Camefficients without online. Application of Theorem

Date: 04.12.2020

Proof of the theorem of the mantle

Let F (x) denotes an arbitrary polynomial of the N-th degree relative to the variable x and let it be divided into bite (Xa) turned out in private Q (x), and in the residue R. It is obvious that Q (x) will be some polynomial ( n-1) -th degree relative to x, and the residue r will be permanent value, i.e. independent of x.

If the residue R was a polynomial at least the first degree relative to X, then this would mean that the division was not fulfilled. So, R from x does not depend.

By definition of division (divided equally, the divider on a private plus residue) I get a identity

f (x) \u003d (x - a) q (x) + r.

This equality is valid at any meaning x, it means that it is valid and at x \u003d a.

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

f (a) \u003d (a - a) q (a) + r. (one)

Here the symbol f (a) is no longer f (x), i.e. Not a polynomial relative to X, and the value of this polynomial at x \u003d a. Q (A) denotes the value Q (x) at x \u003d a.

The residue R remained as it was before, since R from x does not depend.

The product (A-a) Q (A) is zero, since the multiplier (A-a) is zero, and the multiplier Q (a) is a certain number. (The polynomial q (x) does not lose meaning with a certain value of x.)

Therefore, from equality (1) we get:

q.E.D.

Corollary from Theorem

Corollary 1.

The residue from dividing the polynomial F (x) on biccoon (AX + B) is equal to the value

of this polynomial at x \u003d -b / a, i.e. R \u003d F (-b / a).

Evidence:

According to the rule of polynomials:

f (x) \u003d (ax + b) * q (x) + r.

f (-b / a) \u003d (a (-b / a) + b) q (-b / a) + r \u003d r. It means r \u003d f (-b / a),

q.E.D.

Corollary 2:

If the number A is the root of the polynomial F (x), then this polynomial is divided into (x-a) without residue.

Evidence:

By the theorem, the residue from the separation of the polynomial F (x) per (x - a) is equal to f (a), and by condition A is the root f (x), and this means that f (a) \u003d 0, which was required to prove.

From this consequence, the theorem is seen that the problem of solving the equation F (x) \u003d 0 is equivalent to the problem of separating the dividers of a polynomial F having a first degree (linear dividers).

Corollary 3:

If the polynomial f (x) has pairs of various roots a 1, a 2, ..., a n, then it is divided into a piece (x-a 1) ... (x - a n) without a residue.

Evidence:

Conduct proof using mathematical induction by the number of roots. At n \u003d 1, the approval is proved in consequence 2. Let it have already been proven for the case when the number of roots is k, which means that F (x) is divided without a residue

(X - A 1) (X - A 2) ... (X - A K), where A 1, A 2, ..., A K is its roots.

Let f (x) be (k + 1) in pairwise different roots. On the assumption of induction A 1, a 2, a k, ..., (a k + 1) are the roots of the polynomial, and it means that the polynomial is divided into the work (X-A 1) ... (X-a k), where it comes out that

f (x) \u003d (x - a 1) ... (x - a k) q (x).

At the same time (a k + 1) - the root of the polynomial F (X), i.e.

It means that substituting instead of X (a k + 1), we get faithful equality:

f (a k + 1) \u003d (a k + 1 -a 1) ... (a k + 1 -a k) q (a k + 1) \u003d 0.

But (a k + 1) is excellent from numbers a 1, ..., ak, and therefore neither one of the numbers (a k + 1 -a 1), ..., (a k + 1 -ak) is not equal to 0. Therefore, zero Equally q (a k + 1), i.e. (A k + 1) - the root of the polynomial q (x). And from Corollary 2, it turns out that Q (x) is divided into (x - a k + 1) without a residue.

q (x) \u003d (x-a k + 1) q 1 (x), and therefore

f (x) \u003d (x - a 1) ... (x - a k) q (x) \u003d (x - a 1) ... (x - a k) (x - a k + 1) q 1 (x).

This means that f (x) is divided into (x-a 1) ... (x - a k + 1) without a residue.

So, it is proved that the theorem is true for k \u003d 1, and from its justice at n \u003d k it follows that it is true and at n \u003d k + 1. Thus, the theorem is true for any number of roots, which was required to prove.

Corollary 4:

The numerous degree n has no more n different roots.

Evidence:

We use the method from the contrary: if a polynomial F (x) degree n would have more n roots - n + k (a 1, a 2, ..., a n + k - its roots), then on the previously proven consequence of 3 He would shared on the work (Xa 1) ... (Xa N + K), having a degree (n + k), which is impossible.

We have come to contradiction, which means our assumption is incorrect, and the numerous degree n cannot have more than n roots, which was required to prove.

Corollary 5:

For any polynomial F (x) and the number A, the difference (F (X) -F (A)) is divided without a bias (x-a).

Evidence:

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

The polynomial f (x) can be represented as: f (x) \u003d (xa) q (x) + r, where Q (x) is a polynomial, a private in the division of F (X) on (XA), R - the balance of division F (X) on (XA).

And on the theorem, the mouture:

f (x) \u003d (x - a) q (x) + f (a).

f (x) -f (a) \u003d (x-a) q (x),

and this means a division without a residue (f (x) -f (a))

on (x-a), which was required to prove.

Corollary 6:

The number A is the root of a polynomial F (x) degree not lower than the first only when F (x) is divided into (x-a) without a residue.

Evidence:

To prove this theorem you need to consider the need and sufficiency of the formulated condition.

1. The need.

Let A be the root of the polynomial F (X), then in the consequence of 2 f (x) it is divided into (x-a) without a residue.

Thus, the division F (x) per (x - a) is a prerequisite for A to be the root f (x), because It is a consequence of this.

2. Sufficiency.

Let a polynomial F (X) shall be divided without residue on (x-a),

then r \u003d 0, where R is the residue from division F (x) to (xa), but by the theorem R \u003d F (a), where it comes from that f (a) \u003d 0, and this means that A is the root f (x).

Thus, the fraction f (x) per (x-a) is and sufficiently in order to be the root F (x).

The division F (x) per (x-a) is a necessary and sufficient condition so that A is the root f (x), which was required to prove.

Corollary 7:

A polynomial that does not have valid roots, in decomposition of multipliers of linear multipliers does not contain.

Evidence:

We use the method from nasty: Suppose that non-rooted polynomial F (X), when decomposition of multipliers, contains a linear multiplier

then he would have shared on (x-a), but by consequence 6 A would be the root f (x), and by the condition, it does not contain valid roots. We have come to contradiction, which means that our assumption is incorrectly and a polynomial that does not have valid roots, in decomposition of linear multipliers does not contain that it was necessary to prove.

Theorem Bezu, despite the seeming simplicity and evidence, is one of the basic theorems of the theory of polynomials. In this theorem, the algebraic characteristics of polynomials (they allow working with polynomials as with integers) bind to their functional characteristics (which allow us to consider polynomials as functions).

Theorem Bezu It claims that the residue from the division of the polynomial per polynomial is.

The coefficients of the polynomial lie in a certain commutative ring with a unit (for example, in the field of real or complex numbers).

Theorem without the proof.

Delim with a polynomial residue P (X) on polynomial (X-a):

Based on what dEG R (X)< deg (x-a) = 1 - polynomials no higher than zero. We substitute, because we get .

But it is most important that theorem is most important, but a consequence of the theorem:

1. Number - the root of the polynomial P (X) then and only when P (X) shares without a balance on twisted x-A..

Based on this - the set of numerous roots P (X) identically multiple roots of the corresponding equation x-A..

2. The free member of the polynomial is divided into any whole root of a polynomial with integer coefficients (when the older coefficient is equal to one - all rational roots are whole).

3. Suppose that - the whole root of the given polynomial P (X) with whole coefficients. So, for any whole, it is divided into.

The Neua Theorem makes it possible, finding one root of the polynomial, look further the roots of the polynomial, the degree of which is already 1 less: if, then this polynomial P (X) Will look like this:

Immendation theorem Examples:

Find the balance from the division of the polynomial on the bouncer.

Theorem Muzness Examples Solutions:

Based on the mouture theorem, the desired residue corresponds to the value of the polynomial at the point. Then we find, for this, we substitute in the expression for the polynomial instead. We get:

Answer: Residue \u003d 5.

Gorner scheme.

Gorner scheme - This is a division algorithm (division of the Gorner scheme) of polynomials recorded for a particular case, if the private is equal to twisted.

Build this algorithm:

Suppose that - divisible

Private (his degree is likely to be less than), R. - residue (because division is carried out on polynomial 1st degree, then the degree of residue will be less than, i.e. Zero, thus, the residue is a constant).

By definition of division with the residue P (x) \u003d q (x) (x-a) + r. After substitution of polynomial expressions, we get:

We reveal the brackets and equate the coefficients with the same degrees, after which we express the coefficients of private through the division and divider coefficients:

Convenient calculations to reduce such a table:

These cells are highlighted in it, the contents of which are involved in calculations at the next step.

Gorner Scheme Examples:

Let it be necessary to divide the polynomial on the bouncer x-2.

Make a table with two lines. In 1 line, we write out the coefficients of our polynomial. In the second line we will receive the coefficients of incomplete private according to the following scheme: first of all rewrite the senior coefficient of this polynomial, hereinafter, in order to receive another coefficient, we multiply the last found on a \u003d 2. and fold with the appropriate polynomial coefficient F (x). The most recent coefficient will be the residue, and all previous ones - incomplete private coefficients.

The number is the root of the polynomial if and only when it is divided into

Let _ the root of the polynomial, i.e. We divide on where the degree is less than the extent that is equal to the degree equal, i.e. . So. Since, then from the last equality it follows that i.e. .

Back, let it divide, i.e. . Then.

Corollary. The balance of the division of the polynomial is equal.

Mounted first degrees are called linear polynomials. The theorem of the mow shows that the location of the roots of the polynomial is equivalent to the location of its linear divisors with the senior coefficient 1.

The polynomial can be divided into linear polynomial using the separation algorithm with the residue, but there is a more convenient way of dividing, known as the Gunner scheme.

Let and let where. Compare coefficients with the same degrees unknown with the left and right parts of the last equality, we have:

The number is called the corner of the multiplicity of the polynomial, if it divides, but no longer divides.

To believe whether the number of the root of the polynomial and what kind of multiplicity can be used by the Gorner Scheme. It is first divided by then if the residue is zero, the received private is divided into, etc. Before receiving not zero residue.

The number of different roots of the polynomial does not exceed its degree.

The following main theorem is of great importance.

Basic theorem.. Any polynomial with numerical coefficients of nonzero degree has at least one root (may be comprehensive).

Corollary. Every single degree has in C (multiple integrated numbers) so much roots, what its degree, considering every root as many times as its multiplicity.

where _ roots, i.e. In a set C, any polynomial decomposes into the product of linear multipliers. If the same multipliers to collect together, then:

where there are already various roots, _ the multiplicity of the root.

If polynomial, with valid coefficients has a root, then the number is also root

It means that a polynomial with valid coefficients complex roots are included in pairs.

Corollary. The polynomial with valid odd degree coefficients has an odd number of valid roots.

Let both the roots are then divided into and but since there are no common divisors, it is divided into transversion.

Approval 2. The polynomial with valid degree coefficients is always decomposed on the set of real numbers into the product of linear polynomials corresponding to its material roots, and polynomials of the 2nd degree corresponding to the pair of conjugate complex roots.

When calculating the integrals from rational functions, we will need a rational fraction in the form of the sum of the simplest.

The rational fraction is called the fraction where and _ polynomials with valid coefficients, and the polynomial. The rational fraction is called properly, if the degree of numerator is less than the degree of denominator. If a rational fraction is not correct, then, by making a division of the numerator to the denominator according to the rule of polynomial division, it can be represented in the form, where and some polynomials, and the correct rational fraction.

Lemma 1. If it is the right rational fraction, and the number is the real root of the multiplicity of the polynomial, i.e. And, there is a real number and polynomial with real coefficients, such that where the fraction is also correct.

At the same time it is easy to show that the resulting expression is a rational fraction with real coefficients.

Lemma 2. If it is the right rational fraction, and the number (and - real,) is the root of the multiplicity of the polynomial, i.e. And, and if, there are real numbers and the polynomial with real coefficients, such that where the fraction is also correct.

The rational fraction of the species, _ three-stale with valid coefficients, not having valid roots, is called the simplest (or elementary) fractions.

Any correct rational fraction is present in the only way as the sum of the simplest frains.

With the practical preparation of this decomposition, it turns out to be a convenient so-called method of uncertain coefficients. It consists in the following:

· For this fraction is written decomposition in which the coefficients are considered unknown;
· After that, both parts of equality are given to a common denominator and the coefficients obtained in the numeric number in the numerator are equal.

In this case, if the degree of polynomial is equal, then in the numerator after bringing to the general denominator it turns out a polynomial degree, i.e. polynomial with coefficients.

The number of unknowns is also equal :.

Thus, a system of equations with unknowns is obtained. The existence of solving this system follows from the above theorem.

scientific work

Application of Theorem

I will focus on the consideration of some examples of applying the theorem of the MUM to solve practical problems.

It should be noted that when solving equations using the mouture theorem, it is necessary:

· Find all the whole divisters of a free member;

· From these divisors to find at least one root of the equation (A);

· The left part of the equation is divided into (x-a);

· Write on the left side of the equation the product of the divider and private;

· Solve the resulting equation.

Find the balance from dividing the polynomial x 3 -3x 2 + 6x-5

on bicked x-2.

By the theorem, the mant:

R \u003d F (2) \u003d 2 3 -3 * 2 2 + 6 * 2-5 \u003d 3.

Answer: R \u003d 3.

With what value A is a polynomial x 4 + AX \u200b\u200b3 + 3X 2 -4x-4 divided without a residue on twisted x-2?

By the theorem, the mouture: r \u003d f (2) \u003d 16 + 8a + 12-8- 4 \u003d 8a + 16.

But under the condition R \u003d 0, it means 8a + 16 \u003d 0, hence a \u003d -2.

Answer: a \u003d -2.

Under what values \u200b\u200ba and b polynomial AX 3 + BX 2 -73x + 102 is divided into three-fold x 2 -5x + 6 without a residue?

Spatulate the divider on multipliers: x 2 -5x + 6 \u003d (x-2) (x-3).

Since twisted X-2 and X-3 are mutually simple, then this polynomial is divided into X-2 and on X-3, which means that by the mouture theorem:

R 1 \u003d F (2) \u003d 8A + 4B-146 + 102 \u003d 8A + 4B-44 \u003d 0

R 2 \u003d F (3) \u003d 27A + 9B-219 + 102 \u003d 27A + 9B-117 \u003d 0

I will solve the system of equations:

8A + 4B-44 \u003d 0 2A + B \u003d 11

27A + 9B-117 \u003d 0 3A + B \u003d 13

From here we obtain: a \u003d 2, b \u003d 7.

Answer: A \u003d 2, B \u003d 7.

Under what values \u200b\u200ba and b is a polynomial x 4 + AX \u200b\u200b3 -9x 2 + 11x + b

is divided without a speech on the three-tier x 2 -2x + 1?

Imagine the divider as follows: x 2 - 2x + 1 \u003d (x - 1) 2

This polynomial is divided into X-1 without a residue, if on the theorem mouture:

R 1 \u003d F (1) \u003d 1 + A-9 + 11 + B \u003d A + B + 3 \u003d 0.

Find a private from dividing this polynomial to X-1:

X 4 + AX \u200b\u200b3 -9x 2 + 11x-A-3 X-1

x 4 -X 3 x 3 + (A + 1) x 2 + (A-8) X + (A + 3)

(a + 1) x 3 - (a + 1) x 2

(A-8) x 2 - (A-8) x

Private X 3 + (A + 1) x 2 + (A-8) X + (A + 3) is divided into (x-1) without a residue, from where

R 2 \u003d F (1) \u003d 1 + (A + 1) * 1 + (A-8) * 1 + A + 3 \u003d 3A-3 \u003d 0.

I will solve the system of equations:

a + b + 3 \u003d 0 a + b \u003d -3

3A - 3 \u003d 0 a \u003d 1

From the system: a \u003d 1, b \u003d -4

Answer: A \u003d 1, B \u003d -4.

Dispatch a polynomial f (x) \u003d x 4 + 4x 2 -5 on multipliers.

Among the free member divisors, the number 1 is the root of this polynomial F (x), which means that, by consequence, 2 of the theorem F (X) is divided into (x-1) without a residue:

f (x) / (x - 1) \u003d x 3 + x 2 + 5x + 5, it means f (x) \u003d (x - 1) (x 3 + x 2 + 5x + 5).

Among the divisors of the free member of the polynomial x 3 + x 2 + 5x + 5 x \u003d -1 is its root, and this means that by the consequence of 2 of the x 3 + x 2 + 5x + 5 theorem, it is divided into (x + 1) without Remain:

X 4 + 4x 2 -5 x-1 _x 3 + x 2 + 5x + 5 x + 1

x 4 -x 3 x 3 + x 2 + 5x + 5 x 3 + x 2 x 2 +5

X 3 + 4x 2 _5x + 5

(x 3 + x 2 + 5x + 5) / (x + 1) \u003d x 2 +5, which means x 3 + x 2 + 5x + 5 \u003d (x + 1) (x 2 +5).

Hence F (X) \u003d (X - 1) (X + 1) (x 2 +5).

In a consequence of 7 (x 2 +5), there are no multipliers, because There is no valid root, so F (x) is further not exposed to multipliers.

Answer: x 4 + 4x 2 -5 \u003d (x - 1) (x + 1) (x 2 +5).

Eliminate a polynomial f (x) \u003d x 4 +324 on multipliers.

f (X) roots does not have, because X 4 cannot be equal to -324, which means that the investigation of 7 f (x) does not expand on multipliers.

Answer: The polynomials are not laid out.

Create a cubic polynomial having a root 4 of multiplicity 2 and root -2.

By consequence 3, if the polynomial F (x) has a root 4 of multiplicity 2 and root -2, then it is divided without a residue on (x-4) 2 (x + 2), it means:

f (x) / (x-4) 2 (x + 2) \u003d q (x), i.e.

f (x) \u003d (x-4) 2 (x + 2) q (x),

f (x) \u003d (x 2 -8x + 16) (x + 2) q (x),

f (x) \u003d (x 3 -8x 2 + 16x + 2x 2 -16x + 32) Q (x),

f (x) \u003d (x 3 -6x 2 +32) Q (x).

(x 3 -6x 2 +32) - cubic polynomial, but under the condition F (x) - also the cubic polynomial, therefore, Q (x) is some valid number. Let q (x) \u003d 1, then f (x) \u003d x 3 -6x 2 +32.

Answer: x 3 -6x 2 +32.

Solve equation x 4 + 3x 3 -13x 2 -9x + 30 \u003d 0.

301; 2, 3, 5, 6, 10.

(x-2) (x 3 + 5x 2 -3x-15) \u003d 0

(x - 2) (x + 5) (x 2 -3) \u003d 0

X 4 + 3x 3 -13x 2 -9x + 30 x-2

x 4 -2x 3 x 3 + 5x 2 -3x-15

Answer: x 1 \u003d 2, x 2 \u003d -5, x 3,4 \u003d.

Solve equation x 6 + x 5 -7x 4 -5x 3 + 16x 2 + 6x-12 \u003d 0.

Looking at the equation, it can be immediately said that in a consequence of 4 it has no more than 6 roots of the equation.

12 1; 2; 3; 4; 6; 12.

X 6 + x 5 -7x 4 -5x 3 + 16x 2 + 6x-12 x-1

x 6 -x 5 x 5 + 2x 4 -5x 3 -10x 2 + 6x + 12

10x 3 + 16x 2 _x 5 + 2x 4 -5x 3 -10x 2 + 6x + 12 x + 2

10x 3 -10x 2 x 5 + 2x 4 x 4 -5x 2 +6

6x 2 + 6x _ -5x 3 -10x 2

6x 2 -6x -5x 3 -10x 2

x 6 + x 5 -7x 4 -5x 3 + 16x 2 + 6x-12 \u003d (x-1) (x 5 + 2x 4 -5x 3 -10x 2 + 6x + 12) \u003d 0

x 6 + x 5 -7x 4 -5x 3 + 16x 2 + 6x-12 \u003d (x - 1) (x + 2) (x 4 -5x 2 +6) \u003d 0

x 4 -5x 2 + 6 \u003d 0 - Biquetratic equation, x 1.2 \u003d, x 3,4 \u003d.

Answer: x 1.2 \u003d, x 3,4 \u003d, x 5 \u003d 1, x 6 \u003d -2.

Solve equation x 3 -5x 2 + 8x-6 \u003d 0.

X 3 -5x 2 + 8x-6 x-3

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

x 3 -5x 2 + 8x-6 \u003d (x 2 -2x + 2) (x-3) \u003d 0

x 2 -2x + 2 \u003d 0 - a square equation, no roots, because D.<0.

Answer: x \u003d 3.

Solve equation 6x 3 + 11x 2 -3x-2 \u003d 0.

6x 3 + 11x 2 -3x-2 x + 2

6x 3 + 12x 2 6x 2 -X-1

6x 3 + 11x 2 -3x-2 \u003d (6x 2 -X-1) (x + 2) \u003d 0

6x 2 -X-1 \u003d 0 - square equation, x 1 \u003d ѕ, x 2 \u003d -?.

Answer: x 1 \u003d ѕ, x 2 \u003d -?, X 3 \u003d -2.

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City Open Scientific and Practical Conference

schoolchildren and students

Topic: "Studying the theorem of the muddle to solve equationsn.- degree whenn.\u003e 2 "

Performed:

Scientific adviser:

Introduction

Etienne Moul

Theorem Bezu

Proof of Theorem 6.

Corollary of Theorem:

Corollary 1.

Corollary 2.

Corollary 3.

Corollary 4.

Corollary 5.

Corollary 6.

Corollary 7.

Application of Theorem

Conclusion

Sources

Introduction

It is difficult to solve the third degree equations and above. The decomposition of the left part of the factory equation, if the right side is zero, is the most common method of solving a variety of equations. There are no general recipes. Much depends on the skill, intelligence, observation and experience.

But such equations can not always be decomposed on multipliers. One of the methods that helped me solve the high degrees equations is the mouture theorem.

The purpose of my work: the study of the theorem of the manth.

To perform the goal, it was assumed to perform the following tasks:

· To familiarize yourself with Etienne's biography;

· Analyze the definition and proof of the theorem;

· Denote and prove the effect from the moutness theorem;

· Show specific examples of the use of the theorem.

Etienne Moul

Etienne Bezu is a French mathematician, a member of the Paris Academy of Sciences (since 1758).

Since 1763, the math has taught math in the Gardenaryan School, and from 1768 and in the Royal Artillery Corps.

The main works of Etienne Rememne are related to the highest algebra, they are devoted to the creation of the theory of solutions of algebraic equations. In the theory of solutions of systems of linear equations, he facilitated the emergence of the theory of determinants, developed the theory of exclusion of unknowns from the systems of equations of the highest degrees, proved the theorem (first formulated by K. Maclogen) that the two curves of order M and N intersect no more than Mn points.

In France and beyond its border until 1848, his six languid "Mathematics Course" was very popular, which I wrote five years from 1764 to 1769. Also, he developed the method of indefinite factors: in elementary algebra, it is called a method for solving systems of equations based on this method. Part of the work of the mud is devoted to the outer ballistics.

The name of the scientist is called one of the main theorems of the algebra, which will be mentioned below.

Theorem Bezu

When dividing the n-th degree polynomial relative to XNA, the X-A residue is equal to the value of dividery at x \u003d a. (Letter A may designate any valid or imaginary number, i.e. any complex number.)

Before proving the theorem, I will make two explanations.

1. We know that there are such algebraic expressions that lose meaning in some individual values \u200b\u200bof the letters of it. For example, 1 / X pulses meaning at x \u003d 0; The expression 1 / (x 2 -25) loses its meaning at x \u003d 5 and at x \u003d -5.

Note that a polynomial of any positive degree never loses its meaning. With any value of the variable, it takes a certain value.

2. The product of two multipliers, of which one appeals to zero, and the other takes a certain meaning, always equal to zero. If one multiplier addresses to zero, and the other loses its meaning, then this product cannot be said that it is equal to zero. It is impossible to say anything about this work. In each case, a special study is needed.

I will consider the work (1-x) *

. At x \u003d 1, the first factor appeals to zero, and the second loses its meaning. It cannot be argued that this product at x \u003d 1 is zero. ] \u003d Lim \u003d 1/2.

So, with x \u003d 1, the work itself (1-x) *

It makes no sense. But its limit makes sense, namely equal to ½, and not zero, as it was mistaken, it was possible to assume.

Proof of the theorem of the mantle

Let F (x) denotes an arbitrary polynomial of the N-th degree relative to the variable x and let it be divided into bite (Xa) turned out in private Q (x), and in the residue R. It is obvious that Q (x) will be some polynomial ( n-1) -th degree relative to x, and the residue r will be permanent value, i.e. independent of x.

If the residue R was a polynomial at least the first degree relative to X, then this would mean that the division was not fulfilled. So, R from x does not depend.

By definition of division (divided equally, the divider on a private plus residue) I get a identity

f (x) \u003d (x - a) q (x) + r.

This equality is valid at any meaning x, it means that it is valid and at x \u003d a.

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

f (a) \u003d (a - a) q (a) + r. (one)

Here the symbol f (a) is no longer f (x), i.e. Not a polynomial relative to X, and the value of this polynomial at x \u003d a. Q (A) denotes the value Q (x) at x \u003d a.

The residue R remained as it was before, since R from x does not depend.

The product (A-a) Q (A) is zero, since the multiplier (A-a) is zero, and the multiplier Q (a) is a certain number. (The polynomial q (x) does not lose meaning with a certain value of x.)

Therefore, from equality (1) we get:

q.E.D.

Corollary from Theorem

Corollary 1.

The residue from dividing the polynomial F (x) on biccoon (AX + B) is equal to the value

of this polynomial at x \u003d -b / a, i.e. R \u003d F (-b / a).

Evidence:

According to the rule of polynomials:

f (x) \u003d (ax + b) * q (x) + r.

f (-b / a) \u003d (a (-b / a) + b) q (-b / a) + r \u003d r. It means r \u003d f (-b / a),

q.E.D.

Corollary 2:

If the number is the root of the polynomial F (X), then this polynomial is divided into (x-a) without a residue.

Evidence:

By the beam theorem, the residue from the division of the polynomial F (x) per (x - a) is f (a), and by the condition apply the root f (x), which means that f (a) \u003d 0, which was required to prove.

Corollary 3:

If the polynomial f (x) has pairs of various roots a 1, a 2, ..., a n, then it is divided into a piece (x-a 1) ... (x - a n) without a residue.

Evidence:

(X - A 1) (X - A 2) ... (X - A K), wherea 1, a 2, ..., A K - ECCORNII.

Let f (x) be (k + 1) in pairwise different roots. On the assumption of induction A 1, a 2, a k, ..., (a k +1) are the roots of the polynomial, and, it means that the polynomial is divided into the work (X-A 1) ... (X-a k), where it comes out that

f (x) \u003d (x - a 1) ... (x - a k) q (x).

At the same time (a k +1) - the root of the polynomial F (X), i.e.

So, substituting instead of X (a k +1), we get faithful equality:

f (a k + 1) \u003d (a k + 1 -a 1) ... (a k + 1 -a k) q (a k + 1) \u003d 0.

But (AK +1) is excellent on numbers a 1, ..., ak, and therefore neither one of the numbers (AK +1 -a 1), ..., (AK +1 -AK) is not equal to 0. Therefore, zero is q ( AK +1), i.e. (A k +1) - the root of the polynomial q (x). And from Corollary 2, it turns out that Q (x) is divided into (x - a k + 1) without a residue.

q (x) \u003d (x - a k +1) q 1 (x), and therefore

f (x) \u003d (x - a 1) ... (x - a k) q (x) \u003d (x - a 1) ... (x - a k) (x - a k + 1) q 1 (x).

This means that f (x) is divided into (x-a 1) ... (x - a k +1) without a residue.

Corollary 4:

The numerous degree n has no more n different roots.

Evidence:

We have come to contradiction, which means our assumption is incorrect, and the numerous degree n cannot have more than n roots, which was required to prove.

Corollary 5:

For any polynomial F (x) and the number A, the difference (F (X) -F (A)) is divided without a bias (x-a).

Evidence:

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

The polynomial f (x) can be represented as: f (x) \u003d (xa) q (x) + r, where Q (x) is a polynomial, a private in the division of F (X) on (XA), R - the balance of division F (X) on (XA).

And on the theorem, the mouture:

f (x) \u003d (x - a) q (x) + f (a).

f (x) -f (a) \u003d (x-a) q (x),

and this means a division without a residue (f (x) -f (a))

on (x-a), which was required to prove.

Corollary 6:

The number A is the root of a polynomial F (x) degree not lower than the first only when F (x) is divided into (x-a) without a residue.

Evidence:

To prove this theorem you need to consider the need and sufficiency of the formulated condition.

1. The need.

Let A be the root of the polynomial F (X), then in the consequence of 2 f (x) it is divided into (x-a) without a residue.

Thus, the division F (x) per (x - a) is a prerequisite for A to be the root f (x), because It is a consequence of this.

2. Sufficiency.

Let a polynomial F (X) shall be divided without residue on (x-a),

then r \u003d 0, where R is the residue from division F (x) to (xa), but by the theorem R \u003d F (a), where it comes from that f (a) \u003d 0, and this means that A is the root f (x).