Theorem on rational roots of a polynomial. Rational numbers, definition, examples

Etc. is of a general educational nature and is of great importance for studying the ENTIRE course of higher mathematics. Today we will repeat “school” equations, but not just “school” ones - but those that are found everywhere in various vyshmat problems. As usual, the story will be told in an applied way, i.e. I will not focus on definitions and classifications, but will share with you my personal experience of solving it. The information is intended primarily for beginners, but more advanced readers will also find many interesting points for themselves. And, of course, there will be new material that goes beyond high school.

So the equation…. Many remember this word with a shudder. What are the “sophisticated” equations with roots worth... ...forget about them! Because then you will meet the most harmless “representatives” of this species. Or boring trigonometric equations with dozens of solution methods. To be honest, I didn’t really like them myself... Don't panic! – then mostly “dandelions” await you with an obvious solution in 1-2 steps. Although the “burdock” certainly clings, you need to be objective here.

Oddly enough, in higher mathematics it is much more common to deal with very primitive equations like linear equations

What does it mean to solve this equation? This means finding SUCH value of “x” (root) that turns it into a true equality. Let’s throw the “three” to the right with a change of sign:

and drop the “two” to the right side (or, the same thing - multiply both sides by) :

To check, let’s substitute the won trophy into the original equation:

The correct equality is obtained, which means that the value found is indeed the root of this equation. Or, as they also say, satisfies this equation.

Please note that the root can also be written as a decimal fraction:
And try not to stick to this bad style! I repeated the reason more than once, in particular, at the very first lesson on higher algebra.

By the way, the equation can also be solved “in Arabic”:

And what’s most interesting is that this recording is completely legal! But if you are not a teacher, then it’s better not to do this, because originality is punishable here =)

And now a little about

graphical solution method

The equation has the form and its root is "X" coordinate intersection points linear function graph with the graph of a linear function (x axis):

It would seem that the example is so elementary that there is nothing more to analyze here, but one more unexpected nuance can be “squeezed” out of it: let’s present the same equation in the form and construct graphs of the functions:

Wherein, please don't confuse the two concepts: an equation is an equation, and function– this is a function! Functions only help find the roots of the equation. Of which there may be two, three, four, or even infinitely many. The closest example in this sense is the well-known quadratic equation, the solution algorithm for which received a separate paragraph "hot" school formulas. And this is no coincidence! If you can solve a quadratic equation and know Pythagorean theorem, then, one might say, “half of higher mathematics is already in your pocket” =) Exaggerated, of course, but not so far from the truth!

Therefore, let’s not be lazy and solve some quadratic equation using standard algorithm:

, which means the equation has two different valid root:

It is easy to verify that both found values ​​actually satisfy this equation:

What to do if you suddenly forgot the solution algorithm, and there are no means/helping hands at hand? This situation may arise, for example, during a test or exam. We use the graphical method! And there are two ways: you can build point by point parabola , thereby finding out where it intersects the axis (if it crosses at all). But it’s better to do something more cunning: imagine the equation in the form, draw graphs of simpler functions - and "X" coordinates their points of intersection are clearly visible!


If it turns out that the straight line touches the parabola, then the equation has two matching (multiple) roots. If it turns out that the straight line does not intersect the parabola, then there are no real roots.

To do this, of course, you need to be able to build graphs of elementary functions, but on the other hand, even a schoolchild can do these skills.

And again - an equation is an equation, and functions , are functions that only helped solve the equation!

And here, by the way, it would be appropriate to remember one more thing: if all the coefficients of an equation are multiplied by a non-zero number, then its roots will not change.

So, for example, the equation has the same roots. As a simple “proof”, I’ll take the constant out of brackets:
and I’ll remove it painlessly (I will divide both parts by “minus two”):

BUT! If we consider the function, then here we cannot get rid of the constant! It is only permissible to take the multiplier out of brackets: .

Many people underestimate the graphical solution method, considering it something “undignified,” and some even completely forget about this possibility. And this is fundamentally wrong, since plotting graphs sometimes just saves the situation!

Another example: suppose you don’t remember the roots of the simplest trigonometric equation: . The general formula is in school textbooks, in all reference books on elementary mathematics, but they are not available to you. However, solving the equation is critical (aka “two”). There is an exit! – build graphs of functions:


after which we calmly write down the “X” coordinates of their intersection points:

There are infinitely many roots, and in algebra their condensed notation is accepted:
, Where ( – set of integers) .

And, without “going away”, a few words about the graphical method for solving inequalities with one variable. The principle is the same. So, for example, the solution to the inequality is any “x”, because The sinusoid lies almost completely under the straight line. The solution to the inequality is the set of intervals in which the pieces of the sinusoid lie strictly above the straight line (x-axis):

or, in short:

But here are the many solutions to the inequality: empty, since no point of the sinusoid lies above the straight line.

Is there anything you don't understand? Urgently study the lessons about sets And function graphs!

Let's warm up:

Exercise 1

Solve the following trigonometric equations graphically:

Answers at the end of the lesson

As you can see, to study exact sciences it is not at all necessary to cram formulas and reference books! Moreover, this is a fundamentally flawed approach.

As I already reassured you at the very beginning of the lesson, complex trigonometric equations in a standard course of higher mathematics have to be solved extremely rarely. All complexity, as a rule, ends with equations like , the solution of which is two groups of roots originating from the simplest equations and . Don’t worry too much about solving the latter – look in a book or find it on the Internet =)

The graphical solution method can also help out in less trivial cases. Consider, for example, the following “ragtag” equation:

The prospects for its solution look... don’t look like anything at all, but you just have to imagine the equation in the form , build function graphs and everything will turn out to be incredibly simple. There is a drawing in the middle of the article about infinitesimal functions (will open in the next tab).

Using the same graphical method, you can find out that the equation already has two roots, and one of them is equal to zero, and the other, apparently, irrational and belongs to the segment . This root can be calculated approximately, for example, tangent method. By the way, in some problems, it happens that you don’t need to find the roots, but find out do they exist at all?. And here, too, a drawing can help - if the graphs do not intersect, then there are no roots.

Rational roots of polynomials with integer coefficients.
Horner scheme

And now I invite you to turn your gaze to the Middle Ages and feel the unique atmosphere of classical algebra. For a better understanding of the material, I recommend that you read at least a little complex numbers.

They are the best. Polynomials.

The object of our interest will be the most common polynomials of the form with whole coefficients A natural number is called degree of polynomial, number – coefficient of the highest degree (or just the highest coefficient), and the coefficient is free member.

I will briefly denote this polynomial by .

Roots of a polynomial call the roots of the equation

I love iron logic =)

For examples, go to the very beginning of the article:

There are no problems with finding the roots of polynomials of the 1st and 2nd degrees, but as you increase this task becomes more and more difficult. Although on the other hand, everything is more interesting! And this is exactly what the second part of the lesson will be devoted to.

First, literally half the screen of theory:

1) According to the corollary fundamental theorem of algebra, the degree polynomial has exactly complex roots Some roots (or even all) may be particularly valid. Moreover, among the real roots there may be identical (multiple) roots (minimum two, maximum pieces).

If some complex number is the root of a polynomial, then conjugate its number is also necessarily the root of this polynomial (conjugate complex roots have the form ).

The simplest example is a quadratic equation, which was first encountered in 8 (like) class, and which we finally “finished off” in the topic complex numbers. Let me remind you: a quadratic equation has either two different real roots, or multiple roots, or conjugate complex roots.

2) From Bezout's theorem it follows that if a number is the root of an equation, then the corresponding polynomial can be factorized:
, where is a polynomial of degree .

And again, our old example: since is the root of the equation, then . After which it is not difficult to obtain the well-known “school” expansion.

The corollary of Bezout's theorem has great practical value: if we know the root of an equation of the 3rd degree, then we can represent it in the form and from the quadratic equation it is easy to find out the remaining roots. If we know the root of an equation of the 4th degree, then it is possible to expand the left side into a product, etc.

And there are two questions here:

Question one. How to find this very root? First of all, let's define its nature: in many problems of higher mathematics it is necessary to find rational, in particular whole roots of polynomials, and in this regard, further we will be mainly interested in them.... ...they are so good, so fluffy, that you just want to find them! =)

The first thing that comes to mind is the selection method. Consider, for example, the equation . The catch here is in the free term - if it were equal to zero, then everything would be fine - we take the “x” out of brackets and the roots themselves “fall out” to the surface:

But our free term is equal to “three”, and therefore we begin to substitute various numbers into the equation that claim to be “root”. First of all, the substitution of single values ​​suggests itself. Let's substitute:

Received incorrect equality, thus, the unit “did not fit.” Well, okay, let's substitute:

Received true equality! That is, the value is the root of this equation.

To find the roots of a polynomial of the 3rd degree, there is an analytical method (the so-called Cardano formulas), but now we are interested in a slightly different task.

Since - is the root of our polynomial, the polynomial can be represented in the form and arises Second question: how to find a “younger brother”?

The simplest algebraic considerations suggest that to do this we need to divide by . How to divide a polynomial by a polynomial? The same school method that divides ordinary numbers - “column”! I discussed this method in detail in the first examples of the lesson. Complex Limits, and now we will look at another method, which is called Horner scheme.

First we write the “highest” polynomial with everyone , including zero coefficients:
, after which we enter these coefficients (strictly in order) into the top row of the table:

We write the root on the left:

I’ll immediately make a reservation that Horner’s scheme also works if the “red” number Not is the root of the polynomial. However, let's not rush things.

We remove the leading coefficient from above:

The process of filling the lower cells is somewhat reminiscent of embroidery, where “minus one” is a kind of “needle” that permeates the subsequent steps. We multiply the “carried down” number by (–1) and add the number from the top cell to the product:

We multiply the found value by the “red needle” and add the following equation coefficient to the product:

And finally, the resulting value is again “processed” with the “needle” and the upper coefficient:

The zero in the last cell tells us that the polynomial is divided into without a trace (as it should be), while the expansion coefficients are “removed” directly from the bottom line of the table:

Thus, we moved from the equation to an equivalent equation and everything is clear with the two remaining roots (in this case we get conjugate complex roots).

The equation, by the way, can also be solved graphically: plot "lightning" and see that the graph crosses the x-axis () at point . Or the same “cunning” trick - we rewrite the equation in the form , draw elementary graphs and detect the “X” coordinate of their intersection point.

By the way, the graph of any function-polynomial of the 3rd degree intersects the axis at least once, which means the corresponding equation has at least one valid root. This fact is true for any polynomial function of odd degree.

And here I would also like to dwell on important point which concerns terminology: polynomial And polynomial function – it's not the same thing! But in practice they often talk, for example, about the “graph of a polynomial,” which, of course, is negligence.

However, let's return to Horner's scheme. As I mentioned recently, this scheme works for other numbers, but if the number Not is the root of the equation, then a non-zero addition (remainder) appears in our formula:

Let’s “run” the “unsuccessful” value according to Horner’s scheme. In this case, it is convenient to use the same table - write a new “needle” on the left, move the leading coefficient from above (left green arrow), and off we go:

To check, let’s open the brackets and present similar terms:
, OK.

It is easy to see that the remainder (“six”) is exactly the value of the polynomial at . And in fact - what is it like:
, and even nicer - like this:

From the above calculations it is easy to understand that Horner’s scheme allows not only to factor the polynomial, but also to carry out a “civilized” selection of the root. I suggest you consolidate the calculation algorithm yourself with a small task:

Task 2

Using Horner's scheme, find the integer root of the equation and factor the corresponding polynomial

In other words, here you need to sequentially check the numbers 1, –1, 2, –2, ... – until a zero remainder is “drawn” in the last column. This will mean that the “needle” of this line is the root of the polynomial

It is convenient to arrange the calculations in a single table. Detailed solution and answer at the end of the lesson.

The method of selecting roots is good for relatively simple cases, but if the coefficients and/or degree of the polynomial are large, then the process may take a long time. Or maybe there are some values ​​from the same list 1, –1, 2, –2 and there is no point in considering? And, besides, the roots may turn out to be fractional, which will lead to a completely unscientific poking.

Fortunately, there are two powerful theorems that can significantly reduce the search for “candidate” values ​​for rational roots:

Theorem 1 Let's consider irreducible fraction , where . If the number is the root of the equation, then the free term is divided by and the leading coefficient is divided by.

In particular, if the leading coefficient is , then this rational root is an integer:

And we begin to exploit the theorem with just this tasty detail:

Let's return to the equation. Since its leading coefficient is , then hypothetical rational roots can be exclusively integer, and the free term must necessarily be divided into these roots without a remainder. And “three” can only be divided into 1, –1, 3 and –3. That is, we have only 4 “root candidates”. And, according to Theorem 1, other rational numbers cannot be roots of this equation IN PRINCIPLE.

There are a little more “contenders” in the equation: the free term is divided into 1, –1, 2, – 2, 4 and –4.

Please note that the numbers 1, –1 are “regulars” of the list of possible roots (an obvious consequence of the theorem) and the best choice for priority testing.

Let's move on to more meaningful examples:

Problem 3

Solution: since the leading coefficient is , then hypothetical rational roots can only be integer, and they must necessarily be divisors of the free term. “Minus forty” is divided into the following pairs of numbers:
– a total of 16 “candidates”.

And here a tempting thought immediately appears: is it possible to weed out all the negative or all the positive roots? In some cases it is possible! I will formulate two signs:

1) If All If the coefficients of the polynomial are non-negative or all non-positive, then it cannot have positive roots. Unfortunately, this is not our case (Now, if we were given an equation - then yes, when substituting any value of the polynomial, the value of the polynomial is strictly positive, which means that all positive numbers (and irrational ones too) cannot be roots of the equation.

2) If the coefficients for odd powers are non-negative, and for all even powers (including free member) are negative, then the polynomial cannot have negative roots. Or “mirror”: the coefficients for odd powers are non-positive, and for all even powers they are positive.

This is our case! Looking a little closer, you can see that when substituting any negative “X” into the equation, the left-hand side will be strictly negative, which means that negative roots disappear

Thus, there are 8 numbers left for research:

We “charge” them sequentially according to Horner’s scheme. I hope you have already mastered mental calculations:

Luck awaited us when testing the “two”. Thus, is the root of the equation under consideration, and

It remains to study the equation . This is easy to do through the discriminant, but I will conduct an indicative test using the same scheme. Firstly, let us note that the free term is equal to 20, which means Theorem 1 the numbers 8 and 40 drop out of the list of possible roots, leaving the values ​​for research (one was eliminated according to Horner’s scheme).

We write the coefficients of the trinomial in the top row of the new table and We start checking with the same “two”. Why? And because the roots can be multiples, please: - this equation has 10 identical roots. But let's not get distracted:

And here, of course, I was lying a little, knowing that the roots are rational. After all, if they were irrational or complex, then I would be faced with an unsuccessful check of all the remaining numbers. Therefore, in practice, be guided by the discriminant.

Answer: rational roots: 2, 4, 5

In the problem we analyzed, we were lucky, because: a) the negative values ​​immediately fell off, and b) we found the root very quickly (and theoretically we could check the entire list).

But in reality the situation is much worse. I invite you to watch an exciting game called “The Last Hero”:

Problem 4

Find the rational roots of the equation

Solution: By Theorem 1 the numerators of hypothetical rational roots must satisfy the condition (we read “twelve is divided by el”), and the denominators correspond to the condition . Based on this, we get two lists:

"list el":
and "list um": (fortunately, the numbers here are natural).

Now let's make a list of all possible roots. First, we divide the “el list” by . It is absolutely clear that the same numbers will be obtained. For convenience, let's put them in a table:

Many fractions have been reduced, resulting in values ​​that are already in the “hero list”. We add only “newbies”:

Similarly, we divide the same “list” by:

and finally on

Thus, the team of participants in our game is completed:


Unfortunately, the polynomial in this problem does not satisfy the "positive" or "negative" criterion, and therefore we cannot discard the top or bottom row. You'll have to work with all the numbers.

How are you feeling? Come on, get your head up - there is another theorem that can figuratively be called the “killer theorem”…. ...“candidates”, of course =)

But first you need to scroll through Horner's diagram for at least one the whole numbers. Traditionally, let's take one. In the top line we write the coefficients of the polynomial and everything is as usual:

Since four is clearly not zero, the value is not the root of the polynomial in question. But she will help us a lot.

Theorem 2 If for some in general value of the polynomial is nonzero: , then its rational roots (if they are) satisfy the condition

In our case and therefore all possible roots must satisfy the condition (let's call it Condition No. 1). This four will be the “killer” of many “candidates”. As a demonstration, I'll look at a few checks:

Let's check the "candidate". To do this, let us artificially represent it in the form of a fraction, from which it is clearly seen that . Let's calculate the test difference: . Four is divided by “minus two”: , which means that the possible root has passed the test.

Let's check the value. Here the test difference is: . Of course, and therefore the second “subject” also remains on the list.

This polynomial has integer coefficients. If an integer is the root of this polynomial, then it is a divisor of the number 16. Thus, if a given polynomial has integer roots, then these can only be the numbers ±1; ±2; ±4; ±8; ±16. By direct verification, we are convinced that the number 2 is the root of this polynomial, that is, x 3 – 5x 2 – 2x + 16 = (x – 2)Q (x), where Q (x) is a polynomial of the second degree. Consequently, the polynomial is decomposed into factors, one of which is (x – 2). To find the type of polynomial Q (x) we use the so-called Horner scheme. The main advantage of this method is the compactness of notation and the ability to quickly divide a polynomial into a binomial. In fact, Horner's scheme is another form of recording the grouping method, although, unlike the latter, it is completely non-visual. The answer (factorization) is obtained here by itself, and we do not see the process of obtaining it. We will not engage in a rigorous substantiation of Horner's scheme, but will only show how it works.

The number from the cell above it is “moved” into the second cell of the bottom line, that is, 1. Then they do this. The root of the equation (number 2) is multiplied by the last written number (1) and the result is added with the number that is in the top row above the next free cell, in our example we have:

Irrational number- This real number, which is not rational, that is, cannot be represented as a fraction, where are integers, . An irrational number can be represented as an infinite non-periodic decimal fraction.

The set of irrational numbers is usually denoted by a capital Latin letter in bold style without shading. Thus: , i.e. there are many irrational numbers difference between the sets of real and rational numbers.

About the existence of irrational numbers, more precisely segments incommensurable with a segment of unit length were already known to the ancient mathematicians: they knew, for example, the incommensurability of the diagonal and the side of the square, which is equivalent to the irrationality of the number.

Properties

• Any real number can be written as an infinite decimal fraction, while irrational numbers and only they are written as non-periodic infinite decimal fractions.
• Irrational numbers define Dedekind cuts in the set of rational numbers that do not have a largest number in the lower class and do not have a smallest number in the upper class.
• Every real transcendental number is irrational.
• Every irrational number is either algebraic or transcendental.
• The set of irrational numbers is dense everywhere on the number line: between any two numbers there is an irrational number.
• The order on the set of irrational numbers is isomorphic to the order on the set of real transcendental numbers.
• The set of irrational numbers is uncountable and is a set of the second category.

Examples

Irrational are:

Examples of proof of irrationality

Root of 2

Let us assume the opposite: it is rational, that is, it is represented in the form of an irreducible fraction, where is an integer and is a natural number. Let's square the supposed equality:

It follows that even is even and . Let it be where the whole is. Then

Therefore, even means even and . We found that and are even, which contradicts the irreducibility of the fraction . This means that the original assumption was incorrect, and it is an irrational number.

Binary logarithm of the number 3

Let us assume the opposite: it is rational, that is, it is represented as a fraction, where and are integers. Since , and can be chosen to be positive. Then

But even and odd. We get a contradiction.

e

Story

The concept of irrational numbers was implicitly adopted by Indian mathematicians in the 7th century BC, when Manava (c. 750 BC - c. 690 BC) figured out that the square roots of some natural numbers, such as 2 and 61 cannot be expressed explicitly.

The first proof of the existence of irrational numbers is usually attributed to Hippasus of Metapontus (c. 500 BC), a Pythagorean who found this proof by studying the lengths of the sides of the pentagram. At the time of the Pythagoreans, it was believed that there was a single unit of length, sufficiently small and indivisible, which entered any segment an integer number of times. However, Hippasus argued that there is no single unit of length, since the assumption of its existence leads to a contradiction. He showed that if the hypotenuse of an isosceles right triangle contains an integer number of unit segments, then this number must be both even and odd. The proof looked like this:

• The ratio of the length of the hypotenuse to the length of the leg of an isosceles right triangle can be expressed as a:b, Where a And b chosen as the smallest possible.
• According to the Pythagorean theorem: a² = 2 b².
• Because a- even, a must be even (since the square of an odd number would be odd).
• Because the a:b irreducible b must be odd.
• Because a even, we denote a = 2y.
• Then a² = 4 y² = 2 b².
• b² = 2 y², therefore b- even, then b even.
• However, it has been proven that b odd. Contradiction.

Greek mathematicians called this ratio of incommensurable quantities alogos(unspeakable), but according to the legends they did not pay due respect to Hippasus. There is a legend that Hippasus made the discovery while on a sea voyage and was thrown overboard by other Pythagoreans “for creating an element of the universe that denies the doctrine that all entities in the universe can be reduced to integers and their ratios.” The discovery of Hippasus posed a serious problem for Pythagorean mathematics, destroying the underlying assumption that numbers and geometric objects were one and inseparable.

As we have already noted, one of the most important problems in the theory of polynomials is the problem of finding their roots. To solve this problem, you can use the selection method, i.e. take a number at random and check whether it is the root of a given polynomial.

In this case, you can quickly “bump into” the root, or you may never find it. After all, it is impossible to check all the numbers, since there are infinitely many of them.

It would be another matter if we were able to narrow the search area, for example, to know that the roots we are looking for are, say, among the thirty specified numbers. And for thirty numbers you can do a check. In connection with all that has been said above, this statement seems important and interesting.

If the irreducible fraction l/m (l,m are integers) is the root of a polynomial f (x) with integer coefficients, then the leading coefficient of this polynomial is divided by m, and the free term is divided by 1.

Indeed, if f (x) =anxn+an-1xn-1+... +a1x+a0, an?0, where an, an-1,...,a1, a0 are integers, then f (l/m) =0, i.e. аn (l/m) n+an-1 (l/m) n-1+... +a1l/m+a0=0.

Let's multiply both sides of this equality by mn. We get anln+an-1ln-1m+... +a1lmn-1+a0mn=0.

This implies:

anln=m (-an-1ln-1-... - a1lmn-2-a0mn-1).

We see that the integer anln is divisible by m. But l/m is an irreducible fraction, i.e. the numbers l and m are coprime, and then, as is known from the theory of divisibility of integers, the numbers ln and m are also coprime. So, anln is divisible by m and m is coprime to ln, which means an is divisible by m.

The proven topic allows us to significantly narrow the search area for rational roots of a polynomial with integer coefficients. Let's demonstrate this with a specific example. Let's find the rational roots of the polynomial f (x) =6x4+13x2-24x2-8x+8. According to the theorem, the rational roots of this polynomial are among the irreducible fractions of the form l/m, where l is the divisor of the free term a0=8, and m is the divisor of the leading coefficient a4=6. Moreover, if the fraction l/m is negative, then the “-” sign will be assigned to the numerator. For example, - (1/3) = (-1) /3. So we can say that l is a divisor of the number 8, and m is a positive divisor of the number 6.

Since the divisors of the number 8 are ±1, ±2, ±4, ±8, and the positive divisors of the number 6 are 1, 2, 3, 6, then the rational roots of the polynomial in question are among the numbers ±1, ±1/2, ± 1/3, ±1/6, ±2, ±2/3, ±4, ±4/3, ±8, ±8/3. Let us recall that we wrote out only irreducible fractions.

Thus, we have twenty numbers - “candidates” for roots. All that remains is to check each of them and select those that are really roots. But again, you will have to do quite a lot of checks. But the following theorem simplifies this work.

If the irreducible fraction l/m is the root of a polynomial f (x) with integer coefficients, then f (k) is divisible by l-km for any integer k, provided that l-km?0.

To prove this theorem, divide f (x) by x-k with a remainder. We get f (x) = (x-k) s (x) +f (k). Since f (x) is a polynomial with integer coefficients, so is the polynomial s (x), and f (k) is an integer. Let s (x) =bn-1+bn-2+…+b1x+b0. Then f (x) - f (k) = (x-k) (bn-1xn-1+bn-2xn-2+ …+b1x+b0). Let us put x=l/m in this equality. Considering that f (l/m) =0, we get

f (k) = ((l/m) - k) (bn-1 (l/m) n-1+bn-2 (l/m) n-2+…+b1 (l/m) +b0) .

Let's multiply both sides of the last equality by mn:

mnf (k) = (l-km) (bn-1ln-1+bn-2ln-2m+…+b1lmn-2+b0mn-1).

It follows that the integer mnf (k) is divisible by l-km. But since l and m are coprime, then mn and l-km are also coprime, which means f (k) is divisible by l-km. The theorem has been proven.

Let us now return to our example and, using the proven theorem, we will further narrow the circle of searches for rational roots. Let us apply this theorem for k=1 and k=-1, i.e. if the irreducible fraction l/m is the root of the polynomial f (x), then f (1) / (l-m), and f (-1) / (l+m). We easily find that in our case f (1) = -5, and f (-1) = -15. Note that at the same time we excluded ±1 from consideration.

So, the rational roots of our polynomial should be sought among the numbers ±1/2, ±1/3, ±1/6, ±2, ±2/3, ±4, ±4/3, ±8, ±8/3.

Consider l/m=1/2. Then l-m=-1 and f (1) =-5 is divided by this number. Further, l+m=3 and f (1) =-15 is also divisible by 3. This means that the fraction 1/2 remains among the “candidates” for roots.

Let now lm=- (1/2) = (-1) /2. In this case, l-m=-3 and f (1) =-5 is not divisible by - 3. This means that the fraction - 1/2 cannot be the root of this polynomial, and we exclude it from further consideration. Let's check for each of the fractions written above and find that the required roots are among the numbers 1/2, ±2/3, 2, - 4.

Thus, using a fairly simple technique, we have significantly narrowed the search area for rational roots of the polynomial under consideration. Well, to check the remaining numbers, we’ll use Horner’s scheme:

Table 10

We found that the remainder when dividing g (x) by x-2/3 is equal to - 80/9, i.e. 2/3 is not a root of the polynomial g (x), and therefore neither is f (x).

Next, we easily find that - 2/3 is the root of the polynomial g (x) and g (x) = (3x+2) (x2+2x-4). Then f (x) = (2x-1) (3x+2) (x2+2x-4). Further verification can be carried out for the polynomial x2+2x-4, which, of course, is simpler than for g (x) or even more so for f (x). As a result, we find that the numbers 2 and - 4 are not roots.

So, the polynomial f (x) =6x4+13x3-24x2-8x+8 has two rational roots: 1/2 and - 2/3.

Recall that the method described above makes it possible to find only rational roots of a polynomial with integer coefficients. Meanwhile, a polynomial can also have irrational roots. So, for example, the polynomial considered in the example has two more roots: - 1±v5 (these are the roots of the polynomial x2+2x-4). And, generally speaking, a polynomial may not have rational roots at all.

Now let's give some tips.

When testing “candidates” for the roots of the polynomial f (x) using the second of the theorems proved above, the latter is usually used for cases k=±1. In other words, if l/m is a "candidate" root, then check whether f (1) and f (-1) are divisible by l-m and l+m, respectively. But it may happen that, for example, f (1) = 0, i.e. 1 is a root, and then f (1) is divisible by any number, and our check becomes meaningless. In this case, you should divide f (x) by x-1, i.e. obtain f(x) = (x-1)s(x), and test for the polynomial s(x). At the same time, we should not forget that we have already found one root of the polynomial f (x) - x1=1. If, when checking the “candidates” for roots remaining after using the second theorem on rational roots, using Horner’s scheme we find that, for example, l/m is a root, then its multiplicity should be found. If it is equal to, say, k, then f (x) = (x-l/m) ks (x), and further testing can be done for s (x), which reduces the calculations.

Thus, we have learned to find rational roots of a polynomial with integer coefficients. It turns out that by doing this we have learned to find the irrational roots of a polynomial with rational coefficients. In fact, if we have, for example, a polynomial f (x) =x4+2/3x3+5/6x2+3/8x+2, then, bringing the coefficients to a common denominator and putting it out of brackets, we get f (x) =1/24 (24x4+16x3-20x2+9x+48). It is clear that the roots of the polynomial f (x) coincide with the roots of the polynomial in parentheses, and its coefficients are integers. Let us prove, for example, that sin100 is an irrational number. Let's use the well-known formula sin3?=3sin?-4sin3?. Hence sin300=3sin100-4sin3100. Considering that sin300=0.5 and carrying out simple transformations, we get 8sin3100-6sin100+1=0. Therefore, sin100 is the root of the polynomial f (x) =8x3-6x+1. If we look for rational roots of this polynomial, we will be convinced that there are none. This means that the root sin100 is not a rational number, i.e. sin100 is an irrational number.

A polynomial in the variable x is an expression of the form: anxn+an-1 xn-1+. . . +a 1 x+a 0, where n is a natural number; аn, an-1, . . . , a 1, a 0 - any numbers called the coefficients of this polynomial. Expressions anxn, an-1 xn-1, . . . , a 1 x, a 0 are called the terms of the polynomial, and 0 is the free term. an is the coefficient of xn, an-1 is the coefficient of xn-1, etc. A polynomial in which all coefficients are equal to zero is called zero. for example, the polynomial 0 x2+0 x+0 is zero. From the notation of a polynomial it is clear that it consists of several members. This is where the term ‹‹polynomial›› (many terms) comes from. Sometimes a polynomial is called a polynomial. This term comes from the Greek words πολι - many and νομχ - member.

A polynomial in one variable x is denoted: . f (x), g (x), h (x), etc. for example, if the first of the above polynomials is denoted f (x), then we can write: f (x) =x 4+2 x 3+ (- 3) x 2+3/7 x+√ 2. 1. The polynomial h(x) is called the greatest common divisor of the polynomials f(x) and g(x) if it divides f(x), g(x) and each of their common divider. 2. A polynomial f(x) with coefficients from the field P of degree n is said to be reducible over the field P if there exist polynomials h(x), g(x) О P[x] of degree less than n such that f(x) = h( x)g(x).

If there is a polynomial f (x) =anxn+an-1 xn-1+. . . +a 1 x+a 0 and an≠ 0, then the number n is called the degree of the polynomial f (x) (or they say: f (x) - nth degree) and write st. f(x)=n. In this case, an is called the leading coefficient, and anxn is the leading term of this polynomial. For example, if f (x) =5 x 4 -2 x+3, then art. f (x) =4, leading coefficient - 5, leading term - 5 x4. The degree of a polynomial is the largest non-zero number of its coefficients. Polynomials of degree zero are numbers other than zero. , the zero polynomial has no degree; the polynomial f (x) =a, where a is a non-zero number and has degree 0; the degree of any other polynomial is equal to the largest exponent of the variable x, the coefficient of which is equal to zero.

Equality of polynomials. Two polynomials f (x) and g (x) are considered equal if their coefficients for the same powers of the variable x and free terms are equal (their corresponding coefficients are equal). f (x) =g (x). For example, the polynomials f (x) =x 3+2 x 2 -3 x+1 and g(x) =2 x 23 x+1 are not equal, the first of them has a coefficient of x3 equal to 1, and the second has zero ( according to the accepted conventions, we can write: g (x) =0 x 3+2 x 2 -3 x+1. In this case: f (x) ≠g (x). The polynomials are not equal: h (x) =2 x 2 -3 x+5, s (x) =2 x 2+3 x+5, since their coefficients for x are different.

But the polynomials f 1 (x) =2 x 5+3 x 3+bx+3 and g 1 (x) =2 x 5+ax 3 -2 x+3 are equal if and only if a = 3, a b= -2. Let the polynomial f (x) =anxn+an-1 xn-1+ be given. . . +a 1 x+a 0 and some number c. Number f (c) =ancn+an-1 cn-1+. . . +a 1 c+a 0 is called the value of the polynomial f (x) at x=c. Thus, to find f (c), you need to substitute c into the polynomial instead of x and carry out the necessary calculations. For example, if f (x) =2 x 3+3 x 2 -x+5, then f (-2) =2 (-2) 3+ (-2) 2 - (-2) +5=3. A polynomial can take on different values ​​for different values ​​of the variable x. The number c is called the root of the polynomial f (x) if f (c) =0.

Let us pay attention to the difference between two statements: “the polynomial f (x) is equal to zero (or, what is the same, the polynomial f (x) is zero)” and “the value of the polynomial f (x) at x = c is equal to zero.” For example, the polynomial f (x) =x 2 -1 is not equal to zero, it has non-zero coefficients, and its value at x=1 is zero. f (x) ≠ 0, and f (1) =0. There is a close relationship between the concepts of equality of polynomials and the value of a polynomial. If two equal polynomials f (x) and g (x) are given, then their corresponding coefficients are equal, which means f (c) = g (c) for each number c.

Operations on polynomials Polynomials can be added, subtracted and multiplied using the usual rules for opening parentheses and bringing similar terms. The result is again a polynomial. These operations have known properties: f (x) +g (x) =g (x) +f (x), f (x) + (g (x) +h (x)) = (f (x) +g (x)) +h (x), f (x) g (x) =g (x) f (x), f (x) (g (x) h (x)) = (f (x) g ( x)) h (x), f (x) (g (x) +h (x)) =f (x) g (x) +f (x) h (x).

Let two polynomials f(x) =anxn+an-1 xn-1+ be given. . . +a 1 x+a 0, an≠ 0, and g(x)=bmxm+bm-1 xm-1+. . . +b 1 x+bm≠ 0. It is clear that Art. f(x)=n, and art. g(x)=m. If we multiply these two polynomials, we get a polynomial of the form f(x) g(x)=anbmxm+n+. . . +a 0 b 0. Since an≠ 0 and bn≠ 0, then anbm≠ 0, which means st. (f(x)g(x))=m+n. An important statement follows from this.

The degree of the product of two non-zero polynomials is equal to the sum of the degrees of the factors, art. (f (x) g (x)) = st. f (x) +st. g(x). The leading term (coefficient) of the product of two non-zero polynomials is equal to the product of the leading terms (coefficients) of the factors. The free term of the product of two polynomials is equal to the product of the free terms of the factors. The powers of the polynomials f (x), g (x) and f (x) ±g (x) are related by the following relation: art. (f (x) ±g (x)) ≤ max (st. f (x), st. g (x)).

The superposition of polynomials f (x) and g (x) is called. a polynomial denoted f (g (x)), which is obtained if in the polynomial f (x) we substitute the polynomial g (x) instead of x. For example, if f(x)=x 2+2 x-1 and g(x) =2 x+3, then f(g(x))=f(2 x+3)=(2 x+3)2 +2(2 x+3)-1=4 x 2+16 x+14, g(f(x))=g(x 2+2 x-1)=2(x 2+2 x-1)+ 3=2 x 2+4 x+1. It can be seen that f (g (x)) ≠g (f (x)), i.e., the superposition of the polynomials f (x), g (x) and the superposition of the polynomials g (x), f (x) are different. Thus, the superposition operation does not have the commutability property.

, Division algorithm with remainder For any f(x), g(x), there exist q(x) (quotient) and r(x) (remainder) such that f(x)=g(x)q(x)+ r(x), and the degree r(x)

Divisors of a polynomial The divisor of a polynomial f(x) is a polynomial g(x), such that f(x)=g(x)q(x). The greatest common divisor of two polynomials The greatest common divisor of the polynomials f(x) and g(x) is their common divisor d(x) that is divisible by any of their other common divisors.

Euclidean algorithm (sequential division algorithm) for finding the greatest common divisor of the polynomials f(x) and g(x) Then is the greatest common divisor of f(x) and g(x).

Reduce the fraction Solution: Find the gcd of these polynomials using the Euclidean algorithm 1) x3 + 6 x2 + 11 x + 6 x3 + 7 x2 + 14 x + 8 1 – x2 – 3 x – 2 2) x3 + 7 x2 + 14 x + 8 x3 + 3 x2 + 2 x – x2 – 3 x – 2 –x– 4 4 x2 + 12 x + 8 0 Therefore, the polynomial (– x2 – 3 x – 2) is the gcd of the numerator and denominator of a given fraction. The result of dividing the denominator by this polynomial is known.

Let's find the result of dividing the numerator. x 3 + 6 x2 + 11 x + 6 – x2 – 3 x – 2 x3 + 3 x2 + 2 x –x– 3 3 x2 + 9 x + 6 0 Thus, Answer:

Horner's scheme Dividing a polynomial f(x) with a remainder by a nonzero polynomial g(x) means representing f(x) in the form f(x)=g(x) s(x)+r(x), where s(x ) and r(x) are polynomials and either r(x)=0 or st. r(x)

The polynomials on the left and right sides of this relation are equal, which means their corresponding coefficients are equal. Let us equate them by first opening the brackets and bringing similar terms on the right side of this equality. We get: a= bn-1, a-1 = bn-2 - cbn-1, a-2 = bn-3 - cbn-2, a 2 = b 1 - cb 2, a 1 = b 0 - cb 1, a 0 = r - cb 0. Recall that we need to find the incomplete quotient, i.e. its coefficients, and the remainder. Let us express them from the obtained equalities: bn-1 = an, b n-2 = cbn-1 + an-1, b n-3 = cbn-2 + a n-2, b 1 = cb 2 + a 2, b 0 = cb 1 +a 1, r = cb 0 + a 0. We have found formulas that can be used to calculate the coefficients of the partial quotient s (x) and the remainder r. In this case, the calculations are presented in the form of the following table; it is called the Horner scheme.

Table 1. Coefficients f (x) c an bn-1 an-1 bn-2=cbn-1+ an-1 an-2 bn-3 = cbn-2+an-2 … … a 0 r = cb 0 + a 0 Coefficients s (x) remainder In the first row of this table, write all the coefficients of the polynomial f (x) in a row, leaving the first cell free. In the second line, in the first cell, write the number c. The remaining cells of this line are filled in by calculating one by one the coefficients of the incomplete quotient s (x) and the remainder r. In the second cell, write the coefficient bn-1, which, as we have established, is equal to an.

The coefficients in each subsequent cell are calculated according to the following rule: the number c is multiplied by the number in the previous cell, and the number above the cell being filled is added to the result. To remember, say, the fifth cell, that is, to find the coefficient in it, you need to multiply c by the number in the fourth cell, and add the number above the fifth cell to the result. Let us divide, for example, the polynomial f (x) =3 x 4 -5 x 2+3 x-1 by x-2 with remainder, using Horner's scheme. When filling out the first line of this diagram, we must not forget about the zero coefficients of the polynomial. So, the coefficients f (x) are the numbers 3, 0, - 5, 3, - 1. And you should also remember that the degree of an incomplete quotient is one less than the degree of the polynomial f (x).

So, we carry out the division according to Horner’s scheme: Table 2. 2 3 3 0 6 -5 7 3 17 -1 33 We obtain the partial quotient s (x) =3 x 3+6 x 2+7 x+17 and the remainder r=33. Note that at the same time we calculated the value of the polynomial f (2) =33. Let us now divide the same polynomial f (x) by x+2 with a remainder. In this case c=-2. we get: Table 3. -2 3 3 0 -6 -5 7 3 -11 -1 21 As a result, we have f (x) = (x+2) (3 x 3 -6 x 2+7 x-11) +21 .

Roots of polynomials Let c1, c2, …, cm be different roots of the polynomial f (x). Then f (x) is divided by x-c1, i.e. f (x) = (x-c 1) s 1 (x). Let's put x=c2 in this equality. We get f (c 2) = (c 2 -c 1) s 1 (c 2) and, so f (c 2) =0, then (c2 -c1) s 1 (c 2) =0. But с2≠с1, i.e. с2 -с1≠ 0, which means s 1 (c 2) =0. Thus, c2 is the root of the polynomial s 1 (x). It follows that s 1 (x) is divisible by x-c2, i.e. s 1 (x) = (x-c 2) s 2 (x). Let us substitute the resulting expression for s 1 (x) into the equality f (x) = (x-c 1) s 1 (x). We have f (x) = (x-c 1) (x-c 2) s 2 (x). Putting x=c3 in the last equality, taking into account the fact that f (c 3) =0, c3≠c1, c3≠c2, we obtain that c3 is the root of the polynomial s 2 (x). This means s 2 (x) = (x-c 3) s 3 (x), and then f (x) = (x-c 1) (x-c 2) (x-c 3) s 3 (x), etc. Continuing this reasoning for the remaining roots c4, c5, ..., cm, we finally obtain f (x) = (x-c 1) (x-c 2) ... (x-cm) sm (x), i.e. the statement formulated below is proven.

If с1, с2, …, сm are different roots of the polynomial f (x), then f (x) can be represented as f(x)=(x-c 1) (x-c 2)…(x-cm) sm(x). An important corollary follows from this. If c1, c2, ..., cm are different roots of the polynomial f(x), then f(x) is divided by the polynomial (x-c1) (x-c2) ... (x-cm). The number of different roots of a nonzero polynomial f (x) is no greater than its degree. Indeed, if f(x) has no roots, then it is clear that the theorem is true, because art. f(x) ≥ 0. Now let f(x) have m roots с1, с2, …, сm, and all of them are different. Then, by what has just been proven, f (x) is divided into (x-c1) (x -c2)…(x-cm). In this case, Art. f(x)≥st. ((x-c1) (x-c2)…(x-cm))= st. (x-c1)+st. (x-s2)+…+st. (x-cm)=m, i.e. art. f(x)≥m, and m is the number of roots of the polynomial in question. But the zero polynomial has infinitely many roots, because its value for any x is equal to 0. In particular, for this reason it is not prescribed any specific degree. The following statement follows from the theorem just proved.

If a polynomial f(x) is not a polynomial of degree greater than n and has more than n roots, then f(x) is a zero polynomial. In fact, from the conditions of this statement it follows that either f (x) is a zero polynomial, or Art. f (x) ≤n. If we assume that the polynomial f (x) is not zero, then Art. f (x) ≤n, and then f (x) has at most n roots. We arrive at a contradiction. This means f(x) is a non-zero polynomial. Let f (x) and g (x) be nonzero polynomials of degree at most n. If these polynomials take the same values ​​for n+1 values ​​of the variable x, then f (x) =g (x).

To prove this, consider the polynomial h (x) =f (x) - g (x). It is clear that either h (x) =0 or st. h (x) ≤n, i.e. h (x) is not a polynomial of degree greater than n. Now let the number c be such that f (c) = g (c). Then h (c) = f (c) - g (c) = 0, i.e. c is the root of the polynomial h (x). Therefore, the polynomial h (x) has n+1 roots, and when, as just proved, h (x) =0, i.e. f (x) =g (x). If f (x) and g (x) take the same values ​​for all values ​​of the variable x, then these polynomials are equal

Multiple roots of a polynomial If a number c is a 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. by (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 is a natural number), but 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).

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+1 s(x), and this 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+1 s(x). Then (x-c)mg(x)=(x-c)m+1 s (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.

Let's find out, for example, whether the number 2 is the root of the polynomial f (x) =x 5 -5 x 4+3 x 3+22 x 2 -44 x+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. 2 1 1 -5 -3 3 -3 22 16 -44 -12 24 0 As you can see, the remainder when dividing f(x) by x-2 is equal to 0, i.e., it is divided by x-2. This means that 2 is the root of this polynomial. In addition, we got that f(x)=(x-2)(x 4 -3 x 3 -3 x 2+16 x-12). Now let's find out whether f(x) is on (x-2) 2. This depends, as we just proved, on the divisibility of the polynomial g (x) =x 4 -3 x 3 -3 x 2+16 x-12 by x-2.

Let's use Horner's scheme again: Table 5. 1 -3 -3 16 -12 2 1 -1 -5 6 0 We found that g(x) is divisible by x-2 and g(x)=(x-2)(x 3 -x 2 -5 x+6). Then f(x)=(x-2)2(x 3 -x 2 -5 x+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) =x 3 -x 2 -5 x+6 is divisible by x-2: Table 6. 1 -1 -5 6 2 1 1 -3 0 We find that h(x) is divisible by x-2, which means f(x) is divided by (x-2) 3, and f(x)=(x-2)3(x 2+x-3).

Next, we similarly check whether f(x) is divisible by (x-2)4, i.e. whether s(x)=x 2+x-3 is divisible by x-2: Table 7. 2 1 1 1 3 -3 3 We find that the remainder when dividing s(x) by x-2 is equal to 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 a root of multiplicity 3 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. 1 -5 3 22 -44 -24 2 2 1 1 -3 -1 1 3 -3 -5 -3 3 16 6 0 -12 0 0 In other words, according to the scheme Horner's division of 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) =x 4+2 x 3+ax 2+ (a+b)x+2 have the number - 2 as a root of multiple 2? Since the multiplicity of the root - 2 should be equal to 2, then, when dividing by x+2 according to the proposed scheme, we should get a remainder of 0 twice, and the third time - a remainder different from zero. We have: Table 9. -2 -2 -2 1 1 2 0 -2 -4 a a a+4 a+12 a+b -3 a+b-8 2 2 a-2 b+2

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

Rational roots of a polynomial If the irreducible fraction l/m (l, m are integers) is the root of a polynomial f (x) with integer coefficients, then the leading coefficient of this polynomial is divided by m, and the free term is divided by 1. Indeed, if f (x)=anxn+an-1 xn-1+…+a 1 x+a 0, an≠ 0, where an, an-1, . . . , a 1, a 0 are integers, then f(l/m) =0, i.e. аn (l/m) n+an-1 (l/m) n-1+. . . +a 1 l/m+a 0=0. Let's multiply both sides of this equality by mn. We get anln+an-1 ln-1 m+. . . +a 1 lmn-1+a 0 mn=0. This implies anln=m (-an-1 ln-1 -…- a 1 lmn-2 -a 0 mn-1).

We see that the integer anln is divisible by m. But l/m is an irreducible fraction, i.e. the numbers l and m are coprime, and then, as is known from the theory of divisibility of integers, the numbers ln and m are also coprime. So, anln is divisible by m and m is coprime to ln, which means an is divisible by m. Let's find the rational roots of the polynomial f (x) =6 x 4+13 x 2 -24 x 2 -8 x+8. According to the theorem, the rational roots of this polynomial are among the irreducible fractions of the form l/m, where l is the divisor of the free term a 0=8, and m is the divisor of the leading coefficient a 4=6. Moreover, if the fraction l/m is negative, then the “-” sign will be assigned to the numerator. For example, - (1/3) = (-1) /3. So we can say that l is a divisor of the number 8, and m is a positive divisor of the number 6.

Since the divisors of the number 8 are ± 1, ± 2, ± 4, ± 8, and the positive divisors of the number 6 are 1, 2, 3, 6, then the rational roots of the polynomial in question are among the numbers ± 1, ± 1/2, ± 1/3, ± 1/6, ± 2/3, ± 4, ± 4/3, ± 8/3. Let us recall that we wrote out only irreducible fractions. Thus, we have twenty numbers - “candidates” for roots. All that remains is to check each of them and select those that are really roots. the following theorem simplifies this work. If the irreducible fraction l/m is the root of a polynomial f (x) with integer coefficients, then f (k) is divisible by l-km for any integer k, provided that l-km≠ 0.

To prove this theorem, divide f(x) by x-k with a remainder. We get f(x)=(x-k)s(x)+f(k). Since f(x) is a polynomial with integer coefficients, so is the polynomial s(x), and f(k) is an integer. Let s(x)=bn-1+bn-2+…+b 1 x+b 0. Then f(x)-f(k)=(x-k) (bnxn-1+bn-2 xn-2+ … +b 1 x+b 0). Let's put 1 x=l/m in this equality. Considering that f(l/m)=0, we get f(k)=((l/m)-k)(bn-1(l/m)n-1+bn-2(l/m)n- 2+…+b 1(l/m)+b 0). Let's multiply both sides of the last equality by mn: mnf(k)=(l-km)(bn-1 ln-1+bn-2 ln-2 m+…+b 1 lmn-2+b 0 mn-1). It follows that the integer mnf (k) is divisible by l-km. But since l and m are coprime, then mn and l-km are also coprime, which means f(k) is divisible by l-km. The theorem has been proven.

Let's return to our example and, using the proven theorem, we will further narrow the circle of searches for rational roots. Let us apply this theorem for k=1 and k=-1, i.e. if the irreducible fraction l/m is the root of the polynomial f(x), then f(1)/(l-m), and f(-1)/(l +m). We easily find that in our case f(1)=-5, and f(-1)= -15. Note that at the same time we excluded ± 1 from consideration. So, the rational roots of our polynomial should be sought among the numbers ± 1/2, ± 1/3, ± 1/6, ± 2, ± 2/3, ± 4/3, ± 8 /3. Consider l/m=1/2. Then l-m=-1 and f (1) =-5 is divided by this number. Further, l+m=3 and f (1) =-15 is also divisible by 3. This means that the fraction 1/2 remains among the “candidates” for roots.

Let now lm=-(1/2)=(-1)/2. In this case, l-m=-3 and f (1) =-5 is not divisible by - 3. This means that the fraction -1/2 cannot be the root of this polynomial, and we exclude it from further consideration. Let's check for each of the fractions written above and find that the required roots are among the numbers 1/2, ± 2/3, 2, - 4. Thus, using a fairly simple technique, we have significantly narrowed the search area for rational roots of the polynomial in question. Well, to check the remaining numbers, we’ll use Horner’s scheme: Table 10. 6 13 -24 -8 8 1/2 6 16 -16 0

We see that 1/2 is the root of the polynomial f(x) and f(x)= (x-1/2) (6 x 3+16 x 2 -16 x-16) = (2 x-1) (3 x 3+8 x 2 -8 x-8). It is clear that all other roots of the polynomial f (x) coincide with the roots of the polynomial g (x) =3 x 3+8 x 2 -8 x-8, which means that further verification of “candidates” for roots can be carried out for this polynomial. We find: Table 11. 3 8 -8 -8 2/3 3 10 -4/3 -80/9 We found that the remainder when dividing g(x) by x-2/3 is equal to - 80/9, i.e. 2/3 is not a root of the polynomial g(x), and therefore neither is f(x). Next we find that - 2/3 is the root of the polynomial g(x) and g (x) = (3 x+2) (x 2+2 x-4).

Then f(x) = (2 x-1) (3 x+2) (x 2+2 x-4). Further verification can be carried out for the polynomial x 2+2 x-4, which, of course, is simpler than for g (x) or, even more so, for f (x). As a result, we find that the numbers 2 and - 4 are not roots. So, the polynomial f (x) =6 x 4+13 x 3 -24 x 2 -8 x+8 has two rational roots: 1/2 and - 2/3. This method makes it possible to find only rational roots of a polynomial with integer coefficients. Meanwhile, a polynomial can also have irrational roots. So, for example, the polynomial considered in the example has two more roots: - 1±√ 5 (these are the roots of the polynomial x2+2 x-4). a polynomial may not have rational roots at all.

When testing “candidate” roots of the polynomial f(x) using the second of the theorems proved above, the latter is usually used for cases k = ± 1. In other words, if l/m is a “candidate” root, then check whether f( 1) and f (-1) by l-m and l+m, respectively. But it may happen that, for example, f(1) =0, i.e. 1 is a root, and then f(1) is divisible by any number, and our check becomes meaningless. In this case, you should divide f(x) by x-1, i.e. get f(x)=(x-1)s(x), and test for the polynomial s(x). At the same time, we should not forget that we have already found one root of the polynomial f(x)-x 1=1. If we check the “candidates” for roots remaining after using the second theorem on rational roots, using Horner’s scheme, we find that, for example, l/m is a root, then its multiplicity should be found. If it is equal to, say, k, then f(x)=(x-l/m) ks (x), and further testing can be done on s(x), which reduces the computation.

Solution. Having replaced the variable y=2 x, we move on to a polynomial with a coefficient equal to one at the highest degree. To do this, first multiply the expression by 4. If the resulting function has integer roots, then they are among the divisors of the free term. Let's write them down: ± 1, ± 2, ± 3, ± 4, ± 5, ± 6, ± 10, ± 12, ± 15 ±, ± 20, ± 30, ± 60

Let us sequentially calculate the values ​​of the function g(y) at these points until we reach zero. That is, y=-5 is a root and therefore is the root of the original function. Let us divide the polynomial by a binomial using a column (corner)

It is not advisable to continue checking the remaining divisors, since it is easier to factorize the resulting quadratic trinomial. Therefore,

Using Abbreviated Multiplication Formulas and Newton's Binomial to Factor a Polynomial Sometimes the appearance of a polynomial suggests a way to factor it. For example, after simple transformations, the coefficients are lined up in a line from Pascal’s triangle for the coefficients of Newton’s binomial. Example. Factor the polynomial.

Solution. Let's transform the expression to the form: The sequence of coefficients of the sum in brackets clearly indicate that this is Therefore, Now we apply the difference of squares formula: The expression in the second bracket has no real roots, and for the polynomial from the first bracket we once again apply the difference of squares formula

Vieta formulas expressing the coefficients of a polynomial through its roots. These formulas are convenient to use for checking the correctness of finding the roots of a polynomial, as well as for composing a polynomial based on its given roots. Formulation If are the roots of a polynomial, then the coefficients are expressed in the form of symmetric polynomials of the roots, namely

In other words, ak is equal to the sum of all possible products of k roots. If the leading coefficient is a polynomial, then to apply the Vieta formula it is necessary to first divide all coefficients by a 0. In this case, the Vieta formulas give an expression for the ratio of all coefficients to the leading one. From Vieta's last formula it follows that if the roots of a polynomial are integer, then they are divisors of its free term, which is also integer. The proof is carried out by considering the equality obtained by expanding the polynomial by roots, taking into account that a 0 = 1 Equating the coefficients at the same powers of x, we obtain the Vieta formulas.

Solve the equation x 6 – 5 x 3 + 4 = 0 Solution. Let us denote y = x 3, then the original equation takes the form y 2 – 5 y + 4 = 0, solving which we obtain Y 1 = 1; Y 2 = 4. Thus, the original equation is equivalent to a set of equations: x 3 = 1 or x 3 = 4, i.e. X 1 = 1 or X 2 = Answer: 1;

Bezout's Theorem Definition 1. An element is called a root of a polynomial if f(c)=0. Bezout's theorem. The remainder of dividing the polynomial Pn(x) by the binomial (x-a) is equal to the value of this polynomial at x = a. Proof. By virtue of the division algorithm, f(x)=(xc)q(x)+r(x), where either r(x)=0, or, and therefore. So f(x)=(x-c)q(x)+r, therefore f(c)=(c-c)q(c)+r=r, and therefore f(x)=(xc)q(x) +f(c).

Corollary 1: The remainder of dividing the polynomial Pn (x) by the binomial ax+b is equal to the value of this polynomial at x = -b/a, i.e. R=Pn (-b/a). Corollary 2: If the number a is the root of the polynomial P (x), then this polynomial is divisible by (x-a) without a remainder. Corollary 3: If the polynomial P(x) has pairwise distinct roots a 1 , a 2 , ... , an, then it is divided by the product (x-a 1) ... (x-an) without a remainder. Corollary 4: A polynomial of degree n has at most n different roots. Corollary 5: For any polynomial P(x) and number a, the difference (P(x)-P(a)) is divisible by the binomial (x-a) without remainder. Corollary 6: A number a is a root of a polynomial P(x) of degree at least first if and only if P(x) is divisible by (x-a) without a remainder.

Decomposition of a rational fraction into simple fractions Let us show that any proper rational fraction can be decomposed into a sum of simple fractions. Let a proper rational fraction (1) be given.

Theorem 1. Let x=a be the root of the denominator of brevity k, i.e., where f(a)≠ 0, then this proper fraction can be represented as the sum of two other proper fractions as follows: (2) , where A is a constant not equal to zero, and F 1(x) is a polynomial whose degree is lower than the degree of the denominator