Let's find the remainder of the division of the polynomial P(x) to a linear binomial of the form ( x ‑ a), Where a– a certain number. Since the divisor polynomial has a first degree, the remainder must have a zero degree, that is, it must be a certain number r. Then if Q(x) is a quotient polynomial, then the equality holds: P(x) = Q(x)·( x ‑ a) + r. Substituting into the resulting equality instead x number a, we get: P(a) = Q(a)·( a ‑ a) + r = Q(a)·0 + r = r. Thus, it turns out that the remainder of the division of the polynomial P(x) by binomial ( x ‑ a) can be found without performing division by substituting into the dividend polynomial a instead of x. The proven statement, successfully used in solving many non-standard problems, is the essence of Bezout’s theorem (Etienne Bezout, 1730 - 1783, French mathematician, member of the Paris Academy of Sciences).

Bezout's theorem: Remainder r from dividing a polynomial P(x) by binomial ( x ‑ a) is equal to the value of this polynomial at the point a, i.e. r = P(a).

Note 1: Polynomial called given, if its leading coefficient (i.e., the coefficient of the term of the highest degree) is equal to 1. For example, the polynomials , are reduced, but , are not.

Note 2: When dividing a polynomial with integer coefficients by a reduced polynomial with integer coefficients, all coefficients of the quotient polynomial and the remainder polynomial also turn out to be integer (this is easy to understand by remembering how a polynomial is divided by a “corner” polynomial). In particular, when dividing a polynomial with integer coefficients by a binomial ( x ‑ a), Where a– an integer, all coefficients of the quotient polynomial are integers.

Number a called root of the polynomial P(x), If P(a) = 0 (in other words, if the number a- root of the equation P(x) = 0). For example, the numbers 1 and -1 are the roots of the polynomial, the numbers -2 and 5 are the roots of the polynomial, and the polynomial has no roots, since equality is impossible. From Bezout's theorem it follows that if a number a is the root of the polynomial P(x), then the remainder of the division of the polynomial P(x) by binomial ( x ‑ a) is equal P(a) = 0, that is, a polynomial P(x) divided by ( x ‑ a) without remainder. In other words, if a– root of a polynomial P(x), That P(x) will be presented in the form: P(x) = (x ‑ a)· Q(x). This statement is the essence of the corollary to Bezout's theorem.

Corollary to Bezout's theorem: Number a is the root of the polynomial P(x) if and only if P(x) divided by ( x ‑ a) without remainder.

Tasks:

1. Find the remainder when dividing the polynomial by .

2. Find a polynomial of the third degree which, when divided by x gives remainder 1, on x- 2 – remainder 3, and is divisible by without remainder.

3. Prove that the polynomial is divisible by .

4. At what values a And b polynomial is divisible without remainder into the following polynomials:

5. When dividing a polynomial by x- 1 leaves a remainder of 2, and when divided by x- 2 - 1. What is the remainder when this polynomial is divided by ?

6. Find the remainder when dividing the polynomial by .

7. It is known that the remainder when dividing a polynomial by is equal to 2 x+ 1. Find the remainder when dividing this polynomial by:

A) x – 1;

b) 3 x + 2;

8. Find the reduced fourth-degree polynomial if it is known that it is divisible by , and when divided by it leaves a remainder of .

9. What can be the remainders from division of a polynomial? P(x) on x– 1, and the polynomial Q(x) - on x+ 1 if when dividing by x 2 – 1 polynomial leaves a remainder of -6?

10. Prove that the number is divisible by 7.

11. Prove that the remainders of division by 11 of the numbers 100.000 and 1.000.000.000 are equal.

12. Find the remainder when dividing the number by 26.

13. One of the roots of the equation is equal to 3. Find the value of the parameter a and solve the equation.

Gal.: page 111, No. 9.10 (b).

Homework:

14. Without performing division, find the remainder when dividing the polynomial by x + 2.

Etienne Bezout–

French mathematician, member of the Paris Academy of Sciences (since 1758), born in Nemours on March 31, 1730 and died on September 27, 1783.

Since 1763, Bezu taught mathematics at the midshipmen's school, and since 1768 at the Royal Artillery Corps.

Etienne Bezout's main works relate to higher algebra; they are devoted to the creation of a theory for solving algebraic equations. In the theory of solving systems of linear equations, he contributed to the emergence of the theory of determinants, developed the theory of eliminating unknowns from systems of equations of higher degrees, and proved the theorem (first formulated by K. Maclaurin) that two curves of order m and n intersect at most mn points. In France and abroad, until 1848, his six-volume “Course of Mathematics,” written by him in 1764-69, was very popular. Bezou developed the method of indefinite multipliers; in elementary algebra, a method for solving systems of equations based on this method is named after him. Part of Bezout's works is devoted to external ballistics. One of the basic theorems of algebra is named after the scientist.

Bezout's theorem.

Remainder of polynomial division P n ( x )

by binomial ( x - a ) is equal to the value

this polynomial at x = a .

Pn(x) – given polynomial of degree n ,

binomial (x- a) - its divisor,

Qn-1 (x) – quotient of division Pn(x) on x- a(polynomial of degree n-1) ,

R- remainder of the division ( R does not contain a variable x as a first degree divisor with respect to x).

Proof:

According to the rule for dividing polynomials with a remainder, we can write:

Pn(x) = (x-a)Qn-1(x)+R .

Hence, at x = a :

Pn(a) = (a-a)Qn-1(a) + R =0*Qn-1(a)+R=

=0+ R= R .

Means, R = Pn(a) , i.e. remainder when dividing a polynomial by (x- a) equal to the value of this

polynomial at x= a, which was what needed to be proven.

Corollaries from the theorem .

WITH consequence 1 :

Remainder of polynomial division P n ( x )

by binomial ax + b equal to the value

this polynomial at x = - b / a ,

T . e . R=P n (-b/a) .

Proof:

According to the rule for dividing polynomials:

Pn(x)= (ax + b)* Qn-1(x)+R.

Pn (-b/a) = (a(-b/a) + b)Qn-1(-b/a) + R = R. This means that R = Pn (-b/a) , which is what needed to be proved.

Corollary 2 :

If the number a is the root

polynomial P ( x ) , That this

the polynomial is divisible by ( x - a ) without

remainder.

Proof:

According to Bezout's theorem, the remainder of a polynomial is P (x) on x- a equals P (a) , and by condition a is the root P (x) , which means that P (a) = 0 , which is what needed to be proven .

From this corollary of Bezout’s theorem it is clear that the problem of solving the equation P (x) = 0 is equivalent to the problem of identifying divisors of a polynomial P having the first degree (linear divisors).

Corollary 3 :

If a polynomial P ( x ) It has

pairwise distinct roots

a 1 , a 2 , … , a n , then it is divided by

work ( x - a 1 ) … ( x - a n )

without a trace .

Proof:

Let's carry out the proof using mathematical induction on the number of roots. At n=1 the statement is proved in Corollary 2. Suppose it has already been proven for the case when the number of roots is equal k, it means that P(x) divisible without remainder by (x- a1 )(x- a2 ) … (x- ak) , Where

a1 ,a2 , … , ak- its roots.

Let P(x) It has k+1 pairwise different roots. By the induction hypothesis a1 , a2 , ak , … , ak+1 are roots of the polynomial, which means the polynomial is divisible by the product (x- a1 ) … (x- ak) , where does it come from that

P(x) = (x-a1 ) … (x-ak)Q(x).

Wherein ak+1 – root of a polynomial P(x) , i.e. . P(ak+1 ) = 0 .

So, substituting instead xak+1 , we get the correct equality:

P(ak+1) = (ak+1-a1 ) ... (ak+1-ak)Q(ak+1) =

But ak+1 different from numbers a1 , … , ak, and therefore none of the numbers ak+1 - a1 , … , ak+1 - ak not equal to 0. Therefore, zero is equal Q(ak+1 ) , i.e. ak+1 – root of a polynomial Q(x) . And from Corollary 2 it turns out that Q(x) divided by x- ak+ 1 without remainder.

Q(x) = (x- ak+1 ) Q1 (x) , and that's why

P(x) = (x-a1) … (x-ak)Q(x) =

=(x- a1 ) … (x- ak)(x- ak+1 ) Q1 (x) .

This means that P(x) divided by (x- a1 ) … (x- ak+1 ) without a trace.

So, it has been proven that the theorem is true for k =1 , and from its validity at n = k it follows that it is also true when n = k+1 . Thus, the theorem is true for any number of roots, what andneeded to be proven .

Corollary 4 :

Polynomial of degree n has no more

n different roots.

Proof:

Let's use the method by contradiction: if the polynomial Pn(x) degrees n would have more n roots - n+ k (a1 , a2 , … , an+ k- its roots), then, according to the previously proven Corollary 3, it

would be divided by the product (x- a1 ) … (x- an+ k) , having a degree n+ k, which is impossible.

We have come to a contradiction, which means our assumption is incorrect and a polynomial of degree n cannot have more than n roots, Q.E.D.

Corollary 5 :

For any polynomial P ( x )

and numbers a difference

( P ( x )- P ( a )) is divided without

remainder by binomial ( x - a ) .

Proof:

Let P(x) – given polynomial of degree n , a- any number.

Polynomial Pn(x) can be represented as: Pn(x)=(x- a) Qn-1 (x)+ R ,

Where Qn-1 (x) – polynomial, quotient when dividing Pn(x) on (x- a) ,

R- remainder of the division Pn(x) on (x- a) .

Moreover, according to Bezout’s theorem:

R = Pn(a), i.e.

Pn(x)=(x-a)Qn-1(x)+Pn(a) .

Pn(x) - Pn(a) = (x-a)Qn-1(x) ,

and this means divisibility without remainder (Pn(x) – Pn(a))

on (x- a) , which is what needed to be proven .

Corollary 6 :

Number a is the root

polynomial P ( x ) degrees

not lower than the first then and

only when

P ( x ) divided by ( x - a )

without a trace .

Proof:

To prove this theorem, it is necessary to consider the necessity and sufficiency of the formulated condition.

1. Necessity .

Let a– root of a polynomial P(x) , then by Corollary 2 P(x) divided by (x- a) without a trace.

Thus divisibility P(x) on (x- a) is a necessary condition for a was the root P(x) , because is a consequence of this.

2. Adequacy .

Let the polynomial P(x) divisible without remainder by (x- a) ,

Then R = 0 , Where R- remainder of the division P(x) on (x- a) , but according to Bezout’s theorem R = P(a) , where does it come from that P(a) = 0 , which means that a is the root P(x) .

Thus divisibility P(x) on (x- a) is also a sufficient condition for a was the root P(x) .

Divisibility P(x) on (x- a) is necessary and sufficient condition for a was the root P(x) , Q.E.D.

A polynomial that has no real

solid roots, in decomposition

to factors of linear factors

does not contain.

Proof:

Let's use the method by contradiction: suppose that the polynomial without roots P(x) when factored, contains a linear factor (x – a) :

P(x) = (x – a)Q(x),

then it would be divided by (x – a) , but by Corollary 6 a would be the root P(x) , but by condition it does not contain roots. We have come to a contradiction, which means our assumption is incorrect and the polynomial

Proof of Bezout's theorem

Let f(x) denote an arbitrary polynomial of the nth degree with respect to the variable x and let when it is divided by the binomial (x-a) the result is q(x) in the quotient and R in the remainder. Obviously, q(x) will be some polynomial ( n-1)th degree relative to x, and the remainder R will be a constant value, i.e. independent of x.

If the remainder R were a polynomial of at least the first degree with respect to x, then this would mean that the division failed. So R does not depend on x.

By the definition of division (the dividend is equal to the product of the divisor and the quotient plus the remainder), I get the identity

f(x) =(x-a)q(x)+R.

This equality is true for any value of x, which means it is also true for x=a.

Substituting the number a into the left and right sides of the equality instead of the variable x, I get:

f(a)=(a-a)q(a)+R. (1)

Here the symbol f(a) no longer denotes f(x), i.e. not a polynomial for x, but the value of this polynomial at x=a. q(a) denotes the value of q(x) at x=a.

The remainder of R remains the same as it was before, since R does not depend on x.

The product (a-a)q(a) is equal to zero, since the factor (a-a) is equal to zero, and the factor q(a) is a certain number. (The polynomial q(x) does not lose its meaning for any particular value of x.)

Therefore, from equality (1) we obtain:

Q.E.D.

Corollaries from the theorem

Corollary 1.

The remainder of dividing the polynomial f(x) by the binomial (ax+b) is equal to the value

of this polynomial at x=-b/a, i.e. R=f(-b/a).

Proof:

According to the rule for dividing polynomials:

f(x)= (ax+b)*q(x)+R.

f(-b/a)=(a(-b/a)+b)q(-b/a)+R=R. So, R=f(-b/a),

Q.E.D.

Corollary 2:

If the number a is the root of a polynomial f(x), then this polynomial is divisible by (x-a) without a remainder.

Proof:

By Bezout’s theorem, the remainder of dividing the polynomial f(x) by (x-a) is equal to f(a), and by condition, a is the root of f(x), which means that f(a) = 0, which is what needed to be proved.

From this corollary of Bezout’s theorem it is clear that the problem of solving the equation f(x) = 0 is equivalent to the problem of identifying divisors of the polynomial f that have the first degree (linear divisors).

Corollary 3:

If a polynomial f(x) has pairwise different roots a 1 , a 2 ,… ,a n , then it is divided by the product (x-a 1)…(x-a n) without a remainder.

Proof:

Let's carry out the proof using mathematical induction on the number of roots. For n=1, the statement is proven in Corollary 2. Let it already be proven for the case when the number of roots is equal to k, this means that f(x) is divisible without remainder by

(x-a 1)(x-a 2)…(x-a k), where a 1, a 2,…, a k are its roots.

Let f(x) have (k+1) pairwise distinct roots. By the induction hypothesis, a 1, a 2, a k,…, (a k+1) are the roots of the polynomial, which means that the polynomial is divided into the product (x-a 1)…(x-a k), which means that

f(x)=(x-a 1)…(x-a k)q(x).

In this case (a k+1) is the root of the polynomial f(x), i.e.

This means that substituting (a k+1) for x, we obtain the correct equality:

f(a k+1)=(a k+1 -a 1)…(a k+1 -a k)q(a k+1)=0.

But (a k+1) is different from the numbers a 1 ,…, a k , and therefore none of the numbers (a k+1 -a 1),…, (a k+1 -a k) is equal to 0. Therefore, zero is equal to q(a k+1), i.e. (a k+1) is the root of the polynomial q(x). And from Corollary 2 it turns out that q(x) is divisible by (x-a k+1) without a remainder.

q(x)=(x-a k+1)q 1 (x), and therefore

f(x)=(x-a 1)...(x-a k)q(x)=(x-a 1)...(x-a k)(x-a k+1)q 1 (x).

This means that f(x) is divided by (x-a 1)…(x-a k+1) without a remainder.

So, it has been proven that the theorem is true for k=1, and from its validity for n=k it follows that it is also true for n=k+1. Thus, the theorem is true for any number of roots, which is what needed to be proven.

Corollary 4:

A polynomial of degree n has at most n different roots.

Proof:

Let's use the method by contradiction: if a polynomial f(x) of degree n had more than n roots - n+k (a 1 , a 2 ,..., a n+k are its roots), then according to the previously proven corollary 3 it would be divided into the product (x-a 1)...(x-a n+k), which has degree (n+k), which is impossible.

We have arrived at a contradiction, which means our assumption is incorrect, and a polynomial of degree n cannot have more than n roots, which is what we needed to prove.

Corollary 5:

For any polynomial f(x) and number a, the difference (f(x)-f(a)) is divided without remainder by the binomial (x-a).

Proof:

Let f(x) be a given polynomial of degree n, a be any number.

The polynomial f(x) can be represented as: f(x)=(x-a)q(x)+R, where q(x) is the polynomial quotient when dividing f(x) by (x-a), R is the remainder of the division f(x) to (x-a).

Moreover, according to Bezout’s theorem:

f(x)=(x-a)q(x)+f(a).

f(x)-f(a)=(x-a)q(x),

and this means divisibility without remainder (f(x)-f(a))

on (x-a), which is what needed to be proved.

Corollary 6:

A number a is a root of a polynomial f(x) of degree at least first only if f(x) is divisible by (x-a) without a remainder.

Proof:

To prove this theorem, it is necessary to consider the necessity and sufficiency of the formulated condition.

1. Necessity.

Let a be the root of the polynomial f(x), then by Corollary 2 f(x) is divisible by (x-a) without a remainder.

Thus, the divisibility of f(x) by (x-a) is a necessary condition for a to be a root of f(x), because is a consequence of this.

2. Sufficiency.

Let the polynomial f(x) be divided without remainder by (x-a),

then R=0, where R is the remainder of dividing f(x) by (x-a), but by Bezout’s theorem R=f(a), which means that f(a)=0, which means that a is the root f(x).

Thus, the divisibility of f(x) by (x-a) is also a sufficient condition for a to be a root of f(x).

The divisibility of f(x) by (x-a) is a necessary and sufficient condition for a to be a root of f(x), which is what we needed to prove.

Corollary 7:

A polynomial that has no real roots does not contain linear factors when factored.

Proof:

Let's use the method by contradiction: suppose that the unrooted polynomial f(x) when factored contains a linear factor

then it would be divisible by (x-a), but by Corollary 6 a would be a root of f(x), and by condition it does not contain real roots. We have arrived at a contradiction, which means our assumption is incorrect and a polynomial that does not have real roots does not contain linear factors in its factorization, which is what we needed to prove.

1. Divide 5x 4 + 5 x 3 + x 2 − 11 on x − 1 using Horner's scheme.

Solution:

Let's make a table of two lines: in the first line we write the coefficients of the polynomial 5 x 4 +5x 3 +x 2 −11, arranged in descending order of degrees of the variable x. Note that this polynomial does not contain x in the first degree, i.e. coefficient before x to the first power is equal to 0. Since we divide by x−1, then in the second line we write one:

Let's start filling in the empty cells in the second line. In the second cell of the second row we write the number 5 , simply moving it from the corresponding cell of the first line:

Let's fill the next cell according to this principle: 1⋅ 5 + 5 = 10 :

Let's fill in the fourth cell of the second row in the same way: 1⋅ 10 + 1 = 11 :

For the fifth cell we get: 1⋅ 11 + 0 = 11 :

And finally, for the last, sixth cell, we have: 1⋅ 11 + (−11)= 0 :

The problem is solved, all that remains is to write down the answer:

As you can see, the numbers located in the second line (between one and zero) are the coefficients of the polynomial obtained after dividing 5 x 4 +5x 3 +x 2 −11 per x−1. Naturally, since the degree of the original polynomial is 5 x 4 +5x 3 +x 2 −11 was equal to four, then the degree of the resulting polynomial is 5 x 3 +10x 2 +11x+11 is one less, i.e. equals three. The last number in the second line (zero) means the remainder of the division of the polynomial 5 x 4 +5x 3 +x 2 −11 per x−1.
In our case, the remainder is zero, i.e. polynomials are evenly divisible. This result can also be characterized as follows: the value of the polynomial is 5 x 4 +5x 3 +x 2 −11 at x=1 is equal to zero.
The conclusion can also be formulated in this form: since the value of the polynomial is 5 x 4 +5x 3 +x 2 −11 at x=1 is equal to zero, then unity is the root of the polynomial 5 x 4 +5x 3 +x 2 −11.

2. Find the incomplete quotient, remainder of the division of the polynomial

A(X) = X 3 – 2X 2 + 2X– 1 per binomial X – 1.

Solution:

		– 2		– 1
α = 1		– 1

Answer: Q(x) = X 2 – X + 1 , R(x) = 0.

3. Calculate the value of a polynomial A(X) at X = – 1 if A(X) = X 3 – 2 X – 1.

Solution:

			– 2	– 1
α = – 1		– 1	– 1

Answer: A(– 1) = 0.

4. Calculate the value of a polynomialA(X) at X= 3, incomplete quotient and remainder, where

A(X)= 4 X 5 – 7X 4 + 5X 3 – 2 X + 1.

Solution:

		– 7			– 2
α = 3					178	535

Answer: R(x) = A(3) = 535, Q(x) = 4 X 4 + 5X 3 + 20X 2 + 60X +178.

5. Find the roots of the equationX 3 + 4 X 2 + X – 6 = 0.

Solution:

Find the divisors of the free term ±1; ± 2; ± 3; ± 6

Here, a = 1 (x – 1 = x – a), and the coefficients of the dividend polynomial are equal, respectively
1, 4, 1, – 6. We build a table for applying Horner’s scheme:

Class: 11

Presentation for the lesson

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The purpose of the lesson:

• promote the development of skills in dividing a polynomial by a polynomial and the use of Horner's scheme;
• strengthen your skills in OpenOffice.org Calc spreadsheets;
• organize students’ activities to perceive, comprehend and initially memorize new knowledge;
• analyze and prove Bezout’s theorem when solving a problem situation: is it possible to factor a third-degree polynomial;
• consider using Bezout's theorem to solve higher degree equations;
• promote the development of logical thinking, attention, speech and the ability to work independently.

Lesson type: lesson on introducing new material.

Equipment: multimedia projector, lesson presentation, computer class.

“In order to improve the mind, you need to reason more than memorize.”
Descartes (1596 -1650). French mathematician, physicist, philologist, philosopher.

During the classes

I. Organizing time

Our task today in joint activities is to confirm the words of Descartes (slide 1). The topic of our lesson (slide 2) “Bezout’s Theorem” is so significant that it is even used in Unified State Examination tasks and various Olympiads. Bezout's theorem facilitates the solution of many problems containing equations of higher degrees. Unfortunately, it is studied only at the profile level.

II. The emergence of a problematic situation

In this lesson we will learn how to solve equations of higher degrees, and we will derive the solution algorithm ourselves.

Solve the equation: x 3 - 2x 2 - 6x + 4=0(Slide 3). A problem arises: We understand that it would be convenient to represent the left side of the equation as a product, and since the product is equal to zero, then equate each factor to zero. To do this, you need to factor the third-degree polynomial. But how? Is it possible to group or bracket the common factor in our case? (No).

III. Updating of reference knowledge

Let's remember how to factor the polynomial x 2 - 5x - 6? (Slide 4).

(According to the formula for factoring a quadratic trinomial:

ax 2 + bx + c = a(x – x 1)(x-x 2), where x 1 and x 2 are roots of the trinomial).

Find the roots of the trinomial in two ways. Which ones?

(using the formula for the roots of a quadratic equation and Vieta’s theorem).

One student from each group solves on the board. The rest of the students are in their notebooks. We got: x 2 - 5x - 6 = (x - 6) (x + 1).

This means that the trinomial is divisible by each of the binomials: x – 6 and x + 1.

Pay attention to the free term of our trinomial and find its divisors (±1, ±2, ±3, ±6).

Which divisors are the roots of the trinomial? (-1 and 6)

What conclusion can be drawn? (The roots of the trinomial are divisors of the free term).

IV. Proposing a hypothesis

So which monomial will help you find the roots of a polynomial?

P(x) = x 3 - 2x 2 - 6x + 4=0?

(Free member).

Write down its divisors: ±1; ±2; ±4.

Find the polynomial values ​​for each divisor. Using spreadsheets and directly:

1st group calculates in a notebook, the second at computers in OpenOffice.org Calc.

P(1)= -3
Р(-1)=7
Р(2)=-8
Р(-2)=0
Р(4)=12
Р(-4)=-68

(When calculating in spreadsheets, in cell B2, students enter the formula: =A1^3-2*A1^2-6*A1+4. Using the autocomplete marker, they obtain the values ​​of the polynomial in the entire column).

Which divisor is the root of the polynomial? (-2)

Thus, one of the factors in the expansion will be x-(-2) = x + 2.

How to find other multipliers?

(Divide "in a column" by the binomial x + 2)

How else is it possible? (according to Horner's scheme). (Slide 5)

What is Horner's scheme? ( Horner's scheme is an algorithm for dividing polynomials, written for the special case when the divisor is equal to a binomial x–a).

We carry out the division: the first group is “in a column”, the second - according to Horner’s scheme.

Divided without a trace.

Let's return to the equation: x 3 - 2x 2 - 6x + 4= (x 2 -4x+2)(x+ 2)=0

x 2 -2x+2=0 - quadratic equation. Solve it:

D 1 = 4 – 2 = 2;

Answer: -2, .

Could there be a remainder when dividing? We will answer this question later. Now name the value of the polynomial at x = - 2. (The value is zero).

Please note that x = - 2 is the root of the polynomial and the remainder when dividing the polynomial by x-(-2) is 0.

Considerx=1 - is not the root of the equation.

Let's try to divide the polynomial by x-1. The second group performs long division. The first one, according to Horner’s scheme, supplements the table with one more line.

So, x 3 - 2x 2 - 6x + 4 = (x – 1)∙(x 2 - x – 7) – 3.

Note that x=1 is not the root of the polynomial and the remainder when dividing the polynomial by (x-1) is equal to the value of the polynomial at x=1.

Here is the answer to the question about the remainder. Yes, the remainder was obtained for a value of x that is not the root of the polynomial.

Let's continue Horner's scheme for the remaining divisors of the free term. Now let the first group calculate at the computer, and the second in notebooks.

V. Proof of the hypothesis

(Slide 6) You noticed a pattern about the remainder. Which one? (the remainder is obtained for a value of x that is not the root of the polynomial).

Let's write this pattern in general form.

Let P(x) be a polynomial and a be a number.

Let's prove the statement: The remainder when P(x) is divided by (x - a) is equal to P(a).

Proof. Divide P(x) with the remainder by (x - a).

We get P(x)= (x - a)Q(x) + R; By definition of a remainder, the polynomial r is either equal to 0 or has a degree less than the degree (x - a), i.e. less than 1. But the degree of a polynomial is less than 1 only when it is equal to 0, and therefore in both cases R is actually a number - zero or non-zero.

Now substituting the value x = a into the equality P(x)= (x - a)Q(x) + R, we get P(a)= (a - a)Q(x) + R, P(a) = R , so indeed R = P(a).

This pattern was also noted by the mathematician Bezout.

Student's message

(Slide 7) Etienne Bezu - French mathematician, member of the Paris Academy of Sciences (since 1758), was born in Nemours on March 31, 1730 and died on September 27, 1783. Since 1763, Bezu taught mathematics at the midshipmen's school, and since 1768 at the Royal Artillery Corps.

Etienne Bezout's main works relate to higher algebra; they are devoted to the creation of a theory for solving algebraic equations.

In the theory of solving systems of linear equations, he contributed to the emergence of the theory of determinants, developed the theory of eliminating unknowns from systems of equations of higher degrees, and proved the theorem (first formulated by Maclaurin) that two curves of order m and n intersect at most mn points.

In France and abroad, until 1848, his six-volume “Course of Mathematics,” written by him in 1764-69, was very popular.

Bezout developed the method of indefinite multipliers. In elementary algebra, a method for solving systems of equations based on this method is named after him.

Part of Bezout's works is devoted to external ballistics.

One of the basic theorems of algebra is named after the scientist.

Consequence

What must be the remainder for the polynomial P(x) to be divisible by the binomial (x – a)? (equal to 0).

We obtain a corollary from Bezout’s theorem: In order for the polynomial P(x) to be divisible by a binomial (x – a), it is necessary and sufficient that the equality P(a) = 0 hold.

VI. Assimilation of what has been learned

(Slide 8) Solve the equation: x 4 - x 3 - 6x 2 - x + 3 = 0.

The integer roots of the polynomial P(x) = x 4 - x 3 - 6x 2 - x + 3 must be divisors of the free term, so these can be the numbers -1, 1, 3, -3.

Let's select a root according to Horner's scheme:

VII. Result: So, what does Bezout's Theorem give us? (Slide 9) Bezout's theorem makes it possible, having found one root of a polynomial, to further search for the roots of a polynomial whose degree is 1 less: if P(a) = 0, then P(x)= (x - a)Q(x), and all that remains is to solve the equation Q (x) = 0. Sometimes using this technique - it is called reducing the degree - you can find all the roots of a polynomial.