How to solve n-th degree equations. Equations of the highest degrees in mathematics

The use of equations is widespread in our life. They are used in many calculations, building construction, and even sports. Man used equations in antiquity and since then their application has only increased. In mathematics, equations of higher degrees with integer coefficients are quite common. To solve this kind of equation it is necessary:

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 \\

Change variables \\

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

Where can you solve the equation of higher degrees with an online solver?

You can solve the equation on our website https: // site. A free online solver will allow you to solve an equation online of any complexity in a matter of seconds. All you have to do is just enter your data into the solver. You can also watch a video instruction 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.

GORNER'S SCHEME

IN SOLVING EQUATIONS WITH PARAMETERS
FROM GROUP "C" WHEN PREPARING FOR USE

Kazantseva Lyudmila Viktorovna

teacher of mathematics MBOU "Uyarskaya secondary school number 3"

In optional lessons, it is necessary to expand the range of available knowledge by solving tasks of increased complexity of the group "C".

This work covers some of the issues discussed in additional classes.

It is advisable to introduce Horner's scheme after studying the topic "Division of a polynomial by a polynomial". This material allows you to solve higher-order equations not by grouping polynomials, but in a more rational way that saves time.

Lesson plan.

Lesson 1.

1. Explanation of the theoretical material.

2. Solution of examples a B C D).

Lesson 2.

1. Solving equations a B C D).

2. Finding rational roots of a polynomial

Application of Horner's scheme for solving equations with parameters.

Lesson 3.

Tasks a B C).

Lesson 4.

1. Tasks d), e), f), g), h).

Solving equations of higher degrees.

Horner's scheme.

Theorem : Let the irreducible fraction be the root of the equation

a o x n + a 1 x n-1 +… + A n-1 x 1 + a n = 0

with integer coefficients. Then the number ris the divisor of the leading coefficient a about .

Consequence: Any integer root of an equation with integer coefficients is a divisor of its intercept.

Consequence: If the leading coefficient of an equation with integer coefficients is 1 , then all rational roots, if they exist, are whole.

Example 1. 2x 3 - 7x 2 + 5x - 1 \u003d 0

Let an irreducible fraction be a root of the equation, thenr is a divisor of the number1: ± 1

q is a divisor of the highest term: ± 1; ± 2

The rational roots of the equation must be sought among the numbers:± 1; ±.

f (1) \u003d 2 - 7 + 5 - 1 \u003d - 1 ≠ 0

f (–1) \u003d –2 - 7 - 5 - 1 ≠ 0

f () = – + – 1 = – + – = 0

The root is the number .

Division of a polynomial P (x) \u003d a about x p + a 1 x n -1 + … + a n into a binomial ( x - £) it is convenient to perform according to Horner's scheme.

We denote the incomplete quotient P (x)on ( x - £)across Q (x ) = b o x n -1 + b 1 x n -2 + … b n -1 ,

and the remainder after b n

P (x) \u003dQ (x ) (x – £) + b n , then the identity

a about x p + a 1 x n-1 +… + A n \u003d (b o x n-1 + … + b n-1 ) (x - £) +b n

Q (x ) Is a polynomial whose degree is 1 below the degree of the original polynomial. Polynomial coefficients Q (x ) determined according to Horner's scheme.

and oh

a 1

a 2

a n-1

a n

b o \u003d a о

b 1 = a 1 + £· b o

b 2 = a 2 + £· b 1

b n-1 \u003d a n-1 + £· b n-2

b n \u003d a n + £· b n-1

The first line of this table contains the coefficients of the polynomial P (x).

If some degree of the variable is absent, then in the corresponding cell of the table is written 0.

The senior coefficient of the quotient is equal to the senior coefficient of the dividend ( a about = b o ). If £ is the root of the polynomial, then in the last cell we get 0.

Example 2... Factor with integer coefficients

P (x) \u003d 2x 4 - 7x 3 - 3x 2 + 5x - 1

± 1.

Suitable - 1.

Divide P (x) on (x + 1)

– 7

– 3

– 1

– 9

– 1

2x 4 - 7x 3 - 3x 2 + 5x - 1 \u003d (x + 1) (2x 3 - 9x 2 + 6x - 1)

We are looking for whole roots among the free member: ± 1

Since the senior member is 1, then the roots can be fractional numbers: - ; .

Suitable .

– 9

– 1

– 8

2x 3 - 9x 2 + 6x - 1 \u003d (x -) (2x 2 - 8x + 2) \u003d (2x - 1) (x 2 - 4x + 1)

Trinomial x 2 - 4x + 1cannot be decomposed into factors with integer coefficients.

The task:

1. Factor with integer coefficients:

a) x 3 - 2x 2 - 5x + 6

q: ± 1;

p: ± 1; ± 2; ± 3; ± 6

: ± 1; ± 2; ± 3; ± 6

Find the rational roots of the polynomial f (1) = 1 – 2 – 5 + 6 = 0

x \u003d 1

– 2

– 5

– 1

– 6

x 3 - 2x 2 - 5x + 6 \u003d (x - 1) (x 2 - x - 6) \u003d (x - 1) (x - 3) (x + 2)

We define the roots of the quadratic equation

x 2 - x - 6 \u003d 0

x \u003d 3; x \u003d - 2

b) 2x 3 + 5x 2 + x - 2

p: ± 1; ± 2

q: ± 1; ± 2

: ± 1; ± 2; ±

Find the roots of the third degree polynomial

f (1) \u003d 2 + 5 + 1 - 2 ≠ 0

f (–1) \u003d - 2 + 5 - 1 - 2 \u003d 0

One of the roots of the equation x \u003d - 1

– 2

– 1

– 2

2x 3 + 5x 2 + x - 2 \u003d (x + 1) (2x 2 + 3x - 2) \u003d (x + 1) (x + 2) (2x - 1)

We expand the square trinomial 2x 2 + 3x - 2by factors

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

D \u003d 9 + 16 \u003d 25

x 1 \u003d - 2; x 2 \u003d

in) x 3 - 3x 2 + x + 1

p: ± 1

q: ± 1

: ± 1

f (1) \u003d 1 - 3 + 1 - 1 \u003d 0

One of the roots of the third degree polynomial is x \u003d 1

– 3

– 2

– 1

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

Find the roots of the equation x 2 - 2x - 1 \u003d 0

D \u003d 4 + 4 = 8

x 1 = 1 –

x 2 = 1 +

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

d) x 3 - 2x - 1

p: ± 1

q: ± 1

: ± 1

We define the roots of the polynomial

f (1) \u003d 1 - 2 - 1 \u003d - 2

f (–1) \u003d - 1 + 2 - 1 \u003d 0

First root x \u003d - 1

– 2

– 1

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

x 2 - x - 1 \u003d 0

D \u003d 1 + 4 \u003d 5

x 1.2 =

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

2. Solve the equation:

a) x 3 - 5x + 4 \u003d 0

Define the roots of the third degree polynomial

: ± 1; ± 2; ± 4

f (1) \u003d 1 - 5 + 4 \u003d 0

One of the roots is x \u003d 1

– 5

– 4

x 3 - 5x + 4 \u003d 0

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

x 2 + x - 4 \u003d 0

D \u003d 1 + 16 \u003d 17

x 1 =
; x 2 =

Answer: 1;
;

b) x 3 - 8x 2 + 40 = 0

Let us define the roots of the third degree polynomial.

: ± 1; ± 2; ± 4; ± 5; ± 8; ± 10; ± 20; ± 40

f (1) ≠ 0

f (–1) ≠ 0

f (–2) \u003d - 8 - 32 + 40 \u003d 0

One of the roots is x \u003d - 2

– 8

– 2

– 10

Let us factorize the third degree polynomial.

x 3 - 8x 2 + 40 \u003d (x + 2) (x 2 - 10x + 20)

Find the roots of the quadratic equation x 2 - 10x + 20 \u003d 0

D \u003d 100 - 80 \u003d 20

x 1 = 5 –
; x 2 = 5 +

Answer: - 2; 5 –
; 5 +

in) x 3 - 5x 2 + 3x + 1 \u003d 0

We are looking for whole roots among the divisors of the free term: ± 1

f (–1) \u003d - 1 - 5 - 3 + 1 ≠ 0

f (1) \u003d 1 - 5 + 3 + 1 \u003d 0

Suitable x \u003d 1

– 5

– 4

– 1

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

(x - 1) (x 2 - 4x - 1) \u003d 0

Determine the roots of the quadratic equation x 2 - 4x - 1 \u003d 0

D \u003d 20

x \u003d 2 +
; x \u003d 2 -

Answer: 2 –
; 1; 2 +

d) 2x 4 - 5x 3 + 5x 2 – 2 = 0

p: ± 1; ± 2

q: ± 1; ± 2

: ± 1; ± 2; ±

f (1) \u003d 2 - 5 + 5 - 2 \u003d 0

One of the roots of the equation x \u003d 1

– 5

– 2

– 3

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

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

We find the roots of the third degree equation in the same way.

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

p: ± 1; ± 2

q: ± 1; ± 2

: ± 1; ± 2; ±

f (1) \u003d 2 - 3 + 2 + 2 ≠ 0

f (–1) \u003d - 2 - 3 - 2 + 2 ≠ 0

f (2) \u003d 16 - 12 + 4 + 2 ≠ 0

f (–2) \u003d - 16 - 12 - 4 + 2 ≠ 0

f() = – + 1 + 2 ≠ 0

f(–) = – – – 1 + 2 ≠ 0

The next root of the equationx \u003d -

– 3

– 4

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

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

We define the roots of the quadratic equation 2x 2 - 4x + 4 \u003d 0

x 2 - 2x + 2 \u003d 0

D \u003d - 4< 0

Therefore, the roots of the original fourth-degree equation are

1 and –

Answer: –; 1

3. Find the rational roots of the polynomial

a) x 4 - 2x 3 - 8x 2 + 13x - 24

q: ± 1

: ± 1; ± 2; ± 3; ± 4; ± 6; ± 8; ± 12; ± 24

Choose one of the roots of the fourth degree polynomial:

f (1) \u003d 1 - 2 - 8 + 13 - 24 ≠ 0

f (–1) \u003d 1 + 2 - 8 - 13 - 24 ≠ 0

f (2) \u003d 16 - 16 - 32 + 26 - 24 ≠ 0

f (–2) \u003d 16 + 16 - 72 - 24 ≠ 0

f (–3) \u003d 81 + 54 - 72 - 39 - 24 \u003d 0

One of the roots of the polynomial x 0= – 3.

x 4 - 2x 3 - 8x 2 + 13x - 24 \u003d (x + 3) (x 3 - 5x 2 + 7x + 8)

Find the rational roots of the polynomial

x 3 - 5x 2 + 7x + 8

p: ± 1; ± 2; ± 4; ± 8

q: ± 1

f (1) \u003d 1 - 5 + 7 + 8 ≠ 0

f (–1) \u003d - 1 - 5 - 7 - 8 ≠ 0

f (2) \u003d 8 - 20 + 14 + 8 ≠ 0

f (–2) \u003d - 8 - 20 - 14 + 8 ≠ 0

f (–4) \u003d 64 - 90 - 28 + 8 ≠ 0

f (4) ≠ 0

f (–8) ≠ 0

f (8) ≠ 0

Besides the number x 0 = – 3 there are no other rational roots.

b) x 4 - 2x 3 - 13x 2 - 38x - 24

p: ± 1; ± 2; ± 3; ± 4; ± 6; ± 8; ± 12; ± 24

q: ± 1

f (1) \u003d 1 + 2 - 13 - 38 - 24 ≠ 0

f (–1) = 1 – 2 – 13 + 38 – 24 = 39 – 39 = 0, i.e x \u003d - 1polynomial root

– 13

– 38

– 24

– 1

– 14

– 24

x 4 - 2x 3 - 13x 2 - 38x - 24 \u003d (x + 1) (x 3 - x 2 - 14x - 24)

Define the roots of the third degree polynomial x 3 - x 2 - 14x - 24

p: ± 1; ± 2; ± 3; ± 4; ± 6; ± 8; ± 12; ± 24

q: ± 1

f (1) \u003d - 1 + 1 + 14 - 24 ≠ 0

f (–1) \u003d 1 + 1 - 14 - 24 ≠ 0

f (2) \u003d 8 + 4 - 28 - 24 ≠ 0

f (–2) \u003d - 8 + 4 + 28 - 24 ≠ 0

Hence, the second root of the polynomial x \u003d - 2

– 14

– 24

– 2

– 1

– 12

x 4 - 2x 3 - 13x 2 - 38x - 24 \u003d (x + 1) (x 2 + 2) (x 2 - x - 12) \u003d

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

Answer: – 3; – 2; – 1; 4

Applying Horner's scheme to solving equations with a parameter.

Find the largest integer parameter value a,at which the equation f (x) \u003d 0has three distinct roots, one of which x 0 .

a) f (x) \u003d x 3 + 8x 2 + ah +b , x 0 = – 3

So one of the roots x 0 = – 3 , then according to Horner's scheme we have:

– 3

– 15 + a

0 \u003d - 3 (- 15 + a) + b

0 \u003d 45 - 3a + b

b \u003d 3а - 45

x 3 + 8x 2 + ax + b \u003d (x + 3) (x 2 + 5x + (a - 15))

The equation x 2 + 5x + (a - 15) \u003d 0 D > 0

a \u003d 1; b \u003d 5; c \u003d (a - 15),

D \u003d b 2 - 4ac \u003d 25 - 4 (a - 15) \u003d 25 + 60 - 4a\u003e 0,

85 - 4a\u003e 0;

4a< 85;

a< 21

Largest integer parameter value a,at which the equation

f (x) \u003d 0has three roots, a \u003d 21

Answer: 21.

b) f (x) \u003d x 3 - 2x 2 + ax + b, x 0 = – 1

Since one of the roots x 0= – 1, then according to Horner's scheme we have

– 2

– 1

– 3

3 + a

x 3 - 2x 2 + ax + b \u003d (x + 1) (x 2 - 3x + (3 + a))

The equation x 2 – 3 x + (3 + a ) = 0 must have two roots. This is only done when D > 0

a \u003d 1; b \u003d - 3; c \u003d (3 + a),

D \u003d b 2 - 4ac \u003d 9 - 4 (3 + a) \u003d 9 - 12 - 4a \u003d - 3 - 4a\u003e 0,

– 3 - 4a\u003e 0;

– 4a< 3;

a < –

Highest value a \u003d - 1 a \u003d 40

Answer: a \u003d 40

d) f (x) \u003d x 3 - 11x 2 + ax + b, x 0 = 4

Since one of the roots x 0 = 4 , then according to Horner's scheme we have

– 11

– 7

– 28 + a

x 3 - 11x 2 + ax + b \u003d (x - 4) (x 2 - 7x + (a - 28))

f (x ) = 0, if x \u003d 4or x 2 – 7 x + (a – 28) = 0

D > 0, i.e

D \u003d b 2 - 4ac \u003d 49 - 4 (a - 28) \u003d 49 + 112 - 4a \u003d 161 - 4a\u003e 0,

161 - 4a\u003e 0;

– 4a< – 161; f x 0 = – 5 , then according to Horner's scheme we have

– 5

– 40 + a

x 3 + 13x 2 + ax + b \u003d (x +5) (x 2 + 8x + (a - 40))

f (x ) = 0, if x \u003d - 5or x 2 + 8 x + (a – 40) = 0

The equation has two roots if D > 0

D \u003d b 2 - 4ac \u003d 64 - 4 (a - 40) \u003d 64 + 1 60 - 4a \u003d 224 - 4a\u003e 0,

224 - 4a\u003e 0;

a< 56

The equation f (x ) has three roots at the highest value a \u003d 55

Answer: a \u003d 55

g) f (x ) = x 3 + 19 x 2 + ax + b , x 0 = – 6

Since one of the roots – 6 , then according to Horner's scheme we have

– 6

a - 78

x 3 + 19x 2 + ax + b \u003d (x +6) (x 2 + 13x + (a - 78)) \u003d 0

f (x ) = 0, if x \u003d - 6or x 2 + 13 x + (a – 78) = 0

The second equation has two roots if

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

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

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

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

Statements about roots of a polynomial and its divisors:

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

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

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

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

6. For a polynomial of degree 3

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

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

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

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

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

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

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

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

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

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

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

Solution examples

Example 1.

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

Decision.

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

Answer: R \u003d 0.

Example 2.

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

Decision:

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

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

X 2 - 2 x

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

Basic methods for solving equations of higher degrees

1. Introducing a new variable

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

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

Example 1.

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

Decision:

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

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

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

t 2 - 3t + 2 \u003d 0.

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

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

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

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

2. Factorization by grouping and reduced multiplication formulas

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

Example 1.

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

Decision.

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

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

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

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

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

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

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

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

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

3. Factoring by the method of undefined coefficients

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

Example 1.

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

Decision.

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

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

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

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

Having solved the system:

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

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

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

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

Answer: -1; -2.

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

The method is based on the application of theorems:

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

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

Example 1.

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

Decision:

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

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

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

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

Still have questions? Not sure how to solve equations?
To get help from a tutor - register.
The first lesson is free!

site, with full or partial copying of the material, a link to the source is required.

To use the preview of presentations, create yourself a Google account (account) and log into it: https://accounts.google.com

Slide captions:

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

Plan lectures. No. 1. Equations of the highest degrees in the school mathematics course. No. 2. Standard form of a polynomial. # 3. Integral roots of a polynomial. Horner's scheme. № 4. Fractional roots of a polynomial. № 5. Equations of the form: (х + а) (х + в) (х + с)… \u003d А № 6. Return equations. № 7. Homogeneous equations. № 8. The method of undefined coefficients. № 9. Functional - graphic method. № 10. Vieta's formulas for equations of higher degrees. № 11. Non-standard methods for solving equations of higher degrees.

Equations of the highest degrees in the school mathematics course. 7th grade. Standard form of a polynomial. Actions with polynomials. Factoring a polynomial. In the regular class 42 hours, in the special class 56 hours. 8 special class. Integer roots of a polynomial, division of polynomials, recurrent equations, difference and sum of n - th powers of a binomial, method of undefined coefficients. Yu.N. Makarychev "Additional chapters to the school course of 8th grade algebra", MLGalitsky Collection of problems in algebra 8th - 9th grade. 9 special class. Rational roots of a polynomial. Generalized return equations. Vieta's formulas for equations of higher degrees. N. Ya. Vilenkin “Algebra Grade 9 with Advanced 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. The polynomial P (x) \u003d a ⁿ x ⁿ + a n-1 x n-1 +… + a₂x ² + a₁x + a₀. It is called a standard polynomial. a n x ⁿ is the highest term of the polynomial and n is the coefficient of the highest term of the polynomial. For a n \u003d 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.

Integer roots of a polynomial. Horner's scheme. Theorem 1. If an integer a is a 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 \u003d 11x² - 4x - 4 Let us bring the equation to its standard form. X⁴ + 2x³ - 11x² + 4x + 4 \u003d 0. We have the polynomial P (x) \u003d x ⁴ + 2x³ - 11x² + 4x + 4 The divisions of the free term: ± 1, ± 2, ± 4. x \u003d 1 root of the equation because P (1) \u003d 0, x \u003d 2 root of the equation since P (2) \u003d 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 a root of the polynomial P (x), then P (x) is divisible by (x - a). In our equation, P (x) is divisible by (x - 1) and by (x - 2), and hence by (x - 1) (x - 2). When dividing P (x) by (x ² - 3x + 2) in the quotient, we obtain a trinomial x ² + 5x + 2 \u003d 0, which has roots x \u003d (- 5 ± √17) / 2

Fractional roots of a polynomial. Theorem # 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 \u003d 0. Free term divisors: ± 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 make sure that P (4/3) \u003d 0. X \u003d 4/3 is the root of the equation. According to Horner's scheme, we divide P (x) by (x - 4/3).

Examples for an independent solution. Solve the equations: 9x³ - 18x \u003d x - 2, x ³ - x ² \u003d x - 1, x ³ - 3x² -3x + 1 \u003d 0, X ⁴ - 2x³ + 2x - 1 \u003d 0, X⁴ - 3x² + 2 \u003d 0 , x ⁵ + 5x³ - 6x² \u003d 0, x ³ + 4x² + 5x + 2 \u003d 0, X⁴ + 4x³ - x ² - 16x - 12 \u003d 0 4x³ + x ² - x + 5 \u003d 0 3x⁴ + 5x³ - 9x² - 9x + 10 \u003d 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; -one; ± 2, 9) - 5/4 10) -2; - 5/3; one.

Equations of the form (x + a) (x + b) (x + c) (x + d)… \u003d A. Example №3. Solve the equation (x + 1) (x + 2) (x + 3) (x + 4) \u003d 24. a \u003d 1, b \u003d 2, c \u003d 3, d \u003d 4 a + d \u003d b + c. We multiply the first bracket with the fourth and the second with the third. (x + 1) (x + 4) (x + 20 (x + 3) \u003d 24. (x2 + 5x + 4) (x2 + 5x + 6) \u003d 24. Let x2 + 5x + 4 \u003d y , then y (y + 2) \u003d 24, y² + 2y - 24 \u003d 0 y₁ \u003d - 6, y₂ \u003d 4.x² + 5x + 4 \u003d -6 or x² + 5x + 4 \u003d 4.x² + 5x + 10 \u003d 0, D

Examples for an independent solution. (x + 1) (x + 3) (x + 5) (x + 7) \u003d -15, x (x + 4) (x + 5) (x + 9) + 96 \u003d 0, x (x + 3 ) (x + 5) (x + 8) + 56 \u003d 0, (x - 4) (x - 3) (x - 2) (x - 1) \u003d 24, (x - 3) (x -4) ( x - 5) (x - 6) \u003d 1680, (x ² - 5x) (x + 3) (x - 8) + 108 \u003d 0, (x + 4) ² (x + 10) (x - 2) + 243 \u003d 0 (x ² + 3x + 2) (x ² + 9x + 20) \u003d 4, Indication: x + 3x + 2 \u003d (x + 1) (x + 2), x ² + 9x + 20 \u003d (x + 4) (x + 5) Answers: 1) -4 ± √6; - 6; - 2. 6) - 1; 6; (5 ± √97) / 2 7) -7; -one; -4 ± √3.

Reflexive equations. Definition # 1. An equation of the form: ax⁴ + in ³ + cx ² + in + a \u003d 0 is called a return equation of the fourth degree. Definition number 2. An equation of the form: ax⁴ + inx ³ + cx² + bx + k² a \u003d 0 is called a generalized return equation of the fourth degree. k² a: a \u003d k²; kv: v \u003d k. Example No. 6. Solve the equation x ⁴ - 7x³ + 14x² - 7x + 1 \u003d 0. Divide both sides of the equation by x ². x ² - 7x + 14 - 7 / x + 1 / x ² \u003d 0, (x ² + 1 / x ²) - 7 (x + 1 / x) + 14 \u003d 0. Let x + 1 / x \u003d y. We square both sides of the equality. x ² + 2 + 1 / x ² \u003d y², x ² + 1 / x ² \u003d y² - 2. We get the quadratic equation y² - 7y + 12 \u003d 0, y₁ \u003d 3, y₂ \u003d 4.x + 1 / x \u003d 3 or x + 1 / x \u003d 4. We get two equations: x ² - 3x + 1 \u003d 0, x ² - 4x + 1 \u003d 0. Example №7. 3x⁴ - 2x³ - 31x² + 10x + 75 \u003d 0. 75: 3 \u003d 25, 10: (- 2) \u003d -5, (-5) ² \u003d 25. The condition of the generalized return equation is fulfilled k \u003d -5. Solved similarly to example No. 6. Divide both sides of the equation by x ². 3x⁴ - 2x - 31 + 10 / x + 75 / x ² \u003d 0.3 (x ⁴ + 25 / x ²) - 2 (x - 5 / x) - 31 \u003d 0. Let x - 5 / x \u003d y, we square both sides of the equality x ² - 10 + 25 / x ² \u003d y², x ² + 25 / x ² \u003d y² + 10. We have the quadratic equation 3y² - 2y - 1 \u003d 0, y₁ \u003d 1, y₂ \u003d - 1 / 3. x - 5 / x \u003d 1 or x - 5 / x \u003d -1/3. We get two equations: x ² - x - 5 \u003d 0 and 3x² + x - 15 \u003d 0

Examples for an independent solution. 1.78x⁴ - 133x³ + 78x² - 133x + 78 \u003d 0, 2.x - 5x³ + 10x² - 10x + 4 \u003d 0.3x х - x ³ - 10x² + 2x + 4 \u003d 0, 4.6x⁴ + 5x³ - 38x² -10x + 24 \u003d 0.5 x ⁴ + 2x³ - 11x² + 4x + 4 \u003d 0.6 x ⁴ - 5x³ + 10x² -10x + 4 \u003d 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 \u003d 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 \u003d 0 is called a homogeneous equation of the fourth degree with respect to u v. Example No. 8. Solve the equation (x ² - x + 1) ³ + 2x⁴ (x ² - x + 1) - 3x⁶ \u003d 0 Homogeneous equation of the third degree with respect to u \u003d x ²- x + 1, v \u003d x ². We divide both sides of the equation by x ⁶. We checked beforehand that x \u003d 0 is not the root of the equation. (x ² - x + 1 / x ²) ³ + 2 (x ² - x + 1 / x ²) - 3 \u003d 0. (x ² - x + 1) / x ²) \u003d y, y³ + 2y - 3 \u003d 0, y \u003d 1 root of the equation. We divide the polynomial P (x) \u003d 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 an independent solution. 1.2 (x² + 6x + 1) ² + 5 (X² + 6X + 1) (X² + 1) + 2 (X² + 1) ² \u003d 0, 2. (X + 5) ⁴ - 13X² (X + 5) ² + 36X⁴ \u003d 0, 3.2 (X² + X + 1) ² - 7 (X - 1) ² \u003d 13 (X³ - 1), 4.2 (X -1) ⁴ - 5 (X² - 3X + 2) ² + 2 (x - 2) ⁴ \u003d 0, 5. (x ² + x + 4) ² + 3x (x ² + x + 4) + 2x² \u003d 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) No roots.

Undefined coefficient method. Theorem №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 \u003d (y2 + wu + c) (y2 + v₁y + c₁) \u003d y ⁴ + y³ (b₁ + b) + y² (s₁ + s + b₁v) + y (sun ₁ + sv ₁) + ss ₁. According to theorem №3 we have a system of equations: s₁ + s \u003d -4, s₁ + s + s₁v \u003d 5, ss ₁ + sv ₁ \u003d -4, ss ₁ \u003d 1. It is necessary to solve the system in integers. The last equation in integers can have solutions: c \u003d 1, c₁ \u003d 1; c \u003d -1, c₁ \u003d -1. Let с \u003d с ₁ \u003d 1, then from the first equation we have в₁ \u003d -4 –в. Substitute in the second equation of the system в² + 4в + 3 \u003d 0, в \u003d -1, в₁ \u003d -3 or в \u003d -3, в₁ \u003d -1. These values \u200b\u200bfit the third equation in the system. With c \u003d c ₁ \u003d -1 D

Example No. 10. Factor the polynomial y³ - 5y + 2. y³ -5y + 2 \u003d (y + a) (y2 + wu + c) \u003d y³ + (a + b) y² + (ab + c) y + ac. We have a system of equations: a + b \u003d 0, ab + c \u003d -5, ac \u003d 2. Possible integer solutions of the third equation: (2; 1), (1; 2), (-2; -1), (-1 ; -2). Let a \u003d -2, c \u003d -1. From the first equation of the system в \u003d 2, which satisfies the second equation. Substituting these values \u200b\u200binto the desired equality, we get the answer: (y - 2) (y² + 2y - 1). Second way. Y³ - 5y + 2 \u003d y³ -5y + 10 - 8 \u003d (y³ - 8) - 5 (y - 2) \u003d (y - 2) (y² + 2y -1).

Examples for an independent solution. 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 \u003d 0, b) x ⁵ + 5x³ -6x² \u003d 0. Answers: 1) (y² + 2y -2) (y² + 2y +4), 2) (y - 1) ² (y² -2y + 2), 3) (x ² -6x + 18) (x ² + 6x + 18), 4) (y - 1) (y - 3) (y² - 4y + 5), 5a) ± 1; ± √2, 5b) 0; one.

Functional - graphical method for solving equations of higher degrees. Example No. 11. Solve the equation x ⁵ + 5x -42 \u003d 0. Function y \u003d x ⁵ increasing, function y \u003d 42 - 5x decreasing (k

Examples for an independent solution. 1. Using the property of monotonicity of the function, prove that the equation has a single root, and find this root: a) x ³ \u003d 10 - x, b) x ⁵ + 3x³ - 11√2 - x. Answers: a) 2, b) √2. 2. Solve the equation using the functional-graphic method: a) x \u003d ³ √x, b) l x l \u003d ⁵ √x, c) 2 \u003d 6 - x, d) (1/3) \u003d x +4, d ) (x - 1) ² \u003d log₂ x, e) log \u003d (x + ½) ², g) 1 - √x \u003d ln x, h) √x - 2 \u003d 9 / x. Answers: a) 0; ± 1, b) 0; 1, c) 2, d) -1, e) 1; 2, f) 1, g) 1, h) 9.

Vieta's formulas for equations of higher degrees. Theorem 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 the quadratic equation ax² + bx + c \u003d o: x ₁ + x ₂ \u003d -v / a, x₁x ₂ \u003d s / a; For the cubic equation a₃x ³ + a₂x ² + a₁x + a₀ \u003d o: x ₁ + x ₂ + x ₃ \u003d -a₂ / a₃; x₁x ₂ + x₁x ₃ + x₂x ₃ \u003d a₁ / a₃; х₁х₂х ₃ \u003d -а₀ / а₃; ..., for the n-th degree equation: x ₁ + x ₂ + ... x \u003d - a / a, x₁x ₂ + x₁x ₃ + ... + x x \u003d a / a, ..., x₁x ₂ ... · x \u003d (- 1 ) ⁿ a₀ / a. The converse theorem also holds.

Example No. 13. Write a cubic equation, the roots of which are inverse to the roots of the equation x ³ - 6x² + 12x - 18 \u003d 0, and the coefficient at x ³ is 2. 1. By Vieta's theorem for the cubic equation we have: x ₁ + x ₂ + x ₃ \u003d 6, x₁x ₂ + x₁x ₃ + x₂x ₃ \u003d 12, x₁x₂x ₃ \u003d 18. 2. We compose the reciprocals of the given roots and for them we apply the inverse Vieta theorem. 1 / x ₁ + 1 / x ₂ + 1 / x ₃ \u003d (x₂x ₃ + x₁x ₃ + x₁x ₂) / x₁x₂x ₃ \u003d 12/18 \u003d 2/3. 1 / x₁x ₂ + 1 / x₁x ₃ + 1 / x₂x ₃ \u003d (x ₃ + x ₂ + x ₁) / x₁x₂x ₃ \u003d 6/18 \u003d 1/3, 1 / x₁x₂x ₃ \u003d 1/18. We get the equation x ³ + 2 / 3x² + 1 / 3x - 1/18 \u003d 0 · 2 Answer: 2x³ + 4 / 3x² + 2 / 3x -1/9 \u003d 0.

Examples for an independent solution. 1. Write a cubic equation, the roots of which are inverse to the squares of the roots of the equation x ³ - 6x² + 11x - 6 \u003d 0, and the coefficient at x ³ is 8. Answer: 8x³ - 98 / 9x² + 28 / 9x -2/9 \u003d 0. Non-standard methods for solving equations of higher degrees. Example No. 12. Solve the x ⁴ -8x + 63 \u003d 0 equation. Factor the left side of the equation. Let's select the exact squares. X⁴ - 8x + 63 \u003d (x ⁴ + 16x² + 64) - (16x² + 8x + 1) \u003d (x² + 8) ² - (4x + 1) ² \u003d (x ² + 4x + 9) (x ² - 4x + 7) \u003d 0. Both discriminants are negative. Answer: no roots.

Example No. 14. Solve the equation 21x³ + x² - 5x - 1 \u003d 0. If the free term of the equation is ± 1, then the equation is transformed into the reduced equation by replacing x \u003d 1 / y. 21 / у³ + 1 / у² - 5 / у - 1 \u003d 0 · у³, у³ + 5у² -у - 21 \u003d 0. у \u003d -3 is the root of the equation. (y + 3) (y2 + 2y -7) \u003d 0, y \u003d -1 ± 2√2. x ₁ \u003d -1/3, x ₂ \u003d 1 / -1 + 2√2 \u003d (2√2 + 1) / 7, X₃ \u003d 1 / -1 -2√2 \u003d (1-2√2) / 7 ... Example No. 15. Solve the equation 4x³-10x² + 14x - 5 \u003d 0. Multiply both sides of the equation by 2. 8x³ -20x² + 28x - 10 \u003d 0, (2x) ³ - 5 (2x) ² + 14 · (2x) -10 \u003d 0. We introduce a new variable y \u003d 2x, we get the reduced equation y³ - 5y² + 14y -10 \u003d 0, y \u003d 1 is the root of the equation. (y - 1) (y² - 4y + 10) \u003d 0, D

Example No. 16. Prove that the equation x ⁴ + x ³ + x - 2 \u003d 0 has one positive root. Let f (x) \u003d x ⁴ + x ³ + x - 2, f ’(x) \u003d 4x³ + 3x² + 1\u003e o for x\u003e o. The function f (x) is increasing for x\u003e o, and the value f (o) \u003d -2. Obviously, the equation has one positive root ch.d. Example No. 17. Solve the equation 8x (2x² - 1) (8x⁴ - 8x² + 1) \u003d 1. IF Sharygin "Optional course in mathematics for grade 11" .M. Enlightenment 1991 p90. 1. l x l 1 2x² - 1\u003e 1 and 8x⁴ -8x² + 1\u003e 1 2. Make the replacement x \u003d cozy, y € (0; n). For other values \u200b\u200bof y, the values \u200b\u200bof x are repeated, and the equation has no more than 7 roots. 2x² - 1 \u003d 2 cos²y - 1 \u003d cos2y, 8x⁴ - 8x² + 1 \u003d 2 (2x² - 1) ² - 1 \u003d 2 cos²2y - 1 \u003d cos4y. 3. The equation becomes 8 cosycos2ycos4y \u003d 1. Multiply both sides of the equation by siny. 8 sinycosycos2ycos4y \u003d siny. Applying the double angle formula 3 times, we get the equation sin8y \u003d siny, sin8y - siny \u003d 0

Completion of the solution of example No. 17. Apply the sine difference formula. 2 sin7y / 2 cos9y / 2 \u003d 0. Taking into account that y € (0; n), y \u003d 2nk / 3, k \u003d 1, 2, 3 or y \u003d n / 9 + 2nk / 9, k \u003d 0, 1, 2, 3. Returning to the variable x, we obtain answer: Cos2 n / 7, cos4 n / 7, cos6 n / 7, cos n / 9, ½, cos5 n / 9, cos7 n / 9. Examples for an independent solution. Find all values \u200b\u200bof a for which the equation (x ² + x) (x ² + 5x + 6) \u003d a has exactly three roots. The answer is 9/16. Hint: Graph the left side of the equation. F max \u003d f (0) \u003d 9/16. The straight line y \u003d 9/16 intersects the graph of the function at three points. Solve the equation (x ² + 2x) ² - (x + 1) ² \u003d 55. Answer: -4; 2. Solve the equation (x + 3) ⁴ + (x + 5) ⁴ \u003d 16. Answer: -5; -3. Solve the equation 2 (x ² + x + 1) ² -7 (x - 1) ² \u003d 13 (x ³ - 1). Answer: -1; -1/2, 2; 4 Find the number of real roots of the equation x ³ - 12x + 10 \u003d 0 on [-3; 3/2]. Hint: find the derivative and investigate for monot.

Examples for independent solution (continued). 6. Find the number of real roots of the equation x ⁴ - 2x³ + 3/2 \u003d 0. Answer: 2 7. Let x ₁, x ₂, x ₃ be the roots of the polynomial P (x) \u003d x ³ - 6x² -15x + 1. Find X₁² + x ₂² + x ₃². Answer: 66. Direction: Apply Vieta's theorem. 8. Prove that for a\u003e o and arbitrary real in the equation x ³ + ax + b \u003d o has only one real root. Hint: Run proof by contradiction. Apply Vieta's theorem. 9. Solve the equation 2 (x ² + 2) ² \u003d 9 (x ³ + 1). Answer: ½; one; (3 ± √13) / 2. Hint: reduce the equation to a homogeneous one, using the equalities X² + 2 \u003d x + 1 + x² - x + 1, x ³ + 1 \u003d (x + 1) (x ² - x + 1). 10. Solve the system of equations x + y \u003d x ², 3y - x \u003d y². Answer: (0; 0), (2; 2), (√2; 2 - √2), (- √2; 2 + √2). 11. Solve the system: 4y² -3xy \u003d 2x -y, 5x² - 3y² \u003d 4x - 2y. Answer: (o; o), (1; 1), (297/265; - 27/53).

Test. Option 1. 1. Solve the equation (x ² + x) - 8 (x ² + x) + 12 \u003d 0. 2. Solve the equation (x + 1) (x + 3) (x + 5) (x + 7) \u003d - 15 3. Solve the equation 12x² (x - 3) + 64 (x - 3) ² \u003d x ⁴. 4. Solve the equation x ⁴ - 4x³ + 5x² - 4x + 1 \u003d 0 5. Solve the system of equations: x² + 2y² - x + 2y \u003d 6, 1.5x² + 3y² - x + 5y \u003d 12.

Option 2 1. (x ² - 4x) ² + 7 (x ² - 4x) + 12 \u003d 0.2 x (x + 1) (x + 5) (x + 6) \u003d 24.3 x ⁴ + 18 (x + 4) ² \u003d 11x² (x + 4). 4.x ⁴ - 5x³ + 6x² - 5x + 1 \u003d 0. 5.x² - 2xy + y² + 2x²y - 9 \u003d 0, x - y - x²y + 3 \u003d 0.3 option. 1. (x ² + 3x) ² - 14 (x ² + 3x) + 40 \u003d 0 2. (x - 5) (x-3) (x + 3) (x + 1) \u003d - 35.3 x4 + 8x² (x + 2) \u003d 9 (x + 2) ². 4.x ⁴ - 7x³ + 14x² - 7x + 1 \u003d 0. 5.x + y + x ² + y ² \u003d 18, xy + x ² + y² \u003d 19.

Option 4. (x ² - 2x) ² - 11 (x ² - 2x) + 24 \u003d o. (x -7) (x-4) (x-2) (x + 1) \u003d -36. X⁴ + 3 (x -6) ² \u003d 4x² (6 - x). X⁴ - 6x³ + 7x² - 6x + 1 \u003d 0. X² + 3xy + y² \u003d - 1, 2x² - 3xy - 3y² \u003d - 4. Additional task: The remainder of dividing the polynomial P (x) by (x - 1) is 4, the remainder of division by (x + 1) is equal to 2, and when divided by (x - 2) is equal to 8. Find the remainder of division of P (x) by (x ³ - 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; one; 2. Homogeneous equation: u \u003d x -3, v \u003d x² -2; -one; 3; 4. (2; 1); (2/3; 4/3). Indication: 1 · (-3) + 2 · 2 2. -6; -2; -4 ± √6. -3 ± 2√3; - 4; - 2.1 ± √11; 4; - 2. Homogeneous equation: u \u003d x + 4, v \u003d x² 1; 5; 3 ± √13. (2; 1); (0; 3); (- thirty). Indication: 2 · 2 + 1. 3. -6; 2; 4; 12 -3; -2; 4; 12 -6; -3; -one; 2. Homogeneous u \u003d x + 2, v \u003d x² -6; ± 3; 2 (2; 3), (3; 2), (-2 + √7; -2 - √7); (-2 - √7; -2 + √7). Hint: 2 -1. 4. (3 ± √5) / 2 2 ± √3 2 ± √3; (3 ± √5) / 2 (5 ± √21) / 2 (1; -2), (-1; 2). Guideline: 1 4 + 2.

Solution of an additional task. By Bezout's theorem: P (1) \u003d 4, P (-1) \u003d 2, P (2) \u003d 8.P (x) \u003d G (x) (x ³ - 2x² - x + 2) + ax² + inx + from. Substitute 1; - one; 2.P (1) \u003d G (1) · 0 + a + b + c \u003d 4, a + b + c \u003d 4.P (-1) \u003d a - b + c \u003d 2, P (2) \u003d 4a² + 2b + c \u003d 8. Solving the resulting system of three equations, we get: a \u003d b \u003d 1, c \u003d 2. Answer: x ² + 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 through another. 2 points - got one of the solutions. 3 points - one computational error. Additional task: 4 points. 1 point - Bezout's theorem was applied for all four cases. 2 points - made up a system of equations. 3 points - one computational error.

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

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

Equations of the highest degree with integer coefficients

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

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

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

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

Substitute the remaining divisors written out in P n - 1 (x) \u003d 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 as (x - x 1) (x - x 2) P n - 2 (x) \u003d 0. Here P n - 2 (x) will be the quotient of dividing P n - 1 (x) by x - x 2.

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

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

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

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

Example 1

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

Decision

Let's start by finding whole roots.

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

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

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

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

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

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

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

We get that

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

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

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

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

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

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

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

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

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

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

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

Example 2

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

Decision

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

We check them in order:

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

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

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

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

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

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

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

Let's solve the quadratic equation:

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

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

Answer: x \u003d - 3 2 ± i 3 2.

Example 3

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

Decision

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

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

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

Replace the variables y \u003d 2 x:

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

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

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

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