Solving equations of higher degrees. Equations of higher degrees

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Equations 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 (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. With c = c 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. Fmax = 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: x² + 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.

Methods for solving equations: n n n Replacing the equation h(f(x)) = h(g(x)) with the equation f(x) = g(x) Factorization. Introduction of a new variable. Functional - graphic method. Selection of roots. Application of Vieta's formulas.

Replacing the equation h(f(x)) = h(g(x)) with 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 value once. If the function is non-monotonic, then loss of roots is possible.

Solve the equation (3 x + 2)²³ = (5 x – 9)²³ y = x ²³ is an increasing function, so from the equation (3 x + 2)²³ = (5 x – 9)²³ you can go to the equation 3 x + 2 = 5 x – 9, from where 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 definition of the original equation, and discard the rest as extraneous.

Solve the equation x³ – 7 x + 6 = 0 Representing the term 7 x in the form x + 6 x, we obtain 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 a set of equations x – 1 = 0; x² + x – 6 = 0. Answer: 1, 2, – 3.

Introduction of 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 special feature of this equation is the equality of the coefficients of its left side, equidistant from its ends. Such equations are called reciprocal. Because 0 is not a root given equation, dividing by x² we get

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 a homogeneous equation. Let's divide both sides of the equation by (x² – 7 x +12)² (it is clear that the values ​​of x are such that x² – 7 x +12=0 are not solutions). Now we denote We Have From Here Answer:

Functional - graphic 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. Let's transform the equation to the form We notice that the function increases, and the function decreases. This means that the equation has only one root. Answer: 2.

Selection of roots n n n Theorem 1: If an integer m is the 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: – equation with integer Let coefficients. If a number and a fraction where p and q are irreducible integers is the root of an equation, then p is a divisor of the free term an, and q is a divisor of the coefficient of 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 polynomial being divided at x = a. Corollaries of Bezout's theorem n n n n The difference of identical powers of two numbers is divided without a remainder by the difference of the same numbers; The difference between identical even powers of two numbers is divided without a remainder by both the difference of these numbers and their sum; The difference between identical odd powers of two numbers is not divisible by the sum of these numbers; The sum of equal powers of two non-numbers is divided by the difference of these numbers; The sum of identical odd powers of two numbers is divided without a remainder by the sum of these numbers; The sum of identical even powers of two numbers is not divisible either by the difference of these numbers or by their sum; A polynomial is divisible by a binomial (x – a) if and only if the number a is the root of the given polynomial; The number of distinct roots of a nonzero polynomial is no more than 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 are convinced that the number 3 is a root. By corollary to Bezout’s theorem, the polynomial is divisible by (x – 3). Thus, 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, only numbers ± 1 can be integer roots of the equation. Checking shows that these numbers are not roots. Since the equation is not reduced, it can have fractional rational roots. Let's find them. To do this, multiply both sides of the equation by 4: 8 x³ – 20 x² – 4 x + 4 = 0 Substituting 2 x = t, we get t³ – 5 t² – 2 t + 4 = 0. By Theorem 2, all rational roots of this given equation must be intact. They can be found among the divisors of the free term: ± 1, ± 2, ± 4. In this case, t = – 1 is suitable. Therefore, by corollary to Bezout’s 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) Having solved 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 they satisfy the equation. They exhaust all the roots of the equation. Answer:

Find the sum of the squared roots of the equation x³ + 3 x² – 7 x +1 = 0 By Vieta’s theorem Note that where

Indicate how each of these equations can be solved. Solve equations No. 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)) with the equation f(x) = g(x). 4. Factorization. 5. Selection of roots. 6 Functional - graphic method. 7. Application of Vieta formulas. 8. Selection of roots. 9. Replacing the equation h(f(x)) = h(g(x)) with 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. Introduction of a new variable. 18. Factorization.

1. Instruction. 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. Instruction. 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 - 8). Answer:

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

Bibliography. n n n Kolmogorov A. N. “Algebra and the beginnings of analysis, 10 – 11” (M.: Prosveshchenie, 2003). Bashmakov M. I. “Algebra and the beginnings of analysis, 10 – 11” (M.: Prosveshchenie, 1993). Mordkovich A. G. “Algebra and principles of analysis, 10 – 11” (M.: Mnemosyna, 2003). Alimov Sh. A., Kolyagin Yu. M. et al. “Algebra and the beginnings of analysis, 10 – 11” (M.: Prosveshchenie, 2000). Galitsky M. L., Goldman A. M., Zvavich L. I. “Collection of problems in algebra, 8 – 9” (M.: Prosveshchenie, 1997). Karp A. P. “Collection of problems on algebra and principles of analysis, 10 – 11” (M.: Prosveshchenie, 1999). Sharygin I. F. “Optional course in mathematics, problem solving, 10” (M.: Education. 1989). Skopets Z. A. “Additional chapters on the course of mathematics, 10” (M.: Prosveshchenie, 1974). Litinsky G.I. “Mathematics Lessons” (Moscow: Aslan, 1994). Muravin G.K. “Equations, inequalities and their systems” (Mathematics, supplement to the newspaper “First of September”, No. 2, 3, 2003). Kolyagin Yu. M. “Polynomials and equations of higher degrees” (Mathematics, supplement to the newspaper “First of September”, No. 3, 2005).

The use of equations is widespread in our lives. They are used in many calculations, construction of structures and even sports. Man used equations in ancient times, and since then their use has only increased. In mathematics, equations of higher degrees with integer coefficients are quite common. To solve this type of equation you need:

Determine the rational roots of the equation;

Factor the polynomial on the left side of the equation;

Find the roots of the equation.

Let's say we are given an equation of the following form:

Let's find all its real roots. Multiply the left and right sides of the equation by \

Let's perform a change of variables\

Thus, we have the following fourth-degree equation, which can be solved using the standard algorithm: we check the divisors, carry out the division, and as a result we find out that the equation has two real roots\ and two complex ones. We get the following answer to our fourth-degree equation:

Where can I solve higher degree equations online using a solver?

You can solve the equation on our website https://site. The free online solver will allow you to solve online equations of any complexity in a matter of seconds. All you need to do is simply enter your data into the solver. You can also watch video instructions and learn how to solve the equation on our website. And if you still have questions, you can ask them in our VKontakte group http://vk.com/pocketteacher. Join our group, we are always happy to help you.

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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 solving equations of higher degrees and identifying the most accessible of them for practical application.

Based on the goal, the work identifies the following: tasks:

Study literature and Internet resources on this topic;

Get acquainted with historical facts relating 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 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 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 .

Universal formula for finding roots 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, the question of general decision equations of the 3rd degree were studied by Scipio, Dal, Ferro and his students 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 solving cubic equations, 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 There are 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 by the methods considered in school curriculum, 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's transform the equation general view 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. However, 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 that we can write the following equalities into the system:

Formulas derived by Viethe 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 nth polynomial degrees, then the theorem has another interpretation.

Theorem. When dividing a polynomial nth degree relatively 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 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 according to Horner’s scheme are presented in the following table:

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.

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 in various ways factorization of 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 undetermined 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 uncertain 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 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 using the method uncertain coefficients, solving symmetric equations.

During 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 established as a result of a survey of mathematics teachers at MBOU “Secondary School No. 9” and students showing 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 the educational and scientific literature, Internet resources in youth educational forums