Equations of higher degrees, methodological development in algebra (grade 10) on the topic. Equations of higher degrees in mathematics Equations with higher degrees

The text of the work is posted without images and formulas.
The full version of the work is available in the "Work Files" tab in PDF format

Introduction

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

Solving equations of the nth degree is an important task for modern mathematics. There is quite a lot of interest in them, since these equations are closely related to the search for the roots of equations that are not covered in the school mathematics curriculum.

Problem: Students’ lack of skills in solving equations of higher degrees in various ways prevents them from successfully preparing for final certification in mathematics and mathematical Olympiads, and training in a specialized mathematics class.

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

Knowledge of the simplest methods of solving equations of the nth degree reduces the time for completing a task, on which the result of the work and the quality of the learning process depend.

Goal of the work: studying known methods for solving equations of higher degrees and identifying the most accessible of them for practical use.

Based on the goal, the following is defined in the work: tasks:

Study literature and Internet resources on this topic;

Get acquainted with historical facts related to this topic;

Describe different ways to solve higher degree equations

compare the degree of complexity of each of them;

Introduce classmates to ways of solving equations of higher degrees;

Create a selection 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- methods for solving equations of higher degrees.

Hypothesis: There is no general method or single algorithm that allows one to find 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 significance research consists of systematizing methods for solving equations of higher degrees and describing their algorithms.

Practical significance- presented material on this topic and development of a teaching aid for students on this topic.

1. EQUATIONS OF HIGHER DEGREES

1.1 Concept of nth degree equation

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 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 nth degree. The coefficients are distinguished by names: a 0 - senior coefficient; a n is a free member.

Definition 2. Solutions or roots for a given equation are all the values ​​of the variable X, 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 goes to zero. This variable value X also called the root of a polynomial. Solving an equation means finding all its roots or establishing that there are none.

If a 0 = 1, then such an equation is called a reduced integer rational equation n th degrees.

For equations of the third and fourth degree, there are Cardano and Ferrari formulas that express the roots of these equations through 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. However, in many special cases this problem is completely solved. Let's look at some of them.

1.2 Historical facts for solving higher degree equations

Already in ancient times, people realized how important it was to learn to solve algebraic equations. About 4000 years ago, Babylonian scientists knew how to solve a quadratic equation and solved systems of two equations, one of which was 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 varied questions of practice and natural science were reduced to them, since the precise language of mathematics allows one to simply express facts and relationships, which, when stated in ordinary language, may seem confusing and complex .

A universal formula for finding the roots of an algebraic equation nth no degree. Many, of course, had the tempting idea of ​​finding, for any degree n, formulas that would express the roots of the equation through its coefficients, that is, solve the equation in radicals.

Only in the 16th century did Italian mathematicians manage to advance further - to find formulas for n = 3 and n = 4. At the same time, Scipio, Dahl, Ferro and his students Fiori and Tartaglia were studying the question of the general solution of equations of the 3rd degree.

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

A complete presentation of issues related to the solution of equations of the 3rd and 4th degrees was given by F. Viet.

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 knows many ways to solve equations of the nth degree.

The result of the search for methods for solving equations of higher degrees that cannot be solved using the methods considered in the school curriculum were 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 and Ferrari formula for solving cubic and quartic equations.

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

2. SOLUTION OF ENTIRE 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. Let us transform the general equation to the form: x 3 +px 2 +qx+r=0. Let's write down the formula for the cube of the sum; Let's 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, let’s write down the sum cube formula again:

(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 we 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 above 3rd degree equation. It 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

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

But anyone who masters this formula can use it when solving third-degree equations on the Unified State Exam.

2.2 Vieta's theorem

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

Consider the equation:

Let's factor the left side of the equation and divide by ≠ 0.

Let's transform the right side of the equation to the form

; It follows from this that we can write the following equalities into the system:

The formulas derived by Viète for quadratic equations and demonstrated by us for equations of the 3rd degree are also true 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, you must have good computational skills.

2.3 Bezout's theorem

This theorem is named after the 18th century French mathematician 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 non-zero and 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 nth degree, the theorem has another interpretation.

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

Proof: let f(x) denotes an arbitrary polynomial of the nth degree with respect to the variable x and let, when divided by a binomial ( x-a) turned out in private q(x), and the remainder R. It's obvious that q(x) there will be some polynomial (n - 1)th degree relative 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 failed. So, R from x does not depend. By definition of division we obtain the identity: f(x)=(x-a) q(x)+R.

The equality is true for any value of x, which means it is also true 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) stands for value q(x) at x=a. Remainder R remained the same as it was before, because R from x does not depend. Work ( x-a) q(a) = 0, since the factor ( x-a) = 0, and the multiplier q(a) there is a certain number. Therefore, from the equality we get: f(a)= R, etc.

Example 1. Find the remainder of a polynomial x 3 - 3x 2 + 6x- 5 per 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 for its consequences. (Annex 1)

Let us dwell on the consideration of some techniques for 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 down the product of the divisor and the quotient on the left side of the equation;

Solve the resulting equation.

Let's look at the example of solving the equation x 3 + 4X 2 + x - 6 = 0 .

Solution: find the divisors of the free term ±1 ; ± 2; ± 3; ± 6. Let's calculate the values ​​at x= 1, 1 3 + 41 2 + 1- 6=0. Divide the left side of the equation by ( X- 1). Let’s do the division using a “corner” and get:

Conclusion: Bezout’s theorem is one of the methods that we consider in our work, studied in the program of elective classes. It is difficult to understand, because in order to master it, you need to know all the consequences from it, but at the same time, Bezout’s theorem is one of the main assistants for students on the Unified State Exam.

2.4 Horner scheme

To divide a polynomial by a binomial x-α you can use a special simple technique invented by English mathematicians of the 17th century, later called Horner’s scheme. In addition to finding the roots of equations, using Horner's scheme you can more simply calculate their values. To do this, you need 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 dividing 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 through the coefficients of the polynomial Pn( x) and 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 using Horner’s scheme are presented in 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

Because the 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 under bn -1 the result is 0 again, which means α is a multiple root.

Let's look at an example: solve the equation X 3 + 4X 2 + x - 6 = 0.

Let us apply to the left side of the equation the factorization of the polynomial on 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 coefficients of the quotient are the numbers 1, 5, 6, and the remainder r = 0.

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

From here: X- 1 = 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 4th degree equations. Ferrari method

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

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

Stage II: the resulting equations are solved using factorization, but in order to find the required factorization, cubic equations have to be solved.

The idea is to represent the 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: Isolating 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 complete square, you can pick up d, so that the right side of (2) also becomes a complete square. Let's imagine that we have achieved this. Then our equation looks like this:

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

So to find d, we need to solve this 3rd degree equation. This auxiliary equation is called resolvent.

We easily find the whole root of the resolvent: d = 1

Substituting the equation into (1) we get

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

2.6 Method of uncertain coefficients

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

The essence of the method of indefinite 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 powers of the variable.

Example: solve the equation:

Suppose that the left side of our equation can be decomposed into two square trinomials with integer coefficients such that the identical equality is true

Obviously, the coefficients in front of them must be equal to 1, and the free terms must be equal to one + 1, the other - 1.

The coefficients facing the X. 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 coefficients at the same degrees X on 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 polynomial of the fourth degree in the equation can be decomposed into the product of two polynomials of the second degree; two polynomials are identically equal if and only if their coefficients are equal for the same powers X.

2.7 Symmetric equations

Definition. An equation of the form is called symmetric if the first coefficients on the left of 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 a root X= - 1. Next we can lower the degree of the equation by dividing it by ( x+ 1). It turns out that when dividing a symmetric equation by ( x+ 1) a symmetric equation of even degree is obtained. 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)

Let's solve equation (1), divide by X 2 (to medium degree) = 0.

Let us group terms with symmetric

) + 3(x+ . Let's 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 return to replacement x+ = and x+ = 3. We obtain the equations and The first has no solution, and the second has two roots. Answer:.

Conclusion: this type of equation is 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 a full degree

Consider the equation.

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

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

Where is the only root?

RESEARCH RESULTS

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

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

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

− L. Ferrari’s method allows one to reduce the solution to a fourth-degree equation to a cubic one;

− Bezout’s theorem can be used both for cubic equations and for equations of the fourth degree; it is more understandable and visual when applied to solving equations;

Horner's scheme helps to significantly reduce and simplify calculations in solving equations. In addition to finding the roots, using Horner's scheme you can more simply calculate the values ​​of the polynomials on the left side of the equation;

Of particular interest were the solutions of equations by the method of indefinite coefficients and the solution of symmetric equations.

During the research work, it was found that students become familiar with the simplest methods of solving equations of the highest degree in elective mathematics classes, starting in the 9th or 10th grade, as well as in special courses at visiting mathematics schools. This fact was established as a result of a survey of mathematics teachers at MBOU “Secondary School No. 9” and students who showed increased interest in the subject “mathematics”.

The most popular methods for solving equations of higher degrees, which are encountered when solving olympiads, competitive problems and as a result of students preparing 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. methods for solving equations not taught in the school mathematics curriculum interested my classmates.

Conclusion

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

To use presentation previews, create a Google account and log in to it: https://accounts.google.com

Slide captions:

Equations of higher degrees (roots of a polynomial in one variable).

Lecture plan. No. 1. Equations of higher degrees in the school mathematics course. No. 2. Standard form of a polynomial. No. 3. Whole roots of a polynomial. Horner's scheme. No. 4. Fractional roots of a polynomial. No. 5. Equations of the form: (x + a)(x + b)(x + c) ... = A No. 6. Reciprocal equations. No. 7. Homogeneous equations. No. 8. Method of undetermined coefficients. No. 9. Functional - graphic method. No. 10. Vieta formulas for equations of higher degrees. No. 11. Non-standard methods for solving equations of higher degrees.

Equations of higher degrees in the school mathematics course. 7th grade. Standard form of a polynomial. Actions with polynomials. Factoring a polynomial. In a regular class 42 hours, in a special class 56 hours. 8 special class. Integer roots of a polynomial, division of polynomials, reciprocal equations, difference and sum of the nth powers of a binomial, method of indefinite coefficients. Yu.N. Makarychev “Additional chapters to the school algebra course for grade 8”, M.L. Galitsky Collection of problems in algebra for grades 8 – 9.” 9 special class. Rational roots of a polynomial. Generalized reciprocal equations. Vieta formulas for equations of higher degrees. N.Ya. Vilenkin “Algebra 9th grade with in-depth study. 11 special class. Identity of polynomials. Polynomial in several variables. Functional - graphical method for solving equations of higher degrees.

Standard form of a polynomial. Polynomial P(x) = a ⁿ x ⁿ + a p-1 x p-1 + … + a₂x ² + a₁x + a₀. Called a polynomial of standard form. a p x ⁿ is the leading term of the polynomial and p is the coefficient of the leading term of the polynomial. When a n = 1, P(x) is called a reduced polynomial. and ₀ is the free term of the polynomial P(x). n is the degree of the polynomial.

Whole roots of a polynomial. Horner's scheme. Theorem No. 1. If an integer a is the root of the polynomial P(x), then a is a divisor of the free term P(x). Example No. 1. Solve the equation. X⁴ + 2x³ = 11x² – 4x – 4 Let’s bring the equation to standard form. X⁴ + 2x³ - 11x² + 4x + 4 = 0. We have the polynomial P(x) = x ⁴ + 2x³ - 11x² + 4x + 4 Divisors of the free term: ± 1, ± 2, ±4. x = 1 root of the equation because P(1) = 0, x = 2 is the root of the equation because P(2) = 0 Bezout's theorem. The remainder of dividing the polynomial P(x) by the binomial (x – a) is equal to P(a). Consequence. If a is the root of the polynomial P(x), then P(x) is divided by (x – a). In our equation, P(x) is divided by (x – 1) and by (x – 2), and therefore by (x – 1) (x – 2). When dividing P(x) by (x² - 3x + 2), the quotient yields the trinomial x² + 5x + 2 = 0, which has roots x = (-5 ± √17)/2

Fractional roots of a polynomial. Theorem No. 2. If p / g is the root of the polynomial P(x), then p is the divisor of the free term, g is the divisor of the coefficient of the leading term P(x). Example #2: Solve the equation. 6x³ - 11x² - 2x + 8 = 0. Divisors of the free term: ±1, ±2, ±4, ±8. None of these numbers satisfy the equation. There are no whole roots. Natural divisors of the coefficient of the leading term P(x): 1, 2, 3, 6. Possible fractional roots of the equation: ±2/3, ±4/3, ±8/3. By checking we are convinced that P(4/3) = 0. X = 4/3 is the root of the equation. Using Horner’s scheme, we divide P(x) by (x – 4/3).

Examples for independent solutions. Solve the equations: 9x³ - 18x = x – 2, x³ - x² = x – 1, x³ - 3x² -3x + 1 = 0, X⁴ - 2x³ + 2x – 1 = 0, X⁴ - 3x² + 2 = 0 , x ⁵ + 5x³ - 6x² = 0, x ³ + 4x² + 5x + 2 = 0, X⁴ + 4x³ - x ² - 16x – 12 = 0 4x³ + x ² - x + 5 = 0 3x⁴ + 5x³ - 9x² - 9x + 10 = 0. Answers: 1) ±1/3; 2 2) ±1, 3) -1; 2 ±√3, 4) ±1, 5) ± 1; ±√2, 6) 0; 1 7) -2; -1, 8) -3; -1; ±2, 9) – 5/4 10) -2; - 5/3; 1.

Equations of the form (x + a)(x + b)(x + c)(x + d)… = A. Example No. 3. Solve the equation (x + 1)(x + 2)(x + 3)(x + 4) =24. a = 1, b = 2, c = 3, d = 4 a + d = b + c. Multiply the first bracket with the fourth and the second with the third. (x + 1)(x + 4)(x + 20(x + 3) = 24. (x² + 5x + 4)(x² + 5x + 6) = 24. Let x² + 5x + 4 = y , then y(y + 2) = 24, y² + 2y – 24 = 0 y₁ = - 6, y₂ = 4. x ² + 5x + 4 = -6 or x ² + 5x + 4 = 4. x ² + 5x + 10 = 0, D

Examples for independent solutions. (x + 1)(x + 3)(x + 5)(x + 7) = -15, x (x + 4)(x + 5)(x + 9) + 96 = 0, x (x + 3 )(x + 5)(x + 8) + 56 = 0, (x – 4)(x – 3)(x – 2)(x – 1) = 24, (x – 3)(x -4)( x – 5)(x – 6) = 1680, (x² - 5x)(x + 3)(x – 8) + 108 = 0, (x + 4)² (x + 10)(x – 2) + 243 = 0 (x² + 3x + 2)(x² + 9x + 20) = 4, Note: x + 3x + 2 = (x + 1)(x + 2), x² + 9x + 20 = (x + 4)(x + 5) Answers: 1) -4 ±√6; - 6; - 2. 6) - 1; 6; (5± √97)/2 7) -7; -1; -4 ±√3.

Reciprocal equations. Definition No. 1. An equation of the form: ax⁴ + inx ³ + cx ² + inx + a = 0 is called a reciprocal equation of the fourth degree. Definition No. 2. An equation of the form: ax⁴ + inx ³ + cx ² + kinx + k² a = 0 is called a generalized reciprocal equation of the fourth degree. k² a: a = k²; kv: v = k. Example No. 6. Solve the equation x ⁴ - 7x³ + 14x² - 7x + 1 = 0. Divide both sides of the equation by x². x² - 7x + 14 – 7/ x + 1/ x² = 0, (x² + 1/ x²) – 7(x + 1/ x) + 14 = 0. Let x + 1/ x = y. We square both sides of the equation. x² + 2 + 1/ x² = y², x² + 1/ x² = y² - 2. We obtain the quadratic equation y² - 7y + 12 = 0, y₁ = 3, y₂ = 4. x + 1/ x =3 or x + 1/ x = 4. We get two equations: x² - 3x + 1 = 0, x² - 4x + 1 = 0. Example No. 7. 3х⁴ - 2х³ - 31х² + 10х + 75 = 0. 75:3 = 25, 10:(– 2) = -5, (-5)² = 25. The condition of the generalized reciprocal equation is satisfied to = -5. The solution is similar to example No. 6. Divide both sides of the equation by x². 3x⁴ - 2x – 31 + 10/ x + 75/ x² = 0, 3(x⁴ + 25/ x²) – 2(x – 5/ x) – 31 = 0. Let x – 5/ x = y, we square both sides of the equality x² - 10 + 25/ x² = y², x² + 25/ x² = y² + 10. We have a quadratic equation 3y² - 2y – 1 = 0, y₁ = 1, y₂ = - 1/ 3. x – 5/ x = 1 or x – 5/ x = -1/3. We get two equations: x² - x – 5 = 0 and 3x² + x – 15 = 0

Examples for independent solutions. 1. 78x⁴ - 133x³ + 78x² - 133x + 78 = 0. 2. x ⁴ - 5x³ + 10x² - 10x + 4 = 0. 3. x ⁴ - x³ - 10x² + 2x + 4 = 0. 4. 6x⁴ + 5x³ - 38x² -10x + 24 = 0.5. x ⁴ + 2x³ - 11x² + 4x + 4 = 0. 6. x ⁴ - 5x³ + 10x² -10x + 4 = 0. Answers: 1) 2/3; 3/2, 2) 1;2 3) -1 ±√3; (3±√17)/2, 4) -1±√3; (7±√337)/12 5) 1; 2; (-5± √17)/2, 6) 1; 2.

Homogeneous equations. Definition. An equation of the form a₀ u³ + a₁ u² v + a₂ uv² + a₃ v³ = 0 is called a homogeneous equation of the third degree with respect to u v. Definition. An equation of the form a₀ u⁴ + a₁ u³v + a₂ u²v² + a₃ uv³ + a₄ v⁴ = 0 is called a homogeneous equation of the fourth degree with respect to u v. Example No. 8. Solve the equation (x² - x + 1)³ + 2x⁴(x² - x + 1) – 3x⁶ = 0 A homogeneous third-degree equation for u = x²- x + 1, v = x². Divide both sides of the equation by x ⁶. We first checked that x = 0 is not a root of the equation. (x² - x + 1/ x²)³ + 2(x² - x + 1/ x²) – 3 = 0. (x² - x + 1)/ x²) = y, y³ + 2y – 3 = 0, y = 1 root of the equation. We divide the polynomial P(x) = y³ + 2y – 3 by y – 1 according to Horner’s scheme. In the quotient we get a trinomial that has no roots. Answer: 1.

Examples for independent solutions. 1. 2(x² + 6x + 1)² + 5(X² + 6X + 1)(X² + 1) + 2(X² + 1)² = 0, 2. (X + 5)⁴ - 13X²(X + 5)² + 36X⁴ = 0. 3. 2(X² + X + 1)² - 7(X – 1)² = 13(X³ - 1), 4. 2(X -1)⁴ - 5(X² - 3X + 2)² + 2(x – 2)⁴ = 0. 5. (x² + x + 4)² + 3x(x² + x + 4) + 2x² = 0, Answers: 1) -1; -2±√3, 2) -5/3; -5/4; 5/2; 5 3) -1; -1/2; 2;4 4) ±√2; 3±√2, 5) There are no roots.

Method of undetermined coefficients. Theorem No. 3. Two polynomials P(x) and G(x) are identical if and only if they have the same degree and the coefficients of the same degrees of the variable in both polynomials are equal. Example No. 9. Factor the polynomial y⁴ - 4y³ + 5y² - 4y + 1. y⁴ - 4y³ + 5y² - 4y + 1 = (y² + уу + с)(y² + в₁у + с₁) =у ⁴ + у³(в₁ + в) + у² (с₁ + с + в₁в) + у(с₁ + св₁) + сс ₁. According to Theorem No. 3, we have a system of equations: в₁ + в = -4, с₁ + с + в₁в = 5, сс₁ + св₁ = -4, сс₁ = 1. It is necessary to solve the system in integers. The last equation in integers can have solutions: c = 1, c₁ =1; с = -1, с₁ = -1. Let с = с ₁ = 1, then from the first equation we have в₁ = -4 –в. We substitute into the second equation of the system в² + 4в + 3 = 0, в = -1, в₁ = -3 or в = -3, в₁ = -1. These values ​​fit the third equation of the system. When с = с ₁ = -1 D

Example No. 10. Factor the polynomial y³ - 5y + 2. y³ -5y + 2 = (y + a)(y² + vy + c) = y³ + (a + b)y² + (ab + c)y + ac. We have a system of equations: a + b = 0, ab + c = -5, ac = 2. Possible integer solutions to the third equation: (2; 1), (1; 2), (-2; -1), (-1 ; -2). Let a = -2, c = -1. From the first equation of the system in = 2, which satisfies the second equation. Substituting these values ​​into the desired equality, we get the answer: (y – 2)(y² + 2y – 1). Second way. Y³ - 5y + 2 = y³ -5y + 10 – 8 = (y³ - 8) – 5(y – 2) = (y – 2)(y² + 2y -1).

Examples for independent solutions. Factor the polynomials: 1. y⁴ + 4y³ + 6y² +4y -8, 2. y⁴ - 4y³ + 7y² - 6y + 2, 3. x ⁴ + 324, 4. y⁴ -8y³ + 24y² -32y + 15, 5. Solve the equation using the factorization method: a) x ⁴ -3x² + 2 = 0, b) x ⁵ +5x³ -6x² = 0. Answers: 1) (y² +2y -2)(y² +2y +4), 2) (y – 1)²(y² -2y + 2), 3) (x² -6x + 18)(x² + 6x + 18), 4) (y – 1)(y – 3)(y² - 4у + 5), 5a) ± 1; ±√2, 5b) 0; 1.

Functional - graphical method for solving equations of higher degrees. Example No. 11. Solve the equation x ⁵ + 5x -42 = 0. Function y = x ⁵ increasing, function y = 42 – 5x decreasing (k

Examples for independent solutions. 1. Using the property of monotonicity of a function, prove that the equation has a single root and find this root: a) x ³ = 10 – x, b) x ⁵ + 3x³ - 11√2 – x. Answers: a) 2, b) √2. 2. Solve the equation using the functional-graphical method: a) x = ³ √x, b) l x l = ⁵ √x, c) 2 = 6 – x, d) (1/3) = x +4, d ) (x – 1)² = log₂ x, e) log = (x + ½)², g) 1 - √x = ln x, h) √x – 2 = 9/x. Answers: a) 0; ±1, b) 0; 1, c) 2, d) -1, e) 1; 2, f) ½, g) 1, h) 9.

Vieta formulas for equations of higher degrees. Theorem No. 5 (Vieta's theorem). If the equation a x ⁿ + a x ⁿ + … + a₁x + a₀ has n different real roots x ₁, x ₂, …, x, then they satisfy the equalities: For a quadratic equation ax² + bx + c = o: x ₁ + x ₂ = -в/а, x₁х ₂ = с/а; For the cubic equation a₃x ³ + a₂x ² + a₁x + a₀ = o: x ₁ + x ₂ + x ₃ = -a₂/a₃; x₁х ₂ + x₁х ₃ + x₂х ₃ = а₁/а₃; x₁х₂х ₃ = -а₀/а₃; ..., for an equation of the nth degree: x ₁ + x ₂ + ... x = - a / a, x₁x ₂ + x₁x ₃ + ... + x x = a / a, ... , x₁x ₂·… · x = (- 1 ) ⁿ a₀/a. The converse theorem also holds.

Example No. 13. Write a cubic equation whose roots are inverse to the roots of the equation x ³ - 6x² + 12x – 18 = 0, and the coefficient for x ³ is 2. 1. By Vieta’s theorem for the cubic equation we have: x ₁ + x ₂ + x ₃ = 6, x₁x ₂ + x₁х ₃ + x₂х ₃ = 12, x₁х₂х ₃ = 18. 2. We compose the reciprocals of these roots and apply the inverse Vieta theorem for them. 1/ x ₁ + 1/ x ₂ + 1/ x ₃ = (x₂х ₃ + x₁х ₃ + x₁х ₂)/ x₁х₂х ₃ = 12/18 = 2/3. 1/ x₁х ₂ + 1/ x₁х ₃ + 1/ x₂х ₃ = (x ₃ + x ₂ + x ₁)/ x₁х₂х ₃ = 6/18 = 1/3, 1/ x₁х₂х ₃ = 1/18. We get the equation x³ +2/3x² + 1/3x – 1/18 = 0 2 Answer: 2x³ + 4/3x² + 2/3x -1/9 = 0.

Examples for independent solutions. 1. Write a cubic equation whose roots are the inverse squares of the roots of the equation x ³ - 6x² + 11x – 6 = 0, and the coefficient of x ³ is 8. Answer: 8x³ - 98/9x² + 28/9x -2/9 = 0. Non-standard methods for solving equations of higher degrees. Example No. 12. Solve the equation x ⁴ -8x + 63 = 0. Let's factorize the left side of the equation. Let's select the exact squares. X⁴ - 8x + 63 = (x⁴ + 16x² + 64) – (16x² + 8x + 1) = (x² + 8)² - (4x + 1)² = (x² + 4x + 9)(x² - 4x + 7) = 0. Both discriminants are negative. Answer: no roots.

Example No. 14. Solve the equation 21x³ + x² - 5x – 1 = 0. If the dummy term of the equation is ± 1, then the equation is converted to the reduced equation using the substitution x = 1/y. 21/y³ + 1/y² - 5/y – 1 = 0 · y³, y³ + 5y² -y – 21 = 0. y = -3 root of the equation. (y + 3)(y² + 2y -7) = 0, y = -1 ± 2√2. x ₁ = -1/3, x ₂ = 1/ -1 + 2√2 = (2√2 + 1)/7, X₃ = 1/-1 -2√2 = (1-2√2)/7 . Example No. 15. Solve the equation 4x³-10x² + 14x – 5 = 0. Multiply both sides of the equation by 2. 8x³ -20x² + 28x – 10 = 0, (2x)³ - 5(2x)² + 14 (2x) -10 = 0. Let's introduce a new variable y = 2x, we get the reduced equation y³ - 5y² + 14y -10 = 0, y = 1 root of the equation. (y – 1)(y² - 4y + 10) = 0, D

Example No. 16. Prove that the equation x ⁴ + x ³ + x – 2 = 0 has one positive root. Let f (x) = x ⁴ + x ³ + x – 2, f’ (x) = 4x³ + 3x² + 1 > o for x > o. The function f (x) increases for x > o, and the value of f (o) = -2. It is obvious that the equation has one positive root etc. Example No. 17. Solve the equation 8x(2x² - 1)(8x⁴ - 8x² + 1) = 1. I.F. Sharygin “Optional course in mathematics for grade 11.” M. Enlightenment 1991 p.90. 1. l x l 1 2x² - 1 > 1 and 8x⁴ -8x² + 1 > 1 2. Let’s make the replacement x = cozy, y € (0; n). For other values ​​of y, the values ​​of x are repeated, and the equation has no more than 7 roots. 2х² - 1 = 2 cos²y – 1 = cos2y, 8х⁴ - 8х² + 1 = 2(2х² - 1)² - 1 = 2 cos²2y – 1 = cos4y. 3. The equation takes the form 8 cozycos2ycos4y = 1. Multiply both sides of the equation by siny. 8 sinycosycos2ycos4y = siny. Applying the double angle formula 3 times we get the equation sin8y = siny, sin8y – siny = 0

The end of the solution to example No. 17. We apply the difference of sines formula. 2 sin7y/2 · cos9y/2 = 0 . Considering that y € (0;n), y = 2pk/3, k = 1, 2, 3 or y = n/9 + 2pk/9, k =0, 1, 2, 3. Returning to the variable x, we get answer: Cos2 p/7, cos4 p/7, cos6 p/7, cos p/9, ½, cos5 p/9, cos7 p/9. Examples for independent solutions. Find all values ​​of a for which the equation (x² + x)(x² + 5x + 6) = a has exactly three roots. Answer: 9/16. Directions: Graph the left side of the equation. F max = f(0) = 9/16 . The straight line y = 9/16 intersects the graph of the function at three points. Solve the equation (x² + 2x)² - (x + 1)² = 55. Answer: -4; 2. Solve the equation (x + 3)⁴ + (x + 5)⁴ = 16. Answer: -5; -3. Solve the equation 2(x² + x + 1)² -7(x – 1)² = 13(x³ - 1).Answer: -1; -1/2, 2;4 Find the number of real roots of the equation x ³ - 12x + 10 = 0 on [-3; 3/2]. Instructions: find the derivative and investigate the monot.

Examples for independent solutions (continued). 6. Find the number of real roots of the equation x ⁴ - 2x³ + 3/2 = 0. Answer: 2 7. Let x ₁, x ₂, x ₃ be the roots of the polynomial P(x) = x ³ - 6x² -15x + 1. Find X₁² + x ₂² + x ₃². Answer: 66. Directions: Apply Vieta's theorem. 8. Prove that for a > o and an arbitrary real value in the equation x ³ + ax + b = o has only one real root. Hint: Prove by contradiction. Apply Vieta's theorem. 9. Solve the equation 2(x² + 2)² = 9(x³ + 1). Answer: ½; 1; (3 ± √13)/2. Hint: bring the equation to a homogeneous equation using the equalities X² + 2 = x + 1 + x² - x + 1, x³ + 1 = (x + 1)(x² - x + 1). 10. Solve the system of equations x + y = x², 3y – x = y². Answer: (0;0),(2;2), (√2; 2 - √2), (- √2; 2 + √2). 11. Solve the system: 4y² -3y = 2x –y, 5x² - 3y² = 4x – 2y. Answer: (o;o), (1;1),(297/265; - 27/53).

Test. Option 1. 1. Solve the equation (x² + x) – 8(x² + x) + 12 = 0. 2. Solve the equation (x + 1)(x + 3)(x + 5)(x + 7) = - 15 3. Solve the equation 12x²(x – 3) + 64(x – 3)² = x ⁴. 4. Solve the equation x ⁴ - 4x³ + 5x² - 4x + 1 = 0 5. Solve the system of equations: x ² + 2y² - x + 2y = 6, 1.5x² + 3y² - x + 5y = 12.

Option 2 1. (x² - 4x)² + 7(x² - 4x) + 12 = 0. 2. x (x + 1)(x + 5)(x + 6) = 24. 3. x ⁴ + 18(x + 4)² = 11x²(x + 4). 4. x ⁴ - 5x³ + 6x² - 5x + 1 = 0. 5. x² - 2xy + y² + 2x²y – 9 = 0, x – y – x²y + 3 = 0. 3rd option. 1. (x² + 3x)² - 14(x² + 3x) + 40 = 0 2. (x – 5)(x-3)(x + 3)(x + 1) = - 35. 3. x4 + 8x²(x + 2) = 9(x+ 2)². 4. x ⁴ - 7x³ + 14x² - 7x + 1 = 0. 5. x + y + x² + y² = 18, xy + x² + y² = 19.

Option 4. (x² - 2x)² - 11(x² - 2x) + 24 = o. (x -7)(x-4)(x-2)(x + 1) = -36. X⁴ + 3(x -6)² = 4x²(6 – x). X⁴ - 6x³ + 7x² - 6x + 1 = 0. X² + 3xy + y² = - 1, 2x² - 3xy – 3y² = - 4. Additional task: The remainder of dividing the polynomial P(x) by (x – 1) is 4, the remainder when divided by (x + 1) is equal to 2, and when divided by (x – 2) it is equal to 8. Find the remainder when dividing P(x) by (x³ - 2x² - x + 2).

Answers and instructions: option No. 1 No. 2. No. 3. No. 4. No. 5. 1. - 3; ±2; 1 1;2;3. -5; -4; 1; 2. Homogeneous equation: u = x -3, v = x² -2 ; -1; 3; 4. (2;1); (2/3;4/3). Hint: 1·(-3) + 2· 2 2. -6; -2; -4±√6. -3±2√3; - 4; - 2. 1±√11; 4; - 2. Homogeneous equation: u = x + 4, v = x² 1; 5;3±√13. (2;1); (0;3); (- thirty). Hint: 2 2 + 1. 3. -6; 2; 4; 12 -3; -2; 4; 12 -6; -3; -1; 2. Homogeneous u = x+ 2, v = x² -6; ±3; 2 (2;3), (3;2), (-2 + √7; -2 - √7); (-2 - √7; -2 + √7). Instruction: 2 -1. 4. (3±√5)/2 2±√3 2±√3; (3±√5)/2 (5 ± √21)/2 (1;-2), (-1;2). Hint: 1·4 + 2 .

Solving an additional task. By Bezout’s theorem: P(1) = 4, P(-1) = 2, P(2) = 8. P(x) = G(x) (x³ - 2x² - x + 2) + ax² + inx + With. Substitute 1; - 1; 2. P(1) = G(1) 0 + a + b + c = 4, a + b+ c = 4. P(-1) = a – b + c = 2, P(2) = 4a² + 2b + c = 8. Solving the resulting system of three equations, we obtain: a = b = 1, c = 2. Answer: x² + x + 2.

Criterion No. 1 - 2 points. 1 point – one computational error. No. 2,3,4 – 3 points each. 1 point – led to a quadratic equation. 2 points – one computational error. No. 5. – 4 points. 1 point – expressed one variable in terms of another. 2 points – received one of the solutions. 3 points – one computational error. Additional task: 4 points. 1 point – applied Bezout’s theorem for all four cases. 2 points – compiled a system of equations. 3 points – one computational error.

Let's consider solving 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 greatest of the powers 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, because after the operations of opening the brackets and bringing similar ones, we obtain 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. A polynomial of nth degree has a number of roots not exceeding n, and roots of multiplicity m occur exactly m times.

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

3. If α is the 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.

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

6. For a third degree polynomial

P 3 (x) = ax 3 + bx 2 + cx + d one of two things is possible: either it is decomposed into the 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 expanded into the product of two square trinomials.

8. A polynomial f(x) is divisible by a polynomial g(x) without a remainder if there is a polynomial q(x) such that f(x) = g(x) · q(x). To divide polynomials, the “corner division” rule is used.

9. For the polynomial P(x) to be divisible by a binomial (x – c), it is necessary and sufficient that the number c be the 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 .

Solving Examples

Example 1.

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

Solution.

By corollary to Bezout’s theorem: “The remainder of a polynomial divided by a binomial (x – c) is equal to the value of the polynomial of c.” Let's find P(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 incomplete quotient.

Solution:

2x 3 + 3x 2 – 2x + 3| x + 2

2x 3 + 4 x 2 2x 2 – x

X 2 – 2 x

Answer: R = 3; quotient: 2x 2 – x.

Basic methods for solving higher degree equations

1. Introduction of a new variable

The method of introducing a new variable is already familiar from the example of biquadratic 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 through t, obtaining a new equation r(t). Then solving the equation r(t), the roots are found:

(t 1, t 2, …, t n). After this, 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.

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

t 2 – 3t + 2 = 0.

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

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 abbreviated multiplication formulas

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

Example 1.

x 4 – 3x 2 + 4x – 3 = 0.

Solution.

Let's 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. Factorization by the method of undetermined coefficients

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

Example 1.

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

Solution.

A polynomial of degree 3 can be expanded into the product of linear and quadratic factors.

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 selecting a root using the highest and free coefficient

The method is based on the application of theorems:

1) Every 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 number) to be the 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 be a natural divisor of the leading coefficient.

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 will find other roots using corner division, the method of indefinite coefficients or Horner’s scheme.

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

Still have questions? Don't know how to solve equations?
To get help from a tutor -.
The first lesson is free!

blog.site, when copying material in full or in part, a link to the original source is required.

When solving algebraic equations, you often have to factor a polynomial. To factor a polynomial means to represent it as a product of two or more polynomials. We use some methods of decomposing polynomials quite often: taking a common factor, using abbreviated multiplication formulas, isolating a complete square, grouping. Let's look at 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 you somehow select the root of a polynomial of degree, then the polynomial can be represented in the form where is a polynomial of degree

A polynomial can be found either by dividing the polynomial into a binomial in a “column”, or by appropriately grouping the terms of the polynomial and separating the multiplier from them, or by the method of indefinite coefficients.

Example. Factor a polynomial

Solution. Since the coefficient of x4 is equal to 1, then the rational roots of this polynomial exist and are divisors of the number 6, i.e. they can be integers ±1, ±2, ±3, ±6. Let us denote this polynomial by P4(x). Since P P4 (1) = 4 and P4(-4) = 23, the numbers 1 and -1 are not 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 - 3x2 + x - 3 = x2 (x - 3) + (x - 3) = (x - 3)(x2 + 1), then x4 - 5x3 + 7x2 - 5x + 6 = (x - 2) (x - 3)(x2 + 1).

Parameter input method

Sometimes when factoring a polynomial, the method of introducing a parameter helps. We will explain the essence of this method using the following example.

Example. x3 –(√3 + 1) x2 + 3.

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

Since the roots of this trinomial squared with respect to a are a1 = x and a2 = x2 - x, then 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) included in the polynomial Pn(x), through y one can obtain a polynomial with respect to y, which can easily be factorized. Then, after replacing y with f(x), we obtain a factorization of the polynomial Pn(x).

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

Solution. Let's 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's 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's denote x- 4+x+2 = x - 1 by 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[(yg + 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, it is necessary to apply several of the methods discussed above in succession.

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

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

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

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

Finally, applying the difference of squares formula, we get 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, a ≠ 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 the 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 + 4xr + 4x + 3 = 3(x3 + 1) + 4x(x + 1) = (x+ 1)(3x2 - 3x + 3 + 4x) = (x + 1)(3x2 + x + 3) , then equation (2) is equivalent to the 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 by dividing both sides of equation (3) by x2, we obtain an equation equivalent to the original one (3):

Let us rewrite equation (4) as:

Let's make a substitution in this equation, 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 symmetric equation of the fourth degree. Since x = 0 is not its root, then by dividing equation (6) by x2, we obtain an equivalent equation:

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

Putting it, we get an equation that has two roots y1 = 2 and y2 = 3. Consequently, the original equation is equivalent to a 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. Reducing the degree of the equation

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

Example 1: Solve the equation

Solution. Let us denote by, then equation (1) can be rewritten as The last equation has roots and Therefore, equation (1) is equivalent to the set of equations and. The solution to the first equation of this set is and the solution to 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. We rewrite this equation in the form

(3) and denoting by we 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 There are solutions to this set of equations and i.e. equation (2) is equivalent to a set of equations and ( 4)

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

2. Equations of the form

The equation

(5) where - the given 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,t. e. 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 a quadratic equation are and, the solutions to equation (7) are solutions to the set of equations and. This set of equations has two solutions and Therefore, the 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. Let's make a change of unknowns, i.e. y=x+3 or x = y – 3. Then equation (9) can be rewritten as

(y-2)(y-1)(y+1)(y+2)=10, i.e. in the form

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

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

4. Equations of the form

Equation, (11)

Where, x = 0 has no root, therefore, dividing equation (11) by x2, we obtain an equivalent equation

Which, after replacing the unknown, will be rewritten in the form of 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), dividing it by x2, we obtain an equivalent equation

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

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

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

Comment. Equation of the form

Which can always be reduced 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 solving 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 by dividing both sides by x2, we obtain an equivalent original equation. Making a change of variables, we obtain a quadratic equation whose solution is and. Consequently, the original equation (14) is equivalent to the set of equations and.

The solution to the first equation of this set is

The second equation of this set of solutions has no solutions. 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 x = 0 does not have a root, 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. Solving the equation

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

, (17) equivalent to equation (16). Having made the replacement with an unknown, we rewrite equation (17) in the form

Quadratic equation (18) has 2 roots: y1 = 1 and y2 = -1. Therefore, equation (17) is equivalent to the 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 you need to check which of them are not roots of the equation Q(x) = 0. These roots and only they will be solutions to the equation.

Let's consider some methods for solving equations of the form = 0.

1. Equations of the form

The equation

(1) under certain conditions on the numbers can be solved as follows. By 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 replacing the variables, the resulting equation will either have also form (1), but with a smaller number of terms, or 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 type (1), but with a smaller number of terms.

Example. Solve the equation

Solution. Having grouped 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 the terms in each bracket, we rewrite equation (3) in the form

Since there is no solution to equation (4), then, dividing this equation by, we obtain the equation

, (5) equivalent to equation (4). Let us make a replacement for the unknown, then equation (5) will be rewritten in the form

Thus, the solution to equation (2) with five terms on the left side is reduced to the solution to 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. None of these numbers makes the denominator of the rational function on the left side of equation (7) vanish. Consequently, equation (7) has these two roots, and therefore the original equation (2) is equivalent to the set of equations

The solutions to the first equation of this set are

The solutions to 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 numbers can be solved as follows: it is necessary to select the integer part in each of the fractions of the equation, i.e. replace equation (8) with the equation

Reduce it to form (1) and then solve it in the manner 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 brackets, we rewrite equation (10) in the form

By replacing the unknown, we rewrite equation (11) in the form

Summing up the terms on the left side of equation (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: by expanding (if this is, of course, possible) each of the fractions on the left side of equation (14) into a sum of simple fractions

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

Example. Solve the equation

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

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 the set of equations and

To solve the second equation of the set (18), we make a replacement for 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 in the form

Since the equation has no roots, equation (20) also does not have them.

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

4. Equations of the form

The equation

(21) under certain conditions on the numbers and A after representing each term on the left side 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 the 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:. All of them are solutions to the original equation.

5. Equations of the form.

Equation of the form (25)

Under certain conditions on numbers, by replacing the unknown, one can reduce 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

Having made a change of variables, we rewrite equation (27) in the form

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

In general, an equation of degree greater than 4 cannot be solved in radicals. But sometimes we can still find the roots of a polynomial on the left in an equation of the highest degree if we represent it as a product of polynomials to a degree of no more than 4. Solving such equations is based on factoring a polynomial, so we advise you to review this topic before studying this article.

Most often you have 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 so that we can then transform it into a lower degree equation that is easy to solve. In this material we will look at just such examples.

Higher degree equations with integer coefficients

All equations of the form a n x n + a n - 1 x n - 1 + . . . + a 1 x + a 0 = 0 , we can produce an equation of the same degree by multiplying both sides by a n n - 1 and making a variable change of the form y = a n x:

a n x n + a n - 1 x n - 1 + . . . + a 1 x + a 0 = 0 a n n · x n + a n - 1 · a n n - 1 · x n - 1 + … + a 1 · (a n) n - 1 · x + a 0 · (a n) n - 1 = 0 y = a n x ⇒ y n + b n - 1 y n - 1 + … + b 1 y + b 0 = 0

The resulting coefficients will also be integer. Thus, we will need to solve the reduced equation of the nth degree with integer coefficients, having the form x n + a n x n - 1 + ... + a 1 x + a 0 = 0.

We calculate the integer 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's write them down and substitute them into the original equality one by one, checking the result. Once we have obtained the identity and found one of the roots of the equation, we can write it in the form x - x 1 · P n - 1 (x) = 0. Here x 1 is the root of the equation, and P n - 1 (x) is the quotient of x n + a n x n - 1 + ... + a 1 x + a 0 divided by x - x 1 .

We substitute the remaining divisors written out into P n - 1 (x) = 0, starting with x 1, since the roots can be repeated. After obtaining the identity, the root x 2 is considered found, and the equation can be written in the form (x - x 1) (x - x 2) · P n - 2 (x) = 0. Here P n - 2 (x) will be the quotient of dividing P n - 1 (x) by x - x 2.

We continue to sort through the divisors. Let's find all the whole roots and denote their number as m. After this, the original equation can be represented as x - x 1 x - x 2 · … · x - x m · P n - m (x) = 0. Here P n - m (x) is a polynomial of n - m degree. For calculation it is convenient to use Horner's scheme.

If our original equation has integer coefficients, we cannot ultimately obtain fractional roots.

We ended up with the equation P n - m (x) = 0, the roots of which can be found in any convenient way. They can be irrational or complex.

Let us show with a specific example how this solution scheme is used.

Example 1

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

Solution

Let's start by finding whole roots.

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

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

Now let's divide the polynomial x 4 + x 3 + 2 x 2 - x - 3 by (x - 1) in a column:

So x 4 + x 3 + 2 x 2 - x - 3 = x - 1 x 3 + 2 x 2 + 4 x + 3.

1 3 + 2 1 2 + 4 1 + 3 = 10 ≠ 0 (- 1) 3 + 2 (- 1) 2 + 4 - 1 + 3 = 0

We got an identity, which means we 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 = (x - 1) (x 3 + 2 x 2 + 4 x + 3) = = (x - 1) (x + 1) (x 2 + x + 3)

We substitute the next divisor into the equality x 2 + x + 3 = 0, starting from - 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 integer roots.

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

D = 1 2 - 4 1 3 = - 11< 0

It follows from this that this quadratic trinomial has no real roots, but there are complex conjugate ones: x = - 1 2 ± i 11 2.

Let us clarify that instead of dividing into a column, Horner’s scheme can be used. This is done like this: after we have determined the first root of the equation, we fill out the table.

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

After finding the next root, which is - 1, we get the following:

Answer: x = - 1, x = 1, x = - 1 2 ± i 11 2.

Example 2

Condition: solve the equation x 4 - x 3 - 5 x 2 + 12 = 0.

Solution

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

Let's check them in order:

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

This means x = 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) = 0.

2 3 + 2 2 - 3 2 - 6 = 0

This means that 2 will again be the root. Divide x 3 + x 2 - 3 x - 6 = 0 by x - 2:

As a result, we get (x - 2) 2 · (x 2 + 3 x + 3) = 0.

Checking the remaining divisors does not make sense, since the equality x 2 + 3 x + 3 = 0 is faster and more convenient to solve using the discriminant.

Let's solve the quadratic equation:

x 2 + 3 x + 3 = 0 D = 3 2 - 4 1 3 = - 3< 0

We obtain a complex conjugate pair of roots: x = - 3 2 ± i 3 2 .

Answer: x = - 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 = 0.

Solution

x 4 + 1 2 x 3 - 5 2 x - 3 = 0 2 x 4 + x 3 - 5 x - 6 = 0

We multiply 2 3 on both sides of the equation:

2 x 4 + x 3 - 5 x - 6 = 0 2 4 x 4 + 2 3 x 3 - 20 2 x - 48 = 0

Replace variables y = 2 x:

2 4 x 4 + 2 3 x 3 - 20 2 x - 48 = 0 y 4 + y 3 - 20 y - 48 = 0

As a result, we got a standard equation of the 4th degree, which can be solved according to the standard scheme. Let's check the divisors, divide and ultimately get that it has 2 real roots y = - 2, y = 3 and two complex ones. We will not present the entire solution here. Due to the substitution, the real roots of this equation will be x = y 2 = - 2 2 = - 1 and x = y 2 = 3 2.

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

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