Algebraic equations of higher degrees how to solve. Methodical development in algebra (grade 10) on the topic: Equations of higher degrees

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Introduction

Solving algebraic equations of higher degrees in one unknown is one of the most difficult and ancient mathematical problems. The most outstanding mathematicians of antiquity were engaged in these problems.

The solution of equations of the n-th degree is an important problem for modern mathematics. Interest in them is quite large, since these equations are closely related to the search for the roots of equations that are not considered in the school curriculum in mathematics.

Problem: the lack of skills in solving equations of higher degrees in various ways among students prevents them from successfully preparing for the final certification in mathematics and mathematical olympiads, teaching in a specialized mathematical class.

The listed facts determined relevance of our work "Solving equations of higher degrees".

Possession of the simplest methods of solving n-th degree equations reduces the time for completing the task, on which the result of the work and the quality of the learning process depend.

Objective: study of known methods for solving equations of higher degrees and identification of the most accessible of them for practical application.

Based on this goal, the work identified the following tasks:

Study literature and Internet resources on this topic;

Get acquainted with the historical facts regarding this topic;

Describe the different ways to solve higher-degree equations

compare the degree of complexity of each of them;

To acquaint classmates with methods of solving equations of higher degrees;

Create a set of equations for the practical application of each of the considered methods.

Object of study - equations of higher degrees with one variable.

Subject of study - ways to solve equations of higher degrees.

Hypothesis:there is no general method and unified algorithm that allows finding solutions to equations of the nth degree in a finite number of steps.

Research methods:

- bibliographic method (analysis of literature on the research topic);

- classification method;

- method of qualitative analysis.

Theoretical significanceresearch consists in the systematization of methods for solving equations of higher degrees and the description of their algorithms.

Practical significance - submitted material on this topic and the development of a textbook for students on this topic.

1 HIGHER DEGREE EQUATIONS

1.1 Concept of n-th degree equation

Definition 1.An equation of the nth degree is an equation of the form

a 0 xⁿ + a 1 xn -1 + a 2 xⁿ - ² + ... + an -1 x + an \u003d 0, where the coefficients a 0, a 1, a 2…, an -1, an- any real numbers, and , a 0 ≠ 0 .

Polynomial a 0 xⁿ + a 1 xn -1 + a 2 xⁿ - ² + ... + an -1 x + an is called a polynomial of degree n. The odds are distinguished by their names: a 0 - senior coefficient; an is a free member.

Definition 2. Solutions or roots for a given equation are all values \u200b\u200bof the variable x, which turn this equation into a true numerical equality or, for which the polynomial a 0 xⁿ + a 1 xn -1 + a 2 xⁿ - ² + ... + an -1 x + an vanishes. This value of the variable xis also called the root of the polynomial. To solve an equation means to find all its roots or to establish that they do not exist.

If a a 0 \u003d 1, then such an equation is called the reduced whole rational equation n th degree.

For equations of the third and fourth degrees, there are Cardano and Ferrari formulas expressing the roots of these equations in terms of radicals. It turned out that in practice they are rarely used. Thus, if n ≥ 3, and the coefficients of the polynomial are arbitrary real numbers, then finding the roots of the equation is not an easy task. Nevertheless, in many special cases, this problem is solved to the end. Let's dwell on some of them.

1.2 Historical facts of solving equations of higher degrees

Already in ancient times, people realized how important it is to learn how to solve algebraic equations. About 4000 years ago, Babylonian scientists mastered the solution of a quadratic equation and solved systems of two equations, of which one is of the second degree. With the help of equations of higher degrees, various problems of land surveying, architecture and military affairs were solved, many and various questions of practice and natural science were reduced to them, since the exact language of mathematics makes it possible to simply express facts and relationships that, being expressed in ordinary language, may seem confusing and complicated ...

A universal formula for finding the roots of an algebraic equation nth degree no. Many, of course, came up with the tempting idea to find for any power n formulas that would express the roots of the equation in terms of its coefficients, that is, would solve the equation in radicals.

Only in the 16th century, Italian mathematicians managed to advance further - to find formulas for n \u003d 3 and n \u003d 4. At the same time, Scipio, Dal, Ferro and his students Fiori and Tartaglia were engaged in the question of the general solution of equations of the 3rd degree.

In 1545, the book of the Italian mathematician D. Cardano "The Great Art, or the Rules of Algebra" was published, where, along with other questions of algebra, general methods of solving cubic equations were considered, as well as the method for solving equations of the 4th degree discovered by his student L. Ferrari.

F. Viet gave a complete exposition of questions related to the solution of equations of the third and fourth degrees.

In the 20s of the 19th century, the Norwegian mathematician N. Abel proved that the roots of equations of the fifth degree cannot be expressed in terms of radicals.

During the study, it was revealed that modern science knows many ways to solve equations of the n-th degree.

The result of the search for methods for solving equations of higher degrees that cannot be solved by the methods considered in the school curriculum were methods based on the application of Vieta's theorem (for equations of degree n\u003e 2), Bezout's theorems, Horner's schemes, as well as the Cardano and Ferrari formula for solving cubic equations and equations of the fourth degree.

The paper presents methods for solving equations and their types, which became a discovery for us. These include the method of undefined coefficients, the selection of the full degree, symmetric equations.

2. SOLUTION OF INTEGER EQUATIONS OF HIGHER DEGREES WITH INTEGER COEFFICIENTS

2.1 Solution of 3rd degree equations. Formula D. Cardano

Consider equations of the form x 3 + px + q \u003d 0.We transform the general equation to the form: x 3 + px 2 + qx + r \u003d 0. Let's write the formula for the sum cube; We add to the original equality and replace with y... We get the equation: y 3 + (q -) (y -) + (r - \u003d 0. After transformations, we have: y 2 + py + q \u003d 0. Now, again, write the sum cube formula:

(a + b) 3 \u003d a 3 + 3a 2 b + 3ab 2 + b 3 \u003d a 3 + b 3 + 3ab (a + b), replace ( a + b)on the x, we get the equation x 3 - 3abx - (a 3 + b 3) = 0. Now you can see that the original equation is equivalent to the system: and Solving the system, we get:

We have obtained a formula for solving the reduced equation of the 3rd degree. She bears the name of the Italian mathematician Cardano.

Let's look at an example. Solve the equation:.

We have r \u003d 15 and q \u003d 124, then using the Cardano formula we calculate the root of the equation

Conclusion: this formula is good, but not suitable for solving all cubic equations. However, it is cumbersome. Therefore, in practice, it is rarely used.

But those who master this formula can use it to solve third-degree equations on the exam.

2.2 Vieta's theorem

From the course of mathematics, we know this theorem for a quadratic equation, but few people know that it is also used to solve equations of higher degrees.

Consider the equation:

factorize the left side of the equation, divide by ≠ 0.

We transform the right side of the equation to the form

; from this it follows that the following equalities can be written into the system:

The formulas derived by Viet for quadratic equations and demonstrated by us for equations of the third degree are also valid for polynomials of higher degrees.

Let's solve the cubic equation:

Conclusion: this method is universal and easy enough for students to understand, since Vieta's theorem is familiar to them from the school curriculum for n = 2. At the same time, in order to find the roots of equations using this theorem, one must have good computational skills.

2.3 Bezout's theorem

This theorem is named after the French mathematician of the XVIII century J. Bezout.

Theorem.If the equation a 0 xⁿ + a 1 xn -1 + a 2 xⁿ - ² + ... + an -1 x + an \u003d 0, in which all coefficients are integers, and the free term is nonzero, has an integer root, then this root is a divisor of the free term.

Considering that on the left side of the equation there is a polynomial of the nth degree, the theorem has a different interpretation.

Theorem. When dividing a polynomial of degree n with respect to x binomial x - a the remainder is equal to the value of the dividend at x \u003d a... (letter a can denote any real or imaginary number, i.e. any complex number).

Evidence: let be f (x) denotes an arbitrary n-th degree polynomial with respect to the variable x and let, when dividing by a binomial ( x-a) happened in private q (x), and in the remainder R... It's obvious that q (x) there will be some polynomial (n - 1) th degree with respect to xand the remainder R will be a constant value, i.e. independent of x.

If the remainder R was a polynomial of the first degree with respect to x, then this would mean that the division is not satisfied. So, R from x does not depend. By the definition of division, we get the identity: f (x) \u003d (x-a) q (x) + R.

Equality is valid for any value of x, which means that it is also valid for x \u003d a, we get: f (a) \u003d (a-a) q (a) + R... Symbol f (a) denotes the value of the polynomial f (x) at x \u003d a, q (a) denotes the value q (x) at x \u003d a.The remainder R remained the same as it was before, since R from x does not depend. Composition ( x-a) q (a) \u003d 0, since the factor ( x-a) \u003d 0, and the factor q (a) there is a certain number. Therefore, from the equality we get: f (a) \u003d R,ch.d.

Example 1.Find the remainder of a polynomial division x 3 - 3x 2 + 6x-5 for binomial

x-2. By Bezout's theorem : R \u003d f(2) = 23-322 + 62 -5 \u003d 3. Answer: R \u003d3.

Note that Bezout's theorem is important not so much in itself as in its consequences. (Appendix 1)

Let us dwell on some methods of applying Bezout's theorem to solving practical problems. It should be noted that when solving equations using Bezout's theorem, it is necessary:

Find all integer divisors of the free term;

Find at least one root of the equation from these divisors;

Divide the left side of the equation by (Ha);

Write the product of the divisor and the quotient on the left side of the equation;

Solve the resulting equation.

Consider, for example, solving the equation x 3 + 4x 2 + x -6 = 0 .

Solution: find the divisors of the free term ± 1 ; ± 2; ± 3; ± 6. Let us calculate the values \u200b\u200bat x \u003d1, 1 3 + 41 2 + 1- 6 \u003d 0. Divide the left side of the equation by ( x-1). We will perform the division "with a corner", we get:

Conclusion: Bezout's theorem, one of the ways that we consider in our work, is studied in the program of optional classes. It is difficult to understand, because in order to own it, you need to know all the consequences from it, but at the same time Bezout's theorem is one of the main assistants of students on the exam.

2.4 Horner's scheme

To divide a polynomial by a binomial x-α you can use a special simple trick invented by English mathematicians of the 17th century, later called Horner's scheme. In addition to finding the roots of the equations, according to Horner's scheme, it is easier to calculate their values. For this, it is necessary to substitute the value of the variable into the polynomial Pn (x) \u003d a 0 xn + a 1 x n-1 + a 2 xⁿ - ² +… ++ an -1 x + an. (1)

Consider the division of the polynomial (1) by the binomial x-α.

Let us express the coefficients of the incomplete quotient b 0 xⁿ - ¹+ b 1 xⁿ - ²+ b 2 xⁿ - ³+…+ bn -1 and the remainder r in terms of the coefficients of the polynomial Pn ( x) and the number α. b 0 \u003d a 0 , b 1 = α b 0 + a 1 , b 2 = α b 1 + a 2 …, bn -1 =

= α bn -2 + an -1 = α bn -1 + an .

Calculations according to Horner's scheme are presented in the form of the following table:

	and 0	a 1	a 2 ,
	b 0 \u003d a 0	b 1 = α b 0 + a 1	b 2 = α b 1 + a 2		r \u003d αb n-1 + an

Insofar as r \u003d Pn (α), then α is the root of the equation. To check whether α is a multiple root, Horner's scheme can be applied to the quotient b 0 x +b 1 x + ... +bn -1 according to the table. If in the column below bn -1 it turns out 0 again, so α is a multiple root.

Consider an example: Solve the equation x 3 + 4x 2 + x -6 = 0.

Apply the factorization of the polynomial on the left side of the equation to the left side of the equation, Horner's scheme.

Solution: find the divisors of the free term ± 1; ± 2; ±3; ± 6.

				6 ∙ 1 + (-6) = 0

The quotients are numbers 1, 5, 6, and the remainder is r \u003d 0.

Hence, x 3 + 4x 2 + x - 6 = (x - 1) (x 2 + 5x + 6) = 0.

Hence: x - 1 \u003d 0 or x 2 + 5x + 6 = 0.

x = 1, x 1 = -2; x 2 = -3. Answer:1,- 2, - 3.

Conclusion: Thus, on one equation, we have shown the use of two different methods of factoring polynomials. In our opinion, Horner's scheme is the most practical and economical.

2.5 Solving equations of the 4th degree. Ferrari method

Cardano's student Ludovic Ferrari discovered a way to solve the 4th degree equation. The Ferrari method consists of two stages.

Stage I: equations of the form are represented as a product of two square trinomials, this follows from the fact that the equation is of the 3rd degree and at least one solution.

Stage II: the obtained equations are solved using factorization, but in order to find the required factorization, it is necessary to solve cubic equations.

The idea is to represent equations as A 2 \u003d B 2, where A \u003d x 2 + s,

B-linear function of x... Then it remains to solve the equations A \u003d ± B.

For clarity, consider the equation: We solitary the 4th degree, we get: For any dthe expression will be a perfect square. We add to both sides of the equation we obtain

On the left is a full square, you can pick up dso that the right-hand side (2) also becomes a complete square. Let's imagine that we have achieved this. Then our equation looks like this:

Finding the root later will not be difficult. To choose the right one d it is necessary that the discriminant of the right-hand side of (3) vanish, i.e.

So to find d, it is necessary to solve this equation of the 3rd degree. Such an auxiliary equation is called resolution.

We can easily find the whole root of the resolvent: d \u003d1

Substituting the equation into (1), we obtain

Conclusion: Ferrari's method is universal, but complicated and cumbersome. At the same time, if the solution algorithm is clear, then 4th degree equations can be solved by this method.

2.6 Method of undefined coefficients

The success of solving the equation of the 4th degree by the Ferrari method depends on whether we solve the resolvent - the equation of the 3rd degree, which, as we know, is not always possible.

The essence of the method of undefined coefficients is that the type of factors into which a given polynomial is decomposed is guessed, and the coefficients of these factors (also polynomials) are determined by multiplying the factors and equating the coefficients at the same degrees of the variable.

Example: Solve the equation:

Suppose that the left-hand side of our equation can be decomposed into two square trinomials with integer coefficients such that the identity

Obviously, the coefficients in front of the uni should be equal to 1, and the free terms should be equal to one + 1, the other has 1.

The coefficients in front of x... We denote them by and and and to determine them, we multiply both trinomials on the right side of the equation.

As a result, we get:

Equating the coefficients at the same degrees xon the left and right sides of equality (1), we obtain a system for finding and

Having solved this system, we will have

So our equation is equivalent to the equation

Having solved it, we get the following roots:.

The method of indefinite coefficients is based on the following statements: any fourth degree polynomial in the equation can be decomposed into the product of two second degree polynomials; two polynomials are identically equal if and only if their coefficients are equal at the same degrees x.

2.7 Symmetric equations

Definition.An equation of the form is called symmetric if the first coefficients on the left in the equation are equal to the first coefficients on the right.

We see that the first coefficients on the left are equal to the first coefficients on the right.

If such an equation has an odd degree, then it has the root x= - 1. Next, we can lower the degree of the equation by dividing it by ( x +1). It turns out that dividing the symmetric equation by ( x +1) a symmetric equation of even degree is obtained. The proof of the symmetry of the coefficients is presented below. (Appendix 6) Our task is to learn how to solve symmetric equations of even degree.

For example: (1)

Solve equation (1), divide by x 2 (medium) \u003d 0.

Let us group the terms with symmetric

) + 3(x +. We denote at= x +, let's square both sides, hence \u003d at 2 So, 2 ( at 2 or 2 at 2 + 3 solving the equation, we get at = , at \u003d 3. Next, let's return to replacing x + \u003d and x + \u003d 3. We get the equations and The first has no solution, and the second has two roots. Answer:.

Conclusion: this type of equations is not common, but if you come across it, then it can be solved easily and simply without resorting to cumbersome calculations.

2.8 Highlighting the full degree

Consider the equation.

The left side is the cube of the sum (x + 1), i.e.

We extract the root of the third degree from both parts:, then we get

Where is the only root.

RESULTS OF THE STUDY

Based on the results of the work, we came to the following conclusions:

Thanks to the studied theory, we got acquainted with various methods of solving entire equations of higher degrees;

D. Cardano's formula is difficult to apply and gives a high probability of making errors in the calculation;

- L. Ferrari's method makes it possible to reduce the solution of a fourth-degree equation to a cubic one;

- Bezout's theorem can be applied both for cubic equations and for equations of the fourth degree; it is more understandable and clear when applied to solving equations;

Horner's scheme helps to significantly reduce and simplify calculations in solving equations. In addition to finding the roots, according to Horner's scheme, it is easier to calculate the values \u200b\u200bof the polynomials on the left side of the equation;

Of particular interest was the solution of equations by the method of undefined coefficients, the solution of symmetric equations.

In the course of the research work, it was found that students get acquainted with the simplest methods of solving equations of the highest degree in elective classes in mathematics, starting from the 9th or 10th grade, as well as in special courses of visiting mathematical schools. This fact was established as a result of a survey of teachers of mathematics MBOU "Secondary School No. 9" and students showing an increased interest in the subject "mathematics".

The most popular methods for solving equations of higher degrees, which are found in solving Olympiad, competitive problems and as a result of preparing students for exams, are methods based on the application of Bezout's theorem, Horner's scheme and the introduction of a new variable.

Demonstration of research results, i.e. ways to solve equations that are not studied in the school curriculum in mathematics interested classmates.

Conclusion

Having studied educational and scientific literature, Internet resources in youth educational forums

In general, an equation with a degree higher than 4 cannot be solved in radicals. But sometimes we can still find the roots of the polynomial on the left in the equation of the highest degree, if we represent it as a product of polynomials in degree at most 4. The solution to such equations is based on factoring a polynomial, so we advise you to repeat this topic before studying this article.

Most often one has to deal with equations of higher degrees with integer coefficients. In these cases, we can try to find rational roots, and then factor the polynomial in order to then transform it into an equation of a lower degree, which will be easy to solve. Within the framework of this material, we will consider just such examples.

Yandex.RTB R-A-339285-1

Equations of the highest degree with integer coefficients

All equations of the form a n x n + a n - 1 x n - 1 +. ... ... + a 1 x + a 0 \u003d 0, we can reduce to an equation of the same degree by multiplying both sides by a n n - 1 and changing a variable of the form y \u003d a n x:

a n x n + a n - 1 x n - 1 +. ... ... + a 1 x + a 0 \u003d 0 ann xn + an - 1 ann - 1 xn - 1 +… + a 1 (an) n - 1 x + a 0 (an) n - 1 \u003d 0 y \u003d anx ⇒ yn + bn - 1 yn - 1 +… + b 1 y + b 0 \u003d 0

The resulting coefficients will also be whole. Thus, we will need to solve the reduced equation of the nth degree with integer coefficients, which has the form x n + a n x n - 1 +… + a 1 x + a 0 \u003d 0.

Calculate the whole roots of the equation. If the equation has integer roots, you need to look for them among the divisors of the free term a 0. Let us write them down and substitute them into the original equality in turn, checking the result. Once we have obtained an identity and found one of the roots of the equation, we can write it in the form x - x 1 · P n - 1 (x) \u003d 0. Here x 1 is the root of the equation, and P n - 1 (x) is the quotient of dividing x n + a n x n - 1 +… + a 1 x + a 0 by x - x 1.

Substitute the remaining divisors written out into P n - 1 (x) \u003d 0, starting with x 1, since the roots can repeat. After obtaining the identity, the root x 2 is considered found, and the equation can be written as (x - x 1) (x - x 2) P n - 2 (x) \u003d 0. Here P n - 2 (x) will be the quotient of dividing P n - 1 (x) by x - x 2.

We continue to iterate over the divisors. Find all whole roots and denote their number as m. After that, the original equation can be represented as x - x 1 x - x 2 · ... · x - x m · P n - m (x) \u003d 0. Here P n - m (x) is a polynomial of degree n - m. It is convenient to use Horner's scheme for counting.

If our original equation has integer coefficients, we cannot end up with fractional roots.

As a result, we got the equation P n - m (x) \u003d 0, the roots of which can be found in any convenient way. They can be irrational or complex.

Let's show with a specific example how such a solution scheme is applied.

Example 1

Condition: find the solution to the equation x 4 + x 3 + 2 x 2 - x - 3 \u003d 0.

Decision

Let's start by finding whole roots.

We have a free term equal to minus three. It has divisors of 1, - 1, 3, and - 3. Let's substitute them in the original equation and see which of them will result in identities.

With x equal to one, we get 1 4 + 1 3 + 2 · 1 2 - 1 - 3 \u003d 0, which means that one will be the root of this equation.

Now we perform division of the polynomial x 4 + x 3 + 2 x 2 - x - 3 by (x - 1) in a column:

Hence, x 4 + x 3 + 2 x 2 - x - 3 \u003d x - 1 x 3 + 2 x 2 + 4 x + 3.

1 3 + 2 1 2 + 4 1 + 3 \u003d 10 ≠ 0 (- 1) 3 + 2 (- 1) 2 + 4 - 1 + 3 \u003d 0

We got an identity, which means that we have found another root of the equation, equal to - 1.

Divide the polynomial x 3 + 2 x 2 + 4 x + 3 by (x + 1) in a column:

We get that

x 4 + x 3 + 2 x 2 - x - 3 \u003d (x - 1) (x 3 + 2 x 2 + 4 x + 3) \u003d \u003d (x - 1) (x + 1) (x 2 + x + 3)

Substitute the next divisor into the equality x 2 + x + 3 \u003d 0, starting with - 1:

1 2 + (- 1) + 3 = 3 ≠ 0 3 2 + 3 + 3 = 15 ≠ 0 (- 3) 2 + (- 3) + 3 = 9 ≠ 0

The resulting equalities will be incorrect, which means that the equation no longer has integral roots.

The remaining roots will be the roots of the expression x 2 + x + 3.

D \u003d 1 2 - 4 1 3 \u003d - 11< 0

It follows from this that this square trinomial has no real roots, but it has complex conjugate roots: x \u003d - 1 2 ± i 11 2.

Let's clarify that instead of long division, we can use Horner's scheme. This is done like this: after we have determined the first root of the equation, we fill in the table.

In the table of coefficients, we can immediately see the coefficients of the quotient of the division of polynomials, which means that x 4 + x 3 + 2 x 2 - x - 3 \u003d x - 1 x 3 + 2 x 2 + 4 x + 3.

After finding the next root equal to - 1, we get the following:

Answer: x \u003d - 1, x \u003d 1, x \u003d - 1 2 ± i 11 2.

Example 2

Condition: Solve the equation x 4 - x 3 - 5 x 2 + 12 \u003d 0.

Decision

The free term has divisors 1, - 1, 2, - 2, 3, - 3, 4, - 4, 6, - 6, 12, - 12.

We check them in order:

1 4 - 1 3 - 5 1 2 + 12 \u003d 7 ≠ 0 (- 1) 4 - (- 1) 3 - 5 (- 1) 2 + 12 \u003d 9 ≠ 0 2 4 2 3 - 5 2 2 + 12 \u003d 0

Hence, x \u003d 2 will be the root of the equation. Divide x 4 - x 3 - 5 x 2 + 12 by x - 2, using Horner's scheme:

As a result, we get x - 2 (x 3 + x 2 - 3 x - 6) \u003d 0.

2 3 + 2 2 - 3 2 - 6 \u003d 0

Hence, 2 will again be a root. Divide x 3 + x 2 - 3 x - 6 \u003d 0 by x - 2:

As a result, we get (x - 2) 2 (x 2 + 3 x + 3) \u003d 0.

It makes no sense to check the remaining divisors, since the equality x 2 + 3 x + 3 \u003d 0 is faster and more convenient to solve using the discriminant.

Let's solve the quadratic equation:

x 2 + 3 x + 3 \u003d 0 D \u003d 3 2 - 4 1 3 \u003d - 3< 0

We get a complex conjugate pair of roots: x \u003d - 3 2 ± i 3 2.

Answer: x \u003d - 3 2 ± i 3 2.

Example 3

Condition: find the real roots for the equation x 4 + 1 2 x 3 - 5 2 x - 3 \u003d 0.

Decision

x 4 + 1 2 x 3 - 5 2 x - 3 \u003d 0 2 x 4 + x 3 - 5 x - 6 \u003d 0

We perform multiplication 2 3 of both sides of the equation:

2 x 4 + x 3 - 5 x - 6 \u003d 0 2 4 x 4 + 2 3 x 3 - 20 2 x - 48 \u003d 0

Replace the variables y \u003d 2 x:

2 4 x 4 + 2 3 x 3 - 20 2 x - 48 \u003d 0 y 4 + y 3 - 20 y - 48 \u003d 0

As a result, we have a standard 4th degree equation that can be solved using the standard scheme. Let's check the divisors, divide and get in the end that it has 2 real roots y \u003d - 2, y \u003d 3 and two complex roots. We will not present the complete solution here. Due to the replacement, the real roots of this equation will be x \u003d y 2 \u003d - 2 2 \u003d - 1 and x \u003d y 2 \u003d 3 2.

Answer: x 1 \u003d - 1, x 2 \u003d 3 2

If you notice an error in the text, please select it and press Ctrl + Enter

Consider solutions of equations with one variable of degree higher than the second.

The degree of the equation P (x) \u003d 0 is the degree of the polynomial P (x), i.e. the greatest of the degrees of its terms with a coefficient not equal to zero.

So, for example, the equation (x 3 - 1) 2 + x 5 \u003d x 6 - 2 has the fifth degree, since after the operations of opening the brackets and bringing similar ones, we get an equivalent equation x 5 - 2x 3 + 3 \u003d 0 of the fifth degree.

Let us recall the rules that will be needed to solve equations of degree higher than two.

Statements about roots of a polynomial and its divisors:

1. The nth degree polynomial has the number of roots at most n, and the roots of multiplicity m occur exactly m times.

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

3. If α is a root of P (x), then P n (x) \u003d (x - α) Q n - 1 (x), where Q n - 1 (x) is a polynomial of degree (n - 1).

4.

5. The reduced polynomial with integer coefficients cannot have fractional rational roots.

6. For a polynomial of degree 3

P 3 (x) \u003d ax 3 + bx 2 + cx + d one of two things is possible: either it decomposes into a product of three binomials

Р 3 (x) \u003d а (х - α) (х - β) (х - γ), or it can be decomposed into the product of a binomial and a square trinomial Р 3 (x) \u003d a (х - α) (х 2 + βх + γ ).

7. Any polynomial of the fourth degree can be decomposed into the product of two square trinomials.

8. The polynomial f (x) is divisible by the polynomial g (x) without a remainder if there is a polynomial q (x) such that f (x) \u003d g (x) q (x). For dividing polynomials, the "corner division" rule is applied.

9. For the divisibility of the polynomial P (x) into the binomial (x - c), it is necessary and sufficient that the number c be a root of P (x) (Corollary of Bezout's theorem).

10. Vieta's theorem: If x 1, x 2, ..., x n are real roots of the polynomial

P (x) \u003d a 0 x n + a 1 x n - 1 + ... + a n, then the following equalities hold:

x 1 + x 2 + ... + x n \u003d -a 1 / a 0,

x 1 x 2 + x 1 x 3 + ... + x n - 1 x n \u003d a 2 / a 0,

x 1 x 2 x 3 + ... + x n - 2 x n - 1 x n \u003d -a 3 / a 0,

x 1 x 2 x 3 x n \u003d (-1) n a n / a 0.

Solution examples

Example 1.

Find the remainder of dividing P (x) \u003d x 3 + 2/3 x 2 - 1/9 by (x - 1/3).

Decision.

By corollary to Bezout's theorem: "The remainder of dividing a polynomial by a binomial (x - c) is equal to the value of the polynomial in c". Let us find Р (1/3) \u003d 0. Therefore, the remainder is 0 and the number 1/3 is the root of the polynomial.

Answer: R \u003d 0.

Example 2.

Divide with a corner 2x 3 + 3x 2 - 2x + 3 by (x + 2). Find the remainder and the incomplete quotient.

Decision:

2x 3 + 3x 2 - 2x + 3 | x + 2

2x 3 + 4 x 2 2x 2 - x

X 2 - 2 x

Answer: R \u003d 3; private: 2x 2 - x.

Basic methods for solving equations of higher degrees

1. Introducing a new variable

The method of introducing a new variable is already familiar with the example of biquadratic equations. It consists in the fact that to solve the equation f (x) \u003d 0, a new variable (substitution) t \u003d x n or t \u003d g (x) is introduced and f (x) is expressed in terms of t, obtaining a new equation r (t). Then solving the equation r (t), the roots are found:

(t 1, t 2, ..., t n). After that, a set of n equations q (x) \u003d t 1, q (x) \u003d t 2, ..., q (x) \u003d t n are obtained, from which the roots of the original equation are found.

Example 1.

(x 2 + x + 1) 2 - 3x 2 - 3x - 1 \u003d 0.

Decision:

(x 2 + x + 1) 2 - 3 (x 2 + x) - 1 \u003d 0.

(x 2 + x + 1) 2 - 3 (x 2 + x + 1) + 3 - 1 \u003d 0.

Replacement (x 2 + x + 1) \u003d t.

t 2 - 3t + 2 \u003d 0.

t 1 \u003d 2, t 2 \u003d 1. Reverse replacement:

x 2 + x + 1 \u003d 2 or x 2 + x + 1 \u003d 1;

x 2 + x - 1 \u003d 0 or x 2 + x \u003d 0;

Answer: From the first equation: x 1, 2 \u003d (-1 ± √5) / 2, from the second: 0 and -1.

2. Factoring by grouping and reduced multiplication formulas

The basis of this method is also not new and consists in grouping the terms in such a way that each group contains a common factor. To do this, sometimes you have to use some artificial methods.

Example 1.

x 4 - 3x 2 + 4x - 3 \u003d 0.

Decision.

Imagine - 3x 2 \u003d -2x 2 - x 2 and group:

(x 4 - 2x 2) - (x 2 - 4x + 3) \u003d 0.

(x 4 - 2x 2 +1 - 1) - (x 2 - 4x + 3 + 1 - 1) \u003d 0.

(x 2 - 1) 2 - 1 - (x - 2) 2 + 1 \u003d 0.

(x 2 - 1) 2 - (x - 2) 2 \u003d 0.

(x 2 - 1 - x + 2) (x 2 - 1 + x - 2) \u003d 0.

(x 2 - x + 1) (x 2 + x - 3) \u003d 0.

x 2 - x + 1 \u003d 0 or x 2 + x - 3 \u003d 0.

Answer: There are no roots in the first equation, from the second: x 1, 2 \u003d (-1 ± √13) / 2.

3. Factoring by the method of undefined coefficients

The essence of the method is that the original polynomial is decomposed into factors with unknown coefficients. Using the property that the polynomials are equal if their coefficients are equal at the same degrees, the unknown expansion coefficients are found.

Example 1.

x 3 + 4x 2 + 5x + 2 \u003d 0.

Decision.

A polynomial of the 3rd degree can be expanded into a product of a linear and a square factor.

x 3 + 4x 2 + 5x + 2 \u003d (x - a) (x 2 + bx + c),

x 3 + 4x 2 + 5x + 2 \u003d x 3 + bx 2 + cx - ax 2 - abx - ac,

x 3 + 4x 2 + 5x + 2 \u003d x 3 + (b - a) x 2 + (cx - ab) x - ac.

Having solved the system:

(b - a \u003d 4,
(c - ab \u003d 5,
(-ac \u003d 2,

(a \u003d -1,
(b \u003d 3,
(c \u003d 2, i.e.

x 3 + 4x 2 + 5x + 2 \u003d (x + 1) (x 2 + 3x + 2).

The roots of the equation (x + 1) (x 2 + 3x + 2) \u003d 0 are easy to find.

Answer: -1; -2.

4. Method of selection of the root by the highest and the free coefficient

The method is based on the application of theorems:

1) Any integer root of a polynomial with integer coefficients is a divisor of an intercept.

2) In order for the irreducible fraction p / q (p is an integer, q is a natural) to be a root of an equation with integer coefficients, it is necessary that the number p is an integer divisor of the free term a 0, and q is a natural divisor of the leading coefficient.

Example 1.

6x 3 + 7x 2 - 9x + 2 \u003d 0.

Decision:

6: q \u003d 1, 2, 3, 6.

Therefore, p / q \u003d ± 1, ± 2, ± 1/2, ± 1/3, ± 2/3, ± 1/6.

Having found one root, for example - 2, we find the other roots using division by an angle, the method of undefined coefficients or Horner's scheme.

Answer: -2; 1/2; 1/3.

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Methods for solving equations: n n n Replacing the equation h (f (x)) \u003d h (g (x)) by the equation f (x) \u003d g (x) Factoring. Introducing a new variable. Functionally - graphical method. Selection of roots. Application of Vieta formulas.

Replacing the equation h (f (x)) \u003d h (g (x)) by the equation f (x) \u003d g (x). The method can be used only in the case when y \u003d h (x) is a monotonic function that takes each of its values \u200b\u200bonce. If the function is non-monotonic, then the loss of roots is possible.

Solve the equation (3 x + 2) ²³ \u003d (5 x - 9) ²³ y \u003d x ²³ an increasing function, so from the equation (3 x + 2) ²³ \u003d (5 x - 9) ²³ you can go to the equation 3 x + 2 \u003d 5 x - 9, whence we find x \u003d 5, 5. Answer: 5, 5.

Factorization. The equation f (x) g (x) h (x) \u003d 0 can be replaced by a set of equations f (x) \u003d 0; g (x) \u003d 0; h (x) \u003d 0. Having solved the equations of this set, you need to take those roots that belong to the domain of the original equation, and discard the rest as extraneous.

Solve the equation x³ - 7 x + 6 \u003d 0 Representing the term 7 x as x + 6 x, we get sequentially: x³ - x - 6 x + 6 \u003d 0 x (x² - 1) - 6 (x - 1) \u003d 0 x (x - 1) (x + 1) - 6 (x - 1) \u003d 0 (x - 1) (x² + x - 6) \u003d 0 Now the problem is reduced to solving the set of equations x - 1 \u003d 0; x² + x - 6 \u003d 0. Answer: 1, 2, - 3.

Introducing a new variable. If the equation y (x) \u003d 0 can be transformed to the form p (g (x)) \u003d 0, then you need to introduce a new variable u \u003d g (x), solve the equation p (u) \u003d 0, and then solve the set of equations g ( x) \u003d u 1; g (x) \u003d u 2; ...; g (x) \u003d un, where u 1, u 2,…, un are the roots of the equation p (u) \u003d 0.

Solve the equation A feature of this equation is the equality of the coefficients of its left side, equidistant from its ends. Such equations are called recurrent. Since 0 is not a root of this equation, dividing by x² we get

Let's introduce a new variable Then We get a quadratic equation So the root y 1 \u003d - 1 can be ignored. We get the Answer: 2, 0, 5.

Solve the equation 6 (x² - 4) ² + 5 (x² - 4) (x² - 7 x +12) + (x² - 7 x + 12) ² \u003d 0 This equation can be solved as homogeneous. Divide both sides of the equation by (x² - 7 x +12) ² (clearly, the values \u200b\u200bof x such that x² - 7 x + 12 \u003d 0 are not solutions). Now let us denote We have Hence the Answer:

Functionally - graphical method. If one of the functions y \u003d f (x), y \u003d g (x) increases, and the other decreases, then the equation f (x) \u003d g (x) either has no roots or has one root.

Solve the Equation It is fairly obvious that x \u003d 2 is the root of the equation. Let us prove that this is the only root. We transform the equation to the form Notice that the function increases and the function decreases. Hence, the equation has only one root. Answer: 2.

Selection of roots n n n Theorem 1: If an integer m is a root of a polynomial with integer coefficients, then the free term of the polynomial is divisible by m. Theorem 2: The reduced polynomial with integer coefficients has no fractional roots. Theorem 3: - an equation with integers Let the coefficients. If the number and fraction where p and q are integers is irreducible, is the root of the equation, then p is the divisor of the free term an, and q is the divisor of the coefficient at the leading term a 0.

Bezout's theorem. The remainder when dividing any polynomial by a binomial (x - a) is equal to the value of the divisible polynomial at x \u003d a. Consequences of Bezout's theorem n n n n The difference of the same powers of two numbers is divided without remainder by the difference of the same numbers; The difference of the same even powers of two numbers is divided without a remainder both by the difference of these numbers, and by their sum; The difference of the same odd powers of two numbers is not divisible by the sum of these numbers; The sum of the same powers of two non-numbers is divided by the difference of these numbers; The sum of the same odd powers of two numbers is divided without a remainder by the sum of these numbers; The sum of the same even powers of two numbers is not divisible either by the difference of these numbers or by their sum; A polynomial is divisible entirely by a binomial (x - a) if and only if the number a is a root of the given polynomial; The number of distinct roots of a nonzero polynomial is at most its degree.

Solve the equation x³ - 5 x² - x + 21 \u003d 0 The polynomial x³ - 5 x² - x + 21 has integer coefficients. By Theorem 1, its integer roots, if any, are among the divisors of the free term: ± 1, ± 3, ± 7, ± 21. By checking, we make sure that the number 3 is a root. By the corollary to Bezout's theorem, the polynomial is divisible by (x - 3). So x³– 5 x² - x + 21 \u003d (x - 3) (x²– 2 x - 7). Answer:

Solve the equation 2 x³ - 5 x² - x + 1 \u003d 0 By Theorem 1, the integer roots of the equation can only be numbers ± 1. Checking shows that these numbers are not roots. Since the equation is not reduced, it can have fractional rational roots. Let's find them. To do this, multiply both sides of the equation by 4: 8 x³ - 20 x² - 4 x + 4 \u003d 0 Having made the substitution 2 x \u003d t, we get t³ - 5 t² - 2 t + 4 \u003d 0. By Theorem 2, all rational roots of this reduced equation must be whole. They can be found among the divisors of the free term: ± 1, ± 2, ± 4. In this case, t \u003d - 1 is suitable. Therefore, by the corollary of Bezout's theorem, the polynomial 2 x³ - 5 x² - x + 1 is divisible by (x + 0, 5 ): 2 x³ - 5 x² - x + 1 \u003d (x + 0, 5) (2 x² - 6 x + 2) Solving the quadratic equation 2 x² - 6 x + 2 \u003d 0, we find the remaining roots: Answer:

Solve the equation 6 x³ + x² - 11 x - 6 \u003d 0 According to Theorem 3, the rational roots of this equation should be sought among the numbers. Substituting them one by one into the equation, we find that satisfy the equation. They also exhaust all the roots of the equation. Answer:

Find the sum of the squares of the roots of the equation x³ + 3 x² - 7 x +1 \u003d 0 By Vieta's theorem Note that whence

Indicate which method can be used to solve each of these equations. Solve equations # 1, 4, 15, 17.

Answers and directions: 1. Introduction of a new variable. 2. Functional - graphic method. 3. Replacing the equation h (f (x)) \u003d h (g (x)) by the equation f (x) \u003d g (x). 4. Factorization. 5. Selection of roots. 6 Functional - graphic method. 7. Application of Vieta formulas. 8. Selection of roots. 9. Replacing the equation h (f (x)) \u003d h (g (x)) by the equation f (x) \u003d g (x). 10. Introduction of a new variable. 11. Factorization. 12. Introduction of a new variable. 13. Selection of roots. 14. Application of Vieta formulas. 15. Functional - graphic method. 16. Factorization. 17. Introduction of a new variable. 18. Factoring.

1. Indication. Write the equation as 4 (x² + 17 x + 60) (x + 16 x + 60) \u003d 3 x², Divide both sides by x². Enter the variable Answer: x 1 \u003d - 8; x 2 \u003d - 7, 5. 4. Note. Add 6 y and - 6 y to the left side of the equation and write it as (y³ - 2 y²) + (- 3 y² + 6 y) + (- 8 y + 16) \u003d (y - 2) (y² - 3 y - 8). Answer:

14. Indication. By Vieta's theorem Since are integers, the roots of the equation can only be numbers - 1, - 2, - 3. Answer: 15. Answer: - 1. 17. Indication. Divide both sides of the equation by x² and write it as Enter variable Answer: 1; 15; 2; 3.

Bibliography. n n n Kolmogorov A. N. "Algebra and the beginning of analysis, 10 - 11" (Moscow: Education, 2003). Bashmakov M. I. "Algebra and the beginning of analysis, 10 - 11" (Moscow: Education, 1993). Mordkovich A. G. "Algebra and the beginning of analysis, 10 - 11" (M.: Mnemosina, 2003). Alimov Sh. A., Kolyagin Yu. M. et al. "Algebra and the beginning of analysis, 10 - 11" (Moscow: Education, 2000). Galitsky ML, Gol'dman AM, Zvavich LI "Collection of problems in algebra, 8 - 9" (Moscow: Education, 1997). Karp A. P. "Collection of problems in algebra and the principles of analysis, 10 - 11" (Moscow: Education, 1999). Sharygin I. F. "Elective course in mathematics, problem solving, 10" (Moscow: Education. 1989). Skopets Z. A. "Additional chapters on the course of mathematics, 10" (Moscow: Education, 1974). Litinsky G. I. "Lessons in Mathematics" (M.: Aslan, 1994). Muravin G. K. "Equations, inequalities and their systems" (Mathematics, supplement to the newspaper "September First", No. 2, 3, 2003). Kolyagin Yu. M. "Polynomials and equations of higher degrees" (Mathematics, supplement to the newspaper "September 1", No. 3, 2005).

Trifanova Marina Anatolievna
teacher of mathematics, MOU "Gymnasium No. 48 (multidisciplinary)", Talnakh

The triune goal of the lesson:

Educational:
systematization and generalization of knowledge on solving equations of higher degrees.
Developing:
promote the development of logical thinking, the ability to work independently, skills of mutual control and self-control, the ability to speak and listen.
Educational:
developing a habit of constant employment, fostering responsiveness, hard work, and accuracy.

Lesson type:

a lesson in the complex application of knowledge, skills and abilities.

Lesson form:

airing, physical minutes, various forms of work.

Equipment:

basic notes, cards with assignments, lesson monitoring matrix.

DURING THE CLASSES

I. Organizational moment

1. Communicating the lesson goal to students.
2. Homework check (Appendix 1). Work with reference notes (Appendix 2).

Equations and answers are written on the board for each of them. Students check the answers and give a quick analysis of the solution to each equation or answer the teacher's questions (frontal survey). Self-control - students give themselves grades and hand over exercise books to the teacher for correction or approval. The grade school is written on the chalkboard:

“5+” - 6 equations;
“5” - 5 equations;
“4” - 4 equations;
“3” - 3 equations.

Teacher's homework questions:

1 equation

1. What change of variables is made in the equation?
2. What equation is obtained after changing variables?

2 equation

1. What polynomial did both sides of the equation divide into?
2. What change of variables was obtained?

3 equation

1. Which polynomials need to be multiplied to simplify the solution of this equation?

4 equation

1. Name the function f (x).
2. How were the rest of the roots found?

Equation 5

1. How many gaps were obtained to solve the equation?

6 equation

1. In what ways could this equation be solved?
2. Which solution is more rational?

II. Working in groups is the main part of the lesson.

The class is divided into 4 groups. Each group is given a card with theoretical and practical (Appendix 3) questions: "Deconstruct the proposed method for solving the equation and explain it using this example."

1. Group work 15 minutes.
2. Examples are written on the board (the board is divided into 4 parts).
3. The group report takes 2 - 3 minutes.
4. The teacher corrects the reports of the groups and helps in case of difficulty.

Group work continues on cards 5 - 8. Each equation has 5 minutes for group discussion. Then there is a report on this equation at the board - a short analysis of the solution. The equation may not be fully solved - it is being finalized at home, but the sequence of its solution in the classroom is discussed all over.

III. Independent work.Appendix 4.

1. Each student receives an individual assignment.
2. Time work takes 20 minutes.
3. 5 minutes before the end of the lesson, the teacher gives open-ended answers for each equation.
4. Students change notebooks in a circle and check the answers from a friend. Give marks.
5. Notebooks are handed over to the teacher for checking and correcting grades.

IV. Lesson summary.

Homework.

Checkout the solution to unfinished equations. Prepare for a control slice.

Grading.