Definitions and statements to 2.2 can be found in .

The root of a polynomial is the number such that
.

Bezout's theorem. For any function
and numbers
the equality is true:

Where
.

Consequence. Number is a root if and only if
divided by
without a trace.

Convenient for dividing into polynomials of the form (
) is Horner's scheme. We draw a table, in the first row of which we write down all the coefficients
(including zero ones).

- coefficients of incomplete quotient from division
on (
);- the remainder of the division, which, according to Bezout’s theorem, is equal to
. If = 0, then they say that
divided by (
) completely and - root of a polynomial
.

Example 33 Divide by
.

Solution. Let's use Horner's scheme. Let's draw a table and perform calculations.

So, where - coefficients of incomplete quotient. Hence,.

Example 34 Find the value of a function
at the point

x = ‑2.

Solution. Using Horner's scheme we divide
to a polynomial
. When filling out the table, we take into account that the coefficients of the fourth and second powers, as well as the free term in the polynomial, are equal to 0.

As a result of the calculations, we obtained a remainder equal to -8. According to Bezout's theorem, it is equal to the value
at the point x = ‑2.

Answer: (–8).

The division algorithm discussed in 2.1 is applicable for division by a polynomial of any degree, but Horner's scheme is applicable only for division by (
).

1. Irreducible polynomials

Definitions and statements for 2.3 can be found in . Polynomial with real coefficients
is irreducible if there are no polynomials
And
with real coefficients of degree less
, such that
. That is, an irreducible polynomial cannot be expanded into a product of polynomials of lower degrees.

Statement. Irreducible polynomials with real coefficients are polynomials of the 1st or 2nd degree with a negative discriminant, and only these.

Factoring a polynomial is its representation as a product of irreducible polynomials.

Basic methods for factoring polynomials:

1. Taking the common factor out of brackets.

2. Using abbreviated multiplication formulas.

Example 35
.

=. When decomposing, we used the formula.

3. Grouping method.

Example 36 Factor a polynomial
.

We group together the terms containing the factor 5:

=
=
=

= [let’s take the common factor out of brackets] =

Example 37 Factor a polynomial
.

We group the terms starting from the first:

We factor the square trinomial by finding its roots:

. In the end

4. Root selection method.

This method is based on the following statements:

Statement 1. If for a polynomial

numbers
- roots, then the equality is true.

Statement 2. For a polynomial with a leading coefficient equal to 1, only divisors of the free term can have integer roots.

Example 38 Possible integer roots of a polynomial
there may be numbers
. Using the selection method it can be established that
and, therefore, 1 is the root of the polynomial.

Example 39 Factor the polynomial.

Solution. According to Statement 2, the only possible integer roots of a polynomial can be divisors of the number -5. These are numbers
. Let's find the value of the polynomial at the point x = ‑ 1:

Therefore, the root of the polynomial
is x = -1. Divide the polynomial
on ( x + 1). According to Bezout's theorem,
must be divisible by ( x + 1) completely, that is, the remainder of the division must be equal to zero. For division we will use Horner's scheme.

The number obtained in the last column allows you to check the correctness of the calculations. If the result is zero, then all calculations are correct. If the number in the last column is non-zero, it means that either the root was found incorrectly, or the calculations using Horner’s scheme were carried out incorrectly.

So: . Since the polynomial resulting from division
is not irreducible, then the factorization process must be continued. For a polynomial
possible roots are numbers
. We find:. Therefore, 1 is the root of the polynomial
. Let's divide it by ( x - 1) according to Horner’s scheme.

The last column is zero. This means the calculations are correct.

We have: . Let's check if the polynomial is
irreducible. Let's find its roots using the standard formula:

. Since the discriminant of this quadratic trinomial is negative, it is irreducible on the set of real numbers.

Properties

Finding roots

The method of finding the roots of linear and quadratic polynomials, that is, the method of solving linear and quadratic equations, was known back in ancient world. The search for a formula for the exact solution of a general equation of the third degree continued for a long time (the method proposed by Omar Khayyam should be mentioned) until it was crowned with success in the first half of the 16th century in the works of Scipione del Ferro, Niccolò Tartaglia and Gerolamo Cardano. Formulas for the roots of quadratic and cubic equations made it relatively easy to obtain formulas for the roots of equations of the fourth degree.

The fact that the roots of a general equation of the fifth degree and higher cannot be expressed using rational functions and radicals of coefficients was proven by the Norwegian mathematician Niels Abel in 1826. This does not mean at all that the roots of such an equation cannot be found. Firstly, in special cases, for certain combinations of coefficients, the roots of the equation can be determined with some ingenuity. Secondly, there are formulas for the roots of equations of degree 5 and higher, which, however, use special functions - elliptic or hypergeometric (see, for example, the Bring root).

If all the coefficients of a polynomial are rational, then finding its roots leads to finding the roots of a polynomial with integer coefficients. For rational roots of such polynomials, there are algorithms for finding candidates by searching using Horner's scheme, and when finding whole roots, searching can be significantly reduced by cleaning the roots. Also in this case, you can use the polynomial LLL algorithm.

To approximately find (with any required accuracy) the real roots of a polynomial with real coefficients, iterative methods are used, for example, the secant method, the bisection method, Newton's method. The number of real roots of a polynomial on an interval can be estimated using Sturm's theorem.

See also

Notes

Wikimedia Foundation. 2010.

• Sewerage
• Glossary of vexillology terms

See what “Root of a polynomial” is in other dictionaries:

Root of algebraic equation

Root of the equation- The root of a polynomial over the field k is an element that, after substituting it for x, turns the equation into an identity. Properties If c is a root of the polynomial p(x ... Wikipedia

Bring root- Check the information. It is necessary to check the accuracy of the facts and reliability of the information presented in this article. There should be an explanation on the talk page. In algebra, the Bring root or ultraradical is an analytic function such that for... ... Wikipedia

Root (disambiguation)- Root: Wiktionary has an article “root” Root (in botany) is a vegetative axial underground organ of a plant that has sp ... Wikipedia

Root (in mathematics)- Root in mathematics, 1) K. degree n of the number a ≈ number x (denoted), the nth degree of which is equal to a (that is, xn = a). The action of finding K. is called root extraction. For a ¹ 0 there are n different values ​​of K. (generally speaking, ... ...

Root- I Root (radix) is one of the main vegetative organs of leafy plants (with the exception of mosses), serving for attachment to the substrate, absorption of water from it and nutrients, primary transformation of a number of absorbed substances,... ... Great Soviet Encyclopedia

ROOT- 1) K. degree n from the number a number n and the i degree x n to the number is equal to a. 2) The equation of an algebraic equation over a field K, the element which, after substituting it in place, turns the equation into an identity. K. of this equation is called. also K. polynomial If it appears... ... Mathematical Encyclopedia

Multiple root- polynomial f (x) = a0xn + a1xn ​​1 +... + an, a number c such that f (x) is divisible without a remainder by the second or higher power of the binomial (x c). In this case, c is called the root of multiplicity if f (x) is divisible by (x c) k, but not... ... Great Soviet Encyclopedia

Conjugate root- If some irreducible polynomial over a ring is given and some of its roots in the extension are chosen, then the conjugate root for a given root of the polynomial is called any root of the polynomial ... Wikipedia

Square root of 2- equal to the length of the hypotenuse in right triangle with leg length 1. Square root from number 2 is positive ... Wikipedia

If a number c is the root of a polynomial f(x), this polynomial is known to be divisible by x-c. It may happen that f (x) is also divisible by some power of the polynomial x-c, i.e. on (x-c) k, k>1. In this case, c is called a multiple root. Let us formulate the definition more clearly.

A number c is called a root of multiplicity k (k-fold root) of a polynomial f (x), if the polynomial is divisible by (x-c) k, k>1 (k - natural number), but is not divisible by (x-c) k+1. If k=1, then c is called a simple root, and if k>1, then it is called a multiple root of the polynomial f (x).

In the future, when determining the multiplicity of roots, the following sentence will be useful to us.

If the polynomial f (x) is represented as f (x) = (x-c) mg (x), m is a natural number, then it is divisible by (x-c) m+1 if and only if g (x) is divisible on x-s. In fact, if g (x) is divisible by x-c, i.e. g (x) = (x-c) s (x), then f (x) = (x-c) m+1s (x), which means f (x) is divisible by (x-c) m+1.

Conversely, if f (x) is divisible by (x-c) m+1, then f (x) = (x-c) m+1s (x). Then (x-c) mg (x) = (x-c) m+1s (x) and after reduction by (x-c) m we get g (x) = (x-c) s (x). It follows that g(x) is divisible by x-c.

Now let's return to the concept of root multiplicity. Let's find out, for example, whether the number 2 is the root of the polynomial f (x) =x5-5x4+3x3+22x2-44x+24, and if so, find its multiplicity. To answer the first question, let's check using Horner's circuit whether f (x) is divisible by x-2. we have:

Table 4

We found that g (x) is divisible by x-2 and g (x) = (x-2) (x3-x2-5x+6). Then f (x) = (x-2) 2 (x3-x2-5x+6).

So f(x) is divisible by (x-2)2, now we need to find out if f(x) is divisible by (x-2)3.

To do this, let’s check whether h (x) = x3-x2-5x+6 is divisible by x-2:

Table 6

We find that the remainder when dividing s (x) by x-2 is 3, i.e. s(x) is not divisible by x-2. This means f(x) is not divisible by (x-2)4.

Thus, f(x) is divisible by (x-2)3, but not divisible by (x-2)4. Therefore, the number 2 is the multiplicity 3 root of the polynomial f(x).

Typically, checking the root for multiplicity is performed in one table. For this example, this table looks like this:

Table 8

In other words, according to Horner's scheme, dividing the polynomial f (x) by x-2, in the second line we get the coefficients of the polynomial g (x). Then we consider this second line to be the first line of the new Horner system and divide g (x) by x-2, etc. We continue the calculations until we get a remainder that is different from zero. In this case, the multiplicity of the root is equal to the number of zero residues obtained. The line containing the last non-zero remainder also contains the coefficients of the quotient when dividing f (x) by (x-2) 3. Now, using the just proposed scheme for checking the root for multiplicity, we will solve the following problem. For what a and b does the polynomial f (x) =x4+2x3+ax2+ (a+b) x+2 have the number - 2 as a root of multiple 2?

Since the multiplicity of the root - 2 must be equal to 2, then when dividing by x + 2 according to the proposed scheme, we must obtain a remainder of 0 twice, and the third time - a remainder different from zero. We have:

Table 9

Thus, the number - 2 is a root of multiplicity 2 of the original polynomial if and only if

From here we get: a=-7/2, b=-5/2.

Angle division scheme

Division of polynomials

Division with remainder. Theorem. If P(x) and S(x) 0 are two polynomials, then there is a unique pair of polynomials Q(x) and R(x) that satisfies the relations: 1) , 2) or the degree of R(x) is less than or equal to degree S(x), or R(x) = 0.

Q(x) is called the quotient, and R(x) is the remainder.

Example 1. , . Find the quotient and remainder of the polynomial P(x) divided by S(x).

Answer: quotient, remainder.

Example 2. Find the quotient and remainder when dividing by .

Answer: quotient equals ; the remainder is zero.

Theorem. A polynomial P(x) is divisible by a polynomial S(x) if the remainder when dividing P(x) by S(x) is zero.

It follows from the theorem that in order to find out whether the polynomial P(x) is divisible by S(x), you can perform division by angle and find the remainder. If the remainder is zero, then the polynomial P(x) is divided by the polynomial S(x).

Example 3. Determine whether a polynomial is divisible

to a polynomial?

Let us divide the polynomial P(x) by S(x) with a “corner”. As a result, we get that the quotient is equal to , and the remainder is equal to zero. This means that the polynomial P(x) is divisible by the polynomial S(x).

Let c be some real number (in the general case, a complex number). Meaning polynomial P(x) for x = c is the number that is obtained by substituting x into the given polynomial and performing the actions.

If , then the value of this polynomial at x = c is denoted by P(c): .

Example 1. The value of the polynomial P(x) = at x = 2 is:

at x = 0, P(0) = -5; at x = 1, P(1) = 3 - 2 + 4 - 5 = 0.

Thus, at x = 0, the value of the polynomial is equal to the free term:

for x = 1, the value of the polynomial is equal to the sum of its coefficients:

Definition. If at the value of the polynomial is equal to zero, then it is called the root of the polynomial P(x).

Example 1. A polynomial is given. When x = 2, the value of this polynomial is zero, which means x = 2 is the root of the polynomial S(x).

The fact that when x = 1 the value of a polynomial is equal to the sum of its coefficients is used in reverse: if the sum of the coefficients of a polynomial is zero, then x = 1 is the root of that polynomial.

Definition. If the task is to find all values ​​of the variable x for which the polynomial f(x) is equal to zero, then we say that we need to solve the equation f(x) = 0.

Let us especially highlight that decide equation means to find All its roots.

Thus, algebraic equation the equation is called f(x) = 0, where f(x) is some polynomial. If f(x) is a polynomial nth degree, then the equation is called algebraic equation nth degrees .

When solving algebraic equations, the following theorem (called Bezout's theorem) is useful.

Theorem 1. The remainder of the division of the polynomial f(x) by x - a is equal to f(a) (i.e., equal to the value of this polynomial at x = a).

Proof

Let us divide with the remainder of the polynomial f(x) by x - a:

where the remainder r(x), if it is not equal to zero, is a polynomial whose degree is less than the degree of the divisor x - a, i.e., equal to zero. Therefore r(x) = r is number:

To find the number r, let's put x = a in this equality. Then, we get f(a) = r, which proves the theorem.

Consequence. If a is the root of a polynomial f(x), then this polynomial is divisible by .

Example 1. Given a polynomial. It is easy to see that 1 is the root of this polynomial, in fact: therefore, by corollary to the theorem, the polynomial must be divisible by x - 1.

Divide the polynomial by x - 1 with a “corner”:

The remainder is zero, which means the polynomial is divisible by x - 1.

Theorem 2. If all the coefficients of the polynomial

are integers, then every integer root of this polynomial is a divisor of the free term.

Proof

Let c be an integer root of the polynomial f(x), i.e.

Since the number in parentheses is an integer (since all coefficients are integers, by convention), it is divisible by c.

The proven theorem greatly simplifies the search for integer roots of polynomials with integer coefficients.

1 . We need to find and write down all the divisors of the free term (positive and negative).

2 . You can check (by substitution) which of them are roots of a given polynomial.

3 . If no divisor of the free term makes the polynomial vanish, then this polynomial has no integer roots.

Example 1. Solve the equation.

1. Find the divisors of the free term 12: .

2. If the equation has integer roots, then they are among these divisors, let’s check this. We denote the polynomial on the left side of the equation as f(x).

f(1) = 24, which means 1 is not the root of the equation;

f(-1) = -24, which means -1 is not the root of the equation;

f(2) = 0, which means 2 is the root of the equation.

3. According to Bezout’s theorem, the polynomial f(x) is divisible by x - 2. By dividing with a “corner”, we find: .

To find the remaining roots you need to solve the equation

We repeat the previous process again.

1. We write down the divisors of the free term 6: .

2. Check them. The numbers 1 and -1 have already been checked. Let's test other divisors by substituting them one by one into the polynomial.

g(2) = -40, which means 2 is not a root of the polynomial g(x);

g(-2) = 12, -2 is not a root;

g(3) = -48, 3 is not a root;

g(-3) = 0, which means -3 is the root of the polynomial g(x).

According to Bezout's theorem, it is divided by x + 3. As a result of division we get:

To find other roots, if they exist, we solve quadratic equation.

Thus, the original fourth-degree equation has four roots.

Answer: , , , .

Comment. Sometimes it can be difficult to check the supposed roots of a polynomial or calculate its value, especially if the polynomial is of high degree and the numbers being tested are large.

To facilitate this process, there is a Horner scheme.

Lesson objectives:

• teach students to solve equations higher degrees using Horner's scheme;
• develop the ability to work in pairs;
• create, in conjunction with the main sections of the course, a basis for developing students’ abilities;
• help the student assess his potential, develop interest in mathematics, the ability to think, and speak out on the topic.

Equipment: cards for group work, poster with Horner's diagram.

Teaching method: lecture, story, explanation, performing training exercises.

Control form: checking independent solution problems, independent work.

Lesson progress

1. Organizational moment

2. Updating students’ knowledge

What theorem allows you to determine whether a number is the root of a given equation (formulate a theorem)?

Bezout's theorem. The remainder of the division of the polynomial P(x) by the binomial x-c is equal P(c), the number c is called the root of the polynomial P(x) if P(c)=0. The theorem allows, without performing the division operation, to determine whether given number the root of the polynomial.

What statements make it easier to find roots?

a) If the leading coefficient of the polynomial equal to one, then the roots of the polynomial should be sought among the divisors of the free term.

b) If the sum of the coefficients of a polynomial is 0, then one of the roots is 1.

c) If the sum of the coefficients in even places is equal to the sum of the coefficients in odd places, then one of the roots is equal to -1.

d) If all coefficients are positive, then the roots of the polynomial are negative numbers.

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

3. Learning new material

When solving entire algebraic equations, you have to find the values ​​of the roots of polynomials. This operation can be significantly simplified if the calculations are carried out using a special algorithm called the Horner scheme. This circuit is named after the English scientist William George Horner. Horner's scheme is an algorithm for calculating the quotient and remainder of dividing the polynomial P(x) by x-c. Briefly how it works.

Let an arbitrary polynomial P(x) = a 0 x n + a 1 x n-1 + …+ a n-1 x+ a n be given. Dividing this polynomial by x-c is its representation in the form P(x)=(x-c)g(x) + r(x). Partial g(x)=in 0 x n-1 + in n x n-2 +…+in n-2 x + in n-1, where in 0 =a 0, in n =st n-1 +a n , n=1,2,3,…n-1. Remainder r(x)= st n-1 +a n. This calculation method is called the Horner scheme. The word “scheme” in the name of the algorithm is due to the fact that its execution is usually formalized as follows. First, draw table 2(n+2). In the lower left cell write the number c, and in the top line the coefficients of the polynomial P(x). At the same time, the left top cell left empty.

The number that, after executing the algorithm, turns out to be written in the lower right cell is the remainder of the division of the polynomial P(x) by x-c. The other numbers in 0, in 1, in 2,... in the bottom line are the coefficients of the quotient.

For example: Divide the polynomial P(x)= x 3 -2x+3 by x-2.

We get that x 3 -2x+3=(x-2) (x 2 +2x+2) + 7.

4. Consolidation of the studied material

Example 1: Factor the polynomial P(x)=2x4-7x 3 -3x 2 +5x-1 into factors with integer coefficients.

We are looking for whole roots among the divisors of the free term -1: 1; -1. Let's make a table:

X = -1 – root

P(x)= (x+1) (2x 3 -9x 2 +6x -1)

Let's check 1/2.

Therefore, the polynomial P(x) can be represented in the form

P(x)= (x+1) (x-1/2) (x 2 -8x +2) = (x+1) (2x -1) (x 2 - 4x +1)

Example 2: Solve the equation 2x 4 - 5x 3 + 5x 2 - 2 = 0

Since the sum of the coefficients of the polynomial written on the left side of the equation is equal to zero, then one of the roots is 1. Let’s use Horner’s scheme:

We get P(x)=(x-1) (2x 3 -3x 2 =2x +2). We will look for roots among the divisors of free term 2.

We found out that there were no more intact roots. Let's check 1/2; -1/2.

Answer: 1; -1/2.

Example 3: Solve the equation 5x 4 – 3x 3 – 4x 2 -3x+ 5 = 0.

We will look for the roots of this equation among the divisors of the free term 5: 1;-1;5;-5. x=1 is the root of the equation, since the sum of the coefficients is zero. Let's use Horner's scheme:

Let's present the equation as a product of three factors: (x-1) (x-1) (5x 2 -7x + 5) = 0. Solving the quadratic equation 5x 2 -7x+5=0, we got D=49-100=-51, there are no roots.

Card 1

1. Factor the polynomial: x 4 +3x 3 -5x 2 -6x-8
2. Solve the equation: 27x 3 -15x 2 +5x-1=0

Card 2

1. Factor the polynomial: x 4 - x 3 -7x 2 +13x-6
2. Solve the equation: x 4 +2x 3 -13x 2 -38x-24=0

Card 3

1. Factor into: 2x 3 -21x 2 +37x+24
2. Solve the equation: x 3 -2x 2 +4x-8=0

Card 4

1. Factor into: 5x 3 -46x 2 +79x-14
2. Solve the equation: x 4 +5x 3 +5x 2 -5x-6=0

5. Summing up

Testing knowledge when solving in pairs is carried out in class by recognizing the method of action and the name of the answer.

Homework:

Solve the equations:

a) x 4 -3x 3 +4x 2 -3x+1=0

b) 5x 4 -36x 3 +62x 2 -36x+5=0

c) x 4 + x 3 + x + 1 = 4x 2

d) x 4 +2x 3 -x-2=0

Literature

1. N.Ya. Vilenkin et al., Algebra and the beginnings of analysis, grade 10 (in-depth study of mathematics): Enlightenment, 2005.
2. U.I. Sakharchuk, L.S. Sagatelova, Solution of equations of higher degrees: Volgograd, 2007.
3. S.B. Gashkov, Number systems and their application.