Etc. is of a general educational nature and has great importance to study 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 exactly personal experience solutions. The information is intended primarily for beginners, but more advanced readers will also find a lot for themselves. interesting moments. And of course there will be new material, going 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 a root given equation. Or, as they also say, satisfies this equation.
Please note that the root can also be written in the form decimal:
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 you can’t get rid of the constant here! 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 their condensed notation is accepted in algebra:
, 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 better understanding I recommend that you read at least a little of the material complex numbers.
They are the best. Polynomials.
The object of our interest will be the most common polynomials of the form with whole coefficients Natural number 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 that first appeared 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 a 4th degree equation, 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 into the equation different numbers, claiming to be the “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? Same school method, which is used to divide ordinary numbers - in a “column”! This method I in more detail discussed 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 (V 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 3rd degree polynomial function intersects the axis at least once, which means the corresponding equation has at least one valid root. This fact valid 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 the 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 longer. 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 most best choice for priority check.
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, 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 the 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. 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
We were lucky in the problem we analyzed, because: a) they fell off right away negative values, 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 exciting game entitled " Last Hero»:
Problem 4
Find the rational roots of the equation
Solution: By Theorem 1 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.
When solving equations and inequalities, it is often necessary to factor a polynomial whose degree is three or higher. In this article we will look at the easiest way to do this.
As usual, let's turn to theory for help.
Bezout's theorem states that the remainder when dividing a polynomial by a binomial is .
But what is important for us is not the theorem itself, but corollary from it:
If the number is the root of a polynomial, then the polynomial is divisible by the binomial without a remainder.
We are faced with the task of somehow finding at least one root of the polynomial, then dividing the polynomial by , where is the root of the polynomial. As a result, we obtain a polynomial whose degree is one less than the degree of the original one. And then, if necessary, you can repeat the process.
This task breaks down into two: how to find the root of a polynomial, and how to divide a polynomial by a binomial.
Let's take a closer look at these points.
1. How to find the root of a polynomial.
First we check whether the numbers 1 and -1 are roots of the polynomial.
The following facts will help us here:
If the sum of all the coefficients of a polynomial is zero, then the number is the root of the polynomial.
For example, in a polynomial the sum of the coefficients is zero: . It's easy to check what the root of a polynomial is.
If the sum of the coefficients of a polynomial at even powers is equal to the sum of the coefficients at odd powers, then the number is the root of the polynomial. The free term is considered a coefficient for an even degree, since , a is an even number.
For example, in a polynomial the sum of coefficients for even powers is: , and the sum of coefficients for odd powers is: . It's easy to check what the root of a polynomial is.
If neither 1 nor -1 are roots of the polynomial, then we move on.
For a reduced polynomial of degree (that is, a polynomial in which the leading coefficient - the coefficient at - is equal to unity), the Vieta formula is valid:
Where are the roots of the polynomial.
There are also Vieta formulas concerning the remaining coefficients of the polynomial, but we are interested in this one.
From this Vieta formula it follows that if the roots of a polynomial are integers, then they are divisors of its free term, which is also an integer.
Based on this, we need to factor the free term of the polynomial into factors, and sequentially, from smallest to largest, check which of the factors is the root of the polynomial.
Consider, for example, the polynomial
Divisors of the free term: ; ; ;
The sum of all coefficients of the polynomial is equal to , therefore, the number 1 is not the root of the polynomial.
Sum of coefficients for even powers:
Sum of coefficients for odd powers:
Therefore, the number -1 is also not a root of the polynomial.
Let's check whether the number 2 is the root of the polynomial: therefore, the number 2 is the root of the polynomial. This means, according to Bezout’s theorem, the polynomial is divisible by a binomial without a remainder.
2. How to divide a polynomial into a binomial.
A polynomial can be divided into a binomial by a column.
Divide the polynomial by a binomial using a column:
There is another way to divide a polynomial by a binomial - Horner's scheme.
Watch this video to understand how to divide a polynomial by a binomial with a column, and using Horner's scheme.
I note that if, when dividing by a column, some degree of the unknown is missing in the original polynomial, we write 0 in its place - the same way as when compiling a table for Horner’s scheme.
So, if we need to divide a polynomial by a binomial and as a result of the division we get a polynomial, then we can find the coefficients of the polynomial using Horner’s scheme:
We can also use Horner scheme in order to check whether it is given number root of a polynomial: if a number is the root of a polynomial, then the remainder when dividing the polynomial by is equal to zero, that is, in the last column of the second row of Horner’s scheme we get 0.
Using Horner's scheme, we "kill two birds with one stone": we simultaneously check whether the number is the root of a polynomial and divide this polynomial by a binomial.
Example. Solve the equation:
1. Let's write down the divisors of the free term and look for the roots of the polynomial among the divisors of the free term.
Divisors of 24:
2. Let's check whether the number 1 is the root of the polynomial.
The sum of the coefficients of a polynomial, therefore, the number 1 is the root of the polynomial.
3. Divide the original polynomial into a binomial using Horner's scheme.
A) Let’s write down the coefficients of the original polynomial in the first row of the table.
Since the containing term is missing, in the column of the table in which the coefficient should be written we write 0. On the left we write the found root: the number 1.
B) Fill in the first row of the table.
In the last column, as expected, we got zero; we divided the original polynomial by a binomial without a remainder. The coefficients of the polynomial resulting from division are shown in blue in the second row of the table:
It's easy to check that the numbers 1 and -1 are not roots of the polynomial
B) Let's continue the table. Let's check whether the number 2 is the root of the polynomial:
So the degree of the polynomial, which is obtained as a result of division by one, is less than the degree of the original polynomial, therefore, the number of coefficients and the number of columns are one less.
In the last column we got -40 - a number that is not equal to zero, therefore, the polynomial is divisible by a binomial with a remainder, and the number 2 is not the root of the polynomial.
C) Let's check whether the number -2 is the root of the polynomial. Since the previous attempt failed, to avoid confusion with the coefficients, I will erase the line corresponding to this attempt:
Great! We got zero as a remainder, therefore, the polynomial was divided into a binomial without a remainder, therefore, the number -2 is the root of the polynomial. The coefficients of the polynomial that is obtained by dividing a polynomial by a binomial are shown in green in the table.
As a result of division we got quadratic trinomial 
, whose roots can easily be found using Vieta’s theorem:
So, the roots of the original equation are:
{
}
Answer: ( 
}
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.
| 1 | −5 | −2 | 16 | |
| 2 | 1 | −3 | −8 | 0 |
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:
The degree of a polynomial resulting from division is always 1 less than the degree of the original one. So:
It has been proven that to factor a polynomial, you need to find its roots. Formulas for the roots of a square polynomial. Method for finding whole roots. Method of factoring a biquadratic polynomial and reducing it to a quadratic one. Recurrent polynomials.
Basis of the method
Let
- polynomial of degree n ≥ 1
of a real or complex variable z with real or complex coefficients a i. Let us accept the following theorem without proof.
Theorem 1
Equation Pn (z) = 0 has at least one root.
Let us prove the following lemma.
Lemma 1
Let P n (z)- polynomial of degree n, z 1
- root of the equation:
Pn (z 1) = 0.
Then P n (z) can be represented in the only way in the form:
Pn (z) = (z - z 1) P n-1 (z),
where Pn- 1(z)- polynomial of degree n - 1
.
Proof
To prove it, we apply the theorem (see Division and multiplication of a polynomial by a polynomial by a corner and a column), according to which for any two polynomials P n (z) and Q k (z), degrees n and k, with n ≥ k, there is a unique representation in the form:
Pn (z) = P n-k (z) Q k (z) + U k-1 (z),
where Pn-k (z)- polynomial of degree n-k, U k- 1(z)- polynomial of degree not higher than k- 1
.
Let's put k = 1
, Q k (z) = z - z 1, Then
Pn (z) = (z - z 1 ) P n-1 (z) + c,
where c is a constant. Let's substitute z = z here 1
and take into account that P n (z 1) = 0:
Pn (z 1 ) = (z 1 - z 1 ) P n-1 (z 1 ) + c;
0 = 0 + c.
Hence c = 0
. Then
Pn,
Q.E.D.
So, based on Theorem 1, the polynomial P n (z) has at least one root. Let's denote it as z 1
,Pn (z 1) = 0. Then, based on Lemma 1:
Pn (z) = (z - z 1 ) P n-1 (z).
Further, if n > 1
, then the polynomial P n- 1(z) also has at least one root, which we denote as z 2
,Pn- 1 (z 2) = 0. Then
Pn- 1 (z) = (z - z 2 ) P n-2 (z);
Pn (z) = (z - z 1 )(z - z 2 ) P n-2 (z).
Continuing this process, we come to the conclusion that there are n numbers z 1 , z 2 , ... , z n such that
Pn (z) = (z - z 1 )(z - z 2 ) ... (z - z n ) P 0 (z).
But P 0(z)- this is a constant. Equating the coefficients for z n, we find that it is equal to a n. As a result, we obtain the formula for factoring a polynomial:
(1)
Pn (z) = a n (z - z 1 )(z - z 2 ) ... (z - z n ).
The numbers z i are the roots of the polynomial P n (z).
In general, not all z i included in (1)
, are different. Among them there may be the same values. Then factoring the polynomial (1)
can be written as:
(2)
Pn (z) = a n (z - z 1 ) n 1 (z - z 2 ) n 2 ... (z - z k ) n k;
.
Here z i ≠ z j for i ≠ j. If n i = 1
, That root z i called simple. It enters into factorization in the form (z-z i). If n i > 1
, That root z i called the multiple root of the multiplicity n i. It enters into factorization in the form of the product n i prime factors: (z-z i )(z-z i ) ... (z-z i ) = (z-z i ) n i.
Polynomials with real coefficients
Lemma 2
If is a complex root of a polynomial with real coefficients, , then the complex conjugate number is also a root of the polynomial, .
Proof
Indeed, if , and the coefficients of a polynomial are real numbers, then .
Thus, complex roots enter factorization in pairs with their complex conjugate values:
,
where , are real numbers.
Then the decomposition (2)
a polynomial with real coefficients into factors can be represented in a form in which only real constants are present:
(3)
;
.
Methods for factoring a polynomial
Taking into account the above, to factorize a polynomial, you need to find all the roots of the equation P n (z) = 0 and determine their multiplicity. Factors with complex roots must be grouped with complex conjugates. Then the expansion is determined by the formula (3) .
Thus, the method for factoring a polynomial is as follows:
1.
Finding the root z 1
equations Pn (z 1) = 0.
2.1.
If the root z 1
real, then we add the factor to the expansion (z - z 1) (z - z 1) 1
:
.
1(z), starting from point (1)
until we find all the roots.
2.2.
If the root is complex, then the complex conjugate number is also the root of the polynomial. Then the expansion includes the factor
,
where b 1 = - 2 x 1, c 1 = x 1 2 + y 1 2.
In this case, we add the factor to the expansion (z 2 + b 1 z + c 1) and divide the polynomial P n (z) by (z 2 + b 1 z + c 1). As a result, we obtain a polynomial of degree n - 2
:
.
Next we repeat the process for the polynomial P n- 2(z), starting from point (1)
until we find all the roots.
Finding the roots of a polynomial
The main task when factoring a polynomial is to find its roots. Unfortunately, this cannot always be done analytically. Here we will look at several cases when you can find the roots of a polynomial analytically.
Roots of a polynomial of the first degree
A first degree polynomial is a linear function. It has one root. The expansion has only one factor containing the variable z:
.
Roots of a polynomial of the second degree
To find the roots of a polynomial of the second degree, you need to solve a quadratic equation:
P 2 (z) = a 2 z 2 + a 1 z + a 0 = 0.
If the discriminant is , then the equation has two real roots:
, .
Then the factorization has the form:
.
If discriminant D = 0
, then the equation has one double root:
;
.
If discriminant D< 0
, then the roots of the equation are complex,
.
Polynomials of degree higher than two
There are formulas for finding the roots of polynomials of 3rd and 4th degrees. However, they are rarely used because they are bulky. There are no formulas for finding the roots of polynomials of degree higher than 4th. Despite this, in some cases it is possible to factor the polynomial.
Finding whole roots
If it is known that a polynomial whose coefficients are integers has an integer root, then it can be found by searching through all possible values.
Lemma 3
Let the polynomial
,
the coefficients a i of which are integers, has an integer root z 1
. Then this root is a divisor of the number a 0
.
Proof
Let us rewrite the equation P n (z 1) = 0 as:
.
Then the whole
M z 1 = - a 0.
Divide by z 1
:
.
Since M is an integer, then M is an integer. Q.E.D.
Therefore, if the coefficients of the polynomial are integers, then you can try to find the integer roots. To do this, you need to find all divisors of the free term a 0
and, by substituting into the equation P n (z) = 0, check whether they are roots of this equation.
Note. If the coefficients of the polynomial are rational numbers, then multiplying the equation P n (z) = 0 on common denominator numbers a i , we obtain an equation for a polynomial with integer coefficients.
Finding rational roots
If the coefficients of the polynomial are integers and there are no integer roots, then for a n ≠ 1
, you can try to find rational roots. To do this you need to make a substitution
z = y/a n
and multiply the equation by a n n- 1
. As a result, we obtain an equation for a polynomial in the variable y with integer coefficients. Next, we look for the integer roots of this polynomial among the divisors of the free term. If we have found such a root y i, then passing to the variable x, we obtain a rational root
z i = y i /a n .
Useful formulas
We present formulas that can be used to factor a polynomial.
More generally, to expand a polynomial
Pn (z) = z n - a 0,
where a 0
- complex, you need to find all its roots, that is, solve the equation:
z n = a 0
.
This equation can be easily solved by expressing a 0
via modulus r and argument φ:
.
Since a 0
will not change if we add to the argument 2π, then imagine a 0
as:
,
where k is an integer. Then
;
.
Assigning k the values k = 0, 1, 2, ... n-1, we get n roots of the polynomial. Then its factorization has the form:
.
Biquadratic polynomial
Consider the biquadratic polynomial:
.
A biquadratic polynomial can be factorized without finding the roots.
When , we have:
,
Where .
Bicubic and quadratic polynomials
Consider the polynomial:
.
Its roots are determined from the equation:
.
It leads to quadratic equation substitution t = z n :
a 2 n t 2 + a n t + a 0 = 0.
Having solved this equation, we find its roots, t 1
, t 2
. Then we find the expansion in the form:
.
Next, using the method indicated above, we factorize z n - t 1
and z n - t 2
. Finally, we group the factors containing complex conjugate roots.
Recurrent polynomials
The polynomial is called returnable, if its coefficients are symmetric:
An example of a reflexive polynomial:
.
If the degree of a recurrent polynomial n is odd, then such a polynomial has a root z = -1
. Dividing such a polynomial by z + 1
, we obtain a recurrent polynomial of degree n - 1
.
If the degree of a recurrent polynomial n is even, then by substitution , it is reduced to a polynomial of degree n/ 2
. Cm.
The question of finding rational roots of a polynomial f(x)Q[x] (with rational coefficients) reduces to the question of finding rational roots of polynomials k
∙
f(x)
Z[x] (with integer coefficients). Here is the number k is the least common multiple of the denominators of the coefficients of a given polynomial.
Necessary but not sufficient conditions for the existence of rational roots of a polynomial with integer coefficients are given by the following theorem.
Theorem 6.1 (on rational roots of a polynomial with integer coefficients).
If
–
rational root of a polynomialf(x)
=
a n
x n +
+
…+
a 1
x
+
a 0
With
whole
coefficients, and(p,
q)
= 1, then the numerator of the fractionpis a divisor of the free term a 0
, and the denominatorqis a divisor of the leading coefficient a 0
.
Theorem 6.2.If
Q
(
Where
(p,
q)
=
1)
is the rational root of the polynomial
f(x)
with integer coefficients, then
–whole numbers.
Example. Find all rational roots of the polynomial
f(x) = 6 x 4 + x 3 + 2 x 2 – 4 x+ 1.
1. By Theorem 6.1: if
–
rational root of a polynomial f(x),
( Where( p,
q)
= 1),
That a 0
= 1
p,
a n
= 6
q. That's why p 
{
1}, q 
(1, 2, 3, 6), which means
.
2. It is known that (Corollary 5.3) the number A is the root of the polynomial f(x) if and only if f(x) divided by ( x – a).
Therefore, to check whether the numbers 1 and –1 are roots of a polynomial f(x) you can use Horner’s scheme:
f(1)
= 6
0,f(–1)
= 12
0, so 1 and –1 are not roots of the polynomial f(x).
3. To weed out some of the remaining numbers 
, we will use Theorem 6.2. If expressions 
or 
accepts integer values for the corresponding numerator values p and denominator q, then in the corresponding cells of the table (see below) we will write the letter “ts”, otherwise - “dr”.
|
|
|
|
|
|||
|
| ||||||
|
|
=
=
4. Using Horner’s scheme, we check whether the numbers remaining after sifting out will be 
roots f(x). First let's divide f(x) on ( X
–
).
As a result we have: f(x) = (X – )(6 x 3 + 4 x 2 + 4 X - 2) and – root f(x). Private q(x) = 6 x 3 + 4 x 2 + 4 X - divide 2 by ( X + ).
Because q
(–)
= 3
0, then (–) is not a root of the polynomial q(x), and hence the polynomial f(x).
Finally, we divide the polynomial q(x) = 6 x 3 + 4 x 2 + + 4 X - 2 on ( X – ).
Got: q () = 0, i.e. – root q(x), and therefore is the root f (x). So the polynomial f (x) has two rational roots: and.
Liberation from algebraic irrationality in the denominator of a fraction
In the school course, when solving certain types of problems to get rid of irrationality in the denominator of a fraction, it is enough to multiply the numerator and denominator of the fraction by the number conjugate to the denominator.
Examples. 1.t
=
.
Here the abbreviated multiplication formula (difference of squares) works in the denominator, which allows you to free yourself from irrationality in the denominator.
2. Free yourself from irrationality in the denominator of the fraction
t
=
. Expression – incomplete square of the difference of numbers A=
And b= 1. Using the abbreviated multiplication formula A 3
–
b 3
=
(a +b)
· ( a 2
–
ab
+
b 2
), we can determine the multiplier m
= (a +b)
=
+ 1, by which the numerator and denominator of the fraction should be multiplied t to get rid of irrationality in the denominator of the fraction t. Thus,
In situations where abbreviated multiplication formulas do not work, other techniques can be used. Below we will formulate a theorem, the proof of which, in particular, allows us to find an algorithm for getting rid of irrationality in the denominator of a fraction in more complex situations.
Definition 6.1. Number z called algebraic over the field
F, if there is a polynomial f(x)
F[x], whose root is z, otherwise the number z called transcendental over the fieldF.
Definition 6.2.Degree of algebraic over field
F
numbers
z is called the degree of an irreducible over a field F polynomial p(x)
F[x], whose root is the number z.
Example. Let us show that the number z = 
is algebraic over the field Q and find its degree.
Let's find an irreducible over the field Q polynomial p(X), whose root is x
=
. Let's raise both sides of the equality x
=
to the fourth power, we get X 4
= 2 or X 4
–
2
= 0. So, p(X)
= X 4
–
2, and the power of the number z equal to deg
p(X)
= 4.
Theorem 6.3
(on liberation from algebraic irrationality in the denominator of a fraction).Letz– algebraic number over a fieldFdegreesn. Expression of the formt
=
,Where
f(x),
(x)
F[x],
(z) 
0
can only be represented in the form:
t
= With n -1
z n -1
+
c n -2
z n -2
+ … +
c 1
z
+
c 0
,
c i
F.
We will demonstrate the algorithm for getting rid of irrationality in the denominator of a fraction using a specific example.
Example. Free yourself from irrationality in the denominator of a fraction:
t
=
1. The denominator of the fraction is the value of the polynomial 
(X)
= X 2
– X+1 when X
=
. The previous example shows that 
– algebraic number over a field Q degree 4, since it is the root of an irreducible over Q polynomial p(X)
= X 4
–
2.
2. Let us find the linear expansion of GCD ( 
(X),
p(x)) using the Euclidean algorithm.
_x 4 – 2 | x 2 –x + 1
x 4 –x 3 + x 2 x 2 + x = q 1 (x)
_ x 3 –x 2 – 2
x 3 –x 2 + x
x 2 –x + 1 | – x –2 = r 1 (x )
x 2 + 2 x – x + 3 = q 2 (x)
_–3x+ 1
–3 x – 6
_ – x –2 |7 = r 2
– x –2 -x - =q 3 (x)
So, GCD ( 
(X),
p(x))
= r 2
=
7. Let's find its linear expansion.
Let's write down the Euclidean sequence using polynomial notation.
p(x)
=
(x)
· q 1
(x)
+ r 1
(x)
r 1
(x)
=p(x)
–
(x)
· q 1
(x)