Solving equations of higher degrees in various ways. Equations of higher degrees

"Methods for solving equations higher degrees»

( Kiselev readings)

Mathematics teacher L.A. Afanasyeva

MKOU Verkhnekarachanskaya secondary school

Gribanovsky district, Voronezh region

2015 year

Mathematical education received in comprehensive school, is an essential component general education and the general culture of modern man.

The famous German mathematician Courant wrote: “For more than two millennia, the possession of some, not too superficial, knowledge in the field of mathematics was a necessary part of into the intellectual inventory of every educated person. " And among this knowledge, not the last place belongs to the ability to solve equations.

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, one of which is of the second degree. With the help of equations, 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 presented in ordinary language, may seem confusing and complex. Equation is one of the most important concepts in mathematics. Development of methods for solving equations, starting with the birth of mathematics as a science, long time was the main subject of study of algebra. And today in mathematics lessons, starting from the first stage of education, solving equations different types great attention is paid.

Universal formula for finding roots algebraic equation n - no degree. Many, of course, came up with the tempting idea to find for any degree n formulas that would express the roots of an equation in terms of its coefficients, that is, would solve the equation in radicals. However, the “gloomy Middle Ages” turned out to be as gloomy as possible in relation to the problem under discussion - no one found the required formulas for seven whole centuries! Only in the 16th century did Italian mathematicians manage to advance further - to find formulas for n =3 and n =4 ... Simultaneously with the question of general decision equations of the 3rd degree were studied by Scipio Dal Ferro, his student Fiori and Tartaglia. In 1545, the book of the Italian mathematician D Cardano "The Great Art, or On 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 3rd and 4th degrees. And in the 20s of the 19th century, the Norwegian mathematician N. Abel proved that the roots of equations of the 5th and higher degrees cannot be expressed in terms of radicals.

The process of finding solutions to an equation usually consists in replacing the equation with an equivalent one. Replacing an equation with an equivalent is based on the application of four axioms:

1. If equal values are increased by the same number, then the results will be equal.

2. If you subtract the same number from equal values, then the results will be equal.

3. If equal values are multiplied by the same number, then the results will be equal.

4. If equal values are divided by the same number, then the results will be equal.

Since the left side of the equation P (x) = 0 is a polynomial nth degree, it is useful to recall the following statements:

Statements about the 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. An odd degree polynomial has at least one real root.

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

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

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

6. For a polynomial of degree 3

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

Р 3 (x) = а (х - α) (х - β) (х - γ), or decomposes into the product of a binomial and a square trinomial Р 3 (x) = а (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) = g (x) q (x). For dividing polynomials, the “corner division” rule is applied.

9. For the polynomial P (x) to be divisible by the binomial (x - c), it is necessary and sufficient that c is 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) = a 0 x n + a 1 x n - 1 + ... + a n, then the following equalities hold:

x 1 + x 2 + ... + x n = -a 1 / a 0,

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

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

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

Solution examples

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

Solution. 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's find Р (1/3) = 0. Therefore, the remainder is 0 and the number 1/3 is the root of the polynomial.

Answer: R = 0.

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

Solution:

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

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

X 2 - 2x

Answer: R = 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 that to solve the equation f (x) = 0, a new variable (substitution) t = xn or t = 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) = t 1, q (x) = t 2, ..., q (x) = t n is obtained, from which the roots of the original equation are found.

Example;(x 2 + x + 1) 2 - 3x 2 - 3x - 1 = 0.

Solution: (x 2 + x + 1) 2 - 3x 2 - 3x - 1 = 0.

(x 2 + x + 1) 2 - 3 (x 2 + x + 1) + 3 - 1 = 0.

Replacement (x 2 + x + 1) = t.

t 2 - 3t + 2 = 0.

t 1 = 2, t 2 = 1. Reverse replacement:

x 2 + x + 1 = 2 or x 2 + x + 1 = 1;

x 2 + x - 1 = 0 or x 2 + x = 0;

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

The method of introducing a new variable finds application in solving returnable equations, that is, equations of the form a 0 x n + a 1 x n - 1 + .. + a n - 1 x + a n = 0, in which the coefficients of the terms of the equation, equally spaced from the beginning and end, are equal.

2. Factorization by grouping and reduced multiplication formulas

The foundation this method 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: x 4 - 3x 2 + 4x - 3 = 0.

Solution. Imagine - 3x 2 = -2x 2 - x 2 and group:

(x 4 - 2x 2) - (x 2 - 4x + 3) = 0.

(x 4 - 2x 2 +1 - 1) - (x 2 - 4x + 3 + 1 - 1) = 0.

(x 2 - 1) 2 - 1 - (x - 2) 2 + 1 = 0.

(x 2 - 1) 2 - (x - 2) 2 = 0.

(x 2 - 1 - x + 2) (x 2 - 1 + x - 2) = 0.

(x 2 - x + 1) (x 2 + x - 3) = 0.

x 2 - x + 1 = 0 or x 2 + x - 3 = 0.

There are no roots in the first equation, from the second: x 1, 2 = (-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: x 3 + 4x 2 + 5x + 2 = 0.

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

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

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

x 3 + 4x 2 + 5x + 2 = x 3 + (b - a) x 2 + (c - ab) x - ac.

Having solved the system:

get

x 3 + 4x 2 + 5x + 2 = (x + 1) (x 2 + 3x + 2).

The roots of the equation (x + 1) (x 2 + 3x + 2) = 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 the free term.

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 be an integer divisor of the free term a 0, and q is a natural divisor of the leading coefficient.

Example: 6x 3 + 7x 2 - 9x + 2 = 0.

Solution:

2: p = ± 1, ± 2

6: q = 1, 2, 3, 6.

Therefore, p / q = ± 1, ± 2, ± 1/2, ± 1/3, ± 2/3, ± 1/6.

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

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

5. Graphic method.

This method consists in plotting graphs and using the properties of functions.

Example: x 5 + x - 2 = 0

Let's represent the equation in the form x 5 = - x + 2. The function y = x 5 is increasing, and the function y = - x + 2 is decreasing. Hence, the equation x 5 + x - 2 = 0 has a single root -1.

6. Multiplication of an equation by a function.

Sometimes the solution of an algebraic equation becomes much easier if you multiply both sides by some function - a polynomial in the unknown. It should be remembered that unnecessary roots may appear - the roots of the polynomial by which the equation was multiplied. Therefore, one must either multiply by a polynomial that has no roots and get an equivalent equation, or multiply by a polynomial that has roots, and then each of these roots must be substituted into the original equation and establish whether this number is its root.

Example. Solve the equation:

X 8 - X 6 + X 4 - X 2 + 1 = 0. (1)

Solution: Multiplying both sides of the equation by the polynomial X 2 + 1, which has no roots, we get the equation:

(X 2 +1) (X 8 - X 6 + X 4 - X 2 + 1) = 0 (2)
equivalent to equation (1). Equation (2) can be written as:

X 10 + 1 = 0 (3)
It is clear that equation (3) has no real roots, therefore equation (1) does not have them.

Answer: no solutions.

In addition to the above methods for solving equations of higher degrees, there are others. For example, selection of a complete square, Horner's scheme, representation of a fraction in the form of two fractions. From common methods solutions of equations of higher degrees, which are most often found, use: the method of decomposing the left side of the equation into factors;

method of changing a variable (method of introducing a new variable); graphical way. We introduce these methods to 9th grade students when studying the topic "The whole equation and its roots." In the textbook Algebra 9 (authors Makarychev Yu.N., Mindyuk N.G and others) recent years editions in sufficient detail consider the basic methods of solving equations of higher degrees. In addition, in the section "For those who want to know more", in my opinion, the material on the application of theorems about the root of a polynomial and integral roots of an entire equation in solving equations of higher degrees is available. Well-prepared students study this material with interest and then present the solved equations to classmates.

Almost everything that surrounds us is connected to one degree or another with mathematics. And achievements in physics, technology, information technology only confirm this. And what is very important - the solution of many practical problems comes down to solving various types of equations that must be learned to solve.

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

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

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

Factorization. The equation f (x) g (x) h (x) = 0 can be replaced by a set of equations f (x) = 0; g (x) = 0; h (x) = 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 = 0 Representing the term 7 x as x + 6 x, we get sequentially: x³ - x - 6 x + 6 = 0 x (x² - 1) - 6 (x - 1) = 0 x (x - 1) (x + 1) - 6 (x - 1) = 0 (x - 1) (x² + x - 6) = 0 Now the problem is reduced to solving the set of equations x - 1 = 0; x² + x - 6 = 0. Answer: 1, 2, - 3.

Introducing a new variable. If the equation y (x) = 0 can be transformed to the form p (g (x)) = 0, then you need to introduce a new variable u = g (x), solve the equation p (u) = 0, and then solve the set of equations g ( x) = u 1; g (x) = u 2; ...; g (x) = un, where u 1, u 2,…, un are the roots of the equation p (u) = 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 this equation, dividing by x² we obtain

Let's introduce a new variable Then We get a quadratic equation So the root y 1 = - 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 = 0 This equation can be solved as homogeneous. Divide both sides of the equation by (x² - 7 x +12) ² (clearly, the values of x such that x² - 7 x + 12 = 0 are not solutions). Now let us denote We have Hence the Answer:

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

Solve the Equation It is fairly obvious that x = 2 is the root of the equation. Let us prove that this is the only root. We transform the equation to the form Note 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 = 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 = 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 = (x - 3) (x²– 2 x - 7). Answer:

Solve the equation 2 x³ - 5 x² - x + 1 = 0 According to Theorem 1, the integer roots of the equation can only be numbers ± 1. The check 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 = 0 Making the substitution 2 x = t, we get t³ - 5 t² - 2 t + 4 = 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 = - 1 is suitable. Therefore, by the corollary of the Bezout theorem, the polynomial 2 x³ - 5 x² - x + 1 is divisible by (x + 0, 5 ): 2 x³ - 5 x² - x + 1 = (x + 0, 5) (2 x² - 6 x + 2) Solving the quadratic equation 2 x² - 6 x + 2 = 0, we find the remaining roots: Answer:

Solve the equation 6 x³ + x² - 11 x - 6 = 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 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 = 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)) = h (g (x)) by the equation f (x) = g (x). 4. Factorization. 5. Selection of roots. 6 Functional - graphical method. 7. Application of Vieta formulas. 8. Selection of roots. 9. Replacing the equation h (f (x)) = h (g (x)) by the equation f (x) = 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. Introducing a new variable. 18. Factorization.

1. Indication. Write the equation as 4 (x² + 17 x + 60) (x + 16 x + 60) = 3 x², Divide both sides by x². Enter the variable Answer: x 1 = - 8; x 2 = - 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) = (y - 2) (y² - 3 y - eight). 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). Galitskiy 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. "Optional 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 of Mathematics" (M.: Aslan, 1994). Muravin G. K. "Equations, inequalities and their systems" (Mathematics, supplement to the newspaper "September 1", No. 2, 3, 2003). Kolyagin Yu. M. "Polynomials and equations of higher degrees" (Mathematics, supplement to the newspaper "September 1st", No. 3, 2005).

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

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

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

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

Statements about the 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. An odd degree polynomial has at least one real root.

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

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

6. For a polynomial of degree 3

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

Р 3 (x) = а (х - α) (х - β) (х - γ), or decomposes into the product of a binomial and a square trinomial Р 3 (x) = а (х - α) (х 2 + βх + γ ).

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

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) = a 0 x n + a 1 x n - 1 + ... + a n, then the following equalities hold:

x 1 + x 2 + ... + x n = -a 1 / a 0,

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

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

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

Solution examples

Example 1.

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

Solution.

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's find Р (1/3) = 0. Therefore, the remainder is 0 and the number 1/3 is the root of the polynomial.

Answer: R = 0.

Example 2.

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

Solution:

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

2х 3 + 4 x 2 2x 2 - x

X 2 - 2 x

Answer: R = 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 on the example of bi quadratic equations... It consists in the fact that to solve the equation f (x) = 0, a new variable (substitution) t = x n or t = 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) = t 1, q (x) = t 2, ..., q (x) = t n is obtained, from which the roots of the original equation are found.

Example 1.

(x 2 + x + 1) 2 - 3x 2 - 3x - 1 = 0.

Solution:

(x 2 + x + 1) 2 - 3 (x 2 + x) - 1 = 0.

(x 2 + x + 1) 2 - 3 (x 2 + x + 1) + 3 - 1 = 0.

Replacement (x 2 + x + 1) = t.

t 2 - 3t + 2 = 0.

t 1 = 2, t 2 = 1. Reverse replacement:

x 2 + x + 1 = 2 or x 2 + x + 1 = 1;

x 2 + x - 1 = 0 or x 2 + x = 0;

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

2. Factorization 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 = 0.

Solution.

Imagine - 3x 2 = -2x 2 - x 2 and group:

(x 4 - 2x 2) - (x 2 - 4x + 3) = 0.

(x 4 - 2x 2 +1 - 1) - (x 2 - 4x + 3 + 1 - 1) = 0.

(x 2 - 1) 2 - 1 - (x - 2) 2 + 1 = 0.

(x 2 - 1) 2 - (x - 2) 2 = 0.

(x 2 - 1 - x + 2) (x 2 - 1 + x - 2) = 0.

(x 2 - x + 1) (x 2 + x - 3) = 0.

x 2 - x + 1 = 0 or x 2 + x - 3 = 0.

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

3. Factoring by the method of undefined coefficients

Example 1.

x 3 + 4x 2 + 5x + 2 = 0.

Solution.

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

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

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

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

Having solved the system:

(b - a = 4,
(c - ab = 5,
(-ac = 2,

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

x 3 + 4x 2 + 5x + 2 = (x + 1) (x 2 + 3x + 2).

The roots of the equation (x + 1) (x 2 + 3x + 2) = 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.

Example 1.

6x 3 + 7x 2 - 9x + 2 = 0.

Solution:

6: q = 1, 2, 3, 6.

Therefore, p / q = ± 1, ± 2, ± 1/2, ± 1/3, ± 2/3, ± 1/6.

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

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

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When solving algebraic equations, it is often necessary to factor a polynomial. Factoring a polynomial means representing it as a product of two or more polynomials. We use some methods of decomposition of polynomials quite often: removal of a common factor, application of formulas for reduced multiplication, selection of a complete square, grouping. Let's consider some more methods.

Sometimes the following statements are useful when factoring a polynomial:

1) if a polynomial with integer coefficients has a rational root (where is an irreducible fraction, then is the divisor of the free term and the divisor of the leading coefficient:

2) If in some way to choose a root of a polynomial of degree, then the polynomial can be represented in the form where is a polynomial of degree

The polynomial can be found either by dividing the polynomial by the binomial "column", or by the corresponding grouping of the terms of the polynomial and by extracting a factor from them, or by the method of undefined coefficients.

Example. Factor polynomial

Solution. Since the coefficient at x4 is 1, the rational roots of this polynomial, exist, are divisors of 6, that is, they can be integers ± 1, ± 2, ± 3, ± 6. We denote this polynomial by P4 (x). Since Р Р4 (1) = 4 and Р4 (-4) = 23, the numbers 1 and -1 are not the roots of the polynomial PA (x). Since P4 (2) = 0, then x = 2 is the root of the polynomial P4 (x), and, therefore, this polynomial is divisible by the binomial x - 2. Therefore, x4 -5x3 + 7x2 -5x +6 x-2 x4 -2x3 x3 -3x2 + x-3

3x3 + 7x2 -5x +6

3x3 + 6x2 x2 - 5x + 6 x2- 2x

Therefore, P4 (x) = (x - 2) (x3 - 3x2 + x - 3). Since xz - Zx2 + x - 3 = x2 (x - 3) + (x - 3) = (x - 3) (x2 + 1), then x4 - 5x3 + 7x2 - 5x + 6 = (x - 2) (x - 3) (x2 + 1).

Parameter introduction method

Sometimes, when factoring a polynomial into factors, the method of introducing a parameter helps. The essence of this method is illustrated by the following example.

Example. x3 - (√3 + 1) x2 + 3.

Solution. Consider a polynomial with parameter a: x3 - (a + 1) x2 + a2, which for a = √3 turns into a given polynomial. We write this polynomial as a square trinomial with respect to a: a - ax2 + (x3 - x2).

Since the roots of this square trinomial with respect to a are a1 = x and a2 = x2 - x, the equality a2 - ax2 + (xs - x2) = (a - x) (a - x2 + x) is true. Consequently, the polynomial x3 - (√3 + 1) x2 + 3 is decomposed into factors √3 - x and √3 - x2 + x, i.e.

x3 - (√3 + 1) x2 + 3 = (x-√3) (x2-x-√3).

Method of introducing a new unknown

In some cases, by replacing the expression f (x), which appears in the polynomial Pn (x), through y, you can get a polynomial with respect to y, which can already be easily factorized. Then, after replacing y by f (x), we obtain a factorization of the polynomial Pn (x).

Example. Factor the polynomial x (x + 1) (x + 2) (x + 3) -15.

Solution. We transform this polynomial as follows: x (x + 1) (x + 2) (x + 3) -15 = [x (x + 3)] [(x + 1) (x + 2)] - 15 = (x2 + 3x) (x2 + 3x + 2) - 15.

Let us denote x2 + 3x by y. Then we have y (y + 2) - 15 = y2 + 2y - 15 = y2 + 2y + 1 - 16 = (y + 1) 2 - 16 = (y + 1 + 4) (y + 1 - 4) = ( y + 5) (y - 3).

Therefore, x (x + 1) (x + 2) (x + 3) - 15 = (x2 + 3x + 5) (x2 + 3x - 3).

Example. Factor the polynomial (x-4) 4+ (x + 2) 4

Solution. Let us denote x- 4 + x + 2 = x - 1 through y.

(x - 4) 4 + (x + 2) 2 = (y - 3) 4 + (y + 3) 4 = y4 - 12y3 + 54y3 - 108y + 81 + y4 + 12y3 + 54y2 + 108y + 81 =

2y4 + 108y2 + 162 = 2 (y4 + 54y2 + 81) = 2 [(y2 + 27) 2 - 648] = 2 (y2 + 27 - √b48) (y2 + 27 + √b48) =

2 ((x-1) 2 + 27-√b48) ((x-1) 2 + 27 + √b48) = 2 (x2-2x + 28- 18√ 2) (x2- 2x + 28 + 18√ 2 ).

Combining different methods

Often, when factoring a polynomial into factors, it is necessary to apply sequentially several of the methods discussed above.

Example. Factor the polynomial x4 - 3x2 + 4x-3.

Solution. Using grouping, we rewrite the polynomial as x4 - 3x2 + 4x - 3 = (x4 - 2x2) - (x2 -4x + 3).

Applying the method of selecting a full square to the first bracket, we have x4 - 3x3 + 4x - 3 = (x4 - 2 · 1 · x2 + 12) - (x2 -4x + 4).

Applying the full square formula, we can now write that x4 - 3x2 + 4x - 3 = (x2 -1) 2 - (x - 2) 2.

Finally, applying the formula for the difference of squares, we obtain that x4 - 3x2 + 4x - 3 = (x2 - 1 + x - 2) (x2 - 1 - x + 2) = (x2 + x-3) (x2 -x + 1 ).

§ 2. Symmetric equations

1. Symmetric equations of the third degree

Equations of the form ax3 + bx2 + bx + a = 0, and ≠ 0 (1) are called symmetric equations of the third degree. Since ax3 + bx2 + bx + a = a (x3 + 1) + bx (x + 1) = (x + 1) (ax2 + (b-a) x + a), then equation (1) is equivalent to a set of equations x + 1 = 0 and ax2 + (b-a) x + a = 0, which is not difficult to solve.

Example 1. Solve the equation

3x3 + 4x2 + 4x + 3 = 0. (2)

Solution. Equation (2) is a symmetric equation of the third degree.

Since 3x3 + 4xg + 4x + 3 = 3 (x3 + 1) + 4x (x + 1) = (x + 1) (3x2 - Zx + 3 + 4x) = (x + 1) (3x2 + x + 3) , then equation (2) is equivalent to a set of equations x + 1 = 0 and 3x3 + x + 3 = 0.

The solution to the first of these equations is x = -1, the second equation has no solutions.

Answer: x = -1.

2. Symmetric equations of the fourth degree

Equation of the form

(3) is called a symmetric equation of the fourth degree.

Since x = 0 is not a root of equation (3), then, dividing both sides of equation (3) by x2, we obtain an equation equivalent to the original (3):

Let us rewrite equation (4) in the form:

In this equation we make a change, then we get a quadratic equation

If equation (5) has 2 roots y1 and y2, then the original equation is equivalent to a set of equations

If equation (5) has one root y0, then the original equation is equivalent to the equation

Finally, if equation (5) has no roots, then the original equation also has no roots.

Example 2. Solve the equation

Solution. This equation is a fourth degree symmetric equation. Since x = 0 is not its root, then, dividing equation (6) by x2, we obtain an equivalent equation:

Grouping the terms, we rewrite equation (7) in the form or in the form

Putting, we obtain an equation having two roots y1 = 2 and y2 = 3. Therefore, the original equation is equivalent to the set of equations

The solution to the first equation of this set is x1 = 1, and the solution to the second is u.

Therefore, the original equation has three roots: x1, x2 and x3.

Answer: x1 = 1 ,.

§3. Algebraic equations

1. Lowering the degree of the equation

Some algebraic equations by replacing some polynomial in them with one letter can be reduced to algebraic equations, the degree of which is less than the degree of the original equation and the solution of which is simpler.

Example 1. Solve the equation

Solution. Denote by, then equation (1) can be rewritten in the form The last equation has roots and Consequently, equation (1) is equivalent to a set of equations and. The solution of the first equation of this set is and the Solution of the second equation is

The solutions to equation (1) are

Example 2. Solve the equation

Solution. Multiplying both sides of the equation by 12 and denoting by,

We obtain the equation Let us rewrite this equation in the form

(3) and denoting by rewrite equation (3) in the form The last equation has roots and Therefore, we obtain that equation (3) is equivalent to a set of two equations and The solution of this set of equations is, i.e., equation (2) is equivalent to a set of equations and ( 4)

The solutions of the set (4) are and, they are the solutions of the equation (2).

2. Equations of the form

The equation

(5) where are these numbers, can be reduced to a biquadratic equation by replacing the unknown, i.e., replacing

Example 3. Solve the equation

Solution. Let us denote by, i.e. that is, we make a change of variables or Then equation (6) can be rewritten in the form or, using the formula, in the form

Since the roots of the quadratic equation are also the solutions to equation (7) are solutions to the set of equations and. This set of equations has two solutions and Consequently, solutions to equation (6) are and

3. Equations of the form

The equation

(8) where the numbers α, β, γ, δ, and Α are such that α

Example 4. Solve the equation

Solution. We make the change of unknowns, i.e., y = x + 3 or x = y - 3. Then equation (9) can be rewritten in the form

(y-2) (y-1) (y + 1) (y + 2) = 10, that is, in the form

(y2- 4) (y2-1) = 10 (10)

The biquadratic equation (10) has two roots. Therefore, equation (9) also has two roots:

4. Equations of the form

Equation, (11)

Where, does not have a root x = 0, therefore, dividing equation (11) by x2, we obtain the equivalent equation

Which, after replacing the unknown, will be rewritten as a quadratic equation, the solution of which is not difficult.

Example 5. Solve the equation

Solution. Since h = 0 is not a root of equation (12), then, dividing it by x2, we obtain the equivalent equation

Making the substitution unknown, we get the equation (y + 1) (y + 2) = 2, which has two roots: y1 = 0 and y1 = -3. Therefore, the original equation (12) is equivalent to the set of equations

This collection has two roots: x1 = -1 and x2 = -2.

Answer: x1 = -1, x2 = -2.

Comment. An equation of the form,

In which, you can always bring to the form (11) and, moreover, considering α> 0 and λ> 0 to the form.

5. Equations of the form

The equation

, (13) where the numbers α, β, γ, δ, and Α are such that αβ = γδ ≠ 0, can be rewritten by multiplying the first bracket with the second, and the third with the fourth, in the form, i.e., equation (13) is now written in the form (11), and its solution can be carried out in the same way as the solution of equation (11).

Example 6. Solve the equation

Solution. Equation (14) has the form (13), so we rewrite it in the form

Since x = 0 is not a solution to this equation, then, dividing both sides of it by x2, we obtain an equivalent original equation. Making a change of variables, we obtain a quadratic equation, the solution of which is and. Consequently, the original equation (14) is equivalent to a set of equations and.

The solution to the first equation of this set is

The second equation of this set of solutions does not have. So, the original equation has roots x1 and x2.

6. Equations of the form

The equation

(15) where the numbers a, b, c, q, A are such that, does not have a root x = 0, therefore, dividing equation (15) by x2. we obtain an equation equivalent to it, which, after replacing the unknown, will be rewritten in the form of a quadratic equation, the solution of which is not difficult.

Example 7. Solution of the equation

Solution. Since x = 0 is not a root of equation (16), then, dividing both of its parts by x2, we obtain the equation

, (17) which is equivalent to equation (16). Making the change unknown, we rewrite Eq. (17) in the form

Quadratic equation (18) has 2 roots: y1 = 1 and y2 = -1. Therefore, equation (17) is equivalent to a set of equations and (19)

The set of equations (19) has 4 roots:,.

They will be the roots of equation (16).

§4. Rational equations

Equations of the form = 0, where H (x) and Q (x) are polynomials, are called rational.

Having found the roots of the equation H (x) = 0, then it is necessary to check which of them are not the roots of the equation Q (x) = 0. These roots and only they will be the solutions of the equation.

Consider some methods for solving an equation of the form = 0.

1. Equations of the form

The equation

(1) under certain conditions on the numbers can be solved as follows. Grouping the terms of equation (1) by two and summing each pair, it is necessary to obtain in the numerator polynomials of the first or zero degree, differing only in numerical factors, and in the denominators - trinomials with the same two terms containing x, then after changing the variables, the equation will either have also form (1), but with a smaller number of terms, or it will be equivalent to a set of two equations, one of which will be of the first degree, and the second will be an equation of the form (1), but with a smaller number of terms.

Example. Solve the equation

Solution. Grouping on the left side of equation (2) the first term with the last, and the second with the penultimate, we rewrite equation (2) in the form

Summing up the terms in each parenthesis, we rewrite Eq. (3) in the form

Since there is not a solution to equation (4), then, dividing this equation by, we obtain the equation

, (5) equivalent to equation (4). We make the change of the unknown, then the equation (5) will be rewritten in the form

Thus, the solution of equation (2) with five terms on the left side is reduced to the solution of equation (6) of the same form, but with three terms on the left side. Summing up all the terms on the left side of equation (6), we rewrite it in the form

There are solutions to the equation and. None of these numbers zero the denominator rational function on the left side of equation (7). Consequently, equation (7) has these two roots, and therefore the original equation (2) is equivalent to the set of equations

The solutions of the first equation of this set are

The solutions of the second equation from this set are

Therefore, the original equation has roots

2. Equations of the form

The equation

(8) under certain conditions on the numbers can be solved as follows: it is necessary to select the integer part in each of the fractions of the equation, that is, to replace equation (8) by the equation

Reduce it to form (1) and then solve it in the way described in the previous paragraph.

Example. Solve the equation

Solution. Let us write equation (9) in the form or in the form

Summing up the terms in parentheses, we rewrite Eq. (10) as

Replacing the unknown, we rewrite Eq. (11) in the form

Summing up the terms on the left side of Eq. (12), we rewrite it in the form

It is easy to see that equation (13) has two roots: and. Therefore, the original equation (9) has four roots:

3) Equations of the form.

An equation of the form (14) under certain conditions for numbers can be solved as follows: expanding (if it is, of course, possible) each of the fractions on the left side of equation (14) in the sum of the simplest fractions

Reduce equation (14) to form (1), then, having carried out a convenient rearrangement of the terms of the resulting equation, solve it by the method described in paragraph 1).

Example. Solve the equation

Solution. Since and, then, multiplying the numerator of each fraction in equation (15) by 2 and noting that equation (15) can be written in the form

Equation (16) has the form (7). Having rearranged the terms in this equation, we rewrite it in the form or in the form

Equation (17) is equivalent to a set of equations and

To solve the second equation of the set (18), we replace the unknown.Then it will be rewritten in the form or in the form

Summing up all the terms on the left side of equation (19), rewrite it as

Since the equation has no roots, equation (20) also has no roots.

The first equation of the set (18) has a single root.Since this root is included in the GDV of the second equation of the set (18), it is the only root of the set (18), and hence the original equation.

4. Equations of the form

The equation

(21) under certain conditions on the numbers and A after the representation of each term on the left in the form can be reduced to the form (1).

Example. Solve the equation

Solution. Let us rewrite equation (22) in the form or in the form

Thus, equation (23) is reduced to form (1). Now, grouping the first term with the last, and the second with the third, we rewrite equation (23) in the form

This equation is equivalent to a set of equations and. (24)

The last equation of the set (24) can be rewritten as

There are solutions to this equation and, since it is included in the ODZ of the second equation of the set (30), the set (24) has three roots: They are all solutions to the original equation.

5. Equations of the form.

Equation of the form (25)

Under certain conditions on the numbers, replacing the unknown can be reduced to an equation of the form

Example. Solve the equation

Solution. Since it is not a solution to equation (26), then dividing the numerator and denominator of each fraction on the left side by, we rewrite it in the form

Changing variables, we rewrite Eq. (27) in the form

Solving equation (28) is and. Therefore, equation (27) is equivalent to a set of equations and. (29)

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Introduction

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

The solution of equations of the n-th degree is an important task for modern mathematics. Interest in them is quite great, 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: lack of skills in solving equations of higher degrees different ways among students, it 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.

Purpose of work: the study known methods solving equations of higher degrees and identifying the most accessible ones 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 related to 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 methods considered.

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 n-th 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 significance research 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 development study guide for students on this topic.

1 HIGHER DEGREES EQUATIONS

1.1 The concept of an equation of the n-th degree

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

a 0 xⁿ + a 1 x n -1 + a 2 xⁿ - ² + ... + a n -1 x + a n = 0, where the coefficients a 0, a 1, a 2…, a n -1, a n- any real numbers, and , a 0 ≠ 0 .

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

Definition 2. Solutions or roots for a given equation are all values of the variable NS, which turn this equation into a true numerical equality or, for which the polynomial a 0 xⁿ + a 1 x n -1 + a 2 xⁿ - ² + ... + a n -1 x + a n vanishes. This value of the variable NS is 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 0 = 1, then such an equation is called the reduced integer 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 solutions of equations of higher degrees

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 = 3 and n = 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.

The study revealed that modern science there are many ways to solve equations of the n-th degree.

The search for methods for solving equations of higher degrees that cannot be solved by the methods considered in the school curriculum resulted in methods based on the application of Vieta's theorem (for equations of degree n> 2), Bezout's theorems, Horner's schemes, as well as the Cardano-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 Solving 3rd degree equations. Formula D. Cardano

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

(a + b) 3 = a 3 + 3a 2 b + 3ab 2 + b 3 = a 3 + b 3 + 3ab (a + b), replace ( a + b)on 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= 15 and q= 124, then using the Cardano formula we calculate the root of the equation

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

But the one who masters this formula can use it when solving equations of the third degree 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:

factor 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 true for polynomials of higher degrees.

Let's solve the cubic equation:

Output: this way 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 18th century J. Bezout.

Theorem. If the equation a 0 xⁿ + a 1 x n -1 + a 2 xⁿ - ² + ... + a n -1 x + a n = 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 polynomial nth degree, then 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 = a... (letter a can denote any real or imaginary number, i.e. any complex number).

Proof: let be f (x) denotes an arbitrary nth degree polynomial with respect to the variable x, and let, when dividing by the 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 x and 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 obtain the identity: f (x) = (x-a) q (x) + R.

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

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

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

Note that Bezout's theorem is important not so much in itself as in its consequences. (Annex 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 + 4NS 2 + x - 6 = 0 .

Solution: find the divisors of the free term ± 1 ; ± 2; ± 3; ± 6. Let us calculate the values at x = 1, 1 3 + 41 2 + 1- 6 = 0. Divide the left side of the equation by ( NS- 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 equations, according to Horner's scheme, you can more easily calculate their values. For this, it is necessary to substitute the value of the variable into the polynomial Pn (x) = a 0 xn + a 1 x n-1 + a 2 xⁿ - ² +… ++ a n -1 x + a n. (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 = a 0 , b 1 = α b 0 + a 1 , b 2 = α b 1 + a 2 …, bn -1 =

= α bn -2 + a n -1 = α bn -1 + a n .

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

	a 0	a 1	a 2 ,
	b 0 = a 0	b 1 = α b 0 + a 1	b 2 = α b 1 + a 2		r = α b n-1 + a n

Insofar as r = Pn (α), then α is the root of the equation. In order 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, which means that α is a multiple root.

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

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

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 = 0.

Means, NS 3 + 4NS 2 + NS - 6 = (NS - 1) (NS 2 + 5NS + 6) = 0.

Hence: NS- 1 = 0 or NS 2 + 5NS + 6 = 0.

NS = 1, NS 1 = -2; NS 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 steps.

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, however, in order to find the required factorization, it is necessary to solve cubic equations.

The idea is to represent equations in the form A 2 = B 2, where A = x 2 + s,

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

For clarity, consider the equation: Let us isolate the 4th degree, we get: For any d the expression will be a perfect square. Add to both sides of the equation we get

On the left side there is a full square, you can pick up d so 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 afterwards will not be difficult. To choose the right one d it is necessary that the discriminant of the right-hand side (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 = 1

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 NS... Let us denote them by a 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 NS in the left and right sides 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 NS.

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 NS= - 1. Next, we can lower the degree of the equation by dividing it by ( x + 1). It turns out that when the symmetric equation is divided 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)

We solve equation (1), divide by NS 2 (medium) = 0.

Let us group the terms with symmetric

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

Output: given view equations are not often encountered, but if you come across it, then it can be solved easily and simply without resorting to cumbersome calculations.

2.8 Isolation of 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 from.

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 different methods solutions of 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 an equation of the fourth degree 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 the solution of 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 of the polynomials on the left side of the equation;

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

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

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

Demonstration of the results of research work, 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