator, and it can be treated as folder, and not laid out for multipliers. If the numerator, and the denominator declines to multipliers, then among them there may be equal multipliers that can be reduced.
First of all, it is necessary to decompose the numerator on multipliers.
Initially, it is necessary to check whether it is possible to decompose this equation for multipliers, we will find a discriminant. Since, the sign depends on the work (there must be less than 0), in this example, i.e. the specified equation has a root.
To solve, use the Vieta theorem:
In this case, since we are dealing with roots, it's just to choose the roots will be quite difficult. But we see that the coefficients are balanced, i.e., if we assume that, and substitute this value to the equation, the following system is obtained:, i.e. 5-5 \u003d 0. Thus, we picked up one of the roots of this square equation.
We will look for the second root method of substation already known to the system of equations, for example,, i.e. .
Thus, we found both the root of the square equation and can substitute their values \u200b\u200binto the original equation to decompose it on the factors:
Recall the original task, we needed to reduce the fraction.
Let's try to solve the task, substituting instead of a numerator.
It is necessary not to forget that at the same time the denominator cannot be equal to 0, i.e.,.
If these conditions are performed, we reduced the initial fraction to the species.
Task number 3 (task with parameter)
Under what values \u200b\u200bof the parameter the amount of the roots of the square equation
If the roots of this equation exist, then 
, Question: When.
Plan - Lesson Abstract (MBOU "Chernomorskaya Secondary School №2"
Phoe teacherPonomarenko Vladislav Vadimovich
Thing
Algebra
Date of the lesson
19.09.2018
№ lesson
Class
9b
Theme lesson
(in accordance with KTP)
"Decomposition of a square triple to multipliers"
Goaling
- Training: to teach students to lay out square triple to the multipliers, to teach the algorithm of decomposition of the square three decompositions when solving examples, consider the tasks of the GIA database, in which the decomposition algorithm of the square trotter on multipliers is used.
- Ride: To develop the skill with schoolchildren to formulate problems, offer ways to solve them, to promote the development of the skills of the skills to allocate the main thing in the cognitive facility.
- Valid: Help students to realize the value of joint activities, to promote development in children of skills to carry out self-control, self-esteem and self-correction of educational activities.
Type of lesson
studying and primary consolidation of new knowledge.
Equipment:
multimedia Projector, Screen, Computer, Didactic Material, Tutorials, Notebooks, Presentation To the lesson
During the classes
1. Organizing time: the teacher welcomes students, checks the readiness for the lesson.Motivates students:
Today in the lesson in joint activities, we confirm the words understand (Slide 1). ("The task you decisive can be very modest, but if she challenges your curiosity, and if you solve it with your own forces, then you can experience leading to Opening the stress of the mind and enjoy the joy of victory. "Doctor understand.)
Message about Understand (Slide 2)
I want to make a call to your curiosity. Consider the task of GIA. Build a function graph .
Can we enjoy the joy of victory and perform this task? (problem situation).
How to solve this problem?
- Note an action plan to solve this problem.
Corrects the lesson plan, comments on the principle of independent work.
Independent work (classify leaflets with text independent work) (Appendix 1)
Independent work
Spread on multipliers:
x. 2 - 3x;
x. 2 – 9;
x. 2 - 8x + 16;
2a. 2 - 2b. 2 -A + B;
2x. 2 - 7x - 4.
Reduce fraction:
SlideWith answers for self-test.
Question class:
What methods of decomposition of a polynomial to multipliers did you use?
Are you all possible to decompose on multipliers?
Did you change all the fractions?
Problem2:Slide
How to decompose polynomials
2 x. 2 – 7 x. – 4?
How to cut a fraction?
Frontal survey:
What is the polynomials
2 x. 2 – 7 x. - 4 I.x. 2 – 5 x. +6?
Give the definition of square triple.
What do we know about square triple?
How to find his roots?
What does the number of roots depend on?
Match this knowledge with what we have to learn and formulate theme lesson. (Then on the screen the topic of the lesson)Slide
We will put the purpose of the lessonSlide
Note the final resultSlide
Question class: How to solve this problem?
The class works in groups.
Setting Groups:
according to the table of contents to find the right page, with a pencil to read item 4, to highlight the main idea, to make an algorithm for which any square triple can be decomposed on multipliers.
Checking the execution of the task class (Front work):
What is the main idea of \u200b\u200bparagraph 4?Slide (On the screen, the decomposition formula of the square trottelene on multipliers).
Algorithm on the screen.Slide
1. Square the square triple to zero.
2. Invoice discriminant.
3.Night roots square triple.
4. Place the found roots in the formula.
5.If needed, then make a senior coefficient in brackets.
One morelittle problem : If D \u003d 0, then you can decompose the square triple to multipliers, and if possible, how?
(Research work in groups).
Slide (on the screen:
If D \u003d 0, then 
.
If the square threestrate does not have roots,
it is impossible to decompose it.)
Let's return to the task in independent work. Will we be able to decompose on the multipliers of square trothes2 x. 2 – 7 x. - 4 I.x. 2 – 5 x. +6?
The class works independently, lay out on multipliers, I work individually with weak students.
Slide (with the decision)Motion
Can you cut a fraction?
Reduce fraction, causing a strong student to the board.
Let's return to the task Of GIA. Will we can now build a function schedule?
What is a graph of this function?
Build a schedule of a function in my notebook.
Test (fromreproach) Appendix 2.
Self-test and self-esteem Students issued leaflets (Appendix 3) in which the answers must be written. They give the evaluation criteria.
Criteria ratings:
3 tasks - score »4»
4Desses - rating "5"
Reflection: (slide)
1. Today I learned about the lesson ...
2. Today I repeated at the lesson ...
3. I fastened ...
4. I liked it ...
5. I put myself an assessment for the activity in the lesson ...
6. What types of work caused difficulties and demand repeat ...
7. Are we fulfilled the score?
Slide: Thanks for the lesson!
Attachment 1
Independent work
Spread on multipliers:
x. 2 - 3x;
x. 2 – 9;
x. 2 - 8x + 16;
x. 2 + x - 2;
2a. 2 - 2b. 2 -A + B;
2 x. 2 – 7 x. – 4.
Reduce fraction:
Appendix 2.
Test
1 option
azda on multipliers?
x. 2 - 8x+ 7;
x. 2 - 8x+ 16 ;
x. 2 - 8x+ 9;
x. 2 - 8x+ 1 7.
2 x. 2 – 9 x. – 5 = 2( x. – 5)(…)?
Answer:_________ .
Reduce the fraction:
x. – 3;
x. + 3;
x. – 4;
another answer.
Test
Option 2
What square three-shred can notazda on multipliers?
5 x. 2 + X.+ 1;
⅓ x. 2 -8x+ 2;
0,1 x. 2 + 3 x. - 5;
x. 2 + 4 x.+ 5.
What a polynomial must be substituted instead of dots to be equality:2 x. 2 + 5 x. – 3 = 2( x. + 3)(…)?
Answer:_________ .
Reduce the fraction:
3 x. 2 – 6 x. – 15;
0,25(3 x. - 1);
0,25( x. - 1);
another answer.
Appendix 3.
Write down the answers.
Criteria ratings:
True fulfilled: 2 task - rating "3"
3 tasks - score »4»
4Desses - rating "5"
Task number 1
Task number 2.
Task number 3.
1 option
Option 2
In order to decompose the factors, it is necessary to simplify expressions. This is necessary in order to continue to reduce. The decomposition of the polynomial makes sense when its degree is not lower than the second. The polynomial with the first degree is called linear.
The article will reveal all the concepts of decomposition, theoretical foundations and methods of expansions of polynomials to multipliers.
Theory
Theorem 1.When any polynomial with a degree n, having a form p n x \u003d a n x n + a n - 1 x n - 1 +. . . + a 1 x + a 0, represent a product with a constant factor with an older degree of AN and N of linear multipliers (x - xi), i \u003d 1, 2, ..., n, then p n (x) \u003d an (x - XN) (X - XN - 1) ·. . . · (X - x 1), where x i, i \u003d 1, 2, ..., n is the roots of the polynomial.
The theorem is intended for the roots of the complex type x i, i \u003d 1, 2, ..., n and for complex coefficients a k, k \u003d 0, 1, 2, ..., n. This is the basis of any decomposition.
When the coefficients of the form A k, k \u003d 0, 1, 2, ..., n are valid numbers, then complex roots that will meet with pairs. For example, the roots x 1 and x 2 belonging to the polynomial of the form p n x \u003d a n x n + a n - 1 x n - 1 +. . . + A 1 X + A 0 is considered comprehensively conjugate, then the other roots are valid, we obtain from here that the polynomial takes the form p n (x) \u003d a n (x - x n) (x - x n - 1) ·. . . · (X - x 3) x 2 + p x + q, where x 2 + p x + q \u003d (x - x 1) (x - x 2).
Comment
The roots of the polynomial may be repeated. Consider the proof of the theorem of algebra, effect from the theorem of the mant.
The main theorem of algebra
Theorem 2.Any polynomial with a degree n has at least one root.
Theorem Bezu
After the division of the polynomial of the form p n x \u003d a n x n + a n was 1 x n - 1 +. . . + a 1 x + a 0 on (x - s), then we get the residue that is equal to the polynomial at the point s, then we get
P n x \u003d a n x n + a n - 1 x n - 1 +. . . + A 1 X + A 0 \u003d (X - S) · Q n - 1 (x) + p n (S), where Q n - 1 (x) is a polynomial with a degree n - 1.
Consequence of the theorem
When the root of the polynomial P n (x) is considered S, then p n x \u003d a n x n + a n - 1 x n - 1 +. . . + a 1 x + a 0 \u003d (x - s) · q n - 1 (x). This investigation is sufficient when used to describe the solution.
Decomposition for square three-shock multipliers
Square threefold of the form A x 2 + b x + C can be decomposed on linear multipliers. Then we get that a x 2 + b x + c \u003d a (x - x 1) (x - x 2), where x 1 and x 2 are roots (complex or valid).
It can be seen that the decomposition itself is reduced to solving the square equation subsequently.
Example 1.
Determination of square three-shots on multipliers.
Decision
It is necessary to find the roots of the equation 4 x 2 - 5 x + 1 \u003d 0. To do this, it is necessary to find the value of the discriminant according to the formula, then we obtain D \u003d (- 5) 2 - 4 · 4 · 1 \u003d 9. From here we have that
x 1 \u003d 5 - 9 2 · 4 \u003d 1 4 x 2 \u003d 5 + 9 2 · 4 \u003d 1
From here we obtain that 4 x 2 - 5 x + 1 \u003d 4 x - 1 4 x - 1.
To perform checks, you need to reveal brackets. Then we get the expression of the form:
4 x - 1 4 x - 1 \u003d 4 x 2 - x - 1 4 x + 1 4 \u003d 4 x 2 - 5 x + 1
After checking, we arrive at the initial expression. That is, it can be concluded that the decomposition is correct.
Example 2.
Expand on the multipliers of the square three-selected species 3 x 2 - 7 x - 11.
Decision
We obtain that it is necessary to calculate the resulting square equation of the form 3 x 2 - 7 x - 11 \u003d 0.
To find the roots, it is necessary to determine the value of the discriminant. We get that
3 x 2 - 7 x - 11 \u003d 0 d \u003d (- 7) 2 - 4 · 3 · (- 11) \u003d 181 x 1 \u003d 7 + d 2 · 3 \u003d 7 + 181 6 x 2 \u003d 7 - d 2 · 3 \u003d 7 - 181 6
From here we obtain that 3 x 2 - 7 x - 11 \u003d 3 x - 7 + 181 6 x - 7 - 181 6.
Example 3.
Determination of a polynomial 2 x 2 + 1 on multipliers.
Decision
Now you need to solve the square equation 2 x 2 + 1 \u003d 0 and find its roots. We get that
2 x 2 + 1 \u003d 0 x 2 \u003d - 1 2 x 1 \u003d - 1 2 \u003d 1 2 · i x 2 \u003d - 1 2 \u003d - 1 2 · I
These roots are called comprehensively conjugate, it means that the decomposition itself can be depicted as 2 x 2 + 1 \u003d 2 x - 1 2 · i x + 1 2 · i.
Example 4.
Determination of the square three decar x 2 + 1 3 x + 1.
Decision
To begin with, it is necessary to solve the square equation of the form x 2 + 1 3 x + 1 \u003d 0 and find its roots.
x 2 + 1 3 x + 1 \u003d 0 d \u003d 1 3 2 - 4 · 1 · 1 \u003d - 35 9 x 1 \u003d - 1 3 + d 2 · 1 \u003d - 1 3 + 35 3 · i 2 \u003d - 1 + 35 · I 6 \u003d - 1 6 + 35 6 · ix 2 \u003d - 1 3 - d 2 · 1 \u003d - 1 3 - 35 3 · i 2 \u003d - 1 - 35 · i 6 \u003d - 1 6 - 35 6 · I
Having received the roots, write
x 2 + 1 3 x + 1 \u003d x - - 1 6 + 35 6 · i x - - 1 6 - 35 6 · i \u003d x + 1 6 - 35 6 · i x + 1 6 + 35 6 · I
Comment
If the value of the discriminant is negative, then the polynomials will remain polynomials of the second order. It follows that we will not put them on linear multipliers.
Methods of decomposition of polynomials of the degree higher than the second
In decomposition, an universal method is assumed. Most of all cases are based on a consequence of the theorem of the mant. To do this, it is necessary to select the value of the root x 1 and reduce its degree by dividing on a polynomial to 1 division by (x - x 1). The resulting polynomial needs to find the root of X 2, and the search process is cyclically until we receive a complete decomposition.
If the root is not found, then other ways of decomposition of multipliers are applied: grouping, additional terms. This topic believes solving equations with higher degrees and entire coefficients.
Multiplier for brackets
Consider the case when the free member is zero, then the type of polynomial becomes like p n (x) \u003d a n x n + a n - 1 x n - 1 +. . . + A 1 x.
It can be seen that the root of such a polynomial will be x 1 \u003d 0, then the polynomial can be submitted as an expression P n (x) \u003d a n x n + a n - 1 x n - 1 +. . . + a 1 x \u003d x (a n x n - 1 + a n - 1 x n - 2 +... + A 1)
This method is considered to withdraw a common factor for brackets.
Example 5.
Perform a decomposition of a polynomial of a third degree 4 x 3 + 8 x 2 - x on multipliers.
Decision
We see that x 1 \u003d 0 is the root of a given polynomial, then it is possible to make x for brackets of the entire expression. We get:
4 x 3 + 8 x 2 - x \u003d x (4 x 2 + 8 x - 1)
Go to finding the roots of the square three-shredded 4 x 2 + 8 x - 1. We find discriminant and roots:
D \u003d 8 2 - 4 · 4 · (- 1) \u003d 80 x 1 \u003d - 8 + d 2 · 4 \u003d - 1 + 5 2 x 2 \u003d - 8 - d 2 · 4 \u003d - 1 - 5 2
Then it follows that
4 x 3 + 8 x 2 - x \u003d x 4 x 2 + 8 x - 1 \u003d 4 xx - - 1 + 5 2 x - - 1 - 5 2 \u003d 4 xx + 1 - 5 2 x + 1 + 5 2.
To begin with, we will take for consideration a decomposition method containing whole coefficients of the form p n (x) \u003d x n + a n - 1 x n - 1 +. . . + A 1 X + A 0, where the coefficient is one of the senior degree equals 1.
When the polynomial has whole roots, then they are considered free member divisors.
Example 6.
Determination of expression f (x) \u003d x 4 + 3 x 3 - x 2 - 9 x - 18.
Decision
Consider whether there are whole roots. It is necessary to write down the dividers of the number - 18. We obtain that ± 1, ± 2, ± 3, ± 6, ± 9, ± 18. It follows that this polynomial has whole roots. You can check the burner scheme. It is very convenient and allows you to quickly get the plaintiff rates of the polynomial:
It follows that x \u003d 2 and x \u003d - 3 are the roots of the source polynomial, which can be represented as a product of the form:
f (x) \u003d x 4 + 3 x 3 - x 2 - 9 x - 18 \u003d (x - 2) (x 3 + 5 x 2 + 9 x + 9) \u003d \u003d (x - 2) (x + 3) (x 2 + 2 x + 3)
We turn to the decomposition of the square three-selected form x 2 + 2 x + 3.
Since discriminant we get a negative, it means that there are no valid roots.
Answer: f (x) \u003d x 4 + 3 x 3 - x 2 - 9 x - 18 \u003d (x - 2) (x + 3) (x 2 + 2 x + 3)
Comment
It is allowed to use the selection of the root and division of the polynomial to the polynomial instead of the Gunner scheme. We turn to the consideration of the decomposition of a polynomial containing the entire coefficients of the form p n (x) \u003d x n + a n - 1 x n - 1 +. . . + a 1 x + a 0, the eldest of which is equal to one.
This case takes place for fractional rational fractions.
Example 7.
Expand the factors f (x) \u003d 2 x 3 + 19 x 2 + 41 x + 15.
Decision
It is necessary to replace the variable Y \u003d 2 x, you should move to the polynomial with coefficients equal to 1 with a high degree. It is necessary to start with multiplication of expression on 4. We get that
4 f (x) \u003d 2 3 · x 3 + 19 · 2 2 · x 2 + 82 · 2 · x + 60 \u003d \u003d y 3 + 19 y 2 + 82 y + 60 \u003d g (y)
When the resulting function of the form G (y) \u003d y 3 + 19 y 2 + 82 y + 60 has whole roots, then their finding among the free member divisors. The record will take the form:
± 1, ± 2, ± 3, ± 4, ± 5, ± 6, ± 10, ± 12, ± 15, ± 20, ± 30, ± 60
Let us turn to the calculation of the function G (Y) in these dot in order to obtain as a result of zero. We get that
g (1) \u003d 1 3 + 19 · 1 2 + 82 · 1 + 60 \u003d 162 g (- 1) \u003d (- 1) 3 + 19 · (- 1) 2 + 82 · (- 1) + 60 \u003d - 4 G (2) \u003d 2 3 + 19 · 2 2 + 82 · 2 + 60 \u003d 308 g (- 2) \u003d (- 2) 3 + 19 · (- 2) 2 + 82 · (- 2) + 60 \u003d - 36 g (3) \u003d 3 3 + 19 · 3 2 + 82 · 3 + 60 \u003d 504 g (- 3) \u003d (- 3) 3 + 19 · (- 3) 2 + 82 · (- 3) + 60 \u003d - 42 g (4) \u003d 4 3 + 19 · 4 2 + 82 · 4 + 60 \u003d 756 g (- 4) \u003d (- 4) 3 + 19 · (- 4) 2 + 82 · (- 4) + 60 \u003d - 28 g (5) \u003d 5 3 + 19 · 5 2 + 82 · 5 + 60 \u003d 1070 g (- 5) \u003d (- 5) 3 + 19 · (- 5) 2 + 82 · (- 5) + 60.
We obtain that y \u003d - 5 is the root of the equation of the form y 3 + 19 y 2 + 82 y + 60, it means that x \u003d y 2 \u003d - 5 2 is the root of the original function.
Example 8.
It is necessary to divide the column 2 x 3 + 19 x 2 + 41 x + 15 to x + 5 2.
Decision
We write and get:
2 x 3 + 19 x 2 + 41 x + 15 \u003d x + 5 2 (2 x 2 + 14 x + 6) \u003d \u003d 2 x + 5 2 (x 2 + 7 x + 3)
Verification of divisors will take a lot of time, so it is more profitable to take a decomposition on the factors of the resulting square three-starned form x 2 + 7 x + 3. Equating to zero and find discriminant.
x 2 + 7 x + 3 \u003d 0 d \u003d 7 2 - 4 · 1 · 3 \u003d 37 x 1 \u003d - 7 + 37 2 x 2 \u003d - 7 - 37 2 ⇒ x 2 + 7 x + 3 \u003d x + 7 2 - 37 2 x + 7 2 + 37 2
Hence it follows that
2 x 3 + 19 x 2 + 41 x + 15 \u003d 2 x + 5 2 x 2 + 7 x + 3 \u003d 2 x + 5 2 x + 7 2 - 37 2 x + 7 2 + 37 2
Artificial techniques for decomposition of polynomials
Rational roots are not inherent in all polynomials. To do this, use special ways to find multipliers. But not all polynomials can be decomposed or present in the form of a work.
Method of grouping
There are cases when it is possible to group the components of the polynomial to find a common factor and put it for brackets.
Example 9.
Determination of a polynomial x 4 + 4 x 3 - x 2 - 8 x - 2 on multipliers.
Decision
Because the coefficients are integers, then the roots presumably also can be integer. To check, take the value 1, 1, 2 and - 2 in order to calculate the value of the polynomial at these points. We get that
1 4 + 4 · 1 3 - 1 2 - 8 · 1 - 2 \u003d - 6 ≠ 0 (- 1) 4 + 4 · (- 1) 3 - (- 1) 2 - 8 · (- 1) - 2 \u003d 2 ≠ 0 2 4 + 4 · 2 3 - 2 2 - 8 · 2 - 2 \u003d 26 ≠ 0 (- 2) 4 + 4 · (- 2) 3 - (- 2) 2 - 8 · (- 2) - 2 \u003d - 6 ≠ 0
From here it can be seen that there are no roots, it is necessary to use another way of decomposition and solutions.
It is necessary to carry out a grouping:
x 4 + 4 x 3 - x 2 - 8 x - 2 \u003d x 4 + 4 x 3 - 2 x 2 + x 2 - 8 x - 2 \u003d \u003d (x 4 - 2 x 2) + (4 x 3 - 8 x) + x 2 - 2 \u003d \u003d x 2 (x 2 - 2) + 4 x (x 2 - 2) + x 2 - 2 \u003d \u003d (x 2 - 2) (x 2 + 4 x + 1)
After grouping the original polynomial, it is necessary to submit it as a product of two square three-sends. To do this, we need to decompose the factors. We get that
x 2 - 2 \u003d 0 x 2 \u003d 2 x 1 \u003d 2 x 2 \u003d - 2 ⇒ x 2 - 2 \u003d x - 2 x + 2 x 2 + 4 x + 1 \u003d 0 d \u003d 4 2 - 4 · 1 · 1 \u003d 12 x 1 \u003d - 4 - d 2 · 1 \u003d - 2 - 3 x 2 \u003d - 4 - d 2 · 1 \u003d - 2 - 3 ⇒ x 2 + 4 x + 1 \u003d x + 2 - 3 x + 2 + 3.
x 4 + 4 x 3 - x 2 - 8 x - 2 \u003d x 2 - 2 x 2 + 4 x + 1 \u003d x - 2 x + 2 x + 2 - 3 x + 2 + 3
Comment
The simplicity of the group does not mean that it is easy to choose a slighter. A certain way of solving does not exist, so it is necessary to use special theorems and rules.
Example 10.
Determination of multipliers of the polynomial x 4 + 3 x 3 - x 2 - 4 x + 2.
Decision
The specified polynomial does not have whole roots. The grouping of the components should be made. We get that
x 4 + 3 x 3 - x 2 - 4 x + 2 \u003d (x 4 + x 3) + (2 x 3 + 2 x 2) + (- 2 x 2 - 2 x) - x 2 - 2 x + 2 \u003d x 2 (x 2 + x) + 2 x (x 2 + x) - 2 (x 2 + x) - (x 2 + 2 x - 2) \u003d \u003d (x 2 + x) (x 2 + 2 x - 2) - (x 2 + 2 x - 2) \u003d (x 2 + x - 1) (x 2 + 2 x - 2)
After decomposition on multiplies, we get that
x 4 + 3 x 3 - x 2 - 4 x + 2 \u003d x 2 + x - 1 x 2 + 2 x - 2 \u003d x + 1 + 3 x + 1 - 3 x + 1 2 + 5 2 x + 1 2 - 5 2
Using the formulas of abbreviated multiplication and Binome Newton to decompose the polynomial to multipliers
The appearance often does not always make it clear how it is necessary to take advantage of decomposition. After transformations were made, you can build a line consisting of a triangle of Pascal, otherwise they are called Newton's Binom.
Example 11.
Decomposition of a polynomial x 4 + 4 x 3 + 6 x 2 + 4 x - 2 on multipliers.
Decision
It is necessary to perform an expression conversion to the form
x 4 + 4 x 3 + 6 x 2 + 4 x - 2 \u003d x 4 + 4 x 3 + 6 x 2 + 4 x + 1 - 3
The sequence of the coefficients of the amount in brackets indicates the expression X + 1 4.
So, we have x 4 + 4 x 3 + 6 x 2 + 4 x - 2 \u003d x 4 + 4 x 3 + 6 x 2 + 4 x + 1 - 3 \u003d x + 1 4 - 3.
After applying the difference in the squares, we get
x 4 + 4 x 3 + 6 x 2 + 4 x - 2 \u003d x 4 + 4 x 3 + 6 x 2 + 4 x + 1 - 3 \u003d x + 1 4 - 3 \u003d x + 1 4 - 3 \u003d x + 1 2 - 3 x + 1 2 + 3
Consider the expression that is in the second bracket. It is clear that there are no horses there, so it is necessary to apply the formula for the difference of squares again. We get the expression of the view
x 4 + 4 x 3 + 6 x 2 + 4 x - 2 \u003d x 4 + 4 x 3 + 6 x 2 + 4 x + 1 - 3 \u003d x + 1 4 - 3 \u003d x + 1 4 - 3 \u003d x + 1 2 - 3 x + 1 2 + 3 \u003d x + 1 - 3 4 x + 1 + 3 4 x 2 + 2 x + 1 + 3
Example 12.
Determination of multipliers x 3 + 6 x 2 + 12 x + 6.
Decision
We will deal with the transformation of the expression. We get that
x 3 + 6 x 2 + 12 x + 6 \u003d x 3 + 3 · 2 · x 2 + 3 · 2 2 · x + 2 3 - 2 \u003d (x + 2) 3 - 2
It is necessary to apply the formula for the reduced multiplication of the difference of cubes. We get:
x 3 + 6 x 2 + 12 x + 6 \u003d \u003d (x + 2) 3 - 2 \u003d x + 2 - 2 3 x + 2 2 + 2 3 x + 2 + 4 3 \u003d x + 2 - 2 3 x 2 + x 2 + 2 3 + 4 + 2 2 3 + 4 3
Method of replacing a variable when decomposing a polynomial to multipliers
When replacing the variable, a decrease in the degree and decomposition of the polynomial to multipliers.
Example 13.
Determination of polynomial multipliers of the form x 6 + 5 x 3 + 6.
Decision
By condition, it can be seen that it is necessary to replace Y \u003d x 3. We get:
x 6 + 5 x 3 + 6 \u003d y \u003d x 3 \u003d y 2 + 5 y + 6
The roots of the obtained square equation are equal to y \u003d - 2 and y \u003d - 3, then
x 6 + 5 x 3 + 6 \u003d y \u003d x 3 \u003d y 2 + 5 y + 6 \u003d y + 2 y + 3 \u003d x 3 + 2 x 3 + 3
It is necessary to apply the formula for abbreviated multiplication of the amount of cubes. We obtain the expression of the form:
x 6 + 5 x 3 + 6 \u003d y \u003d x 3 \u003d y 2 + 5 y + 6 \u003d y + 2 y + 3 \u003d x 3 + 2 x 3 + 3 \u003d x + 2 3 x 2 - 2 3 x + 4 3 x + 3 3 x 2 - 3 3 x + 9 3
That is, they got the desired decomposition.
The cases discussed above will help in consideration and decomposition of polynomials into multipliers in different ways.
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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 him? 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.
The decomposition of square three-stakes on multipliers refers to school tasks, with which everyone faces early or later. How to execute it? What is the formula for decomposition of a square three-melan for multipliers? We will understand step by step using examples.
General formula
The decomposition of square three-stakes on multipliers is carried out by solving the square equation. This is a simple task that can be solved by several methods - finding a discriminant, with the help of the Vieta Theorem, there is a graphical solution of the solution. The first two ways are studied in high school.
The general formula looks like this:lX 2 + KX + N \u003d L (x - x 1) (x - x 2) (1)
The assignment algorithm
In order to decompose square three-stakes on multipliers, you need to know the theorem of the Vita, have a program at hand to solve, be able to find a decision graphically or search for the roots of the second degree equation through the discriminant formula. If the square is triggered and it is necessary to decompose on multipliers, the algorithm of actions is:
1) equate the initial expression to zero to get the equation.
2) lead similar terms (if there is such a need).
3) Find the roots in any known way. The graphic method is better to apply if it is known in advance that the roots are integer and small numbers. It must be remembered that the number of roots is equal to the maximum degree of equation, that is, the square equation of the roots of two.
4) substitute the value h. In the expression (1).
5) Record the decomposition of square three-stakes on multipliers.
Examples
Finally understand how this task is performed, the practice allows. Illustrate the expansion of the multipliers of the square three decreases. Examples:
it is necessary to decompose the expression:
We resort to our algorithm:
1) x 2 -17x + 32 \u003d 0
2) similar components are reduced
3) According to the Vieta formula, find the roots for this example is difficult, therefore it is better to use the expression for the discriminant:
D \u003d 289-128 \u003d 161 \u003d (12,69) 2
4) Substitut the roots found by us in the main formula for decomposition:
(x-2,155) * (x-14,845)
5) Then the answer will be like this:
x 2 -17x + 32 \u003d (x-2,155) (x-14,845)
Check whether the solutions found by the discriminant correspond to the formulas of the Vieta:
14,845 . 2,155=32
For these roots, the Vieta Theorem is used, they were found correctly, and therefore the factors received by us are also correct.
Similarly, decompose 12x 2 + 7x-6.
x 1 \u003d -7 + (337) 1/2
x 2 \u003d -7- (337) 1/2
In the previous case, the solutions were not meant, but in valid numbers, which are easy to have a calculator. Now consider a more complex example in which the roots will be comprehensive: decompose on multipliers x 2 + 4x + 9. According to the wine formula, the roots will not be found, and the discriminant is negative. The roots will be on the complex plane.
D \u003d -20
Based on this, we obtain the roots of the roots --4 + 2i * 5 1/2 and -4-2i * 5 1/2, because (-20) 1/2 \u003d 2i * 5 1/2.
We get the desired decomposition, substituting the roots into the general formula.
Another example: you need to decompose the expression 23x 2 -14x + 7 expression.
We have an equation 23x 2 -14x + 7 =0
D \u003d -448.
So, roots 14 + 21,166i and 14-21,166i. The answer will be like this:
23x 2 -14x + 7 \u003d 23 (x- 14-21,166I )*(x- 14 + 21,166I ).
Let us give an example, to solve which is possible without the help of the discriminant.
Let it be necessary to decompose the square equation x 2 -32x + 255. Obviously, it can be solved and discriminant, but faster in this case pick up the roots.
x 1 \u003d 15
x 2 \u003d 17
So x 2 -32x + 255 \u003d (x-15) (x-17).
