slide 3

Gorner Williams George (1786-22 September 1837) was an English mathematician. Born in Bristol. He studied and worked there, then in the schools of Bath. Basic works on algebra. In 1819 published a method for the approximate calculation of the real roots of a polynomial, which is now called the Ruffini-Horner method (this method was known to the Chinese as early as the 13th century). The scheme for dividing a polynomial by a binomial x-a is named after Horner.

slide 4

HORNER SCHEME

A method of dividing a polynomial of the nth degree by a linear binomial - a, based on the fact that the coefficients of the incomplete quotient and the remainder r are related to the coefficients of the divisible polynomial and to a by the formulas:

slide 5

Calculations according to the Horner scheme are placed in a table:

Example 1 Divide The incomplete quotient is x3-x2+3x - 13 and the remainder is 42=f(-3).

slide 6

The main advantage of this method is the compactness of the notation and the ability to quickly divide a polynomial into a binomial. In fact, the Horner scheme is another form of recording the grouping method, although, unlike the latter, it is completely non-descriptive. The answer (factorization) here turns out by itself, and we do not see the very process of obtaining it. We will not deal with a rigorous justification of Horner's scheme, but only show how it works.

Slide 7

Example2.

We prove that the polynomial P(x)=x4-6x3+7x-392 is divisible by x-7, and find the quotient. Decision. Using Horner's scheme, we find Р(7): Hence we obtain Р(7)=0, i.e. the remainder when dividing the polynomial by x-7 is zero and, therefore, the polynomial P (x) is a multiple of (x-7). In this case, the numbers in the second row of the table are the coefficients of the quotient from dividing P (x) by (x-7), therefore P(x)=(x-7)(x3+x2+7x+56).

Slide 8

Factor the polynomial x3 - 5x2 - 2x + 16.

This polynomial has integer coefficients. If an integer is the root of this polynomial, then it is a divisor of 16. Thus, if the given polynomial has integer roots, then these can only be numbers ±1; ±2; ±4; ±8; ±16. By direct verification, we make sure that the number 2 is the root of this polynomial, that is, x3 - 5x2 - 2x + 16 = (x - 2)Q(x), where Q(x) is a polynomial of the second degree

Slide 9

The resulting numbers 1, −3, −8 are the coefficients of the polynomial, which is obtained by dividing the original polynomial by x - 2. Hence, the result of the division: 1 x2 + (-3)x + (-8) = x2 - 3x - 8. The degree of the polynomial obtained as a result of division is always 1 less than the degree of the original one. So: x3 - 5x2 - 2x + 16 = (x - 2)(x2 - 3x - 8).

When solving equations and inequalities, it often becomes necessary to factor a polynomial whose degree is equal to 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 powers is equal to the sum of the coefficients at odd powers, 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 - the coefficient of - is 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 if 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 absent, we write 0 in the column of the table in which the coefficient at should be written. 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 that is not equal to zero, therefore, the polynomial is divisible by a binomial with a remainder, and the number 2 is not a polynomial root.

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:

Fine! 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: ( }

The site "professional tutor in mathematics" continues the series of methodological articles on teaching. I publish descriptions of the methods of my work with the most complex and problematic topics of the school curriculum. This material will be useful to teachers and tutors in mathematics, working with students in grades 8-11, both in the regular program and in the program of mathematical classes.

A math tutor cannot always explain material that is poorly presented in a textbook. Unfortunately, there are more and more such topics, and presentation errors, following the authors of the manuals, are made en masse. This applies not only to novice tutors in mathematics and part-time tutors (tutors - students and university tutors), but also to experienced teachers, tutors - professionals, tutors with experience and qualifications. Far from all mathematics tutors have the talent of a competent corrector of the roughness of school textbooks. Not everyone also understands that these corrections (or additions) are necessary. Only a few are engaged in adapting the material for its qualitative perception by children. Unfortunately, the time has passed when mathematics teachers, together with methodologists and authors of publications, massively discussed each letter of the textbook. In the past, before a textbook was introduced into schools, serious analyzes and studies of learning outcomes were carried out. The time has come for dilettantes who strive to make manuals universal, fitting them to the standards of strong mathematical classes.

The race for increasing the amount of information only leads to a decrease in the quality of its assimilation and, as a result, a decrease in the level of real knowledge in mathematics. But no one pays attention to this. And our children are forced to study already in the 8th grade what we went through at the institute: probability theory, solving equations of high degrees, and something else. The adaptation of the material in books for its full perception by the child leaves much to be desired and the math tutor is forced to somehow deal with this.

Let's talk about the methodology for teaching such a specific topic as "dividing a corner of a polynomial by a polynomial", better known in adult mathematics as "Bezout's theorem and Horner's scheme." Just a couple of years ago, the question was not so acute for a math tutor, because he was not included in the main school curriculum. Now the respected authors of the textbook, edited by Telyakovsky, have made changes to the latest edition of the best, in my opinion, textbook, and, having completely spoiled it, only added unnecessary worries to the tutor. Teachers of schools and classes that do not have the status of mathematics, focusing on the innovations of the authors, began to include additional paragraphs in their lessons more often, and inquisitive children, looking at the beautiful pages of their mathematics textbook, increasingly ask the tutor: “What is this division by a corner? Are we going through this? How to share a corner? There is no hiding from such direct questions. The tutor will have to tell the child something.

But as? Probably, I would not describe the method of working with the topic if it was correctly presented in textbooks. How is everything going on with us? Textbooks need to be printed and sold. And for this they need to be updated regularly. Do university teachers complain that children come to them with empty heads, without knowledge and skills? Are the requirements for mathematical knowledge growing? Fine! Let's remove some of the exercises, and instead insert topics that are studied in other programs. Why is our textbook worse? Let's include some additional chapters. Schoolchildren do not know the rule of division by a corner? This is elementary mathematics. We should make such a paragraph optional, heading it "for those who want to know more." Tutors against? And what do we care about tutors in general? Methodists and school teachers are also against it? We will not complicate the material and consider the simplest part of it.

And this is where it starts. The simplicity of the topic and the quality of its assimilation lies, first of all, in the understanding of its logic, and not in the fact that, according to the prescription of the authors of the textbook, to perform a certain set of operations that are not clearly related to each other. Otherwise, the fog in the head of the student will be provided. If the authors are counting on relatively strong students (but studying according to the regular program), then you should not submit the topic in a team form. What do we see in the textbook? Children, it is necessary to divide according to this rule. Get the polynomial at the corner. Thus, the original polynomial will be factorized. However, it is not clear why the terms under the corner are chosen in this way, why they need to be multiplied by a polynomial over the corner, and then subtracted from the current remainder - it is not clear. And most importantly, it is not clear why the chosen monomials must be added in the end and why the resulting brackets will be the expansion of the original polynomial. Any competent mathematician will put a bold question mark over the explanations that are given in the textbook.

I bring to the attention of tutors and teachers of mathematics my solution to the problem, which practically makes everything that is stated in the textbook obvious to the student. In fact, we will prove Bezout's theorem: if the number a is the root of a polynomial, then this polynomial can be decomposed into factors, one of which is x-a, and the second is obtained from the original one in one of three ways: by extracting a linear factor through transformations, dividing by a corner, or according to Horner's scheme. It is with such a formulation that it will be easier for a math tutor to work.

What is a teaching methodology? First of all, it is a clear order in the sequence of explanations and examples, on the basis of which mathematical conclusions are drawn. This topic is no exception. It is very important for a math tutor to introduce the child to Bezout's theorem before the corner division is performed. It is very important! The best way to understand is with a concrete example. Let's take some polynomial with a chosen root and show the technique of its factorization using the method of identical transformations familiar to the student from the 7th grade. With appropriate accompanying explanations, accents and tips from a math tutor, it is quite possible to convey the material without any general mathematical calculations, arbitrary coefficients and degrees.

Important tips for math tutors- follow the instructions from beginning to end and do not change this sequence.

So, let's say we have a polynomial. If we substitute the number 1 instead of its x, then the value of the polynomial will be zero. Hence x=1 is its root. Let's try to decompose into two terms so that one of them is the product of a linear expression and some monomial, and the second would have a degree one less than . That is, we represent it in the form

We choose the monomial for the red field so that when it is multiplied by the leading term, it completely coincides with the leading term of the original polynomial. If the student is not the weakest, then he will be quite capable of giving the tutor in mathematics the desired expression:. The tutor should immediately be asked to insert it into the red box and show what will happen when they are opened. It is best to sign this virtual temporary polynomial under the arrows (under the photo), highlighting it with some color, for example, blue. This will help you to choose the summand for the red field, called the residual from the selection. I would advise tutors to point out here that this remainder can be found by subtraction. Performing this operation, we get:

A tutor in mathematics should draw the student's attention to the fact that by substituting a unit in this equality, we are guaranteed to get zero on its left side (since 1 is the root of the original polynomial), and on the right, obviously, we will also set the first term to zero. So, without any verification, we can say that the unit is the root of the "green residue".

Let's deal with it in the same way as we did with the original polynomial, extracting from it the same linear factor . The math tutor draws two boxes in front of the student and asks them to fill in from left to right.

The student selects for the tutor the monomial for the red field so that when multiplied by the highest term of the linear expression, it gives the highest term of the expanded polynomial. We enter it in the frame, immediately open the bracket and highlight in blue the expression that needs to be subtracted from the expanded one. Performing this operation, we get

And finally, doing the same with the last remainder

finally get

Now we take the expression out of the bracket and we will face the decomposition of the original polynomial into factors, one of which is “x minus the chosen root”.

In order for the student not to think that the last “green residue” was randomly decomposed into the necessary factors, the math tutor should point out an important property of all green residues - each of them has a root 1. Since the degrees of these residues decrease, then no matter what degree of the initial no polynomial was given to us, sooner or later, we will get a linear "green residue" with a root of 1, and therefore it must be decomposed into the product of a certain number and an expression.

After such preparatory work, it will not be difficult for a math tutor to explain to the student what happens when dividing a corner. This is the same process, only in a shorter and more compact form, without equal signs and without rewriting the same selected terms. We write the polynomial from which the linear multiplier is allocated to the left of the corner, collect the selected red monomials at an angle (now it becomes clear why they should add up), to get the “blue polynomials”, you need to multiply the “red” by x-1, and then subtract from the current selected how it is done in the usual division of numbers in a column (here it is an analogy with the previously studied one). The resulting "green residues" are subjected to a new selection and selection of "red monomials". And so on until a zero "green residue" is obtained. The most important thing is that the further fate of the written polynomials above and below the corner becomes clear to the student. Obviously, these are brackets, the product of which is equal to the original polynomial.

The next stage in the work of a tutor in mathematics is the formulation of Bezout's theorem. In fact, its formulation with this approach of the tutor becomes obvious: if the number a is the root of the polynomial, then it can be decomposed into factors, one of which, and the other is obtained from the original one in one of three ways:

• direct decomposition (analogous to the grouping method)
• dividing by a corner (in a column)
• via Horner's scheme

I must say that far from all math tutors show students the horner's scheme, and not all school teachers (fortunately for the tutors themselves) go so deep into the topic in the lessons. However, for a math class student, I see no reason to stop at long division. Moreover, the most convenient and fast The decomposition technique is based precisely on Horner's scheme. In order to explain to the child where it comes from, it is enough to trace the appearance of higher coefficients in green residues using the example of dividing by a corner. It becomes clear that the highest coefficient of the initial polynomial is demolished into the coefficient of the first "red monomial", and further from the second coefficient of the current upper polynomial subtracted the result of multiplying the current "red monomial" coefficient by . Therefore, you can add the result of multiplication by . After focusing the student's attention on the specifics of actions with coefficients, a math tutor can show how these actions are usually performed without writing down the variables themselves. To do this, it is convenient to enter the root and coefficients of the original polynomial in order of precedence into the following table:

If any degree is missing in the polynomial, then its zero coefficient is forcibly entered into the table. The coefficients of the "red polynomials" are alternately entered into the bottom line according to the "hook" rule:

The root is multiplied by the last demolished "red coefficient", added to the next coefficient of the top row and the result is demolished to the bottom line. In the last column, we are guaranteed to get the highest coefficient of the last “green balance”, that is, zero. After the process is completed, the numbers sandwiched between a matched root and zero remainder turn out to be the coefficients of the second (nonlinear) factor.

Since the root a gives zero at the end of the bottom row, then Horner's scheme can be used to check numbers for the rank of the root of a polynomial. If a special theorem on the selection of a rational root. All candidates for this title obtained with its help are simply inserted in turn from the left into Horner's scheme. As soon as we get zero, the tested number will be the root, and at the same time we will get the coefficients of the expansion of the original polynomial into factors. Very comfortably.

In conclusion, I would like to note that for the accurate introduction of the Horner scheme, as well as for the practical consolidation of the topic, a math tutor must have a sufficient number of hours at his disposal. A tutor working with the “once a week” mode should not be engaged in dividing a corner. On the Unified State Exam in mathematics and on the GIA in mathematics, it is unlikely that in the first part there will ever be an equation of the third degree, solved by such means. If a tutor prepares a child for an exam in mathematics at Moscow State University, the study of the topic becomes mandatory. University teachers are very fond of, unlike the compilers of the Unified State Examination, to check the depth of knowledge of the applicant.

Kolpakov Alexander Nikolaevich, mathematics tutor Moscow, Strogino

Lesson Objectives:

• teach students to solve equations of higher degrees using Horner's scheme;
• develop the ability to work in pairs;
• to create, together with the main sections of the course, a basis for developing the abilities of students;
• help the student assess his potential, develop interest in mathematics, the ability to think, speak on the topic.

Equipment: cards for work in groups, a poster with Horner's scheme.

Teaching method: lecture, story, explanation, performance of training exercises.

Form of control: verification of problems of independent solution, independent work.

During the classes

1. Organizational moment

2. Actualization of students' knowledge

What theorem allows you to determine whether the number is the root of a given equation (to formulate a theorem)?

Bezout's theorem. The remainder of dividing the polynomial P(x) by the binomial x-c is equal to P(c), the number c is called the root of the polynomial P(x) if P(c)=0. The theorem allows, without performing the division operation, to determine whether a given number is a root of a polynomial.

Which statements make it easier to find roots?

a) If the leading coefficient of the polynomial is equal to one, then the roots of the polynomial should be sought among the divisors of the free term.

b) If the sum of the coefficients of a polynomial is 0, then one of the roots is 1.

c) If the sum of the coefficients in even places is equal to the sum of the coefficients in odd places, then one of the roots is equal to -1.

d) If all coefficients are positive, then the roots of the polynomial are negative numbers.

e) A polynomial of odd degree has at least one real root.

3. Learning new material

When solving entire algebraic equations, one has to find the values ​​of the roots of polynomials. This operation can be greatly simplified if calculations are carried out according to a special algorithm called Horner's scheme. This scheme is named after the English scientist William George Horner. Horner's scheme is an algorithm for calculating the quotient and remainder of dividing a polynomial P(x) by x-c. Briefly, how it works.

Let an arbitrary polynomial P(x)=a 0 x n + a 1 x n-1 + ...+ a n-1 x+ a n be given. The division of this polynomial by x-c is its representation in the form P(x)=(x-c)g(x) + r(x). Private g (x) \u003d at 0 x n-1 + at n x n-2 + ... + at n-2 x + at n-1, where at 0 \u003d a 0, at n \u003d sv n-1 + a n , n=1,2,3,…n-1. Remainder r (x) \u003d St n-1 + a n. This calculation method is called the Horner scheme. The word "scheme" in the name of the algorithm is due to the fact that usually its execution is formalized as follows. First draw table 2(n+2). The number c is written in the lower left cell, and the coefficients of the polynomial P (x) are written in the upper line. In this case, the upper left cell is left empty.

	at 0 = a 0	in 1 \u003d sv 1 + a 1	in 2 \u003d sv 1 + a 2		in n-1 \u003d sv n-2 +a n-1	r(x)=f(c)=sv n-1 +a n

The number, which after the execution of the algorithm turns out to be written in the lower right cell, is the remainder of dividing the polynomial P(x) by x-c. The other numbers at 0 , at 1 , at 2 ,… of the bottom row are the coefficients of the quotient.

For example: Divide the polynomial P (x) \u003d x 3 -2x + 3 by x-2.

We get that x 3 -2x + 3 \u003d (x-2) (x 2 + 2x + 2) + 7.

4. Consolidation of the studied material

Example 1: Factorize the polynomial P(x)=2x4-7x 3 -3x 2 +5x-1 with integer coefficients.

We are looking for integer roots among the divisors of the free term -1: 1; -one. Let's make a table:

X \u003d -1 - root

P (x) \u003d (x + 1) (2x 3 -9x 2 + 6x -1)

Let's check 1/2.

					X=1/2 - root

Therefore, the polynomial P(x) can be represented as

P (x) \u003d (x + 1) (x-1/2) (x 2 -8x +2) \u003d (x + 1) (2x -1) (x 2 - 4x +1)

Example 2: Solve the equation 2x 4 - 5x 3 + 5x 2 - 2 = 0

Since the sum of the coefficients of the polynomial written on the left side of the equation is zero, then one of the roots is 1. Let's use Horner's scheme:

						X=1 - root

We get P (x) \u003d (x-1) (2x 3 -3x 2 \u003d 2x +2). We will look for roots among the divisors of the free term 2.

We found out that there are no more whole roots. Let's check 1/2; -1/2.

					X \u003d -1/2 - root

Answer: 1; -1/2.

Example 3: Solve the equation 5x 4 - 3x 3 - 4x 2 -3x + 5 = 0.

We will look for the roots of this equation among the divisors of the free term 5: 1; -1; 5; -5. x=1 is the root of the equation, since the sum of the coefficients is zero. Let's use Horner's scheme:

we represent the equation as a product of three factors: (x-1) (x-1) (5x 2 -7x + 5) \u003d 0. Solving the quadratic equation 5x 2 -7x+5=0, we got D=49-100=-51, there are no roots.

Card 1

1. Factor the polynomial: x 4 +3x 3 -5x 2 -6x-8
2. Solve the equation: 27x 3 -15x 2 +5x-1=0

Card 2

1. Factor the polynomial: x 4 -x 3 -7x 2 + 13x-6
2. Solve the equation: x 4 +2x 3 -13x 2 -38x-24=0

Card 3

1. Factorize: 2x 3 -21x 2 + 37x + 24
2. Solve the equation: x 3 -2x 2 +4x-8=0

Card 4

1. Factorize: 5x 3 -46x 2 + 79x-14
2. Solve the equation: x 4 +5x 3 +5x 2 -5x-6=0

5. Summing up

Testing knowledge when solving in pairs is carried out in the lesson by recognizing the method of action and the name of the answer.

Homework:

Solve the equations:

a) x 4 -3x 3 +4x 2 -3x + 1 \u003d 0

b) 5x 4 -36x 3 +62x 2 -36x+5=0

c) x 4 + x 3 + x + 1 \u003d 4 x 2

d) x 4 + 2x 3 -x-2 \u003d 0

Literature

1. N.Ya. Vilenkin et al., Algebra and the Beginnings of Analysis Grade 10 (in-depth study of mathematics): Enlightenment, 2005.
2. U.I. Sakharchuk, L.S. Sagatelova, Solution of equations of higher degrees: Volgograd, 2007.
3. S.B. GashkovNumber systems and their application.

Etc. is of a general nature and great importance to study the WHOLE course of higher mathematics. Today we will repeat the "school" equations, but not just the "school" ones - but those of them that are found everywhere in various tasks of the vyshmat. As usual, the story will go in an applied way, i.e. I will not focus on definitions, classifications, but will share with you my personal experience of solving. The information is intended primarily for beginners, but more prepared readers will also find many interesting points for themselves. And, of course, there will be new material that goes beyond high school.

So the equation... Many people remember this word with a shudder. What are the "fancy" equations with roots... ...forget about them! Because further you will meet the most harmless "representatives" of this species. Or boring trigonometric equations with dozens of methods for solving. To be honest, I didn't really like them either... No panic! - then you are expected mainly by "dandelions" with an obvious solution in 1-2 steps. Although the "burdock", of course, clings - here you need to be objective.

Oddly enough, in higher mathematics it is much more common to deal with very primitive equations like linear equations.

What does it mean to solve this equation? This means - to find SUCH value of "x" (root), which turns it into a true equality. Let's flip the "troika" to the right with a change of sign:

and drop the "two" to the right side (or, the same thing - multiply both parts by) :

To check, we substitute the won trophy into the original equation:

The correct equality is obtained, which means that the found value is indeed the root of this equation. Or, as they say, satisfies this equation.

Note that the root can also be written as a decimal fraction:
And try not to stick to this nasty style! I repeated the reason many times, in particular, at the very first lesson on higher algebra.

By the way, the equation can also be solved "in Arabic":

And what is most interesting - this record is completely legal! But if you are not a teacher, then it is better not to do this, because originality is punishable here =)

And now a little about

graphical solution method

The equation has the form and its root is "x" coordinate intersection points linear function graph with linear function graph (abscissa axis):

It would seem that the example is so elementary that there is nothing more to analyze here, but one more unexpected nuance can be “squeezed” out of it: we represent the same equation in the form and plot the function graphs:

Wherein, please don't confuse the two: an equation is an equation, and function is a function! Functions only help find the roots of the equation. Of which there may be two, three, four, and even infinitely many. The closest example in this sense is everyone knows quadratic equation, whose solution algorithm was awarded a separate item "hot" school formulas. And this is no accident! If you can solve a quadratic equation and know the Pythagorean theorem, then, one might say, “the floor of higher mathematics is already in your pocket” =) Exaggerated, of course, but not so far from the truth!

And therefore, we are not too lazy and solve some quadratic equation according to standard algorithm:

, so the equation has two different valid root:

It is easy to verify that both found values ​​really satisfy this equation:

What to do if you suddenly forgot the solution algorithm, and there are no tools / helping hands at hand? Such a situation may arise, for example, in a test or exam. We use the graphic method! And there are two ways: you can pointwise build parabola , thereby finding out where it intersects the axis (if it crosses at all). But it is better to act more cunningly: we present the equation in the form, draw graphs of simpler functions - and "x" coordinates their points of intersection, at a glance!


If it turns out that the line touches the parabola, then the equation has two coinciding (multiple) roots. If it turns out that the line does not intersect the parabola, then there are no real roots.

To do this, of course, you need to be able to build graphs of elementary functions, but on the other hand, these skills are within the power of even a schoolboy.

And again - an equation is an equation, and functions , are functions that only helped solve the equation!

And here, by the way, it would be appropriate to remember one more thing: if all the coefficients of the equation are multiplied by a non-zero number, then its roots will not change.

So, for example, the equation has the same roots. As the simplest “proof”, I will take the constant out of brackets:
and painlessly remove it (I will divide both parts into “minus two”):

BUT! If we consider the function , then here it is already impossible to get rid of the constant! It is only possible to take the multiplier out of brackets: .

Many underestimate the graphical solution method, considering it to be something "undignified", and some even completely forget about this possibility. And this is fundamentally wrong, because plotting sometimes just saves the day!

Another example: suppose you do not remember the roots of the simplest trigonometric equation:. The general formula is in school textbooks, in all reference books on elementary mathematics, but they are not available to you. However, solving the equation is critical (otherwise "two"). There is an exit! - we build graphs of functions:


after which we calmly write down the "x" coordinates of their intersection points:

There are infinitely many roots, and their folded notation is accepted in algebra:
, where ( – set of integers) .

And, without "departing from the cash desk", a few words about the graphical method for solving inequalities with one variable. The principle is the same. So, for example, any "x" is the solution to the inequality, because the sinusoid lies almost entirely under the straight line. The solution to the inequality is the set of intervals on which the pieces of the sinusoid lie strictly above the straight line (abscissa):

or, in short:

And here is the set of solutions to the inequality - empty, since no point of the sinusoid lies above the straight line.

Anything not clear? Urgently study the lessons about sets and function graphs!

Warm up:

Exercise 1

Solve graphically the following trigonometric equations:

Answers at the end of the lesson

As you can see, to study the exact sciences, it is not at all necessary to cram formulas and reference books! Moreover, this is a fundamentally vicious approach.

As I already reassured you at the very beginning of the lesson, complex trigonometric equations in the standard course of higher mathematics have to be solved extremely rarely. All complexity, as a rule, ends with equations like , the solution of which is two groups of roots, derived from the simplest equations and . Don’t worry too much about the solution of the latter - look in a book or find it on the Internet =)

The graphical method of solving can also help out in less trivial cases. Consider, for example, the following "motley" equation:

The prospects for its solution look ... they don’t look at all, but one has only to present the equation in the form , construct function graphs and everything will be incredibly simple. The drawing is in the middle of the article about infinitesimal functions (opens in next tab).

Using the same graphical method, you can find out that the equation already has two roots, and one of them is equal to zero, and the other, apparently, irrational and belongs to the segment . This root can be calculated approximately, for example, tangent method. By the way, in some tasks, it happens that it is required not to find the roots, but to find out do they exist at all. And here, too, a drawing can help - if the graphs do not intersect, then there are no roots.

Rational roots of polynomials with integer coefficients.
Horner's scheme

And now I suggest you turn your eyes to the Middle Ages and feel the unique atmosphere of classical algebra. For a better understanding of the material, I recommend at least a little familiarization with complex numbers.

They are the most. Polynomials.

The object of our interest will be the most common polynomials of the form with whole coefficients . The natural number is called polynomial degree, number - coefficient at the highest degree (or just the highest coefficient), and the coefficient is free member.

I will denote this polynomial folded by .

Polynomial roots called the roots of the equation

I love iron logic =)

For examples, we go to the very beginning of the article:

There are no problems with finding the roots of polynomials of the 1st and 2nd degrees, but as you increase this task becomes more and more difficult. But on the other hand, everything is more interesting! And this is what the second part of the lesson will be devoted to.

First, literally half a screen of theory:

1) According to the corollary fundamental theorem of algebra, the degree polynomial has exactly integrated roots. Some roots (or even all) can be in particular valid. Moreover, among the real roots there can be identical (multiple) roots (minimum two, maximum pieces).

If some complex number is a root of a polynomial, then conjugate its number is also necessarily the root of this polynomial (conjugate complex roots have the form ).

The simplest example is the quadratic equation, which was first encountered in 8 (like) class, and which we finally “finished off” in the topic complex numbers. I remind you: a quadratic equation has either two different real roots, or multiple roots, or conjugate complex roots.

2) From Bezout's theorems it follows that if the number is the root of the equation, then the corresponding polynomial can be factorized:
, where is a polynomial of degree .

And again, our old example: since is the root of the equation , then . After that, it is easy to get the well-known "school" decomposition .

The consequence of Bezout's theorem is of great practical value: if we know the root of the 3rd degree equation, then we can represent it in the form and from the quadratic equation it is easy to find out the remaining roots. If we know the root of the 4th degree equation, then it is possible to expand the left side into a product, etc.

And there are two questions here:

Question one. How to find this root? First of all, let's define its nature: in many problems of higher mathematics it is required to find rational, in particular whole the roots of polynomials, and in this regard, further we will be interested mainly in them .... …they are so good, so fluffy, that you just want to find them! =)

The first thing that suggests itself is the selection method. Consider, for example, the equation . The catch here is in the free term - if it were equal to zero, then everything would be in openwork - we put the "x" out of brackets and the roots themselves "fall out" to the surface:

But our free term is equal to the “three”, and therefore we begin to substitute various numbers into the equation that claim to be called the “root”. First of all, the substitution of single values ​​suggests itself. Substitute :

Received wrong equality, thus, the unit "did not fit." Okay, let's put it in:

Received correct equality! That is, the value is the root of this equation.

To find the roots of a polynomial of the 3rd degree, there is an analytical method (the so-called Cardano formulas), but now we are interested in a slightly different problem.

Since - is the root of our polynomial, then the polynomial can be represented in the form and arises Second question: how to find the "younger brother"?

The simplest algebraic considerations suggest that for this you need to divide by. How to divide a polynomial by a polynomial? The same school method that divides ordinary numbers - a "column"! I discussed this method in detail in the first examples of the lesson. Complex Limits, and now we will consider another method, which is called Horner's scheme.

First, we write the "senior" polynomial with everyone , including zero coefficients:
, after which we enter these coefficients (strictly in order) in the top row of the table:

On the left we write the root:

I’ll immediately make a reservation that Horner’s scheme also works if the “red” number not is the root of the polynomial. However, let's not rush things.

We take down the senior coefficient from above:

The process of filling in the lower cells is somewhat reminiscent of embroidery, where “minus one” is a kind of “needle” that permeates the subsequent steps. We multiply the "demolished" number by (-1) and add the number from the top cell to the product:

We multiply the found value by the “red needle” and add the following equation coefficient to the product:

And, finally, the resulting value is again “processed” with a “needle” and an upper coefficient:

Zero in the last cell tells us that the polynomial has divided into without a trace (as it should be), while the expansion coefficients are "removed" directly from the bottom row of the table:

Thus, we moved from the equation to an equivalent equation, and everything is clear with the two remaining roots (in this case conjugate complex roots are obtained).

Equation, by the way, can also be solved graphically: build "zipper" and see that the graph crosses the x-axis () at point . Or the same "cunning" trick - we rewrite the equation in the form , draw elementary graphs and detect the "x" coordinate of their intersection point.

By the way, the graph of any polynomial function of the 3rd degree crosses the axis at least once, which means that the corresponding equation has at least one valid root. This fact is true for any polynomial function of odd degree.

And here I also want to stop at important point regarding terminology: polynomial and polynomial function – it's not the same! But in practice, they often talk, for example, about the “polynomial graph”, which, of course, is negligent.

But let's get back to Horner's scheme. As I recently mentioned, this scheme works for other numbers as well, but if the number not is the root of the equation, then a non-zero additive (remainder) appears in our formula:

Let's "drive" the "unsuccessful" value according to Horner's scheme. At the same time, it is convenient to use the same table - we write down a new “needle” on the left, we demolish the highest coefficient from above (left green arrow), and away we go:

To check, we open the brackets and give like terms:
, OK.

It is easy to see that the remainder (“six”) is exactly the value of the polynomial at . And in fact - what is it:
, and even nicer - like this:

From the above calculations, it is easy to understand that Horner's scheme allows not only to factorize the polynomial, but also to carry out a "civilized" selection of the root. I suggest that you independently fix the calculation algorithm with a small task:

Task 2

Using Horner's scheme, find the whole root of the equation and factorize the corresponding polynomial

In other words, here you need to sequentially check the numbers 1, -1, 2, -2, ... - until a zero remainder is “drawn” in the last column. This will mean that the "needle" of this line is the root of the polynomial

Calculations are conveniently arranged in a single table. Detailed solution and answer at the end of the lesson.

The method of selecting roots is good for relatively simple cases, but if the coefficients and / or the degree of the polynomial are large, then the process can be delayed. Or maybe some values ​​​​from the same list 1, -1, 2, -2 and it makes no sense to consider? And, besides, the roots may turn out to be fractional, which will lead to a completely non-scientific poke.

Fortunately, there are two powerful theorems that can significantly reduce the enumeration of “candidate” values ​​for rational roots:

Theorem 1 Consider irreducible fraction , where . If the number is the root of the equation, then the free term is divisible by and the leading coefficient is divisible by.

In particular, if the leading coefficient is , then this rational root is integer:

And we begin to exploit the theorem just from this tasty particular:

Let's go back to the equation. Since its leading coefficient is , the hypothetical rational roots can be exclusively integer, and the free term must be divisible by these roots without a remainder. And the "three" can only be divided into 1, -1, 3 and -3. That is, we have only 4 "candidates for the roots." And, according to Theorem 1, other rational numbers cannot be roots of this equation IN PRINCIPLE.

There are a little more “applicants” in the equation: the free term is divided into 1, -1, 2, -2, 4 and -4.

Please note that the numbers 1, -1 are "regulars" of the list of possible roots (an obvious consequence of the theorem) and the best choice for the first inspection.

Let's move on to more meaningful examples:

Task 3

Decision: since the leading coefficient , then the hypothetical rational roots can only be integers, while they must be divisors of the free term. "Minus forty" is divided into the following pairs of numbers:
- a total of 16 "candidates".

And here a tempting thought immediately appears: is it possible to weed out all negative or all positive roots? In some cases you can! I will formulate two signs:

1) If all If the coefficients of a polynomial are non-negative, then it cannot have positive roots. Unfortunately, this is not our case (Now, if we were given an equation - then yes, when substituting any value of the polynomial is strictly positive, which means that all positive numbers (and irrational too) cannot be roots of the equation.

2) If the coefficients for odd powers are non-negative, and for all even powers (including free member) are negative, then the polynomial cannot have negative roots. This is our case! Looking closely, you can see that when any negative “x” is substituted into the equation, the left side will be strictly negative, which means that the negative roots disappear

Thus, 8 numbers are left for research:

Consistently "charge" them according to the Horner scheme. I hope you have already mastered the mental calculations:

Luck was waiting for us when testing the "deuce". Thus, is the root of the equation under consideration, and

It remains to investigate the equation . It is easy to do this through the discriminant, but I will conduct an exponential test in the same way. First, note that the free term is equal to 20, which means that according to Theorem 1 the numbers 8 and 40 drop out of the list of possible roots, and the values ​​remain for research (one was eliminated according to the Horner scheme).

We write the coefficients of the trinomial in the top row of the new table and we start checking with the same "two". Why? And because the roots can be multiples, please: - this equation has 10 identical roots. But let's not digress:

And here, of course, I was a little cunning, knowing that the roots are rational. After all, if they were irrational or complex, then I would have an unsuccessful check of all the remaining numbers. Therefore, in practice, be guided by the discriminant.

Answer: rational roots: 2, 4, 5

In the analyzed problem, we were lucky, because: a) negative values ​​fell off immediately, and b) we found the root very quickly (and theoretically we could check the entire list ).

But in reality the situation is much worse. I invite you to watch an exciting game called "The Last Hero":

Task 4

Find rational roots of an equation

Decision: on Theorem 1 numerators of hypothetical rational roots must satisfy the condition (read "twelve is divisible by ale"), and the denominators to the condition . Based on this, we get two lists:

"list el":
and "list em": (fortunately, here the numbers are natural).

Now let's make a list of all possible roots. First, we divide the “list of ale” by . It is quite clear that the same numbers will turn out. For convenience, let's put them in a table:

Many fractions have been reduced, resulting in values ​​that are already in the "list of heroes". We add only "newcomers":

Similarly, we divide the same "list of ale" by:

and finally on

Thus, the team of participants in our game is staffed with:


Unfortunately, the polynomial of this problem does not satisfy the "positive" or "negative" criterion, and therefore we cannot discard the top or bottom row. You have to work with all the numbers.

How is your mood? Come on, turn your nose up - there is another theorem that can be figuratively called the “killer theorem” .... ... "candidates", of course =)

But first you need to scroll through Horner's diagram for at least one the whole numbers. Traditionally, we take one. In the upper line we write the coefficients of the polynomial and everything is as usual:

Since four is clearly not zero, the value is not the root of the polynomial in question. But she will help us a lot.

Theorem 2 If for some in general value of the polynomial is nonzero: , then its rational roots (if they are) satisfy the condition

In our case and therefore all possible roots must satisfy the condition (let's call it Condition #1). This four will be the "killer" of many "candidates". As a demonstration, I'll look at a few checks:

Let's check the candidate. To do this, we artificially represent it as a fraction , from which it is clearly seen that . Let's calculate the check difference: . Four is divided by "minus two": which means that the possible root has passed the test.

Let's check the value. Here, the test difference is: . Of course, and therefore the second "test subject" also remains on the list.