Solving equations of higher degrees. Equations of higher degrees Equations of higher degrees definition

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).

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.

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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.

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

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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 be easily 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 are 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)

Class: 9

Basic goals:

Reinforce the concept of an entire rational equation of the th degree.
Formulate the basic methods for solving equations of higher degrees (n > 3).
Teach basic methods for solving higher-order equations.
Learn to use the type of equation to determine the most effective way to solve it.

Forms, methods and pedagogical techniques used by the teacher in the classroom:

Lecture-seminar teaching system (lectures - explanation of new material, seminars - problem solving).
Information and communication technologies (frontal survey, oral work with the class).
Differentiated learning, group and individual forms.
Using a research method in teaching aimed at developing the mathematical apparatus and thinking abilities of each individual student.
Printed material – an individual brief summary of the lesson (basic concepts, formulas, statements, lecture material condensed in the form of diagrams or tables).

Lesson plan:

Organizing time.
The purpose of the stage: to include students in educational activities, to determine the content of the lesson.
Updating students' knowledge.
The purpose of the stage: to update students’ knowledge on previously studied related topics
Studying a new topic (lecture). Goal of the stage: to formulate the basic methods for solving equations of higher degrees (n > 3)
Summarizing.
The purpose of the stage: to once again highlight the key points in the material studied in the lesson.
Homework.
The purpose of the stage: to formulate homework for students.

Lesson summary

1. Organizational moment.

Formulation of the lesson topic: “Equations of higher powers. Methods for solving them.”

2. Updating students' knowledge.

Theoretical survey - conversation. Repetition of some previously studied information from the theory. Students formulate basic definitions and formulate the necessary theorems. Give examples to demonstrate the level of previously acquired knowledge.

The concept of an equation with one variable.
The concept of the root of an equation, the solution of an equation.
The concept of a linear equation with one variable, the concept of a quadratic equation with one variable.
The concept of equivalence of equations, equations-consequences (the concept of extraneous roots), transition not by consequence (the case of loss of roots).
The concept of a whole rational expression with one variable.
The concept of a whole rational equation n th degree. Standard form of a whole rational equation. Reduced whole rational equation.
Transition to a set of equations of lower degrees by factoring the original equation.
The concept of a polynomial n th degree from x. Bezout's theorem. Corollaries from Bezout's theorem. Root theorems ( Z-roots and Q-roots) of an entire rational equation with integer coefficients (reduced and unreduced, respectively).
Horner's scheme.

3. Studying a new topic.

We will consider the whole rational equation n-th power of standard form with one unknown variable x:Pn(x)= 0, where P n (x) = a n x n + a n-1 x n-1 + a 1 x + a 0– polynomial n th degree from x, a n ≠ 0. If a n = 1 then such an equation is called a reduced integer rational equation n th degree. Let us consider such equations for various values n and list the main methods for solving them.

n= 1 – linear equation.

n= 2 – quadratic equation. Discriminant formula. Formula for calculating roots. Vieta's theorem. Selecting a complete square.

n= 3 – cubic equation.

Grouping method.

Example: x 3 – 4x 2 – x+ 4 = 0 (x – 4)(x 2– 1) = 0 x 1 = 4 , x 2 = 1,x 3 = -1.

Reciprocal cubic equation of the form ax 3 + bx 2 + bx + a= 0. We solve by combining terms with the same coefficients.

Example: x 3 – 5x 2 – 5x + 1 = 0 (x + 1)(x 2 – 6x + 1) = 0 x 1 = -1, x 2 = 3 + 2, x 3 = 3 – 2.

Selection of Z-roots based on the theorem. Horner's scheme. When applying this method, it is necessary to emphasize that the search in this case is finite, and we select the roots using a certain algorithm in accordance with the theorem Z-roots of the given whole rational equation with integer coefficients.

Example: x 3 – 9x 2 + 23x– 15 = 0. The equation is given. Let us write down the divisors of the free term ( + 1; + 3; + 5; + 15). Let's apply Horner's scheme:

	x 3	x 2	x 1	x 0	conclusion
	1	-9	23	-15
1	1	1 x 1 – 9 = -8	1 x (-8) + 23 = 15	1 x 15 – 15 = 0	1 – root
	x 2	x 1	x 0

We get ( x – 1)(x 2 – 8x + 15) = 0 x 1 = 1, x 2 = 3, x 3 = 5.

Equation with integer coefficients. Selection of Q-roots based on the theorem. Horner's scheme. When applying this method, it is necessary to emphasize that the search in this case is finite and we select the roots using a certain algorithm in accordance with the theorem about Q-roots of an unreduced integer rational equation with integer coefficients.

Example: 9 x 3 + 27x 2 – x– 3 = 0. The equation is unreduced. Let us write down the divisors of the free term ( + 1; + 3). Let us write down the divisors of the coefficient at the highest power of the unknown. ( + 1; + 3; + 9) Consequently, we will look for roots among the values ( + 1; + ; + ; + 3). Let's apply Horner's scheme:

	x 3	x 2	x 1	x 0	conclusion
	9	27	-1	-3
1	9	1 x 9 + 27 = 36	1 x 36 – 1 = 35	1 x 35 – 3 = 32 ≠ 0	1 – not a root
-1	9	-1 x 9 + 27 = 18	-1 x 18 – 1 = -19	-1 x (-19) – 3 = 16 ≠ 0	-1 – not a root
	9	x 9 + 27 = 30	x 30 – 1 = 9	x 9 – 3 = 0	root
	x 2	x 1	x 0

We get ( x – )(9x 2 + 30x + 9) = 0 x 1 = , x 2 = - , x 3 = -3.

For ease of calculation when selecting Q -roots It can be convenient to make a change of variable, go to the given equation and select Z -roots.

If the dummy term is 1

.

If you can use a replacement of the form y = kx

.

Cardano formula. There is a universal method for solving cubic equations - this is the Cardano formula. This formula is associated with the names of Italian mathematicians Gerolamo Cardano (1501–1576), Nicolo Tartaglia (1500–1557), Scipione del Ferro (1465–1526). This formula is beyond the scope of our course.

n= 4 – equation of the fourth degree.

Grouping method.

Example: x 4 + 2x 3 + 5x 2 + 4x – 12 = 0 (x 4 + 2x 3) + (5x 2 + 10x) – (6x + 12) = 0 (x + 2)(x 3 + 5x – 6) = 0 (x + 2)(x– 1)(x 2 + x + 6) = 0 x 1 = -2, x 2 = 1.

Variable replacement method.

Biquadratic equation of the form ax 4 + bx 2 + s = 0 .

Example: x 4 + 5x 2 – 36 = 0. Replacement y = x 2. From here y 1 = 4, y 2 = -9. That's why x 1,2 = + 2 .

Reciprocal equation of the fourth degree of the form ax 4 + bx 3+c x 2 + bx + a = 0.

We solve by combining terms with the same coefficients by replacing the form

ax 4 + bx 3 + cx 2 – bx + a = 0.

Generalized recurrent equation of the fourth degree of the form ax 4 + bx 3 + cx 2 + kbx + k 2 a = 0.

General replacement. Some standard replacements.

Example 3 . General view replacement(follows from the type of specific equation).

n = 3.

Equation with integer coefficients. Selection of Q-roots n = 3.

General formula. There is a universal method for solving fourth degree equations. This formula is associated with the name of Ludovico Ferrari (1522–1565). This formula is beyond the scope of our course.

n > 5 – equations of the fifth and higher degrees.

Equation with integer coefficients. Selection of Z-roots based on the theorem. Horner's scheme. The algorithm is similar to that discussed above for n = 3.

Equation with integer coefficients. Selection of Q-roots based on the theorem. Horner's scheme. The algorithm is similar to that discussed above for n = 3.

Symmetric equations. Any reciprocal equation of odd degree has a root x= -1 and after factoring it into factors we find that one factor has the form ( x+ 1), and the second factor is a reciprocal equation of even degree (its degree is one less than the degree of the original equation). Any reciprocal equation of even degree together with a root of the form x = φ also contains the root of the species. Using these statements, we solve the problem by lowering the degree of the equation under study.

Variable replacement method. Use of homogeneity.

There is no general formula for solving entire equations of the fifth degree (this was shown by the Italian mathematician Paolo Ruffini (1765–1822) and the Norwegian mathematician Niels Henrik Abel (1802–1829)) and higher degrees (this was shown by the French mathematician Evariste Galois (1811–1832) )).

Let us recall once again that in practice it is possible to use combinations the methods listed above. It is convenient to pass to a set of equations of lower degrees by factoring the original equation.
Outside the scope of our discussion today are those widely used in practice. graphical methods solving equations and approximate solution methods equations of higher degrees.

There are situations when the equation does not have R-roots.

Using the property of monotonicity of functions

4. Summing up.

Summary: Now we have mastered the basic methods for solving various equations of higher degrees (for n > 3). Our task is to learn how to effectively use the algorithms listed above. Depending on the type of equation, we will have to learn to determine which method of solution in a given case is the most effective, as well as correctly apply the chosen method.

5. Homework.

: paragraph 7, pp. 164–174, nos. 33–36, 39–44, 46.47.

: №№ 9.1–9.4, 9.6–9.8, 9.12, 9.14–9.16, 9.24–9.27.

Possible topics for reports or abstracts on this topic:

Cardano formula
Graphical method for solving equations. Examples of solutions.
Methods for approximate solution of equations.

Analysis of student learning and interest in the topic:

Experience shows that students’ interest is primarily aroused by the possibility of selecting Z-roots and Q-roots of equations using a fairly simple algorithm using Horner’s scheme. Students are also interested in various standard types of substitution of variables, which can significantly simplify the type of problem. Graphical solution methods are usually of particular interest. In this case, you can additionally analyze problems using a graphical method for solving equations; discuss the general form of the graph for a polynomial of degree 3, 4, 5; analyze how the number of roots of equations of degree 3, 4, 5 is related to the appearance of the corresponding graph. Below is a list of books where you can find additional information on this topic.

Bibliography:

Vilenkin N.Ya. and others. “Algebra. Textbook for 9th grade students with in-depth study of mathematics” - M., Prosveshchenie, 2007 - 367 p.
Vilenkin N.Ya., Shibasov L.P., Shibasova Z.F.“Behind the pages of a mathematics textbook. Arithmetic. Algebra. 10-11 grades” – M., Education, 2008 – 192 p.
Vygodsky M.Ya.“Handbook of Mathematics” – M., AST, 2010 – 1055 p.
Galitsky M.L.“Collection of problems in algebra. Textbook for grades 8-9 with in-depth study of mathematics” - M., Prosveshchenie, 2008 - 301 p.
Zvavich L.I. and others. “Algebra and the beginnings of analysis. 8–11 grades A manual for schools and classes with in-depth study of mathematics” - M., Drofa, 1999 - 352 p.
Zvavich L.I., Averyanov D.I., Pigarev B.P., Trushanina T.N.“Mathematics assignments for preparing for the written exam in 9th grade” - M., Prosveshchenie, 2007 - 112 p.
Ivanov A.A., Ivanov A.P.“Thematic tests for systematizing knowledge in mathematics” part 1 – M., Fizmatkniga, 2006 – 176 p.
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Ivanov A.P.“Tests and tests in mathematics. Tutorial". – M., Fizmatkniga, 2008 – 304 p.
Leibson K.L.“Collection of practical tasks in mathematics. Part 2–9 grades” – M., MTSNM, 2009 – 184 p.
Makarychev Yu.N., Mindyuk N.G."Algebra. Additional chapters for the 9th grade school textbook. A textbook for students in schools and classes with in-depth study of mathematics.” – M., Education, 2006 – 224 p.
Mordkovich A.G."Algebra. In-depth study. 8th grade. Textbook” – M., Mnemosyne, 2006 – 296 p.
Savin A.P.“Encyclopedic Dictionary of a Young Mathematician” - M., Pedagogy, 1985 - 352 p.
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Chulkov P.V.“Equations and inequalities in the school mathematics course. Lectures 1–4” – M., September 1st, 2006 – 88 p.
Chulkov P.V.“Equations and inequalities in the school mathematics course. Lectures 5–8” – M., September 1, 2009 – 84 p.