This service is designed for decomposing fractions of the form:
For the sum of simple fractions. This service will be useful for solving integrals. see example.
Instructions. Enter the numerator and denominator of the fraction. Click the Solve button.
Note: For example, x 2 is written as x^2, (x-2) 3 is written as (x-2)^3. Between the factors we put a multiply sign (*).Rules for entering a function
This field is intended for entering the numerator of the expressionThe general variable x must first be taken out of brackets. For example, x 3 + x = x(x 2 + 1) or x 3 - 5x 2 + 6x = x(x 2 - 5x + 6) = x(x-3)(x-2).
Rules for entering a function
This field is intended for entering the denominator of the expression. For example, x 2 is written as x^2, (x-2) 3 is written as (x-2)^3. Between the factors we put a multiply sign (*).The general variable x must first be taken out of brackets. For example, x 3 + x = x(x 2 + 1) or x 3 - 5x 2 + 6x = x(x 2 - 5x + 6) = x(x-3)(x-2).
Algorithm for the method of uncertain coefficients
- Factoring the denominator.
- Decomposition of a fraction as a sum of simple fractions with undetermined coefficients.
- Grouping the numerator with the same powers of x.
- Obtaining a system of linear algebraic equations with undetermined coefficients as unknowns.
- Solution of SLAE: Cramer method, Gauss method, inverse matrix method or method of eliminating unknowns.
Example. We use the method of decomposition into simplest ones. Let's break down the function into its simplest terms:
Let us equate the numerators and take into account that the coefficients at the same powers X, standing on the left and right must match
2x-1 = A(x+2) 2 (x-4) + Bx(x+2) 2 (x-4) + Cx(x-4) + Dx(x+2) 2
A+B=0
-12A -8B -4C + 4D = 2
-16A = -1
0A -2B + C + 4D = 0
Solving it, we find:
A = 1/16 ;B = - 1/9 ;C = - 5/12 ;D = 7/144 ;
MINISTRY OF SCIENCE AND EDUCATION OF THE REPUBLIC OF BASHKORTO STAN
SAOU SPO Bashkir College of Architecture and Civil Engineering
Khaliullin Askhat Adelzyanovich,
mathematics teacher at Bashkirsky
College of Architecture and Civil Engineering
UFA
2014
Introduction ___________________________________________________3
Chapter I. Theoretical aspects of using the method of uncertain coefficients_____________________________________________4
Chapter II. Searches for solutions to problems with polynomials using the method of indefinite coefficients_________________________________7
2.1.Factoring a polynomial_____________________ 7
2.2. Problems with parameters_________________________________ 10
2.3. Solving equations__________________________________________14
2.4. Functional equations______________________________19
Conclusion_________________________________________________23
List of used literature__________________________________________24
Application ________________________________________________25
Introduction.
This work is devoted to the theoretical and practical aspects of introducing the method of indefinite coefficients into the school mathematics course. The relevance of this topic is determined by the following circumstances.
No one will argue that mathematics as a science does not stand in one place, it is constantly evolving, new tasks of increased complexity appear, which often causes certain difficulties, since these tasks are usually associated with research. In recent years, such problems have been proposed at school, district and republican mathematical Olympiads, and they are also available in the Unified State Exam versions. Therefore, a special method was required that would allow at least some of them to be solved most quickly, efficiently and affordably. This work clearly presents the content of the method of indefinite coefficients, which is widely used in a wide variety of areas of mathematics, ranging from questions included in the general education course to its most advanced parts. In particular, applications of the method of indefinite coefficients in solving problems with parameters, fractional rational and functional equations are especially interesting and effective; they can easily interest anyone interested in mathematics. The main purpose of the proposed work and selection of problems is to provide ample opportunities to hone and develop the ability to find short and non-standard solutions.
This work consists of two chapters. The first discusses the theoretical aspects of using
method of uncertain coefficients, and secondly, practical and methodological aspects of such use.
The appendix to the work provides conditions for specific tasks for independent solution.
Chapter I . Theoretical aspects of use method of uncertain coefficients
“Man... was born to be a master,
ruler, king of nature, but wisdom,
with which he must rule is not given to him
from birth: it is acquired by learning"
N.I.Lobachevsky
There are various ways and methods for solving problems, but one of the most convenient, most effective, original, elegant and at the same time very simple and understandable to everyone is the method of indefinite coefficients. The method of undetermined coefficients is a method used in mathematics to find the coefficients of expressions whose form is known in advance.
Before considering the application of the method of indefinite coefficients to solving various types of problems, we present a number of theoretical information.
Let them be given
A n (x) = a 0 x n + a 1 x n-1 + a 2 x n-2 + ··· + a n-1 x + a n
B m (x ) = b 0 x m + b 1 x m -1 + b 2 x m -2 + ··· + b m-1 x + b m ,
polynomials relative X with any odds.
Theorem. Two polynomials depending on one and the same argument are identically equal if and only ifn = m and their corresponding coefficients are equala 0 = b 0 , a 1 = b 1 , a 2 = b 2 ,··· , a n -1 = b m -1 , a n = b m And T. d.
Obviously, equal polynomials take for all values X same values. Conversely, if the values of two polynomials are equal for all values X, then the polynomials are equal, that is, their coefficients at the same degreesX match up.
Therefore, the idea of applying the method of indefinite coefficients to solving problems is as follows.
Let us know that as a result of some transformations an expression of a certain type is obtained and only the coefficients in this expression are unknown. Then these coefficients are designated by letters and considered as unknowns. A system of equations is then constructed to determine these unknowns.
For example, in the case of polynomials, these equations are made from the condition that the coefficients are equal for the same powers X for two equal polynomials.
We will demonstrate what was said above using the following specific examples, and let’s start with the simplest.
So, for example, based on theoretical considerations, the fraction
can be represented as a sum
, Where
a
,
b
And
c
-
coefficients to be determined. To find them, we equate the second expression to the first:
=
and freeing ourselves from the denominator and collecting terms with the same powers on the left X, we get:
(a + b + c )X 2 + ( b - c )x - a = 2X 2 – 5 X– 1
Since the last equality must be true for all values X, then the coefficients at the same powersX right and left should be the same. Thus, three equations are obtained to determine the three unknown coefficients:
a+b+c = 2
b - c = - 5
A= 1, whence a = 1 , b = - 2 , c = 3
Hence,
=
,
the validity of this equality is easy to verify directly.
Suppose you also need to represent a fraction
as a
+
b 
+
c 
+ d
, Where a
,
b
,
c
And
d- unknown rational coefficients. We equate the second expression to the first:
a
+
b
+
c
+ d
= or, Freeing ourselves from the denominator, removing, where possible, rational factors from under the signs of the roots and bringing similar terms on the left side, we obtain:
(a-
2
b
+
3
c
) + (-
a+b
+3
d
)
+ (a+c
- 2
d
)
+
+ (b - c
+
d
)
=
1 +
-
.
But such equality is possible only in the case when the rational terms of both parts and the coefficients of the same radicals are equal. Thus, four equations are obtained for finding the unknown coefficients a , b , c And d :
a- 2b+ 3c = 1
- a+b +3 d = 1
a+c - 2 d = - 1
b
-
c
+
d= 0, whence a
= 0 ;
b
= - ;
c
= 0
;
d= , that is = -
+
.
Chapter II. Searches for solutions to problems with polynomials method of undetermined coefficients.
“Nothing contributes to the mastery of a subject better than
the way to act with him in different situations"
Academician B.V. Gnedenko
2. 1. Factoring a polynomial.
Methods for factoring polynomials:
1) placing the common factor out of brackets; 2) grouping method; 3) application of basic multiplication formulas; 4) introduction of auxiliary terms; 5) preliminary transformation of a given polynomial using certain formulas; 6) expansion by finding the roots of a given polynomial; 7) method of entering the parameter; 8)method of undetermined coefficients.
Problem 1. Factor the polynomial into real factors X 4 + X 2 + 1 .
Solution. There are no roots among the divisors of the free term of this polynomial. We cannot find the roots of the polynomial by other elementary means. Therefore, it is not possible to perform the required expansion by first finding the roots of this polynomial. It remains to look for a solution to the problem either by introducing auxiliary terms or by the method of undetermined coefficients. It's obvious that X 4 + X 2 + 1 = X 4 + X 3 + X 2 - X 3 - X 2 - X + X 2 + X + 1 =
= X 2 (X 2 + X + 1) - X (X 2 + X + 1) + X 2 + X + 1 =
= (X 2 + X + 1)(X 2 - X + 1).
The resulting quadratic trinomials have no roots and are therefore indecomposable into real linear factors.
The described method is technically simple, but difficult due to its artificiality. Indeed, it is very difficult to come up with the required auxiliary terms. Only a guess helped us find this decomposition. But
There are more reliable ways to solve such problems.
One could proceed like this: assume that the given polynomial decomposes into the product
(X 2 + A X + b )(X 2 + c X + d )
two square trinomials with integer coefficients.
Thus, we will have that
X 4 + X 2 + 1 = (X 2 + A X + b )(X 2 + c X + d )
It remains to determine the coefficientsa , b , c And d .
Multiplying the polynomials on the right side of the last equality, we get:X 4 + X 2 + 1 = X 4 +
+ (a + c ) X 3 + (b + A c + d ) X 2 + (ad + bc ) x + bd .
But since we need the right side of this equality to turn into the same polynomial that is on the left side, we will require the following conditions to be met:
a + c = 0
b + A c + d = 1
ad + bc = 0
bd = 1 .
The result is a system of four equations with four unknownsa , b , c And d . It is easy to find the coefficients from this systema = 1 , b = 1 , c = -1 And d = 1.
Now the problem is completely solved. We got:
X 4 + X 2 + 1 = (X 2 + X + 1)(X 2 - X + 1).
Problem 2. Factor the polynomial into real factors X
3
– 6
X
2
+ 14
X
– 15 .
Solution. Let us represent this polynomial in the form
X
3
– 6
X
2 + 14
X
– 15 = (X
+
A
)(X
2
+
bx
+
c) , Where a
,
b
And With
- coefficients not yet determined. Since two polynomials are identically equal if and only if the coefficients of the same powersX
are equal, then, equating the coefficients respectively forX
2
,
X
and free terms, we obtain a system of three equations with three unknowns:
a+b= - 6
ab+c = 14
ac = - 15 .
The solution to this system will be significantly simplified if we take into account that the number 3 (divisor of the free term) is the root of this equation, and, therefore,a = - 3 ,
b = - 3 And With = 5 .
Then X
3
– 6
X
2 + 14
X
– 15 = (X
– 3)(X
2
– 3
x
+
5).
The applied method of indefinite coefficients, in comparison with the above method of introducing auxiliary terms, does not contain anything artificial, but it requires the application of many theoretical principles and is accompanied by rather large calculations. For polynomials of higher degree, this method of undetermined coefficients leads to cumbersome systems of equations.
2.2.Tasks and with parameters.
In recent years, versions of the Unified State Exam have offered tasks with parameters. Their solution often causes certain difficulties. When solving problems with parameters, along with other methods, you can quite effectively use the method of indefinite coefficients. It is this method that allows you to greatly simplify their solution and quickly get an answer.
Task 3. Determine at what values of the parameter A equation 2 X 3 – 3 X 2 – 36 X + A – 3 = 0 has exactly two roots.
Solution. 1 way. Using derivative.
Let's represent this equation in the form of two functions
2x 3 – 3 X 2 – 36 X – 3 = – A .
f (x) = 2x 3 – 3 X 2 – 36 X– 3 and φ( X ) = – A .
Let's explore the functionf (x) = 2x 3 – 3 X 2 – 36 X – 3 using the derivative and schematically construct its graph (Fig. 1.).
f(
–
x
)
f
(x
) ,
f
(–
x
) –
f
(x
).
The function is neither even nor odd.
3. Let's find the critical points of the function, its intervals of increase and decrease, extrema. f / (x ) = 6 x 2 – 6 X – 36. D (f / ) = R , therefore we will find all critical points of the function by solving the equation f / (x ) = 0 .
6(X 2 – X– 6) = 0 ,
X 2 – X– 6 = 0 ,
X 1 = 3 , X 2 = – 2 by the theorem inverse to Vieta’s theorem.
f / (x ) = 6(X – 3)(X + 2).
+ max - min +
2 3 x
f
/
(x) > 0 for all X<
– 2 and X
> 3 and the function is continuous at pointsx =– 2 and X
= 3, therefore, it increases on each of the intervals (- 
; - 2] and [ 3 ; ).
f / (x ) < 0 at - 2 < X< 3, therefore, it decreases on the interval [- 2; 3 ].
X = - 2nd maximum point, because at this point the sign of the derivative changes from"+" to "-".
f (– 2) = 2· (– 8) – 3·4 – 36·(– 2) – 3 = – 16 – 12 + 72 – 3 == 72 – 31 = 41 ,
x = 3 minimum point, since at this point the sign of the derivative changes"-" to "+".
f (3) = 2·27 – 3·9 – 36·3 – 3 = 54 – 27 – 108 – 3 = – 138 + +54 = – 84.
Graph of the function φ(X ) = – A is a straight line parallel to the x-axis and passing through the point with coordinates (0; – A ). The graphs have two common points at –A= 41, i.e. a =– 41 and – A= – 84, i.e. A = 84 .
at
41φ( X)
2 3 X
3 f ( x ) = 2x 3 – 3 X 2 – 36 X – 3
Method 2. Method of undetermined coefficients.
Since, according to the conditions of the problem, this equation must have only two roots, the equality is obvious:
2X 3 – 3 X 2 – 36 X + A – 3 = (x + b ) 2 (2 x + c ) ,
2X 3 – 3 X 2 – 36 X + A – 3 = 2 x 3 + (4 b + c ) x 2 + (2 b 2 + +2 bc ) x + b 2 c ,
Now equating the coefficients at the same degrees X, we obtain a system of equations
4 b + c = - 3
2b 2 + 2bc = - 36
b 2 c = a – 3 .
From the first two equations of the system we findb 2 + b – 6 = 0, whence b 1 = - 3 or b 2 = 2 . Corresponding valuesWith 1 and With 2 easy to find from the first equation of the system:With 1 = 9 or With 2 = - 11 . Finally, the desired value of the parameter can be determined from the last equation of the system:
A = b 2 c + 3 , a 1 = - 41 or a 2 = 84.
Answer: this equation has exactly two different
root at A= - 41 and A= 84 .
Task 4. Find the largest value of the parameterA , for which the equationX 3 + 5 X 2 + Oh + b = 0
with integer coefficients has three different roots, one of which is equal to – 2.
Solution. 1 way. Substituting X= - 2 to the left side of the equation, we get
8 + 20 – 2 A + b= 0, which means b = 2 a – 12 .
Since the number - 2 is a root, we can take out the common factor X + 2:
X 3 + 5 X 2 + Oh + b = X 3 + 2 X 2 + 3 X 2 + Oh + (2 a – 12) =
= x 2 (X + 2) + 3 x (X + 2) – 6 x + Oh + (2 a – 12) =
= x 2 (X + 2) + 3 x (X + 2) + (a – 6)(x +2) - 2(a – 6)+ (2 a – 12) =
= (X + 2)(X 2 + 3 x + (a – 6) ) .
By condition, there are two more roots of the equation. This means that the discriminant of the second factor is positive.
D =3 2 - 4 (a – 6) = 33 – 4 a > 0, that is A < 8,25 .
It would seem that the answer would be a = 8 . But when we substitute the number 8 into the original equation we get:
X 3 + 5 X 2 + Oh + b = X 3 + 5 X 2 + 8 X + 4 = (X + 2)(X 2 + 3 x + 2 ) =
= (X + 1) (X + 2) 2 ,
that is, the equation has only two different roots. But when a = 7 actually produces three different roots.
Method 2. Method of undetermined coefficients.
If the equation X 3 + 5 X 2 + Oh + b = 0 has a root X = - 2, then you can always pick up the numbersc And d so that in front of everyoneX equality was true
X 3 + 5 X 2 + Oh + b = (X + 2)(X 2 + With x + d ).
To find numbersc And d Let's open the brackets on the right side, add similar terms and get
X 3 + 5 X 2 + Oh + b = X 3 + (2 + With ) X 2 +(2 s + d ) X + 2 d
Equating the coefficients at the corresponding powers X we have a system
2 + With = 5
2 With + d = a
2 d = b , where c = 3 .
Hence, X 2 + 3 x + d = 0 , D = 9 – 4 d > 0 or
d
<
2.25, so d
(- ; 2 ].
The problem conditions are satisfied by the value d = 1 . The final desired value of the parameterA = 7.
ANSWER: when a = 7 this equation has three different roots.
2.3. Solving equations.
“Remember that by solving small problems you
prepare yourself to tackle big and difficult
new tasks.”
Academician S.L. Sobolev
When solving some equations, you can and should show resourcefulness and wit, and use special techniques. Mastery of a variety of transformation techniques and the ability to carry out logical reasoning is of great importance in mathematics. One of these tricks is to add and subtract some well-chosen expression or number. The stated fact itself, of course, is well known to everyone - the main difficulty is to see in a specific configuration those transformations of equations to which it is convenient and expedient to apply it.
Using a simple algebraic equation, we will illustrate one non-standard technique for solving equations.
Problem 5. Solve the equation
= 
.
Solution. Let's multiply both sides of this equation by 5 and rewrite it as follows
= 0 ; X 0; -
; 
= 0 ,
= 0 ,
= 0 or 
= 0
Let us solve the resulting equations using the method of undetermined coefficients
X 4 - X 3 –7 X – 3 = (X 2 + ah + b )(x 2 + cx + d ) = 0
X 4 - X 3 –7 X – 3 = X 4 + (a + c ) X 3 + (b + A c + d ) X 2 + (ad + bc ) x+ + bd
Equating the coefficients at X 3 , X 2 , X and free terms, we get the system
a + c = -1
b + A c + d = 0
ad + bc = -7
bd = -3, from where we find:A = -2 ; b = - 1 ;
With = 1 ; d = 3 .
So X 4 - X 3 –7X– 3 = (X 2 – 2 X – 1)(X 2 + X + 3) = 0 ,
X 2 – 2 X– 1 = 0 or X 2 + X + 3 = 0
X 1,2 = 
no roots.
Similarly we have
X 4 – 12X – 5 = (X 2 – 2 X – 1)(X 2 + 2X + 5) = 0 ,
where X 2 + 2 X + 5 = 0 , D = - 16 < 0 , нет корней.
Answer: X 1,2 =
Problem 6. Solve the equation
= 10.
Solution. To solve this equation you need to select numbersA And b so that the numerators of both fractions are the same. Therefore, we have the system:
= 0 , X 0; -1 ; -
= - 10
So the task is to find the numbersA And b , for which equality holds
(a + 6) X 2 + ah – 5 = X 2 + (5 + 2 b ) x + b
Now, according to the theorem on the equality of polynomials, it is necessary that the right side of this equality turns into the same polynomial that is on the left side.
In other words, the relations must be satisfied
a + 6 = 1
A = 5 + 2 b
– 5 = b , from where we find the valuesA = - 5 ;
b = - 5 .
At these valuesA And b equality A + b = - 10 is also fair.
= 0 , X 0; -1 ; -
= 0 ,
= 0 ,
(X 2 – 5X– 5)(X 2 + 3X + 1) = 0 ,
X 2 – 5X– 5 = 0 or X 2 + 3X + 1 = 0 ,
X 1,2 = 
, X 3,4 = 
Answer: X 1,2 = , X 3,4 =
Problem 7. Solve the equation
= 4
Solution. This equation is more complex than the previous ones and therefore we will group it this way: X 0;-1;3;-8;12
0 ,
= - 4.
From the condition of equality of two polynomials
Oh 2 + (a + 6) X + 12 = X 2 + (b + 11) x – 3 b ,
we obtain and solve a system of equations for unknown coefficientsA And b :
A = 1
a + 6 = b + 11
12 = – 3 b , where a = 1 , b = - 4 .
Polynomials - 3 – 6X + cx 2 + 8 cx And X 2 + 21 + 12 d – dx are equal to each other identically only when
With = 1
8 With - 6 = - d
3 = 21 + 12 d , With = 1 , d = - 2 .
With valuesa = 1 , b = - 4 , With = 1 , d = - 2
equality = - 4 is correct.
As a result, this equation takes the following form:
= 0 or 
= 0 or 
= 0 ,
= - 4 , 
= - 3 , 
= 1 ,
= - 
.
From the examples considered, it is clear how the skillful use of the method of indefinite coefficients,
helps to simplify the solution of a rather complex, unusual equation.
2.4. Functional equations.
“The highest purpose of mathematics... is
is to find the hidden order in
chaos that surrounds us"
N. Viner
Functional equations are a very general class of equations in which the unknown function is a certain function. A functional equation in the narrow sense of the word is understood as equations in which the desired functions are related to known functions of one or more variables using the operation of forming a complex function. A functional equation can also be considered as an expression of a property characterizing a particular class of functions
[for example, functional equation f ( x ) = f (- x ) characterizes the class of even functions, the functional equationf (x + 1) = f (x ) – class of functions having period 1, etc.].
One of the simplest functional equations is the equationf (x + y ) = f (x ) + f (y ). Continuous solutions of this functional equation have the form
f (x ) = Cx . However, in the class of discontinuous functions this functional equation has other solutions. Associated with the considered functional equation are
f (x + y ) = f (x ) · f (y ), f (x y ) = f (x ) + f (y ), f (x y ) = f (x )· f (y ),
continuous solutions, which, respectively, have the form
e cx , WITHlnx , x α (x > 0).
Thus, these functional equations can be used to define exponential, logarithmic and power functions.
The most widely used equations are those in complex functions in which the required functions are external functions. Theoretical and practical applications
It was precisely these equations that prompted outstanding mathematicians to study them.
For example, at alignment
f 2 (x) = f (x - y)· f (x + y)
N.I.Lobachevskyused when determining the angle of parallelism in my geometry.
In recent years, problems related to solving functional equations are quite often offered at mathematical Olympiads. Their solution does not require knowledge beyond the scope of the mathematics curriculum in secondary schools. However, solving functional equations often causes certain difficulties.
One of the ways to find solutions to functional equations is the method of indefinite coefficients. It can be used when the general form of the desired function can be determined by the appearance of the equation. This applies, first of all, to those cases when solutions to equations should be sought among integer or fractional rational functions.
Let us outline the essence of this technique by solving the following problems.
Task 8. Functionf
(x
) is defined for all real x and satisfies for allX
R
condition
3 f(x) - 2 f(1- x) = x 2 .
Findf (x ).
Solution. Since on the left side of this equation over the independent variable x and the values of the functionf Only linear operations are performed, and the right side of the equation is a quadratic function, then it is natural to assume that the desired function is also quadratic:
f (X) = ax 2 + bx + c , Wherea, b, c – coefficients to be determined, that is, uncertain coefficients.
Substituting the function into the equation, we arrive at the identity:
3(ax 2 + bx+ c) – 2(a(1 – x) 2 + b(1 – x) + c) = x 2 .
ax 2 + (5 b + 4 a) x + (c – 2 a – 2 b) = x 2 .
Two polynomials will be identically equal if they are equal
coefficients for the same powers of the variable:
a = 1
5b + 4a = 0
c– 2 a – 2 b = 0.
From this system we find the coefficients
a = 1 , b = - , c = , Alsosatisfiesequality
3 f (x ) - 2 f (1- x ) = x 2 on the set of all real numbers. At the same time, there is suchx 0 Task 9. Functiony =f(x) for all x is defined, continuous and satisfies the conditionf (f (x)) – f(x) = 1 + 2 x . Find two such functions.
Solution. Two actions are performed on the desired function - the operation of composing a complex function and
subtraction. Considering that the right side of the equation is a linear function, it is natural to assume that the desired function is also linear:f(x) = ah +b , WhereA Andb – uncertain coefficients. Substituting this function intof (f ( (x ) = - X - 1 ;
f 2 (x ) = 2 X+ , which are solutions to the functional equationf (f (x)) – f(x) = 1 + 2 x .
Conclusion.
In conclusion, it should be noted that this work will certainly contribute to the further study of an original and effective method for solving a variety of mathematical problems, which are problems of increased difficulty and require deep knowledge of the school mathematics course and a high logical culture. Anyone who wants to independently deepen their knowledge of mathematics will also find This work contains material for reflection and interesting tasks, the solution of which will bring benefit and satisfaction.
The work, within the framework of the existing school curriculum and in a form accessible for effective perception, sets out the method of indefinite coefficients, which helps deepen the school course in mathematics.
Of course, all the possibilities of the method of indefinite coefficients cannot be demonstrated in one work. In fact, the method still requires further study and research.
List of used literature.
Glazer G.I..History of mathematics in school.-M.: Education, 1983.
Gomonov S.A. Functional equations in a school mathematics course // Mathematics at school. – 2000. –№10 .
Dorofeev G.V., Potapov M.K., Rozov N.H.. A manual on mathematics. - M.: Nauka, 1972.
Kurosh A.G.. Algebraic equations of arbitrary degrees. - M.: Nauka, 1983.
Likhtarnikov L.M.. Elementary introduction to functional equations. – St. Petersburg. : Lan, 1997.
Manturov O.V., Solntsev Yu.K., Sorokin Yu.I., Fedin N.G.. Explanatory dictionary of mathematical terms.-M.: Education, 1971
Modenov V.P.. A manual on mathematics. Part 1.-M.: Moscow State University, 1977.
Modenov V.P.. Problems with parameters. - M.: Exam, 2006.
Potapov M.K., Aleksandrov V.V., Pasichenko P.I.. Algebra and analysis of elementary functions. - M.: Nauka, 1980.
Khaliullin A.A.. You can solve it easier // Mathematics at school. – 2003 . - №8 .
Khaliullin.
4. Expand polynomial 2X 4 – 5X 3 + 9X 2 – 5X+ 3 for multipliers with integer coefficients.
5. At what value A X 3 + 6X 2 + Oh+ 12 per X+ 4 ?
6. At what value of the parameterA the equationX 3 +5 X 2 + + Oh + b = 0 with integer coefficients has two different roots, one of which is 1 ?
7. Among the roots of the polynomial X 4 + X 3 – 18X 2 + Oh + b with integer coefficients there are three equal integers. Find the value b .
8. Find the largest integer value of the parameter A, at which the equation X 3 – 8X 2 + ah +b = 0 with integer coefficients has three different roots, one of which is equal to 2.
9. At what values A And b division is performed without remainder X 4 + 3X 3 – 2X 2 + Oh + b on X 2 – 3X + 2 ?
10. Factor polynomials:
A)X 4 + 2 X 2 – X + 2 V)X 4 – 4X 3 +9X 2 –8X + 5 d)X 4 + 12X – 5
b)X 4 + 3X 2 + 2X + 3 G)X 4 – 3X –2 e)X 4 – 7X 2 + 1 .
11. Solve the equations:
A) 
= 2
= 2
f
(1 –
X
) =
X
2
.
Find f (X) .
13. Function at= f (X) in front of everyone X defined, continuous and satisfies the condition f ( f (X)) = f (X) + X. Find two such functions.