Polynomials - Masaryk University
Polynomials
Text for Students of
Mathematics Teaching
Irena Budínová
Masaryk University
Faculty of Education
P
OLYNOMIALS
T
EXT FOR
S
TUDENTS OF
M
ATHEMATICS
T
EACHING
M
GR
.
I
RENA
B
UDÍNOVÁ
, P
H
.D.
2013
1 Table of Contents
3
2 Polynomials
Chapter Objectives
Having studied this chapter you will be able to:
define polynomial
divide polynomials
perform Taylor series expansion
find the greatest common divisor of polynomials
find roots of polynomials
use Vieta’s formulas for searching for roots of polynomials
Time to Study
Circa 3 hours
Study Guide
The notion of polynomial must be already known to you from a high school. A polynomial (also known as a multinomial) was seen as a mathematical expression, which is expressed as a sum of several monomials. A monomial consists of several parts – a coefficient (which can be any real number) and one or more variables with a positive integer exponent.
In a mathematical analysis course you have encountered a polynomial as a special case of a (real) function (of a single real variable), id est as a mapping from the set of real numbers into the set of real numbers. These functions are very pleasant to work with due to their properties: their domain of definition is always the whole set of real numbers, they are continuous on this whole set and have derivations of any order which are polynomials again, etc.
At elementary school you have learned to add polynomials and to multiply them. At high school your skills have been extended by dividing polynomials.
From the algebraic point of view it is especially interesting that the result of adding and multiplying polynomials is again a polynomial. Thence polynomials form an algebraic structure with two (binary) operations.
Let us remind the form in which you are used to writing polynomials. Any polynomial can be transformed into the form
5
𝑎 𝑛 𝑥 𝑛 + 𝑎 𝑛−1 𝑥 𝑛−1 + ⋯ + 𝑎
1 𝑥 + 𝑎
0
, where 𝑛 ∈ 𝑵, 𝑎
0
, 𝑎
1
, … , 𝑎 𝑛
∈ 𝑹 .
In this text we will extend your knowledge in the field of polynomials by important concept definitions and by other possibilities of polynomials manipulation.
We will use concepts and knowledge you have already encountered in previous algebra courses. Proofs of theorems will be included only in case the proof is not overly demanding and there can be a certainty that the student will be able to follow the line of the proof. End of a proof will be marked by a square.
When constructing polynomials we have basically two ways we can proceed with.
We can regard them as functions, which you have come across with in mathematical analysis. In this field a polynomial is perceived as a function of a single variable 𝑥 , which can be in the whole domain written as 𝑎 𝑛 𝑥 𝑛 + 𝑎 𝑛−1 𝑥 𝑛−1 + ⋯ + 𝑎
1 𝑥 + 𝑎
0
, where 𝑎
0
, 𝑎
1
, … , 𝑎 𝑛
are constants. This definition states that every polynomial can be expressed as a finite sum of monomials in the form 𝑎 𝑘 𝑥 𝑘
, where the variable is exponentiated to a nonnegative integer.
Yet we can also look at them from the point of view that a polynomial is defined by its coefficients and that the polynomials equality, addition and multiplication can be described by equalities and operations of their coefficients. We will use this second approach and will base the study on the algebraic definition of polynomial.
2.1
Algebraic Definition of Polynomial
Definition 1
By a polynomial over a field T
1
we mean a sequence (𝑎
0
, 𝑎
1
, 𝑎
2
, … ) of elements of the field T, in which there is at most a finite count of nonzero elements. The set of all polynomials over T is denoted T[x].
The polynomial 𝑜̅ = (0,0, … ,0) is called a zero polynomial .
Note:
For instance a polynomial 5𝑥 7 + 13𝑥 6 − 2𝑥 3 + 2 can be identified with the sequence (2, 0, 0,-2 ,0 ,0 ,13 ,5).
1 By a numerical field we mean an ordered triple
T ,
,
, where T is a subset of the set of complex numbers C, such that
0 T , 1 T
and there applies:
(
x , y
T )( x
y
T
x .
y
T ) (it is closed for both addition and multiplication),
( x T )( 1 ).
x T
(it is closed for inverse elements),
(
x
T )( x
0
1
T ) (it is closed for inverse values of nonzero elements). x
Definition 2
Let us define over the set T[x] that for 𝑓 = (𝑎
0, 𝑎
1
, … ) , 𝑔 = (𝑏
0, 𝑏
1
, … ) ∈
𝑇[𝑥] :
1) 𝑓 = 𝑔
∀𝑘 ∈ 𝑁
0
: 𝑎 𝑘
= 𝑏 𝑘
2) 𝑓 + 𝑔 = (𝑎
0
+ 𝑏
0
, 𝑎
1
+ 𝑏
1
, … , 𝑎 𝑘
+ 𝑏 𝑘
, … ) , 𝑓 + 𝑔 ∈ 𝑇[𝑥]
3) 𝑓 ∙ 𝑔 = (𝑐
0
, 𝑐
1
, 𝑐
2
, … ) , 𝑓 ∙ 𝑔 ∈ 𝑇[𝑥] , where 𝑐
0 𝑎
0 𝑏
2
+ 𝑎
1 𝑏
1
+ 𝑎
2 𝑏
0
, etc.
= 𝑎
0 𝑏
0
, 𝑐
1
= 𝑎
0 𝑏
1
+ 𝑎
1 𝑏
0
, 𝑐
2
=
Note:
E.g. the polynomials 𝑓 = 2𝑥 3 + 𝑥 2 − 𝑥 + 3 and 𝑔 = 3𝑥 2 + 2𝑥 + 1 are multiplied in this way: 𝑓 ∙ 𝑔 = 2𝑥 3 ∙ 3𝑥 2 + 2𝑥 3
− 𝑥 ∙ 1 + 3 ∙ 3𝑥 2
∙ 2𝑥 + 2𝑥 3 ∙ 1 + 𝑥
+ 3 ∙ 2𝑥 + 3 ∙ 1
2
= 6𝑥 5
= 6𝑥 5
+ 4𝑥
+ 7𝑥
4
4
+ 2𝑥 3
+ 𝑥 3
+ 3𝑥
+ 8𝑥 2
4 + 2𝑥 3 + 𝑥
+ 5𝑥 + 3,
2
∙ 3𝑥 2
− 3𝑥
+ 𝑥
3
2 ∙ 2𝑥 + 𝑥
− 2𝑥 2
2 ∙ 1 − 𝑥 ∙ 3𝑥
− 𝑥 + 9𝑥 2
2 − 𝑥 ∙ 2𝑥
+ 6𝑥 + 3 hence 𝑐
0
= 3 = 3 ∙ 1, 𝑐
1 the form (3, 5, 8, 1, 7, 6) .
= 5 = 3 ∙ 2 − 1 ∙ 1 , etc. The polynomial 𝑓 ∙ 𝑔 can be written in
Theorem 1 (𝑇[𝑥], +, . ) is an integral domain
2
for each field 𝑇 .
Proof:
Addition of polynomials is commutative, i.e. 𝑓 + 𝑔 = 𝑔 + 𝑓 .
Addition of polynomials is associative, i.e. 𝑓 + (𝑔 + ℎ) = (𝑓 + 𝑔) + ℎ .
The zero element is the zero polynomial.
The inverse element for a polynomial 𝑓 = (𝑎
0
, 𝑎
1
, … ) is a polynomial −𝑓 =
(−𝑎
0
, −𝑎
1
, … ) .
Multiplication of polynomials is commutative.
Multiplication of polynomials is associative.
The identity element is the polynomial 𝑖 = (1,0,0, … ,0) .
Zero divisors: it is obvious that if polynomials f and g are nonzero then their product is nonzero too.
Note:
Following is the demonstration how a polynomial represented using a sequence can be turned into the expression familiar from high school.
Let the polynomial 𝑓 = (𝑎, 0,0, … ) identify with an element 𝑎 ∈ 𝑇 ,
(𝒂, 𝟎, 𝟎, … ) ≡ 𝒂 .
2 Let us remind that an integral domain is a commutative ring that contains the identity element and where the zero element does not have any non-trivial divisors.
7
Next let the polynomial (0,1,0,0, … ) be denoted as 𝑥 , (0,0,1,0 … ) as 𝑥 2
etc. Then there applies
(𝑎
0
, 𝑎
2
, … , 𝑎 𝑘
) = (𝑎
0
, 0,0, … ) + (0, 𝑎
1
, 0, … ) + ⋯ + (0, … , 𝑎 𝑘
, 0, … ) =
= 𝑎
0
(1,0, … ) + 𝑎
1
(0,1,0, … ) + 𝑎
2
(0,0,1,0, … ) + ⋯ + 𝑎 𝑘
(0,0, … ,1,0, … ) =
= 𝒂
𝟎
+ 𝒂
𝟏 𝒙 + 𝒂
𝟐 𝒙 𝟐 + ⋯ + 𝒂 𝒌 𝒙 𝒌
.
Definition 3
Let 𝑓 ∈ 𝑇[x] . Then the degree of a polynomial ( 𝑑𝑔 𝑓 ) is defined in this way: if 𝑓 ≠ 𝑜̅ , 𝑓 = 𝑎
0
+ 𝑎
1 𝑥 + ⋯ + 𝑎 𝑛 𝑥 𝑛
, 𝑎 𝑛
≠ 0 , then it is defined 𝑑𝑔 𝑓 = 𝑛.
Theorem 2
For degrees of polynomials 𝑓, 𝑔 ∈ 𝑇[𝑥] there applies:
1) 𝑑𝑔(𝑓 + 𝑔) ≤ 𝑚𝑎𝑥{𝑑𝑔 𝑓, 𝑠𝑡 𝑔} ,
2) 𝑑𝑔 (𝑓𝑔) = 𝑑𝑔 𝑓 + 𝑑𝑔 𝑔
Proof:
The proof of the first part is obvious, thus part 2: 𝑓 = 𝑎
0 𝑏
0
+ 𝑏
1 𝑥 + ⋯ + 𝑏 𝑙 𝑥 𝑙
, 𝑑𝑔 𝑓 = 𝑘 , 𝑑𝑔 𝑔 = 𝑙 , 𝑓𝑔 = 𝑎
0 𝑏
0
+ 𝑎
0 𝑏
1
+ 𝑎
1 𝑥 + ⋯ + 𝑎 𝑥 + ⋯ + 𝑎 𝑘 𝑏 𝑙 𝑥 𝑘+𝑙 𝑘 𝑥 𝑘
, 𝑔 =
and it is apparent that the product 𝑎 𝑘
𝑏 𝑙
is nonzero. Hence the degree of the polynomial 𝑓𝑔 is 𝑘 + 𝑙 .
Note:
Let us consider the polynomials 𝑓 = 𝑥 3
3, 𝑑𝑔 𝑔 = 3 , a polynomial 𝑓 + 𝑔 = (1 − 1)𝑥 3
+ 2𝑥 2 + 𝑥 + 1
+ (2 + 1)𝑥 2
, 𝑔 = −𝑥 3 + 𝑥 2 − 1 , 𝑑𝑔 𝑓 =
+ (1 + 0)𝑥 + 1 − 1 = 3𝑥 2 + 𝑥 , 𝑑𝑔(𝑓 + 𝑔) = 2 , 𝑓 ∙ 𝑔 = −𝑥 6 − 𝑥 5 + 𝑥 4 − 𝑥 3 − 𝑥 2 − 𝑥 − 1 , 𝑑𝑔 (𝑓 ∙ 𝑔) = 6 .
2.2
Polynomial Division
For polynomial division the knowledge of integer division will be applied. Surely, you can remember the algorithm to divide for instance a five-digit number by a two-digit one. The process of polynomial division is very similar.
Just as when dividing two integers over the integer domain the result can be either without or with remainder, the result of division of two polynomials can likewise be a polynomial without a remainder, or so-called partial quotient and a remainder.
Note:
Let us remind the theorem regarding integer division with a remainder:
For each pair of integers 𝑎, 𝑏 ∈ 𝑍 , where 𝑏 ≠ 0 , there exists exactly one pair of integers 𝑞, 𝑟 ∈ 𝑍 such that there applies:
1.
𝑎 = 𝑏𝑞 + 𝑟 ,
2.
0 ≤ 𝑟 < b
.
For instance, we can consider numbers 𝑎 = 51, 𝑏 = 5 . We know that 51: 5 = 10 +
1
5
, thus the partial quotient is 10 and the remainder 1. This fact can be expressed as
51 = 5 ∙ 10 + 1.
The form 51 = 5 ∙ 9 + 6 could also be chosen to be used, but according to the previous theorem, the remainder of division by five can attain values 0, 1, 2, 3 or 4.
The theorem regarding polynomial division is analogical. Before it is introduced let us remind the algorithm for dividing two polynomials, well known from high school.
Note (Algorithm for Dividing Two Polynomials):
Before dividing two polynomials 𝑓 and 𝑔 , it is first necessary to check whether both of them are ordered by their power 𝑥 descending. We divide the first term of the polynomial 𝑓 by the first term of the polynomial 𝑔 , with the result of 𝜑
1
. Then we multiply the polynomial 𝜑
1
with the polynomial 𝑔 and subtract the resulting polynomial 𝑔𝜑
1
from the polynomial 𝑓 . The polynomial 𝑓 − 𝑔 ∙ 𝜑
1
will be denoted as 𝑓
1
. If 𝑑𝑔 𝑓
1
≥ 𝑠𝑡 𝑔 we continue with dividing.
We then divide the first term of the polynomial 𝑓
1
by the first term of the polynomial 𝑔 again and proceed analogically. The succession of steps will be illustrated on the following example.
Example 1 Divide polynomials 𝑓, 𝑔 , when 𝑓 = 2𝑥 4 − 3𝑥 3 + 𝑥 2 − 5𝑥 + 2 , 𝑔 = 𝑥 2
3𝑥 + 2 .
−
Solution: 𝜑
1
𝜑
2
𝜑
3
(2𝑥 4 − 3𝑥 3 + 𝑥 2 − 5𝑥 + 2): (𝑥 2
↓ ↓ ↓
− 3𝑥 + 2) = 2𝑥 2 + 3𝑥 + 6 + 𝑟 𝑔 𝑔. 𝜑
1
: 2𝑥 4 − 6𝑥 3 + 4𝑥 2 𝑓
1
= 𝑓 − 𝑔. 𝜑
1
: 3𝑥 3 − 3𝑥 2 𝑔. 𝜑
2
: 3𝑥 3 − 9𝑥 2
− 5𝑥 + 2
+ 6𝑥 𝑓
2
= 𝑓 − 𝑔. 𝜑
1
− 𝑔. 𝜑
2
: 6𝑥 2 − 11𝑥 + 2 𝑔. 𝜑
3
: 6𝑥 2 − 18𝑥 + 12 𝑟: 7𝑥 − 10
Hence the result of division is a partial quotient 𝑞(𝑥) = 2𝑥 2 + 3𝑥 + 6 and a remainder 𝑟(𝑥) = 7𝑥 − 10 .
Theorem 3
(Polynomial Long Division) For each pair of polynomials 𝑓, 𝑔 ∈ 𝑇[𝑥] , where 𝑔 ≠ 0 , there exists exactly one pair of polynomials 𝑞, 𝑟 ∈ 𝑇[𝑥] for which there applies:
1) 𝑓 = 𝑔 ∙ 𝑞 + 𝑟 ,
9
2) 𝑑𝑔 𝑟 < 𝑑𝑔 𝑔 .
Proof:
Firstly, the existence of polynomials 𝑞, 𝑟 will be proved: a) When 𝑑𝑔 𝑓 < 𝑠𝑡 𝑔 , then 𝑞 = 0, 𝑟 = 𝑓 . b) When 𝑑𝑔 𝑓 ≥ 𝑠𝑡 𝑔 , then we follow the division algorithm. 𝑓 − 𝑔 ∙ 𝜑
1
= 𝑓
1,
𝑑𝑔 𝑓
1
< 𝑑𝑔 𝑓 . If 𝑑𝑔 𝑓
1
≥ 𝑑𝑔 𝑔 , then 𝑓
1
− 𝑔 ∙ 𝜑
2
= 𝑓
2,
𝑑𝑔 𝑓
2
< 𝑑𝑔 𝑓 ,
⋮ 𝑓 𝑘−1
− 𝑔 ∙ 𝜑 𝑘
= 𝑓 𝑘,
𝑑𝑔 𝑓 𝑘
< 𝑑𝑔 𝑔 .
By adding these equations together we get 𝑓 − 𝑔 ∙ (𝜑
1 denoting (𝜑
1
+ ⋯ + 𝜑 𝑘
) = 𝑞 and 𝑓 𝑘
+ ⋯ + 𝜑 𝑘
) = 𝑓 𝑘
, and by
= 𝑟 , we come to 𝑓 = 𝑔 ∙ 𝑞 + 𝑟 . By this we have proven the first part of the theorem.
Secondly, there will come the proof that there exists exactly one such pair of polynomials. The proof will be performed by contradiction, let us therefore assume that there could exist two different pairs of these polynomials: 𝑓 = 𝑔 ∙ 𝑞 + 𝑟 , 𝑑𝑔 𝑟 < 𝑑𝑔 𝑔, 𝑓 = 𝑔 ∙ 𝑞̅ + 𝑟̅ , 𝑑𝑔 𝑟̅ < 𝑑𝑔 𝑔 .
By subtraction we get 0 = 𝑔 ∙ (𝑞 − 𝑞̅) + (𝑟 − 𝑟̅) 𝑟 . It is then apparent that 𝑑𝑔 𝑔 > 𝑑𝑔 (𝑟̅ − 𝑟) = 𝑑𝑔 𝑔 + 𝑑𝑔 (𝑞 − 𝑞̅) and since 𝑑𝑔 𝑔 = 𝑑𝑔 𝑔 , there must apply 𝑞 − 𝑞̅ = 0 , thus 𝑞̅ = 𝑞 and from the validity of the equation 𝑟̅ − 𝑟 = 𝑔(𝑞 − 𝑞̅) there must also apply 𝑟̅ = 𝑟 . We have proven that it is the same pair of polynomials, which is contradictory with the initial assumption. By this the proof has been performed.
Example 2 Solve individually: Compute 𝑞(𝑥), 𝑟(𝑥) , for 𝑓(𝑥) = 𝑥 5 − 𝑥 3 + 3𝑥 − 5 , a) 𝑔(𝑥) = 𝑥 2 + 7 , b) 𝑔(𝑥) = 𝑥 − 2 .
Result: a) 𝑞(𝑥) = 𝑥 3 − 8𝑥, 𝑟(𝑥) = 59𝑥 − 5 , b) 𝑞(𝑥) = 𝑥 4 + 2𝑥 3 + 3𝑥 2 + 6𝑥 +
15, 𝑟(𝑥) = 25 .
2.3
Horner’s Method
Let us say that our task is to find the value of the polynomial 5𝑥 4 − 4𝑥 3 + 7𝑥 2 − 𝑥 + 1 for 𝑥 = 4 . We can certainly substitute 4 for 𝑥 , but let us assume we are indolent to exponentiate 4 4
, and therefore we need some other approach with only simpler expressions involved.
The procedure will be as follows: Factoring out 𝑥 3
from the first two polynomial terms, we get (5𝑥 − 4)𝑥 3
. Then, from this newly formed expression and the third term of the polynomial, we factor out 𝑥 2
, resulting into [(5𝑥 − 4)𝑥 + 7]𝑥 2
. From this expression
and the fourth term, when factoring out 𝑥 , we get {[(5𝑥 − 4)𝑥 + 7]𝑥 + 1}𝑥 . Altogether the original polynomial will gain the form
{[(5𝑥 − 4)𝑥 + 7]𝑥 + 1}𝑥 + 1 and by substituting 𝑥 with 4 we get {[(16)4 + 7]4 + 1}4 + 1 = 1141 .
The previous example can be generalised. When 𝑑𝑔 𝑓 = 𝑛 ≥ 1, 𝑓 = 𝑎
0 𝑥 𝑛 𝑎
1 𝑥 𝑛−1 𝑏
0 𝑥 𝑛−1
+ ⋯ + 𝑎 𝑛
+ 𝑏
1 𝑥 𝑛−2
, 𝑐 ∈ 𝑇, 𝑔 = 𝑥 − 𝑐
+ ⋯ + 𝑏 𝑛−1
then there exists a polynomial
, 𝑟 = 𝑏 𝑛
and there applies
+ 𝑞, 𝑑𝑔 𝑞 = 𝑛 − 1, 𝑞 = 𝒂
𝟎 𝒙 𝒏 + 𝒂
𝟏 𝒙 𝒏−𝟏 + ⋯ + 𝒂 𝒏
= (𝒙 − 𝒄)(𝒃
𝟎 𝒙 𝒏−𝟏 + ⋯ + 𝒃 𝒏−𝟏
) + 𝒃 𝒏
.
This situation can be expressed in a table called a Horner scheme : 𝑎
0 𝑎
1
… 𝑎 𝑛−1 𝑐 𝑏
0 𝑏
1
… 𝑏 𝑛−1 𝑎 𝑏 𝑛 𝑛
When expanding the right side of the previous equation (you can perform by yourselves to ascertain the validity of the statement), we get: 𝑎
0
= 𝑏
0 𝑎
1
= 𝑏
1
− 𝑐𝑏
0 𝑎
2
= 𝑏
2
− 𝑐𝑏
1
⋮ 𝑎 𝑛
= 𝑏 𝑛
− 𝑐𝑏 𝑛−1
We are interested in the number 𝑏 𝑛
, which is the value of the given polynomial at the point 𝑐 . Therefore we transform the previous sequence into 𝑏
0
= 𝑎
0 𝑏
1
= 𝑐𝑏
0
+ 𝑎
1 𝑏
2
= 𝑐𝑏
1
+ 𝑎
2
⋮ 𝒃 𝒏
= 𝒄𝒃 𝒏−𝟏
+ 𝒂 𝒏
.
Example 3 Determine the value of a polynomial 𝑥 4 − 3𝑥 3 + 2𝑥 2 − 2𝑥 − 16 for 𝑥 = 5 .
Solution:
The value of the polynomial at the given point will be found by using Horner’s method. We will proceed according to the fore-mentioned sequence 𝑏
0
= 𝑎
0 𝑏
1
= 𝑐𝑏
0
+ 𝑎
1 𝑏
2
= 𝑐𝑏
1
+ 𝑎
2
11
⋮ 𝑏 𝑛
= 𝑐𝑏 𝑛−1
+ 𝑎 𝑛
.
In our case, we place the polynomial coefficients into the first row of the table and the value 𝑥 = 5 into the first column. Then we write the first coefficient 𝑏
0
= 𝑎
0
= 1 into the second row and calculate on: 5 ∙ 1 − 3 = 2; 5 ∙ 2 + 2 = 12; 5 ∙ 12 − 2 = 58; 5 ∙ 58 +
16 = 274 .
1 -3 2 -2 16
5 1 2 12 58 274
That means we have found out that the value of the polynomial at the point 5 is 274. Yet the result of the table can be interpreted in another way, too. When dividing the polynomial 𝑥 4 𝑥 3
− 3𝑥
+ 2𝑥
3
2
+ 2𝑥 2 − 2𝑥 − 16
+ 12𝑥 + 58
by the polynomial 𝑥 − 5
together with the remainder
, we get the partial quotient 𝑟(𝑥) = 274 𝑞(𝑥) =
. This fact can be verified by dividing them. The result can be expressed as 𝑥 4
2𝑥 2 + 12𝑥 + 58)(𝑥 − 5) + 274 .
− 3𝑥 3 + 2𝑥 2 − 2𝑥 − 16 = (𝑥 3 +
Hence Horner’s method can be used for simple division of a polynomial by a linear polynomial, as demonstrated in the following example.
Example 4
Divide the polynomials from the example 2 b) using Horner’s method.
Solution:
1 0 -1 0 3 -5
2 1 2 3 6 15 25
The result is therefore 𝑞(𝑥) = 𝑥 4 + 2𝑥 3 + 3𝑥 2 + 6𝑥 + 15, 𝑟(𝑥) = 25 .
Example 5 Solve individually: Determine the value of the polynomial 𝑓(𝑥) = 𝑥 5
4𝑥 3 + 2𝑥 2 − 7 at the point 𝑥 = 6 . Write in the form 𝑓(𝑥) = 𝑞(𝑥)𝑔(𝑥) + 𝑟(𝑥) .
−
Result: 𝑓(6) = 6977, 𝑥 5 − 4𝑥 3 + 2𝑥 2 − 7 = (𝑥 4 + 6𝑥 3 + 32𝑥 2 + 194𝑥 + 1164)(𝑥 −
6) + 6977
2.4
Application of Horner Scheme for Taylor Expansion of Polynomial
You certainly can recall Taylor series that could be used in mathematical analysis to replace some functions in the neighbourhood of a given point. Let us remind Taylor’s theorem, which states that a function 𝑓(𝑥) , having in the neighbourhood of a point 𝑎
derivations up to an order of (𝑛 + 1) , can be in the neighbourhood of the point 𝑎 expressed as 𝑓(𝑥) = 𝑓(𝑎) + 𝑓 ′ (𝑎)
1!
(𝑥 − 𝑎) + 𝑓 ′ ′(𝑎)
2!
(𝑥 − 𝑎) 2 + ⋯ + 𝑓 (𝑛) (𝑎) 𝑛!
(𝑥 − 𝑎) 𝑛 + 𝑅, where 𝑓(𝑎) + 𝑓
′ (𝑎)
(𝑥 − 𝑎) + 𝑓
′′ (𝑎)
(𝑥 − 𝑎) 2 + ⋯ + 𝑓
(𝑛) (𝑎)
(𝑥 − 𝑎) 𝑛
is the Taylor series
1!
2!
𝑛!
and 𝑅 is the Taylor series remainder. This way the approximate value of the given function in the point 𝑎 can be found.
In case the function 𝑓(𝑥) is a polynomial, there exists a much easier way of finding
Taylor series, which is by the application of Horner’s method.
Definition 4
Let 𝑓 ∈ 𝑇[𝑥], 𝑓(𝑥) = 𝑎
0 𝑥 𝑛
𝑇 . Then if
+ 𝑎
1 𝑥 𝑛−1 + ⋯ + 𝑎 𝑛
, 𝑠𝑡 𝑓 = 𝑛, 𝑐, 𝑐
1
, … , 𝑐 𝑛
∈ 𝑓 = 𝑑
0
+ 𝑑
1
(𝑥 − 𝑐) + 𝑑 where 𝑑
0
, … , 𝑑 𝑛
(Taylor expansion) centred at 𝒄 .
2
(𝑥 − 𝑐) 2 … + 𝑑 𝑛
(𝑥 − 𝑐) 𝑛 ,
∈ 𝑇 , then the expression on the right side is called a Taylor series
Theorem 4
For each 𝑐 ∈ 𝑇, 𝑓 ∈ 𝑇[𝑥] there exists a Taylor series of a polynomial 𝑓 centred at 𝑐 .
Proof:
When 𝑑𝑔 𝑓 = 0 , then 𝑓 = 𝑑
0
. When 𝑠𝑡 𝑓 ≥ 1 , then after dividing the polynomial 𝑓 by the expression (𝑥 − 𝑐) we get 𝑓 = (𝑥 − 𝑐)𝑓
1
+ 𝑑
0
, 𝑑𝑔 𝑓
1
= 𝑛 − 1 𝑓
1
= (𝑥 − 𝑐)𝑓
2
+ 𝑑
1
, 𝑑𝑔 𝑓
2
= 𝑛 − 2
⋮ 𝑓 𝑛−1
= (𝑥 − 𝑐)𝑓 𝑛
+ 𝑑 𝑛−1
, 𝑑𝑔 𝑓 𝑛
= 𝑛 − 𝑛 = 0 which means that 𝑓 𝑛
is a constant polynomial. Now when we leave the polynomial 𝑓 as it is, then multiply the polynomial 𝑓
1
by the expression (𝑥 − 𝑐) , the polynomial 𝑓
2
by the expression (𝑥 − 𝑐) 2
, etc., up to multiplying the polynomial 𝑓 𝑛
by the expression (𝑥 − 𝑐) 𝑛 and when we sum up all the resulting polynomials, we come to the Taylor series centred at 𝑐 , i.e. 𝑓 = 𝑑
0
+ 𝑑
1
(𝑥 − 𝑐) + 𝑑
2
(𝑥 − 𝑐) 2 … + 𝑑 𝑛
(𝑥 − 𝑐) 𝑛 .
Example 6 Given the polynomial 𝑓(𝑥) = 𝑎
0 𝑥 𝑛 method to find a Taylor series centred at 𝑐 .
+ 𝑎
1 𝑥 𝑛−1 + ⋯ + 𝑎 𝑛
, apply the Horner’s
Solution: We are to rearrange the polynomial 𝑓 in the following way: 𝑓 = (𝑥 − 𝑐)𝑓
1 𝑑
0
= (𝑥 − 𝑐)[(𝑥 − 𝑐)𝑓
2
+ 𝑑
1
] + 𝑑
0
= ⋯
+
. The coefficients of the polynomial will be written into the first line, and the centre c into the first column. In the first step we proceed like we did in Horner scheme higher above. The last number in the second row is 𝑑
0
. In the
13
second step we do not work with coefficients of the original polynomial anymore, but with the coefficients of the polynomial 𝑓
1
, instead, i.e. with the numbers from the second row.
The last number of the row is left out, processing only up to 𝑑
1
. In this way we advance onwards. The table has the following form: 𝑎
0 𝑎
1
… 𝑎 𝑛 c 𝑎
0
… 𝒅
𝟎 c 𝑎
0 𝒅
𝟏 c 𝑎
0 𝒅
𝟐 c ⋮ c 𝒅 𝒏
The numbers 𝑑
0
, … , 𝑑 𝑛
are the sought Taylor series coefficients.
Example 7 For the polynomial 𝑓(𝑥) = 𝑥 5
find a Taylor series centred at 𝑐 = 1 .
Solution:
1 0 0 0 0 0
1 1 1 1 1 1 1
1 1 2 3 4 5
1 1 3 6 10
1 1 4 10
1 1 5
1 1
Therefore the result is 𝑥 5 = 1 + 5(𝑥 − 1) + 10(𝑥 − 1) 2 + 10(𝑥 − 1) 3 + 5(𝑥 − 1) 4 + (𝑥 − 1) 5
Note:
There is a noticeable similarity of Taylor series coefficients from the previous example with the Pascal’s triangle coefficients:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
Examine for yourselves whether for polynomials 𝑥 3
or 𝑥 6
their coefficients of a Taylor series centred at 𝑐 = 1 are also equal to the Pascal’s triangle coefficients. Examine further what do Taylor series with different centres look like, for instance for 𝑐 = −1 , 𝑐 = 2 .
Example 8 Solve individually: For the polynomial 𝑓(𝑥) = 𝑥 4 − 8𝑥 3 + 24𝑥 2 − 50𝑥 + 90 find out a Taylor series centred at 𝑐 = 2 .
Result: 𝑥 4 − 8𝑥 3 + 24𝑥 2 − 50𝑥 + 90 = 38 − 18(𝑥 − 2) + (𝑥 − 2) 4
.
Definition 5
Let 𝑓 ∈ 𝑇[𝑥], 𝑓(𝑥) = 𝑎
0 𝑥 𝑛
𝑇 . If
+ 𝑎
1 𝑥 𝑛−1 + ⋯ + 𝑎 𝑛
, 𝑑𝑔 𝑓 = 𝑛, 𝑐, 𝑐
1
, … , 𝑐 𝑓 = 𝑑
0
+ 𝑑
1
(𝑥 − 𝑐
1
) + 𝑑
2
(𝑥 − 𝑐
1
)(𝑥 − 𝑐
2
) … + 𝑑 𝑛
(𝑥 − 𝑐
1
)(𝑥 − 𝑐 then the right side is called a Gaussian interpolation polynomial .
2
) … (𝑥 − 𝑐 𝑛
), 𝑛
∈
Theorem 5
Let 𝑓 ∈ 𝑇[𝑥], 𝑑𝑔 𝑓 = 𝑛 ≥ 1, 𝑐
1
, 𝑐
2
, … , 𝑐 𝑛 𝑑
1
, 𝑑
2
, … , 𝑑 𝑛
∈ 𝑇 such, that there applies 𝑓 = 𝑑
0
∈ 𝑇 . Then there exist numbers
+ 𝑑
1
(𝑥 − 𝑐
1
) + 𝑑
2
(𝑥 − 𝑐
1
)(𝑥 − 𝑐
2
) +
⋯ + 𝑑 𝑛
(𝑥 − 𝑐
1
)(𝑥 − 𝑐
2
) … (𝑥 − 𝑐 𝑛
) . The polynomial 𝑓 can therefore be expressed in the form of a Gaussian interpolation polynomial with chosen numbers 𝑐
1
, 𝑐
2
, … , 𝑐 𝑛
.
Proof:
When dividing the polynomial 𝑓 by the expression (𝑥 − 𝑐
1
), then by the expression
(𝑥 − 𝑐
2
) , etc., we get: 𝑓 = (𝑥 − 𝑐
1
)𝑓
1
+ 𝑑
0
. 1 𝑓
1
= (𝑥 − 𝑐
2
)𝑓
2
⋮
+ 𝑑
1
. (𝑥 − 𝑐
1
) 𝑓 𝑛−1
= (𝑥 − 𝑐 𝑛 𝑓 𝑛
)𝑓 𝑛
= 𝑑 𝑛
+ 𝑑 𝑛−1
. (𝑥 − 𝑐
1
)(𝑥 − 𝑐
2
) … (𝑥 − 𝑐 𝑛−1
)
. (𝑥 − 𝑐
1
)(𝑥 − 𝑐
2
) … (𝑥 − 𝑐 𝑛
)
When we multiply the resulting polynomials back again with the expressions in the right column and sum them up, we come to 𝑓 = 𝑑
0 𝑑 𝑛
(𝑥 − 𝑐
1
)(𝑥 − 𝑐
2
) … (𝑥 − 𝑐 𝑛
) .
+ 𝑑
1
(𝑥 − 𝑐
1
) + 𝑑
2
(𝑥 − 𝑐
1
)(𝑥 − 𝑐
2
) + ⋯ +
Example 9 Express the polynomial 𝑓 = 2𝑥 5 − 3𝑥 4 − 5𝑥 3 + 2𝑥 2 + 7𝑥 − 9 in the form of
Gaussian interpolation polynomial, given 𝑐
1
= 1, 𝑐
2
= 2, 𝑐
3
= −3, 𝑐
4
= 2, 𝑐
5
= −1 .
Solution: We will use similar procedure as when seeking a Taylor series by application of
Horner’s method. Into the left column we will put values 𝑐
1
, … , 𝑐
5
successively.
2 -3 -5 2 7 -9
1 2 -1 -6 -4 3 -6
15
2
-3
2
2
2
2
3
-3
1
0
9
11
-4
-31
-5
-1 2
2
-1
The polynomial is then of the form 𝑓 = −6 − 5(𝑥 − 1) − 31(𝑥 − 1)(𝑥 − 2) + 11(𝑥 − 1)(𝑥 − 2)(𝑥 + 3) −
(𝑥 − 1)(𝑥 − 2)(𝑥 + 3)(𝑥 − 2) + 2(𝑥 − 1)(𝑥 − 2)(𝑥 + 3)(𝑥 − 2)(𝑥 + 1) .
Note:
The Gaussian interpolation polynomial can be used to find a polynomial of order 𝑛 specified by values at 𝑛 + 1 points, as will be demonstrated in the next example.
Let us merely remind that two points uniquely specify a straight line, three points specify a parabola, and so forth. For determining of a first-degree polynomial we therefore need values of two points, for a polynomial of the second degree we need values of three points, and generally for finding out of a polynomial of degree (𝑛 + 1) there are 𝑛 points needed.
Example 10 Find a polynomial 𝑓 ∈ 𝑅[𝑥] 3 for which there applies 𝑓(−2) = 3, 𝑓(−1) =
1, 𝑓(2) = −2, 𝑓(3) = 5 .
Solution: Since the sought polynomial will be apparently of the third degree, we can express it in the form of the Gaussian interpolation polynomial, where we will choose to substitute 𝑐
1
, 𝑐
2
, 𝑐
3
with values of independent variable −2; −1; 2 ,
4
resulting to 𝑓 = 𝑑
0 𝑑
1
(𝑥 + 2) + 𝑑
2
(𝑥 + 2)(𝑥 + 1) + 𝑑
3
(𝑥 + 2)(𝑥 + 1)(𝑥 − 2) .
+
In this form of the polynomial we substitute 𝑥 with 𝑐
1
= −2 and as according to the assignment there applies 𝑓(−2) = 3 , we can replace 𝑓 with 3, thus 3 = 𝑑
0
+ 𝑑
1
(−2 + 2) + 𝑑
2
(−2 + 2)(−2 + 1) + 𝑑
3
(−2 + 2)(−2 + 1)(−2 − 2) , and we find out that 𝑑
0
= 3 . 𝑑
1
Similarly continuing for 𝑐
2
(−1 + 2) , and then 𝑑
1
= −2 .
= −1 , 𝑓(−1) = 1 , after rearrangement we get 1 = 3 +
Next we set 𝑥 to 𝑐
3
3, 𝑓(3) = 5 , the result is 𝑑
3
=
= 2, 𝑓(2) = −2 , which results into 𝑑
2
=
7
20
.
1
4
. Lastly 𝑐
4
=
3 The symbol 𝑅[𝑥] stands for a set of all polynomials over the field of real numbers.
4 Any three values of an independent variable can be arbitrarily chosen in any order, but this decision must be respected in the rest of the process and suitable function values must be substituted for 𝑓 respectively.
The form of the polynomial is hence 𝑓 = 3 − 2(𝑥 + 2) +
1
4
7
20
(𝑥 + 2)(𝑥 + 1)(𝑥 − 2) . After expansion we get
(𝑥 + 2)(𝑥 + 1) + 𝑓 =
7
20 𝑥 3 +
3
5 𝑥 2 −
53
20 𝑥 −
19
10
.
Now you can check by yourselves whether the values from the assignment comply with the resulting polynomial.
Example 11 Solve individually: By an application of the Gaussian interpolation polynomial, determine 𝑓 ∈ 𝑅[𝑥] such that there applies a) 𝑓(2) = 2, 𝑓(−2) = −4 , b) 𝑓(−1) =
9
2
, 𝑓(0) = 4, 𝑓(2) = 12 .
Result: a) 𝑓 =
3
2 𝑥 − 1 , b) 𝑓 = 3𝑥 2 −
1
2 𝑥 + 1
Definition 6
Let 𝑓 ∈ 𝑇[𝑥], 𝑓(𝑥) = 𝑎
0 𝑥 𝑛
𝑇 . Considering 𝑓 = 𝐿
1
(𝑥)𝑦
1
+ 𝐿
2
(𝑥)𝑦
2
+ 𝑎
1
+ ⋯ + 𝐿 𝑥 𝑛 𝑛−1 + ⋯ + 𝑎
(𝑥)𝑦 𝑛 𝑛
, 𝑑𝑔 𝑓 = 𝑛, 𝑐, 𝑐
, where 𝑦
1
, … , 𝑦 𝑛
∈ 𝑇 ,
1
, … , 𝑐 𝑛
∈
𝐿
1
(𝑥) =
(𝑥 − 𝑐
2
)(𝑥 − 𝑐
3
(𝑐
1
− 𝑐
2
)(𝑐
1
− 𝑐
3
) … (𝑥 − 𝑐
) … (𝑐
1 𝑛
)
− 𝑐 𝑛
)
,
⋮
𝐿 𝑛
(𝑥) =
(𝑐
(𝑥 − 𝑐
1 𝑛
− 𝑐
1
)(𝑥 − 𝑐
)(𝑐 𝑛
2
− 𝑐
2
) … (𝑥 − 𝑐
) … (𝑐 𝑛 𝑛−1
)
− 𝑐 𝑛−1
)
.
If 𝑐 𝑖
≠ 𝑐 𝑗
for 𝑖 ≠ 𝑗 , then the right side is called a Lagrange (interpolation) polynomial.
Note:
Lagrange polynomial can be, similarly as Gaussian interpolation polynomial, applied to find a polynomial passing through given points on a certain interval.
Example 12 Use the Lagrange polynomial to find the polynomial that passes through the points [−1; 9], [1; 1], [2; 6] .
Solution: The form of the polynomial will be 𝑓 = 9𝐿
1
(𝑥) + 𝐿
2
(𝑥) + 6𝐿
3
(𝑥) , where
𝐿
1
(𝑥) =
(𝑥 − 1)(𝑥 − 2)
(−1 − 1)(−1 − 2)
=
1
6
(𝑥 2 − 3𝑥 + 2),
𝐿
2
(𝑥) =
(𝑥 + 1)(𝑥 − 2)
(1 + 1)(1 − 2)
= −
1
2
(𝑥 2 − 𝑥 − 2),
𝐿
3
(𝑥) =
(𝑥 + 1)(𝑥 − 1)
(2 + 1)(2 − 1)
=
1
3
(𝑥 2 − 1).
17
After substitution we get 𝑓 =
9
(𝑥 2 − 3𝑥 + 2) −
1
(𝑥 2 − 𝑥 − 2) +
6
(𝑥 2 − 1) = 3𝑥 2 −
6 2 3
4𝑥 + 2 . You can check by yourselves that the polynomial conforms to the assignment requirements.
Example 13 Solve individually: Use the Lagrange polynomial to solve Example 11.
Example 14 Solve individually: By an application of the Lagrange polynomial find a polynomial that passes through the points [−1; 0], [0; 1], [1; 3], [2; −2] .
Result: 𝑓 = −
4
3 𝑥 3 +
1
2 𝑥 2 +
17 𝑥 + 1
6
2.5
Divisibility of Polynomials
Definition 7
Let 𝑓, 𝑔 ∈ 𝑇[𝑥] . We say that a polynomial 𝒈 divides a polynomial 𝒇 if and only if there exists any such polynomial ℎ ∈ 𝑇[𝑥] for which there applies 𝑓 = 𝑔. ℎ . Then we write 𝑔 𝑓 .
Theorem 6
(Basic Properties of Divisibility of Polynomials)
Let 𝑓, 𝑔 ∈ 𝑇[𝑥] , 𝑓 ≠ 0, 𝑔 ≠ 0 . Then there applies:
1) 𝑓 𝑓
2) For all 𝑢, 𝑣 ∈ 𝑇[𝑥] and all 𝑓
1
, 𝑓
2
∈ 𝑇[𝑥] there applies: If 𝑔|𝑓
1
and at the same time 𝑔|𝑓
2
, then also 𝑔|(𝑓
1 𝑢 + 𝑓
2 𝑣) .
3) Let ℎ ∈ 𝑇[𝑥] , then there applies: If ℎ|𝑔 and also 𝑔|𝑓 , then ℎ|𝑓 as well.
4) 𝑔|𝑓 𝑑𝑔 𝑔 ≤ 𝑑𝑔 𝑓
5) 𝑔|𝑓 and 𝑓|𝑔 at the same time if and only if there exists such nonzero element 𝑐 ∈
𝑇 , for which there applies 𝑔 = 𝑐𝑓 .
Proof:
Re 1) 𝑓 = 1. 𝑓 = 𝑓. 1
Re 2) 𝑔|𝑓
1
∃ℎ
1
∈ 𝑇(𝑥): 𝑓
1
= 𝑔. ℎ
1
, 𝑔|𝑓
2
∃ℎ
2
∈ 𝑇(𝑥): 𝑓
2
= 𝑔. ℎ
2
. Then 𝑓
1
. 𝑢 + 𝑓
2
. 𝑣 = 𝑔. ℎ
1
. 𝑢 + 𝑔. ℎ
2
. 𝑣 = 𝑔. (ℎ
1 𝑢 + ℎ
2 𝑣) , upon denoting ℎ
1 𝑢 + ℎ
2 𝑣 = ℎ , ℎ ∈ 𝑇[𝑥] , we get 𝑓
1
. 𝑢 + 𝑓
2
. 𝑣 = ℎ .
Re 3) If ℎ|𝑔 then 𝑔 = ℎ𝜑
1
( ℎ𝜑
1
)𝜑
2
= ℎ(𝜑
1 𝜑
2
and similarly if 𝑔|𝑓 , then 𝑓 = 𝑔𝜑
2
) , where 𝜑
1 𝜑
2
∈ 𝑇(𝑥) . Hence ℎ|𝑓 .
. Therefore 𝑓 = 𝑔𝜑
2
=
Re 4) If 𝑔|𝑓 , then 𝑓 = 𝑔. ℎ . For the degrees of these polynomials there applies 𝑑𝑔 𝑓 = 𝑑𝑔 𝑔 + 𝑑𝑔 ℎ , where 𝑑𝑔 𝑓 ≠ −∞ , 𝑑𝑔 𝑔 ≠ −∞ , 𝑑𝑔 ℎ ≥ 0 , so then 𝑑𝑔 𝑓 ≥ 𝑑𝑔 𝑔 .
Re 5) If 𝑔|𝑓 and at the same time 𝑓|𝑔 then 𝑓 = 𝑔𝜑
1
and also 𝑔 = 𝑓𝜑
2
. From the previous part of the proof we know that 𝑑𝑔 𝑔 ≤ 𝑑𝑔 𝑓 and also 𝑑𝑔 𝑓 ≤ 𝑑𝑔 𝑔 , which means 𝑑𝑔 𝑓 =
𝑑𝑔 𝑔 . Since 𝑑𝑔 𝑔 = 𝑑𝑔 𝑓 + 𝑑𝑔 𝜑
2
, it must be 𝑑𝑔 𝜑
2
𝑇 .
= 0 , so 𝜑
2
is a nonzero constant 𝑐 ∈
Definition 8
We say that a polynomial ℎ ∈ 𝑇[𝑥] is a common divisor of polynomials 𝑓, 𝑔 ∈ 𝑇[𝑥] if and only if there apply at the same time both ℎ|𝑔 and ℎ|𝑓 .
Definition 9
Let 𝑓, 𝑔, 𝑑 ∈ 𝑇[𝑥] . We call the polynomial 𝑑 a greatest common divisor of polynomials 𝑓, 𝑔 , if and only if there applies:
1.
𝑑|𝑓 𝑑|𝑔 ,
2.
For all polynomials ℎ ∈ 𝑇[𝑥] there applies: if ℎ|𝑓 and also ℎ|𝑔 then ℎ|𝑑 .
Note:
The definition above implies that a greatest common divisor is the common divisor of both polynomials which is divisible by all of the common divisors.
Note:
Following there is presented the sequence of equations called Euclidean division of polynomials, which can be used to determine a greatest common divisor of two polynomials. Earlier you have encountered the Euclidean algorithm for finding the greatest common divisor of two integers, which is similar in principle and which will be demonstrated on an example.
Example 15 Use the Euclidean algorithm to find out the greatest common divisor of numbers 450 and 294.
450 = 294 ∙ 1 + 156
294 = 156 ∙ 1 + 138
156 = 138 ∙ 1 + 18
138 = 18 ∙ 7 + 12
18 = 12 ∙ 1 + 6
12 = 6 ∙ 1 + 0
The greatest common divisor of two numbers is the last nonzero remainder in the
Euclidean algorithm, which is 𝐷(450, 294) = 6 .
Euclidean Algorithm for Polynomials:
Let 𝑓, 𝑔 ∈ 𝑇[𝑥], 𝑑𝑔 𝑓 ≥ 𝑑𝑔 𝑔, 𝑔 ≠ 0 . We start by dividing the polynomial 𝑓 by the polynomial 𝑔 and then we repeatedly divide the divisor from the previous step by the remainder, continuing until there is a result with a zero remainder. 𝑓 = 𝑔. 𝑞
1
+ 𝑟
1
, 𝑑𝑔 𝑟
1
< 𝑑𝑔 𝑔, 𝑟
1
≠ 0 𝑔 = 𝑟
1
. 𝑞
2
+ 𝑟
2
, 𝑑𝑔 𝑟
2
< 𝑑𝑔 𝑟
1
, 𝑟
2
≠ 0
19
𝑟
1
= 𝑟
2
. 𝑞
3
+ 𝑟
3
, 𝑑𝑔 𝑟
3
< 𝑑𝑔 𝑟
2
, 𝑟
3
≠ 0
⋮ 𝑟 𝑘−2
= 𝑟 𝑘−1
. 𝑞 𝑘
+ 𝒓 𝒌
, 𝑟 𝑘
≠ 0 𝑟 𝑘−1
= 𝑟 𝑘
. 𝑞 𝑘+1
+ 0
Theorem 7
A greatest common divisor can be found for each pair of polynomials 𝑓, 𝑔 ∈
𝑇[𝑥], 𝑔 ≠ 0 . One of these greatest common divisors is the last non-zero Euclidean algorithm remainder for the polynomials 𝑓 and 𝑔 .
Proof: F rom the previous equation it is apparent from the previous equation that 𝑟 𝑘
|𝑟 𝑘−1
, then from the next-to-last equation 𝑟 𝑘
|𝑟 𝑘−2
, and so onwards to the second equation from which 𝑟 𝑘
|𝑔 , and from the first equation then 𝑟 𝑘
|𝑓 , hence 𝑟 𝑘
is a common divisor of the polynomials 𝑓, 𝑔 .
Now it only remains to be proven that every common divisor of polynomials 𝑓, 𝑔 divides 𝑟 𝑘
. If ℎ is a common divisor of both polynomials then it is evident that ℎ|𝑟
1
from the first equation, ℎ|𝑟
2
from the second one, and so forth to ℎ|𝑟 𝑘
from the last one.
Note:
In accordance with the previous theorem, no greatest common divisor there exists only when 𝑓 = 𝑔 = 0 .
Note:
If 𝑑
1
, 𝑑
2
are two different common divisors of polynomials 𝑓, 𝑔 , then according to the property 2) in the Theorem 6 there applies 𝑑
1
|𝑑
2
𝑑
2
|𝑑
1
and according to 5) there exists a non-zero number 𝑐 ∈ 𝑇 , for which it applies 𝑑
2
= 𝑐. 𝑑
1
.
There also applies that when 𝑑
1
is a greatest common divisor of two polynomials, then 𝑑
2
= 𝑐. 𝑑
1
is their greatest common divisor as well.
Theorem 8
When 𝑑 ∈ 𝑇[𝑥] is a greatest common divisor of polynomials 𝑓, 𝑔 ∈ 𝑇[𝑥] , then the set {𝑐𝑑|𝑑 ∈ 𝑇, 𝑐 ≠ 0} is the set of all greatest common divisors of the polynomials 𝑓, 𝑔 . We say these common divisors are associated with each other.
Among the greatest common divisors of the polynomials 𝑓, 𝑔 there exists exactly one whose coefficient of the highest degree term is equal to 1. This polynomial is called a normalized greatest common divisor and denoted as 𝐺𝐶𝐷(𝑓, 𝑔) .
Example 16 Find the greatest common divisor of following polynomials over 𝑄[𝑥] : 𝑓(𝑥) = 𝑥 4 − 2𝑥 3 − 2𝑥 2 + 7𝑥 − 6 , 𝑔(𝑥) = 2𝑥 3 − 4𝑥 2 − 𝑥 + 2 .
Solution: Euclidean algorithm will be applied. Since 𝐺𝐶𝐷(𝑓, 𝑔) will not change when replacing the polynomials 𝑓, 𝑔 with the polynomials associated with them, we will first multiply 𝑓 by 2. Then we will divide.
(2𝑥 4 − 4𝑥 3 − 4𝑥 2 + 14𝑥 − 12): (2𝑥 3 − 4𝑥 2 − 𝑥 + 2) = 𝑥 +
−3𝑥 2 + 12𝑥 − 12
2𝑥 3 − 4𝑥 2 − 𝑥 + 2
, thus 𝑞
1
(𝑥) = 𝑥, 𝑟
1
(𝑥) = −3𝑥 2 + 12𝑥 − 12 . Normalizing the polynomial 𝑟
1
(𝑥) into 𝑥 2
4𝑥 + 3 we proceed to the next step, this time dividing 𝑔(𝑥) by 𝑟
1
(𝑥) .
(2𝑥 3 − 4𝑥 2 − 𝑥 + 2): ( 𝑥 2 − 4𝑥 + 4) = 2𝑥 + 4 + 𝑥 2
7𝑥 − 14
− 4𝑥 + 3
,
− therefore 𝑞
2
(𝑥)2𝑥 + 4, 𝑟
2
(𝑥) = 7𝑥 − 14 . After replacing the polynomial 𝑟
2
(𝑥) with the normalized polynomial again we can proceed to the division, getting
(𝑥 2 − 4𝑥 + 4): (𝑥 − 2) = 𝑥 − 2, i.e. 𝑞
3
(𝑥) = 𝑥 − 2, 𝑟
3
(𝑥) = 0 . Hence 𝐺𝐶𝐷(𝑓, 𝑔) = 𝑟
2
(𝑥) = 𝑥 − 2 . The discovered greatest common divisor is really over the complex numbers field, thus satisfying the assignment requirement.
Note:
In some cases it is also possible to find a greatest common divisor of two polynomials by their decomposition. Let us, for this purpose, revise some basic polynomial decompositions:
𝑎 2 − 𝑏 2 = (𝑎 − 𝑏)(𝑎 + 𝑏) ,
𝑎 3 − 𝑏 3 = (𝑎 − 𝑏)(𝑎 2 + 𝑎𝑏 + 𝑏 2 ) ,
𝑎 3 + 𝑏 3 = (𝑎 + 𝑏)(𝑎 2 − 𝑎𝑏 + 𝑏 2 ) ,
𝑎 2 ± 2𝑎𝑏 + 𝑏 2 = (𝑎 ± 𝑏) 2
, etc.
Definition 10
A polynomial 𝑓 ∈ 𝑇[𝑥] is called irreducible 5 over a field 𝑇 if and only if 𝑑𝑔 𝑓 ≥ 1 and there can be found no polynomials 𝑔, ℎ ∈ 𝑇[𝑥] such that there applies 𝑓 = 𝑔. ℎ, 𝑑𝑔 𝑔 ≥ 1, 𝑑𝑔 ℎ ≥ 1 .
Definition 11
A polynomial 𝑓 ∈ 𝑇[𝑥] is called reducible over a field 𝑇 if and only if there exist polynomials 𝑔, ℎ ∈ 𝑇[𝑥] for which there applies 𝑓 = 𝑔. ℎ, 𝑑𝑔 𝑔 ≥ 1, 𝑑𝑔 ℎ ≥ 1 .
Note:
The role irreducible polynomials play in the integral domain 𝑇[𝑥] is analogical to the role of prime numbers in the ring 𝑍 .
Theorem 9
For each non-constant polynomial 𝑓 ∈ 𝑇[𝑥] there exists a number 𝑎 ∈ 𝑇 and irreducible polynomials 𝜑
1
, … , 𝜑 𝑘
∈ 𝑇[𝑥] over 𝑇 such that there applies 𝑓 = 𝑎𝜑
1
… 𝜑 𝑘
, where 𝜑 𝑖
, 𝑖 = 1, … , 𝑘 are normalized polynomials. This decomposition is unambiguous, except for the ordering of the polynomials 𝜑
1
, … , 𝜑 𝑘
.
5 Irreducible stands for one that cannot be decomposed into a product of non-constant polynomials.
21
The theorem could be also formulated in a different way: Every non-constant polynomial 𝑓 over a field 𝑇 can be decomposed into the product of a constant and normalized irreducible polynomials.
Proof:
Let 𝑓 ∈ 𝑇[𝑥] .
The polynomial 𝑓 is irreducible over 𝑇 , then we normalize it into the form 𝑓 = 𝑎𝜑
1
.
Or the polynomial 𝑓 is reducible. Then there can be found its decomposition into polynomials irreducible over 𝑇: 𝑓 =
1
…
𝑘
. Upon rearranging them into the normalized form, we get 𝑓 = 𝑎𝜑
1
… 𝜑 𝑘
.
Now there still remains to prove the unambiguity of the decomposition. The proof will be performed by contradiction. So let us assume that there exist two distinct decompositions of the polynomial 𝑓 : 𝑓 = 𝑎𝜑
1
… 𝜑 𝑘
= 𝑐
1
…
𝑘
The polynomials 𝑓 𝜑 𝑖
,
𝑖
are irreducible over
, then there must also apply 𝜑 𝑖
|(
1
…
𝑘
)
𝑇 . If the polynomial
. Then either 𝜑
1
=
1 𝜑
1
divides the polynomial
, or the polynomials are coprime. In the case they are coprime, then among polynomials
2
, … ,
𝑘 𝜑
1
,
1
there must be one that is equal to 𝜑
1
. Analogically, we can reason for the other polynomials 𝜑
2
, … , 𝜑 𝑘
.
Thus we reach the contradiction with the assumption that the decompositions are distinct.
Note:
When the same irreducible polynomials in the decomposition are multiplied by each other, then 𝑓 = 𝑎𝜑 𝛼
1
1
… 𝜑 𝛼 𝑟 𝑟 , 𝛼 𝑖
∈ 𝑁.
This decomposition is called a canonical decomposition .
An analogy can be found in the ring of integers. From now on we will consider only non-negative integers, negative integer could be obtained by multiplication 𝑎 =
(−1)𝑛, 𝑛 ∈ 𝑁 . Every non-negative integer 𝑛 greater than 1 can be decomposed into a product of prime factors, 𝑛 = 𝑝
1
∙ … ∙ 𝑝 𝑘
, i.e. 𝑛 = 𝑝
1 𝛼
1 … 𝑝 𝛼 𝑟 𝑟
.
For instance, number 20 580 can be expressed in the form of a canonical decomposition as
2 2 ∙ 3 ∙ 5 ∙ 7 3
.
Theorem 10
Over the field of complex numbers, all polynomials of the 1 st
degree are irreducible. All polynomials of a degree greater than 1 are reducible.
Theorem 11
Over the field of real numbers, there are irreducible just all polynomials of the 1 st degree and those of the 2 nd degree that have a negative discriminant.
Theorem 12
For each 𝑛 ∈ 𝑁 there exists a polynomial 𝑓 ∈ 𝑄[𝑥], 𝑑𝑔 𝑓 = 𝑛 such that 𝑓 is irreducible over the field 𝑄 .
Example 17 Find decompositions of the following polynomials over 𝑄[𝑥], 𝑅[𝑥], 𝐶[𝑥] . a) 𝑥 2 + 𝑥 + 1 , b) 2𝑥 3 + 2𝑥 2 + 𝑥 + 1 , c) 𝑥 4 + 𝑥 3 + 𝑥 + 1 , d) 𝑥 6 − 2 .
Solution: a) The polynomial can be decomposed using the procedure well known from a high school, i.e. 𝑥
1,2
=
−𝑏 ± √𝑏 2
2𝑎
− 4𝑎𝑐
=
−1 ± √1 − 4
=
2
−1 ± √−3
.
2
We can clearly see by now that over both 𝑄[𝑥] and 𝑅[𝑥] the polynomial is irreducible and over 𝐶[𝑥] its decomposition is (𝑥 −
−1−𝑖√3
) (𝑥 −
2
−1+𝑖√3
) .
2 b) 2𝑥 3 + 2𝑥 2 + 𝑥 + 1 = 2𝑥 2 (𝑥 + 1) + (𝑥 + 1) = (𝑥 + 1)(2𝑥 2 + 1) , which is the decomposition in both 𝑄[𝑥] and 𝑅[𝑥] . Over 𝐶[𝑥] we can continue with rearrangement
(𝑥 + 1)(2𝑥 2 + 1) = (𝑥 + 1)(2𝑥 2 − 𝑖 2 ) = (𝑥 + 1)(√2𝑥 − 𝑖)(√2𝑥 + 𝑖) . c) 𝑥 4 + 𝑥 3 + 𝑥 + 1 = 𝑥 3 (𝑥 + 1) + (𝑥 + 1) = (𝑥 + 1)(𝑥 3 𝑥 + 1) = (𝑥 + 1) 2 (𝑥 2 − 𝑥 + 1)
+ 1) = (𝑥 + 1)(𝑥 + 1)(𝑥
, which is the decomposition in both 𝑄[𝑥] and
2
𝑅[𝑥]
−
. In
𝐶[𝑥] the expression in the last parentheses can be decomposed just like in the part a), thus getting (𝑥 + 1) 2 (𝑥 −
1−𝑖√3
) (𝑥 −
2
1+𝑖√3
) .
2 d) 𝑥 6 − 2 = (𝑥 3 − √2)(𝑥 3 + √2) . It is apparent that the polynomial is irreducible over
𝑄[𝑥] . In 𝑅[𝑥] it is possible to continue, (𝑥 3 − √2)(𝑥 3
3
+ √2) = (𝑥 − √√2 ) (𝑥 2 +
3
√√2 3 𝑥 + √2
3
) (𝑥 + √√2 ) (𝑥 2
3
− √√2
3 𝑥 + √2 ) . In 𝐶[𝑥] further decomposition of the polynomial is possible but we will not perform it due to the form of the expression.
Example 18 Determine a greatest common divisor of polynomials 𝑓, 𝑔 when 𝑓 = 𝑥 4
2𝑥 2 + 1 , 𝑔 = 𝑥 3 + 3𝑥 2 − 𝑥 − 3
− a) using the Euclidean algorithm, b) by decomposition.
Solution: a) (𝑥 4 − 2𝑥 2 + 1): ( 𝑥 3 + 3𝑥 2 − 𝑥 − 3) = 𝑥 − 3 + 𝑥 3
8𝑥
+3𝑥
2 −8
2 −𝑥−3
.
23
Now we normalize the remainder 8𝑥 2 − 8 into the polynomial 𝑥 2 − 1 and continue with the algorithm.
(𝑥 3 + 3𝑥 2 − 𝑥 − 3): ( 𝑥 2 − 1) = 𝑥 + 3 remainder. 0.
The last non-zero remainder is 8𝑥 2 and 𝐺𝐶𝐷(𝑓, 𝑔) = 𝑥 2 − 1 .
− 8 , which is one of greatest common divisors, b) In some cases it is possible to find out common divisors and also greatest common divisors by decomposition of the polynomials. 𝑓 = 𝑥 4 − 2𝑥 2 + 1 = (𝑥 2 − 1) 2 = (𝑥 − 1) 2 (𝑥 + 1) 2
, 𝑔 = 𝑥 3
(𝑥 2
+ 3𝑥 2 − 𝑥 − 3 = 𝑥 3 − 𝑥 + 3𝑥 2 − 3 = 𝑥(𝑥
− 1)(𝑥 + 3) = (𝑥 − 1)(𝑥 + 1)(𝑥 + 3) ,
2 − 1) + 3(𝑥 2 − 1) =
𝐺𝐶𝐷(𝑓, 𝑔) = (𝑥 − 1)(𝑥 + 1) = (𝑥 2 − 1) .
Example 19 Solve individually: Apply the Euclidean algorithm to determine 𝐺𝐶𝐷(𝑓, 𝑔) when 𝑓 = 𝑥 5 − 𝑥 4 + 𝑥 3 − 𝑥 2 + 𝑥 − 1 , 𝑔 = 𝑥 4 − 2𝑥 2 − 3𝑥 − 2.
Result: 𝐺𝐶𝐷(𝑓, 𝑔) = 𝑥 2 + 𝑥 + 1
Definition 12
Polynomials 𝑓, 𝑔 ∈ 𝑇(𝑥) are called coprime if and only if 𝐺𝐶𝐷(𝑓, 𝑔) = 1 .
Theorem 13
Let 𝑓, 𝑔 ∈ 𝑇[𝑥], 𝑔 ≠ 0 . Then there exist 𝑢, 𝑣 ∈ 𝑇[𝑥] such that 𝑓𝑢 + 𝑔𝑣 =
𝐺𝐶𝐷(𝑓, 𝑔) .
Furthermore, if 𝑑𝑔 𝑓 ≥ 1, 𝑑𝑔 𝑔 ≥ 1 , then polynomials 𝑢, 𝑣 ∈ 𝑇(𝑥) can be found such that 𝑑𝑔 𝑢 < 𝑑𝑔 𝑔 𝑑𝑔 𝑣 < 𝑑𝑔 𝑓 .
Example 20 Use the Euclidean algorithm to determine polynomials 𝑢, 𝑣 such that
𝐺𝐶𝐷(𝑓, 𝑔) = 𝑓𝑢 + 𝑔𝑣 , je-li 𝑓 = 𝑥 4 + 2𝑥 3 − 𝑥 2 − 4𝑥 − 2 , 𝑔 = 𝑥 4 + 𝑥 3 − 𝑥 2 − 2𝑥 − 2 .
Solution: The first step will be to find 𝐺𝐶𝐷(𝑓, 𝑔) .
(𝑥 4 + 2𝑥 3 − 𝑥 2 − 4𝑥 − 2): (𝑥 4 + 𝑥 3 − 𝑥 2 − 2𝑥 − 2) = 1 + 𝑥 3 − 2𝑥 𝑥 4 + 𝑥 3 − 𝑥 2 − 2𝑥 − 2
We will write this result in the form 𝑓 = 𝑔 ∙ 𝑞
1
+ 𝑟
1
, 𝑞
1
= 1, 𝑟
1
= 𝑥 3 − 2𝑥.
Continuing with division,
(𝑥 4 + 𝑥 3 − 𝑥 2 − 2𝑥 − 2): (𝑥 3 − 2𝑥) = 𝑥 + 1 + 𝑥 𝑥 2
3
− 2
− 2𝑥
,
We will again rewrite it in the form 𝑔 = 𝑟
1 𝑞
2
+ 𝑟
2
, 𝑞
2
= 𝑥 + 1, 𝑟
2
= 𝑥 2 − 2.
Since 𝑥 3 − 2𝑥 = 𝑥(𝑥 2
𝐺𝐶𝐷(𝑓, 𝑔) = 𝑥 2 − 2 .
− 2) , the last non-zero remainder is the polynomial 𝑟
2
and hence
To determine the sought form of the greatest common divisor 𝑟
2 proceed backwards in the equation sequence
= 𝑓𝑢 + 𝑔𝑣 , we will 𝑓 = 𝑔 ∙ 𝑞
1
+ 𝑟
1 𝑔 = 𝑟
1 𝑞
2
+ 𝑟
2
.
When expressing 𝑟
2
= 𝑔 − 𝑟
1 𝑞
2 equation for 𝑟
1
, we get
from the second equation and substituting the first 𝑟
2
= 𝑔 − (𝑓 − 𝑔𝑞
1
)𝑞
2
= (−𝑞
2
)𝑓 + (1 + 𝑞
1 𝑞
2
)𝑔, whence it is already apparent that 𝑢 = −𝑞
2
= −𝑥 − 1 , 𝑣 = 1 + 𝑞
1 𝑞
2
= 𝑥 + 2 .
Overall 𝑥 2 − 2 = (−𝑥 − 1)(𝑥 4 + 2𝑥 3 − 𝑥 2 − 4𝑥 − 2) + (𝑥 + 2)(𝑥 4 + 𝑥 3 − 𝑥 2 − 2𝑥 − 2).
Verify the validity of the decomposition by yourselves by expanding the right side of the equation.
Note:
When searching for a greatest common divisor of polynomials in the form 𝑓𝑢 + 𝑔𝑣 using the Euclidean algorithm, it is not possible to normalize the polynomials or use associated polynomials for the purpose of simplicity.
Theorem 14
Let 𝑓, 𝑔, ℎ ∈ 𝑇[𝑥].
Then there applies
1) If 𝑓|𝑔ℎ and at the same time 𝐺𝐶𝐷(𝑓, 𝑔) = 1 , then 𝑓|ℎ .
2) If 𝑓|ℎ , 𝑔|ℎ and 𝐺𝐶𝐷(𝑓, 𝑔) = 1 , then (𝑓𝑔)|ℎ .
Proof: Re
1: We know from the previous theorem that there exist polynomials 𝑢, 𝑣 ∈ 𝑇[𝑥] such that 𝑓𝑢 + 𝑔𝑣 = 𝐺𝐶𝐷(𝑓, 𝑔) , then obviously 𝑓𝑢 + 𝑔𝑣 = 1.
When multiplying this equation with the polynomial ℎ , we get 𝑓ℎ𝑢 + 𝑔ℎ𝑣 = ℎ.
It is apparent that 𝑓|𝑓 and according to the initial assumption 𝑓|𝑔ℎ , therefore from the previous equation there must also apply that 𝑓|ℎ .
Re 2: If 𝑓|ℎ , then there exists ℎ
1
∈ 𝑇[𝑥] such that ℎ = 𝑓ℎ
1
. Since by the initial assumption also 𝑔|ℎ , then after the substitution for ℎ from the previous step, there also applies 𝑔|𝑓ℎ
1
. Furthermore, we assume that 𝐺𝐶𝐷(𝑓, 𝑔) = 1 , so we can employ the first part of the theorem, resulting into 𝑔|ℎ
1
∃ℎ
2
∈ 𝑇[𝑥]: ℎ
1
= 𝑔ℎ
1
ℎ = 𝑓𝑔ℎ
2
(𝑓𝑔)|ℎ
25
Exercises for Check
1.
Define polynomial using the algebraic definition. Demonstrate how from this form there can be obtained the form you are familiar with from high school.
2.
State the theorem on polynomial division with a remainder.
3.
Define the Taylor polynomial.
4.
Describe the Euclidean algorithm for division of polynomials.
5.
Define common divisor of two polynomials and greatest common divisor of two polynomials.
6.
While calculating the value of the polynomial, a student pressed the following buttons on a calculator:
7 × 6 =× 6 = −2 =× 6 = −3 =× 6 = +1 =× 6 = +2 =× 6 =
7.
Determine the polynomial the student was evaluating and the value of 𝑥 .
Verify that there applies 𝑡 4 + 𝑡 2 − 3𝑡 + 7 = (𝑡 3 + 3𝑡 2 + 10𝑡 + 27)(𝑡 − 3) + 88 .
8.
Assemble a Horner scheme corresponding to this equation.
Divide the polynomial −5𝑥 4 + 4𝑥 2 − 3𝑥 + 4 by the polynomial 𝑥 3 + 2𝑥 2 − 4 .
9.
Find out a polynomial that passes through the points
[−1; −7], [0; −3], [1; −1], [2; 11] .
10.
Find a greatest common divisor of the polynomials 𝑓 = 𝑥 4 𝑥 3 − 𝑥 2 − 𝑥 + 1
− 𝑥 3
by both Euclidean algorithm and decomposition.
− 2𝑥 2 + 𝑥 + 1 , 𝑔 =
Results:
6.
7𝑥 6 − 2𝑥 4 − 3𝑥 3 + 𝑥 2 + 2𝑥 , 𝑐 = 6
7.
1 0
3 1 3
8.
9.
−5𝑥 + 10 +
2𝑥 3 − 𝑥 2
−16𝑥
2 𝑥 3
−23𝑥+44
+2𝑥 2 −4
+ 𝑥 − 3
10.
𝑥 2 − 1
1
10
-3
27
7
88
3 Algebraic Equations
Chapter Objectives
Having studied this chapter you will be able to:
define algebraic equation
find roots of a polynomial
review the procedures for solving of linear and quadratic equation
generically solve equations of 3 rd
degree
solve binomial and reciprocal equations
Time to Study
Circa 3 hours
Study Guide
Definition 13
By an algebraic equation we mean an equation in the form 𝑎 𝑛 𝑥 𝑛 where 𝑎 𝑖
𝑹 for 𝑖 = 0, 1, . . . , 𝑛 .
+ 𝑎 𝑛−1 𝑥 𝑛−1 … 𝑎
1 𝑥 + 𝑎
0
= 0 ,
Note:
Solving an algebraic equation stands for searching for roots of the polynomial on the left side of the equation. In the following text there will be demonstrated procedures for solving of selected cases of algebraic equations as well as algebraic inequalities.
This text is focused solely on the equations whose coefficients are real numbers.
There also exist equations with complex coefficients, for which the polynomial on their left side is over the field of complex numbers. However, since you will not usually encounter these polynomials in your common practice, this text is limited to polynomials over the field of real numbers.
Yet it is worth reminding that a polynomial with real coefficients can have both real and complex roots.
27
3.1
Roots of Polynomial
Definition 14
A number 𝑐 ∈ 𝑇 is called a root of a polynomial 𝑓 ∈ 𝑇[𝑥] , if and only if 𝑓(𝑐) = 0 .
Theorem 15
(Bézout’s Theorem) : Number 𝑐 ∈ 𝑇 is a root of a polynomial 𝑓 ∈ 𝑇[𝑥] if and only if
(𝑥 − 𝑐)|𝑓 .
Proof:
For any 𝑐 ∈ 𝑇 and an arbitrary 𝑓 ∈ 𝑇[𝑥] there applies 𝑓 = (𝑥 − 𝑐)𝑔 + 𝑟 .
1) 𝑐 is a root, then 𝑓(𝑐) = 0 𝑟 = 0 𝑓 = (𝑥 − 𝑐)𝑔 (𝑥 − 𝑐)|𝑓
2) (𝑥 − 𝑐)|𝑓 𝑓 = (𝑥 − 𝑐)ℎ, ℎ ∈ 𝑇[𝑥] 𝑓(𝑐) = 0 𝑐 is a root.
Definition 15
A number 𝑐 ∈ 𝑇 is called a k-fold multiple root of a polynomial 𝑓 ∈
𝑇[𝑥], 𝑘 ∈ 𝑁 if and only if there applies:
1) (𝑥 − 𝑐) 𝑘 |𝑓
2) (𝑥 − 𝑐) 𝑘+1
does not divide 𝑓 .
Example 21 What is the root multiplicity of number 2 for the polynomial 𝑓(𝑥) = 𝑥 5
5𝑥 4 + 7𝑥 3 − 2𝑥 2 + 4𝑥 − 8 ?
−
Solution: The Horner scheme will be used:
1 -5 7 -2 4 -8
2 1 -3 1 0 4 0
2 1 -1 -1 -2 0
2 1 1 1 0
2 1 3 7
Thus number 2 is a 3-fold multiple root of the polynomial 𝑓 and there applies 𝑓(𝑥) = 𝑥 5 − 5𝑥
= (𝑥 − 2) 2
4 + 7𝑥 3
(𝑥 3
− 2𝑥 2
− 𝑥 2
+ 4𝑥 − 8 = (𝑥 − 2)(𝑥
− 𝑥 − 2) = (𝑥 − 2) 3 (𝑥 2
4 − 3𝑥
+ 𝑥 + 1).
3 + 𝑥 2 + 4)
Example 22 Find the value of 𝑎 in the polynomial 𝑓 = 𝑥 5 − 𝑎𝑥 2 − 𝑎𝑥 + 1 such that the number 𝑐 = −1 is its at least a 2-fold multiple root.
Solution:
1 0 0 a a 1
-1 1 -1 1 -(1+ a ) 1 0
-1 1 -2 3 -(4+ a ) 5+ a
There must apply that 5 + 𝑎 = 0 , hence 𝑎 = −5 . When actually dividing the polynomial 𝑥 5 + 5𝑥 2 + 5𝑥 + 1 by the polynomial (𝑥 + 1) 2 = 𝑥 2 + 2𝑥 + 1 , we then get 𝑥 5 + 5𝑥 2 + 5𝑥 + 1 = (𝑥 + 1) 2 (𝑥 3 − 2𝑥 2 + 3𝑥 + 1).
Note:
When searching for roots of algebraic equations, it might be useful to apply the following theorem, which is valid for algebraic equations with real coefficients.
Theorem 16
Let 𝑎 𝑛 𝑥 𝑛 + 𝑎 𝑛−1 𝑥 𝑛−1 … 𝑎
1 𝑥 + 𝑎
0
= 0 be an algebraic equation. Let 𝑥
1
, . . . , 𝑥 𝑛 be its roots (both real and complex).
Let 𝐴 = 𝑚𝑎𝑥 { 𝑎 𝑖
; 𝑖 = 0, . . . , 𝑛
1} . Then there applies:
1.
For each index 𝑖 , 𝑖 = 1, 2, . . . , 𝑛 there applies the inequality |𝑥 𝑖
| < 1 +
𝐴
|𝑎 𝑛
|
.
2.
The number of positive real roots is equal to the number of sign changes in the sequence of non-zero coefficients 𝑎 𝑛
, 𝑎 𝑛−1
, … , 𝑎
0 or it is lesser by an even number.
Implication:
When all coefficients of a polynomial are positive real numbers, this polynomial has no positive roots.
Example 23 Determine intervals in which there lie real roots of the equation 2𝑥 5
3𝑥 4 − 7𝑥 3 + 6𝑥 2 − 11𝑥 + 5 = 0 .
+
Solution: 𝑎 𝑛
= 2, 𝐴 = 11 equation lie in the interval 𝑥
, thus 𝑖
∈ (− 𝑥 𝑖
𝐴
∈ 1 +
|𝑎 𝑛
|
13
2
,
13
) .
2
= 1 +
11
2
=
13
2
. Hence all roots of the
There are four sign changes in the sequence of coefficients 2, 3, -7, 6, -11, 5, thus in the interval (0,
13
) there lie either 4 or 2 roots.
2
Definition 16
We call a field 𝑇 to be algebraically closed if and only if every polynomial 𝑓 ∈ 𝑇[𝑥] with the degree of 1 or greater has at least one root over the field 𝑇 .
Example 24 The field of rational numbers is not algebraically closed. For example the polynomial 𝑓 = 𝑥 2 − 2 has rational coefficients, its degree is 2, but its roots 𝑘 = ±√2 do not belong into the set of rational numbers. Similarly, the field of real numbers is not
29
algebraically closed. For instance, the polynomial 𝑓 = 𝑥 2 degree is 2, but it does not have real roots because 𝑘 = ±𝑖 .
+ 1 has real coefficients, its
Theorem 17
(Fundamental Theorem of Algebra) : Every polynomial with complex coefficients and with the degree of 1 or greater has at least one complex root.
The fundamental theorem of algebra can be also formulated as follows: The field of complex numbers is algebraically closed.
Note:
All known proofs of the Fundamental theorem of algebra are based on methods of mathematical analysis. Performing the proof would require application of some theorems from mathematical analysis, whose implication is the fundamental theorem of algebra.
Therefore no proof will be included.
Theorem 18
Let 𝑓 ∈ 𝑇[𝑥], 𝑑𝑔 𝑓 = 𝑛 ≥ 1 . Then the polynomial 𝑓 has 𝑛 roots over 𝐶 , provided that a 𝑘 − fold multiple root is counted as 𝑘 roots. Therefore for each polynomial 𝑓 ∈ 𝑇[𝑥], 𝑑𝑔 𝑓 ≥ 1 there exist numbers 𝑎, 𝛼
1
, … , 𝛼 𝑛 𝛼
2
) … (𝑥 − 𝛼 𝑛
).
∈ 𝐶 such that 𝑓 = 𝑎(𝑥 − 𝛼
1
)(𝑥 −
Proof:
According to the Fundamental theorem of algebra, there exists a root 𝛼
1
such that 𝑓 = (𝑥 − 𝛼
1
)𝑓
1
, 𝑑𝑔 𝑓
1
= 𝑛 − 1 . If 𝑛 − 1 ≥ 1 , then there exists a root 𝛼
2
of a polynomial 𝑓
1
such that 𝑓
1
= (𝑥 − 𝛼
2
)𝑓
2
(𝑥 − 𝛼
1
) … (𝑥 − 𝛼 𝑛
)𝑓 𝑛
, 𝑑𝑔 𝑓 𝑛
, so 𝑓 = (𝑥 − 𝛼
1
)(𝑥 − 𝛼
2
)𝑓
2
= 0 , thus the polynomial 𝑓 𝑛
, and so on. Eventually 𝑓 =
is a constant one, 𝑓 𝑛
= 𝑎 .
Theorem 19
Let 𝑓 ∈ 𝑅[𝑥], 𝑐 ∈ 𝐶, 𝑓(𝑐) = 0 . Then the number 𝑐̅ ∈ 𝐶 is also a root of the polynomial 𝑓 and its multiplicity is the same as of the root 𝑐 .
6
Proof:
1) 𝑎
Let 𝑛 𝑐 be a root of a polynomial
= 0 𝑓 = 𝑎
0 𝑥 𝑛 + ⋯ + 𝑎
. There only remains to be proven that 𝑓(𝑐) 𝑛
. Then
̅̅̅̅̅̅ = 𝑓(𝑐̅): 𝑓(𝑐) = 𝑎
0 𝑐 𝑛
̅̅̅̅̅̅ = 𝑎
0 𝑐 𝑛 + ⋯ + 𝑎 𝑛 0 𝑐 𝑛 ̅̅̅ = 𝑎
0 𝑛 ̅̅̅ =
= 𝑎
0
(𝑐̅) 𝑛 + ⋯ + 𝑎 𝑛
= 𝑓(𝑐̅) = 0.
+ ⋯ +
Hence 𝑐̅ is a root of the polynomial 𝑓 as well.
2) Let 𝑐 be a 𝑘 − fold multiple root of a polynomial 𝑓, 𝑘 ∈ 𝑁, 𝑐
𝑅, 𝑐̅ ≠ 𝑐 . We will prove that 𝑐̅ is also a 𝑘 − fold multiple root. 𝑓 = (𝑥 − 𝑐)𝑓
1
= (𝑥 − 𝑐)(𝑥 − 𝑐̅)𝑓
2
As 𝑐 + 𝑐̅ = 2𝑎 and 𝑐𝑐̅ = 𝑎 2 + 𝑏 2
= [𝑥 2 − (𝑐 + 𝑐̅)𝑥 + 𝑐𝑐̅]𝑓
, the polynomial 𝑥 2
2
− (𝑐 + 𝑐̅)𝑥 + 𝑐𝑐̅ and thus also the polynomial 𝑓
2
has real coefficients.
6 Let us remind what a number 𝑐̅ , called a complex conjugate of the number 𝑐 , stands for. When 𝑐 = 𝑎 + 𝑏𝑖, 𝑎, 𝑏 ∈ 𝑅 , then 𝑐̅ = 𝑎 − 𝑏𝑖 . Furthermore, for numbers 𝑐
1
, 𝑐
2
∈ 𝐶 there applies: 𝑐
1
± 𝑐
2
̅ ± 𝑐 ̅
2
, 𝑐
1
. 𝑐
2
̅ . 𝑐 ̅
2
, for 𝑘 ∈ 𝑁 : 𝑧 𝑘
1
̅ ) 𝑘
, for 𝑎 ∈ 𝑅: 𝑎̅ = 𝑎
𝑓 = (𝑥 − 𝑐)(𝑥 − 𝑐̅)𝑓
2
= (𝑥 − 𝑐) 𝑘
= (𝑥 − 𝑐)(𝑥 − 𝑐̅)(𝑥 − 𝑐)(𝑥 − 𝑐̅)𝑓
(𝑥 − 𝑐̅) 𝑘 ℎ,
4
= ⋯ where the polynomial ℎ cannot have roots 𝑐, 𝑐̅ . Therefore 𝑐̅ is a 𝑘 − fold multiple root.
3.2
Polynomial Derivative
Definition 17
Let 𝑓 ∈ 𝑇(𝑥), 𝑓 = 𝑎
0 𝑥 𝑛 polynomial 𝑓 ′ = 𝑛𝑎
0 𝑥 𝑛−1 + (𝑛 − 1)𝑎
1 𝑥
+ 𝑎 𝑛−2
1 𝑥 𝑛−1 + ⋯ + 𝑎
+ ⋯ + 𝑎 𝑛−1 𝑛−1 𝑥 + 𝑎 𝑛
is called a
. Then derivative the
of the polynomial 𝑓 .
7
Theorem 20
For the derivative of polynomials 𝑓, 𝑔 ∈ 𝑇[𝑥] there applies:
1) (𝑓 ± 𝑔) ′ = 𝑓 ′ ± 𝑔 ′
2) (𝑓. 𝑔) ′ = 𝑓 ′
3) 𝑘 ∈ 𝑁, (𝑓 𝑘 ) ′
. 𝑔 + 𝑔 ′ . 𝑓
= 𝑘. 𝑓 𝑘−1 . 𝑓′
4) 𝑐 ′ = 0 for 𝑐 ∈ 𝑇
Note:
The proof of this theorem has been performed in detail in mathematical analysis.
Theorem 21
Let 𝑘 ∈ 𝑁, 𝑘 ≥ 2, 𝑐 ∈ 𝑇, 𝑓 ∈ 𝑇[𝑥] . Then there applies: When the number 𝑐 is a 𝑘 -fold multiple root of the polynomial 𝑓 , then the number 𝑐 is a (𝑘 − 1) - fold multiple root of the polynomial 𝑓′ . When 𝑐 is a simple root of the polynomial 𝑓 then 𝑐 is not a root of the polynomial 𝑓′ .
Proof:
Let 𝑐 be a 𝑘 -fold multiple root of a polynomial 𝑓 . Then there applies
(𝑥 − 𝑐) 𝑘 |𝑓
𝑓 = (𝑥 − 𝑐) 𝑘 . ℎ, ℎ ∈ 𝑇[𝑥] .
For the derivative of the polynomial 𝑓 there applies 𝑓 ′ = 𝑘(𝑥 − 𝑐) 𝑘−1 . ℎ + (𝑥 − 𝑐) 𝑘 . ℎ ′ = (𝑥 − 𝑐) 𝑘−1 [𝑘ℎ + (𝑥 − 𝑐)ℎ′]
(𝑥 − 𝑐) 𝑘−1 |𝑓 .
We will prove by contradiction that 𝑐 is not a 𝑘 -fold multiple root of the polynomial 𝑓′ . If
(𝑥 − 𝑐) 𝑘
divided 𝑓′ , then there would apply 𝑓 ′ = (𝑥 − 𝑐) 𝑘 . ℎ
1 assumption (𝑥 − 𝑐) 𝑘 |𝑓 for the derivative there would have to apply
. From the previous
(𝑥 − 𝑐) 𝑘 ℎ
1
= (𝑥 − 𝑐) 𝑘−1 [𝑘ℎ + (𝑥 − 𝑐)ℎ′] .
After cancellation we get
7 You have already encountered the derivative during your study of mathematical analysis. The derivative was defined generically for a function and could be applied to analyze various functions’ properties, such as local extremes of function, convexity and concavity, examination of its behavior and so on.
31
(𝑥 − 𝑐)ℎ
1
= 𝑘ℎ + (𝑥 − 𝑐)ℎ′ .
Choosing 𝑥 = 𝑐 we get after substitution
0 = 𝑘. ℎ(𝑐) .
Since 𝑘 is independent on 𝑐 then ℎ(𝑐) = 0 , which means that ℎ = (𝑥 − 𝑐)𝑔, 𝑔 ∈ 𝑇[𝑥] and 𝑓 = (𝑥 − 𝑐) 𝑘 ℎ = (𝑥 − 𝑐) 𝑘 (𝑥 − 𝑐)𝑔 = (𝑥 − 𝑐) 𝑘+1 𝑔 .
That would imply that the number 𝑐 is a (𝑘 + 1) -fold multiple root of the polynomial 𝑓 , which is a contradiction with the initial assumption that 𝑐 is only a 𝑘 -fold multiple root of the polynomial 𝑓 .
Note:
The theorem is not valid in the opposite way: If a number 𝑐 is a root of the polynomial 𝑓′ , then it does not necessarily have to be a root of the polynomial 𝑓 . For example. 𝑓 = (𝑥 − 3) 5 + 7 , 𝑓(3) ≠ 0 , 𝑓 ′ = 5(𝑥 − 3) 4
and number 3 is a 4-fold multiple root of the polynomial 𝑓′ .
Example 25 Given the polynomial 𝑓 = 𝑎𝑥 4 + 𝑏𝑥 3 + 1 , find coefficients 𝑎, 𝑏 such that 1 is at least a double root.
Solution: If 1 is a double root of the polynomial 𝑓 then it is a simple root of the polynomial 𝑓´ . Therefore we will find the derivative of the polynomial 𝑓 and determine the values of the polynomials 𝑓, 𝑓´ at the point 1: 𝑓´(𝑥) = 4𝑎𝑥 3 + 3𝑏𝑥 2 𝑓(1) = 𝑎 + 𝑏 + 1 𝑓´(1) = 4𝑎 + 3𝑏
Now all that remains is to solve the following system of equations: 𝑎 + 𝑏 + 1 = 0
4𝑎 + 3𝑏 = 0
Having solved this system of equations, we learn that 𝑎 = 3, 𝑏 = −4 , so the form of the polynomial 𝑓 is 𝑓 = 3𝑥 4 − 4𝑥 3 + 1.
Example 26 Solve Example 22 using the derivative.
Solution: 𝑓´(𝑥) = 5𝑥 4 − 2𝑎𝑥 − 𝑎 , 𝑓(−1) = −1 − 𝑎 + 𝑎 + 1 = 0 , 𝑓´(−1) = 5 + 2𝑎 − 𝑎 = 5 + 𝑎 , 𝑓´(−1) = 0 𝑎 = −5 . The result is the polynomial 𝑓 = 𝑥 5 + 5𝑥 2 + 5𝑥 + 1 .
Theorem 22
Let 𝑘 ∈ 𝑁, 𝑘 ≥ 2, 𝑐 ∈ 𝑇, 𝑓 ∈ 𝑇[𝑥] . Then the following propositions are equivalent to each other:
1) 𝑐 is a 𝑘 − fold multiple root of the polynomial f.
2) (𝑥 − 𝑐) 𝑘 |𝑓 and at the same time (𝑥 − 𝑐) 𝑘−1
does not divide 𝑓 .
3) There exists a polynomial 𝑔 ∈ 𝑇[𝑥] such that 𝑓 = (𝑥 − 𝑐) 𝑘 . 𝑔 and at the same time 𝑔(𝑐) ≠ 0 .
4) 𝑓(𝑐) = 0, 𝑓´(𝑐) = 0, 𝑓´´(𝑐) = 0, … , 𝑓 (𝑘−1) (𝑐) = 0, 𝑓 (𝑘) ≠ 0 .
5) 𝑐 is a (𝑘 − 1) − fold multiple root of the polynomial 𝑑 = 𝑁𝑆𝐷(𝑓, 𝑓´) .
Note:
When searching for multiple roots of a polynomial 𝑓 , we can adopt the following procedure:
1.
Find 𝐺𝐶𝐷(𝑓, 𝑓´) = 𝑑 .
2.
If 𝑑 = 1 , then 𝑓 does not have any multiple roots.
3.
If 𝑑 ≠ 1 , then each root of the polynomial 𝑑 is also a root of the polynomial 𝑓 , and moreover it is its multiple root: its multiplicity is greater by 1 than its multiplicity for the polynomial 𝑑 .
Example 27 Demonstrate that the polynomial 𝑓(𝑥) = 𝑥 3 + 𝑥 2 − 5𝑥 − 2 ∈ 𝑅[𝑥] has only simple roots.
Solution: We will proceed according to the procedure in the previous note. Seeking the derivative of the polynomial 𝑓 , we find it to be 𝑓´(𝑥) = 3𝑥 2 + 2𝑥 − 5 . You can verify for yourselves by application of the Euclidean algorithm that 𝐺𝐶𝐷(𝑓, 𝑓´) = 1 . Thus the polynomial 𝑓(𝑥) has truly only simple roots.
Theorem 23
Let 𝑓 ∈ 𝑇[𝑥], 𝑑 = 𝐺𝐶𝐷(𝑓, 𝑓´), 𝑓 = 𝑑. 𝑔 . Then there applies: The polynomial 𝑔 has the same roots as the polynomial 𝑓 and all of the roots of the polynomial 𝑔 are simple.
Example 28 Let the polynomials for example be 𝑓 = (𝑥 − 1) 3 (𝑥 + 2) 5 (𝑥 + 7) , 𝑑 =
(𝑥 − 1) 2 (𝑥 + 2) 4
. Then the polynomial 𝑔 = (𝑥 − 1)(𝑥 + 2)(𝑥 + 7) and 𝑓 = 𝑑. 𝑔 .
3.3
Polynomials with Integer Coefficients
Theorem 24
Let 𝑝 ∈ 𝑍, 𝑝 ≠ 0, 𝑞 ∈ 𝑁, 𝐺𝐶𝐷(𝑝, 𝑞) = 1, 𝑐 = 𝑝 𝑞
, 𝑓 = 𝑎 𝑛 𝑥 𝑛
𝑍[𝑥] . When 𝑐 is a root of the polynomial 𝑓 , then for any 𝑚 ∈ 𝑍 there applies:
+ ⋯ + 𝑎
0
∈
1) 𝑝|𝑎
0
,
2) 𝑞|𝑎 𝑛
,
3) (𝑝 − 𝑚𝑞)|𝑓(𝑚) .
33
Proof: 𝑐 = 𝑝 𝑞 is a root of the polynomial, i.e. 𝑓(𝑐) = 0 . Hence for the polynomial 𝑓 we can write 𝑎 𝑛
( 𝑝 𝑞
) 𝑛
+ 𝑎 𝑛−1
( 𝑝 𝑞
) 𝑛−1
+ ⋯ + 𝑎
0
= 0.
When multiplying this equation by 𝑞 𝑛
, we get 𝑎 𝑛 𝑝 𝑛 + 𝑎 𝑛−1 𝑝 𝑛−1 𝑞 + ⋯ 𝑎
1 𝑝𝑞 𝑛−1 + 𝑎
0 𝑞 𝑛 = 0.
1) Having transposed the last term 𝑎
0 𝑞 𝑛
to the right side of the equation and factored out 𝑝 from all previous terms, we get: 𝑝[𝑎 𝑛 𝑝 𝑛−1 + ⋯ + 𝑎
1 𝑞 𝑛−1 ] = −𝑎
0 𝑞 𝑛 , thus 𝑝|𝑎
0 𝑞 𝑛
, and as both 𝑝 and 𝑞 are coprime, there must apply 𝑝|𝑎
0
.
2) Similarly as in the previous step we can transpose the term 𝑎 𝑛 𝑝 𝑛
to the right side of the equation and factor out 𝑞 from all following terms, so getting: 𝑞[𝑎 𝑛−1 𝑝 𝑛−1 + ⋯ + 𝑎
0 𝑞 𝑛−1 ] = −𝑎 𝑛 𝑝 𝑛 , reasoning similarly as in step 1), we can conclude that 𝑞|𝑎 𝑛
.
3) The polynomial 𝑓 can be expressed in the form of the Taylor polynomial centred at 𝑚 : 𝑓 = 𝑐
0
+ 𝑐
1
(𝑥 − 𝑚) + ⋯ + 𝑐 𝑛−1
(𝑥 − 𝑚) 𝑛−1 + 𝑐 𝑛
(𝑥 − 𝑚) 𝑛 .
It is obvious that 𝑓(𝑚) = 𝑐
0 𝑐 = 𝑝 𝑞
, we get
.
When we now perform substitution of 𝑥 with the root 𝑓(𝑐) = 𝑐
0
+ 𝑐
1
( 𝑝 − 𝑚𝑞 𝑞
) + ⋯ + 𝑐 𝑛−1
( 𝑝 − 𝑚𝑞 𝑞
) 𝑛−1
+ 𝑐 𝑛
( 𝑝 − 𝑚𝑞 𝑞
) 𝑛
= 0, after multiplication of the equation by the expression 𝑞 𝑛
we get 𝑐
0 𝑞 𝑛 + 𝑐
1
(𝑝 − 𝑚𝑞)𝑞 𝑛−1 + ⋯ + 𝑐 𝑛−1
(𝑝 − 𝑚𝑞) 𝑛−1 𝑞 + 𝑐 𝑛
(𝑝 − 𝑚𝑞) 𝑛 = 0.
Reasoning the same way as in the steps 1) and 2), we can come to the conclusion that (𝑝 − 𝑚𝑞)|𝑐
0 𝑞 𝑛
, and since that is (𝑝 − 𝑚𝑞)|𝑓(𝑚) .
𝐺𝐶𝐷(𝑝 − 𝑚𝑞, 𝑞) = 1 , there applies (𝑝 − 𝑚𝑞)|𝑐
0
,
Note:
The last theorem can be used to find all rational roots of polynomials with integer coefficients. In fact, the application of the part 3) of the last theorem consists in choosing certain values for 𝑚 , which are numbers 1 and −1 , thus
(𝑝 − 𝑞)|𝑓(1) (𝑝 + 𝑞)|𝑓(−1).
Example 29 Find all rational roots of the polynomial 𝑓 = 2𝑥 5 + 𝑥 4 − 12𝑥 3
19𝑥 − 6 .
− 20𝑥 2 −
Solution: The roots being searched for will have the form 𝑐 = 𝑝 𝑞
where 𝑝|(−6) , 𝑞|2 .
Divisors of number −6: 1,2,3,6, −1, −2, −3, −6 .
Positive divisors of number 2: 1,2 .
Next we will create a table where in the first row there will be all possible values of 𝑝 𝑞
, corresponding values of 𝑝 + 𝑞 will be in the second row, and 𝑝 − 𝑞 in the third one.
Since 𝑓(−1) = 4 , we will highlight all divisors of number 4 in the second row, and because of 𝑓(1) = −54 , we will similarly highlight all divisors of number −54 in the third row. Candidates for roots will then be those numbers 𝑝 𝑞
which will have both second and third row values highlighted. 𝑝 𝑞
1
1
1
2
2
1
3
1
3
2
6
1
−1
1
−1
2
−2
1
−3
1
−3
2
−6
1 𝑝 + 𝑞 2 3 3 4 5 7 0 1 -1 -2 -1 -5 𝑝 − 𝑞 0 -1 1 2 1 5 -2 -3 -3 -4 -5 -7
Therefore the candidates for roots are the numbers: 3,
−1
, −2 . We will verify these roots
2 using Horner’s method:
2 1 -12 -20 -19 -6
−1
2
2
3 2
−2 2
0
6
2
-12
6
2
-14
4
0
-12
0
0
We can clearly see in the Horner scheme that 𝑓 = (𝑥 +
1
2
) . (2𝑥 4 − 12𝑥 2 − 14𝑥 − 12) = (2𝑥 + 1)(𝑥
= (2𝑥 + 1)(𝑥 − 3)(𝑥 3 + 3𝑥 2
4 − 6𝑥 2 − 7𝑥 − 6)
+ 3𝑥 + 2) = (2𝑥 + 1)(𝑥 − 3)(𝑥 + 2)(𝑥 2 + 𝑥 + 1)
The polynomial 𝑥 2 numbers 3,
−1
2
, −2 .
+ 𝑥 + 1 has no real roots and all sought rational roots are therefore
Example 30 Solve individually: Find all rational roots of the polynomial 𝑓 = 6𝑥 4
19𝑥 3 − 7𝑥 2 − 26𝑥 + 12 .
+
Result: Candidates for roots: 2,
1
2 roots are −3,
1
2
.
, −3,
−1
, 𝑓 = (𝑥 + 3)(2𝑥 − 1)(3𝑥
3
2 + 2𝑥 − 4) , rational
Note:
The previous procedure can be employed to solving certain algebraic inequalities as well, as shown in the following example.
Example 31 Solve the inequality 𝑥 3 + 4𝑥 2 + 𝑥 − 6 ≤ 0 .
35
Solution: A number 𝑝 is a divisor of number −6 , a number 𝑞 is a divisor of 1 . When substituting number 1 into the expression on the left side of the inequality, we find out that the value is 0 . We will term the left side of the inequality as 𝑓(𝑥) and the discovered fact as 𝑓(1) = 0 , which means that 1 is a root. For this reason we will not include into a table a row for 𝑝 − 𝑞 , because divisors of zero cannot be found. Further on 𝑓(−1) = 4 . 𝑝 𝑞
1 𝑝 + 𝑞 2
2
3
3
4
6
7
-1
0
-2
-1
-3
-2
-6
-5
After checking the candidates −3, −2, 1, 3 using Horner’s method, we find that the numbers −3, −2, 1 are roots of the polynomial, and the assigned inequality can be rewritten into the form
(𝑥 − 1)(𝑥 + 2)(𝑥 + 3) ≤ 0.
Now we can proceed on using the number line and zero points (perform by yourselves).
The result are all numbers 𝑥 ∈ (−∞; −3⟩ ∪ ⟨−2; 1⟩ .
3.4
Vieta’s Formulas
Certain problems can be solved by application of relationships between roots and coefficients of a polynomial that are presented in the following theorem.
Theorem 25
Let 𝑟
1
, 𝑟
2
, … , 𝑟 𝑛 𝑎 𝑛 𝑥 𝑛 + ⋯ 𝑎
1 𝑥 + 𝑎
0
∈ 𝑇 be all roots of a polynomial
. Then there apply so-called Vieta’s formulas: 𝑓 ∈ 𝑇[𝑥], 𝑑𝑔 𝑓 = 𝑛, 𝑓 =
− 𝑎 𝑎 𝑛 𝑛−1 𝑎 𝑛 𝑎 𝑛−2
= 𝑟
= 𝑟
1 𝑟
2
1
+ 𝑟
+ 𝑟
2
1 𝑟
+ ⋯ + 𝑟
3 𝑛
+ ⋯ + 𝑟 𝑛−1 𝑟 𝑛
⋮
(−1) 𝑛−1
(−1) 𝑛 𝑎
0 𝑎 𝑛 𝑎
1 𝑎 𝑛
= 𝑟
= 𝑟
1 𝑟
2
1 𝑟 𝑟
3
2
… 𝑟 𝑛−1
+ 𝑟
… 𝑟 𝑛−1 𝑟 𝑛
1 𝑟
2
… 𝑟 𝑛−2 𝑟 𝑛
+ ⋯ + 𝑟
2 𝑟
3
… 𝑟 𝑛−1 𝑟 𝑛
Proof:
When 𝑟
1
, 𝑟
2
, … , 𝑟 𝑛
are roots of the polynomial 𝑓 , there applies 𝑎 𝑛 𝑥 𝑛 + ⋯ 𝑎
1 𝑥 + 𝑎
0
= 𝑎 𝑛
(𝑥 − 𝑟
1
)(𝑥 − 𝑟
2
) … (𝑥 − 𝑟 𝑛
) . As we expand the right side and compare the coefficients for the same powers of 𝑥 , we get Vieta’s formulas.
Example 32
Apply Vieta’s formulas to find a polynomial whose roots are
−1, −1, −1,3,4 .
Solution: There are five roots, thus we know that 𝑛 = 5 . The coefficient 𝑎
5
can be arbitrarily chosen, so we choose 𝑎
5
− 𝑎
4
1
= 1 . Then
= −1 − 1 − 1 + 3 + 4 = 4 𝑎
3
= 1 + 1 − 3 − 4 + 1 − 3 − 4 − 3 − 4 + 12 = −6
1
(−1)
(−1)
3
4 𝑎
2 𝑎
1
1
= −1 + 3 + 4 + 3 + 4 − 12 + 3 + 4 − 12 − 12 = −16
= −3 − 4 + 12 + 12 + 12 = 29
(−1) 5
1 𝑎
0
1
= (−1). (−1). (−1). 3.4 = −12
After rearrangement we get 𝑎
0 polynomial 𝑓 has the form
= 12, 𝑎
1
= 29, 𝑎
2
= 16, 𝑎
3
= −6, 𝑎
4
= −4, 𝑎
5
= 1 . The 𝑓 = 𝑥 5 − 4𝑥 4 − 6𝑥 3 + 16𝑥 2 + 29𝑥 + 12.
We can verify the validity by expanding: 𝑓 = 𝑎 𝑛
(𝑥 + 1) 3
= 𝑥 5
. (𝑥 − 3). (𝑥 − 4) = 1. (𝑥 + 1)
− 4𝑥 4 − 6𝑥 3 + 16𝑥 2 + 29𝑥 + 12.
3 . (𝑥 − 3). (𝑥 − 4)
Example 33 Solve individually: Apply Vieta’s formulas to find a polynomial whose roots are 𝑖, 𝑖, −1 − 𝑖 .
Result: 𝑓 = 𝑥 3 + (1 − 𝑖)𝑥 2 + (1 − 2𝑖)𝑥 − 1 − 𝑖
Note: Vieta’s formulas can be utilised for simple in-memory solving of some quadratic equations with integer roots. For an equation
− 𝑏 𝑎
, 𝑟
1
. 𝑟
2
= 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0 there applies 𝑟
1 𝑐 𝑎
. For example, in case of the polynomial 𝑥 2 + 5𝑥 + 6
+ 𝑟
2
=
, for its roots there applies 𝑟
1
+ 𝑟
2
= −5 and 𝑟
1
. 𝑟
2
= 6 . Upon solving this system of equations, we get 𝑟
1
=
−2, 𝑟
2
= −3 .
Example 34 Find out the coefficients and the form of a quadratic polynomial 𝑎𝑥 2 𝑏𝑥 + 𝑐 knowing that the product of its roots is equal to −
5
2
and its sum to −
3
2
.
+
Solution: According to Vieta’s formulas there applies:
1) 𝑥
1
2) 𝑥
1 𝑥
+ 𝑥
2
2
=
= − 𝑏 𝑎 𝑐 𝑎
37
Hence we know that − 𝑏 𝑎
= −
3
2
and 𝑐 𝑎
= −
5
2
. The form of the fractions implies that 𝑎 = 2 , then 𝑏 = 3 and 𝑐 = −5 . Thus the form of the sought polynomial is 2𝑥 2 + 3𝑥 − 5 .
Example 35 There is given the quadratic equation 𝑥 2 + 11𝑥 + 19 = 0 . Let us mark its two distinct real roots with the symbols 𝑥
1 𝑥 2
2
.
, 𝑥
2
. Compute the value of the expression 𝑥 2
1
+
Solution: We will apply the knowledge of Vieta’s formulas
1) 𝑥
1
2) 𝑥
1 𝑥
+ 𝑥
2
2
=
= − 𝑏 𝑎
, 𝑐 𝑎
, thus 𝑥
1
2𝑥
1 𝑥
2
+ 𝑥 2
2
(−11) 2
+ 𝑥
2
= −11 ,
, we get
− 2.19 = 83 . 𝑥 𝑥
1
2
1 𝑥
2
= 19
+ 𝑥 2
2
. When rearranging the expression
= (𝑥
1
+ 𝑥
2
) 2 − 2𝑥
1 𝑥
2
(𝑥
1
+ 𝑥
, and after substitution
2
) 𝑥
2
2
1
= 𝑥 2
1
+ 𝑥 2
2
+
=
It was possible to solve the example also without Vieta’s formulas, computing the roots of the quadratic equation using the standard formula, but considering that the roots are not rational, the process of computation could prove to be rather difficult indeed, as you can check by yourselves.
Example 36 Find all roots of the polynomial 𝑥 4 − 4𝑥 3 + 5𝑥 2 − 2𝑥 − 6 given the knowledge that for the roots of the polynomial there applies the relation 𝑥
1 𝑥
4
.
+ 𝑥
2
= 𝑥
3
+
Solution: According to Vieta’s formulas there applies:
1) 𝑥
1
+ 𝑥
2
+ 𝑥
3
+ 𝑥
4
= 4
2) 𝑥
1 𝑥
2
+ 𝑥
1 𝑥
3
+ 𝑥
1 𝑥
4
+ 𝑥
2 𝑥
3
+ 𝑥
2 𝑥
4
+ 𝑥
3 𝑥
4
= 5
3) 𝑥
1 𝑥
2 𝑥
3
+ 𝑥
1 𝑥
2 𝑥
4
+ 𝑥
1 𝑥
3 𝑥
4
+ 𝑥
2 𝑥
3 𝑥
4
= 2
4) 𝑥
1 𝑥
2 𝑥
3 𝑥
4
= −6
When substituting the relation 𝑥
1
2(𝑥
1
+ 𝑥
2
= 𝑥
3
+ 𝑥
2
) = 4 𝑥
1
+ 𝑥
4
into the first equation, we get
+ 𝑥
2
= 2, 𝑥
3
+ 𝑥
4
= 2
From the second equation we get by rearrangement 𝑥
1 𝑥
2
+ 𝑥
1
(𝑥
3
= 𝑥
1 𝑥
2
+ 𝑥
4
) + 𝑥
2
+ 2.2 + 𝑥
3 𝑥
(𝑥
3
4
+ 𝑥
4
) + 𝑥
3
= 5
𝑥
1 𝑥
2 𝑥
4
+ 𝑥
= 𝑥
3 𝑥
4
1 𝑥
2
+ (𝑥
= 1
3
+ 𝑥
4
). (𝑥
1
+ 𝑥
2
) + 𝑥
3 𝑥
4
Combining this rearranged equation with the fourth equation:
2) 𝑥
1 𝑥
2
+ 𝑥
3 𝑥
4
= 1
4) (𝑥
1 𝑥
2
). (𝑥
3 𝑥
4
) = −6
Now we perform the substitution 𝑥
1 𝑥
2
= 𝑢 𝑥
3 𝑥
4
= 1 − 𝑢 𝑢(1 − 𝑢) = −6 𝑢 2 − 𝑢 − 6 = 0
𝑢
1
= 3, 𝑢
2
= −2
Returning to the original variables, we get 𝑥
1 𝑥
2
= 3, 𝑥
3 𝑥
4
= −2 .
Now we will solve the system of two equations 𝑥
1 substitution 𝑥
1
= 𝑣 𝑥
2
= 2 − 𝑣 𝑥
2
= 3 and 𝑥
, and having solved the equation
1
+ 𝑥
2
= 2 , perform the 𝑣 2 − 2𝑣 + 3 = 0 we come to two roots 𝑥
1
= 1 + 𝑖√2, 𝑥
2
= 1 − 𝑖√2 .
When we use an analogical process for the equations 𝑥
3 𝑥
4 another two roots 𝑥
3
= 1 + √3, 𝑥
4
= 1 − √3 .
= −2 and 𝑥
3
+ 𝑥
4
= 2 , we get
As we are dealing with a polynomial of the fourth degree, which has four roots, we have found all of its roots.
Example 37 Solve individually: Find all roots of the polynomial 𝑥 4 + 2𝑥 3 + 2𝑥
10𝑥 + 25 = 0 , when among the roots of the polynomial there applies the relation 𝑥
1 𝑥
2 𝑥
3 𝑥
4
.
2 +
=
Result: 𝑥
1
= −2 − 𝑖, 𝑥
2
= −2 + 𝑖, 𝑥
3
= 1 + 2𝑖, 𝑥
4
= 1 − 2𝑖
Following are briefly described basic methods of solving some selected types of algebraic equations.
3.5
Linear Equation
The term stands for an equation in the form 𝒙 + 𝒃 = 𝟎, 𝑎 ≠ 0 . This equation has always a solution in the form 𝑥 = − 𝑏
.
𝑎
3.6
Quadratic Equation
It is an equation in the form 𝒂𝒙 𝟐 + 𝒃𝒙 + 𝒄 = 𝟎, 𝑎 ≠ 0 . You have already encountered this type of equation at elementary school where you were solving quadratic equations in a special form. You learned to generically solve quadratic equations at high school. The following text reminds how quadratic equations are being solved.
For the calculation of roots a quadratic equation is rearranged using the method of the complement to a full square.
39
𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 𝑎[𝑥 2 + 𝑝𝑥 + 𝑞] = 𝑎 [(𝑥 + 𝑝
2
)
2
− 𝑝 2
+ 𝑞]
4
Now if we mark the expression 𝑝
2
4 squares we get
− 𝑞 with the letter 𝑑 , from the formula for difference of 𝑎 [(𝑥 + 𝑝
2
)
2
− 𝑑] = 𝑎 (𝑥 + 𝑝
2
− √𝑑) (𝑥 + 𝑝
2
+ √𝑑) = 0
Discussing the last term: if the expression 𝑎 (𝑥 + 𝑝
2
− √𝑑) (𝑥 + 𝑝
2
+ √𝑑) is to be zero, then either one or another of the expressions in parentheses must be zero, thus
(𝑥 + 𝑝
2
− √𝑑) = 0
(𝑥 + 𝑝
2
+ √𝑑) = 0.
As 𝑝 = 𝑏 𝑎
, 𝑑 = known formula 𝑝 2
4
− 𝑞 = 𝑏 2
4𝑎 2
− 𝑐 𝑎
= 𝑏 2 −4𝑎𝑐
4𝑎 2
, for roots of quadratic equation we get the well𝑥
1,2
=
−𝑏 ± √𝑏 2
2𝑎
− 4𝑎𝑐
.
The expression 𝑏 2 − 4𝑎𝑐 = 𝐷 is called a discriminant and according to it being positive or negative the decision can be promptly made whether the equation has real or complex roots.
Let us just remind that in some simple cases with integer coefficients Vieta’s formulas are more suitable to be used to determine roots faster. For instance, for the equation 𝑥 2 + 7𝑥 + 10 = 0 there applies 𝑥
1
. 𝑥
2
= 10, 𝑥
1
+ 𝑥
2
= −7 , so the roots are 𝑥
1
=
−2, 𝑥
2
= −5 .
3.7
Cubic Equation
The cubic equation is an equation in the form 𝒂
𝟑 𝒙 𝟑 + 𝒂
𝟐 𝒙 𝟐 + 𝒂
𝟏 𝒙 + 𝒂
𝟎
= 𝟎, 𝑎
3
≠ 0 . The first step in its solution procedure consists of dividing the equation by the number 𝑎
3
and after re-labelling of the coefficients we get the following equation 𝑥 3 + 𝑎 𝑥 2 + 𝑏 𝑥 + 𝑐 = 0
We proceed by using substitution 𝑥 = 𝑡
a
3
. After rearrangement, we get the equation in the so-called reduced form 𝑡 3 + 𝑝 𝑡 + 𝑞 = 0
Let us mark 𝜀 = cos
2𝜋
3
+ 𝑖 sin
2𝜋
3
=
1
2
+ 𝑖
√3
2
. Furthermore, let 𝐾 = √ 𝑞 2
4
+ 𝑝 3
27 denote one
(fixedly chosen) of the two values of the second square root. Next, let 𝑢 denote any (fixed)
3 of the three cube roots √− 𝑞
2
3
+ 𝐾 , and lastly, let v denote one of cube roots √− 𝑞
2
− 𝐾 for
which there applies the relationship 3𝑢𝑣 =
𝑝 . Then roots of the equation 𝑡 3 + 𝑝 𝑡 +
𝑞 = 0 are 𝑡
1
= 𝑢 + 𝑣, 𝑡
2
=
. 𝑢 +
2 . 𝑣, 𝑡
3
=
2 . 𝑢 +
. 𝑣 .
The previous procedure is called Cardan’s method .
A cubic equation can have both real and complex roots. Their nature can be determined from the value of a discriminant which has the form 𝐷 =
4 𝑝 3
27 𝑞 2
.
When being limited to a cubic equation 𝑡 3 + 𝑝 𝑡 + 𝑞 = 0 with real non-zero coefficients (otherwise, when 𝑝 = 0 or 𝑞 = 0 , the solution is trivial). a) 𝐷 = 0 : 𝑡
1
3
= −2√ 𝑞
2
, 𝑡
2
= 𝑡
3
3
= √ 𝑞
2
.
b) 𝐷 0 : one real root and two imaginary complex conjugate roots, specified by the
Cardan formula . c) 𝐷 0 : three real roots, however in this case it is not possible to find them using
Cardan’s method (so-called “cassus irreducibilis”). The goniometrical solution procedure needs to be applied:
Firstly, we compute the value of the angle 𝜑 (one fixedly chosen) from the equation cos 𝜑 =
− 𝑞
2
√(
|𝑝|
3
)
3
. The sought roots are then specified by the following relationships: 𝑡
1
= 2√(
|𝑝|
3
) cos 𝜑
3
, 𝑡
2
= −2√(
|𝑝|
3
) cos ( 𝜑
3
− 𝜋
) , 𝑡
3
3
= −2√(
|𝑝|
3
) cos ( 𝜑
3
+ 𝜋
3
).
It is obvious that both Cardan’s method and the lastly-mentioned relationships are very laborious and cumbersome for practical calculations. Moreover, it is rather difficult to rearrange the acquired results into a usable form. For our purposes it is therefore more suitable to find the solution of a cubic equation in the way described in Chapter 3.3 which allows determining roots of an equation with integer coefficients.
3.8
Fourth Degree Equation
It is an equation in the form 𝒂
𝟒 𝒙 𝟒 + 𝒂
𝟑 𝒙 𝟑 + 𝒂
𝟐 𝒙 𝟐 + 𝒂
𝟏 𝒙 + 𝒂
𝟎
= 𝟎, 𝑎
4
≠ 0 . This equation is of the last degree that is generally solvable. Nevertheless, the general solution procedure for a 4 th
degree equation is toilsome and tedious to that extent that it is hardly ever used in practice, and thus it will not be presented in this text. For equations in the school practice some simpler ways of solving are sufficient (transformation of an biquadratic equation in the form 𝑎𝑥 4 + 𝑏𝑥 2 + 𝑐 = 0 by the substitution to a quadratic equation, procedures for solving binomial and reciprocal equations).
41
3.9
Higher Degree Equations
For equations of the 5 th
and higher degree there does not exist any generic solving algorithm anymore 8 . According to the theory created by French mathematician Galois there exists for every n
5 an algebraic equation of the degree n that is not solvable by algebraic methods. Even though many equations of higher degrees can be solved, they must be in some special form. Either it is possible to “guess” the roots one at a time, or the equation is in the form that allows it to be solved as a binomial or reciprocal equation. Both these methods are being covered at high school and will be reminded in the following chapters.
Another method how to obtain roots of an algebraic equation, is to use numerical methods. Advances in information and computation technology contribute to the widespread use of these methods. The general procedure consists of the following three steps: assessing bounds of polynomial roots (according to Theorem 16), their separation, and approximation. Searching for roots of polynomials with real coefficients by applying these numerical methods relies on the observation that a polynomial is a continuous function in the domain of real numbers. Among contemporarily used methods there belong the Bisection method, the Fixed-point iteration, Newton’s method and Halley’s method.
These problems will not be dealt with in the following text.
3.10
Selected Types of Algebraic Equations
Binomial Equation
A binomial equation stands for an equation of the type 𝒙 𝒏 = 𝒂, 𝑛 ∈ 𝑁, 𝑎 ∈ 𝐶 . According to the Fundamental theorem of algebra there exist 𝑛 solutions of this equation. These solutions are in the form 𝑛 𝑥 = √|𝑎| [cos 𝛼 + 2𝑘𝜋 𝑛
+ 𝑖 sin 𝛼 + 2𝑘𝜋
] 𝑛 and they form vertices of a regular planar 𝑛 − gon. This 𝑛 − gon lies in the Gauss plane on 𝑛 a circle that has its centre in the origin of coordinate axes and its radius 𝑟 = √|𝑎| . When searching for these roots, we proceed as follows:
1.
Write down the sought root in the form 𝑥 𝑘
= 𝑟(cos 𝜑 + 𝑖 sin 𝜑) . 𝑛
2.
Find the radius 𝑟 = √|𝑎| ∈ 𝑅 .
3.
Find the angle 𝜑 for which there applies 𝑛𝜑 = 𝛼 + 2𝑘𝜋, 𝑘 ∈ 𝑍 , thus 𝜑 = 𝛼 𝑛
+
2𝑘𝜋
, 𝑛 where 𝑘 = 0, 1, … , 𝑛 − 1 . The angle 𝛼 can be then found from the position of the
8 No universal formula can be found that would contain only coefficients, a finite number of arithmetical operations and a finite number of n th root operations.
number 𝑎 in the Gauss plane. It is the angle that the position vector of the number 𝑎 contains with the real axis.
Example 38 Solve the equation 𝑥 3 − 2 = 0 in 𝐶 . Depict its roots in the Gauss plane of coordinates.
Solution: First we rewrite the equation in the form 𝑥 3 = 2 .
1) We will write the sought roots in the form 𝑥 𝑘
= 𝑟(cos 𝜑 + 𝑖 sin 𝜑) .
3
2) 𝑟 = √2 .
3) 𝛼 = 0 or 𝛼 = 2𝜋 , since number 2 lies in the Gauss plane on the positive part of the real axis. Then 𝜑 =
0
3
+
2𝑘𝜋
3
=
2𝑘𝜋
, 𝑘 = 0, 1, 2 .
3
The solution are three roots in the following form: 𝑥
0
3
= √2
3
(cos 0 + 𝑖 sin 0) = √2 𝑥
1 𝑥
2
3
= √2 (cos
2𝜋
3
+ 𝑖 sin
2𝜋
)
3
3
= √2 (cos
4𝜋
3
+ 𝑖 sin
4𝜋
)
3
We can see that the first root is a real one while the other two are complex. They form vertices of a regular triangle in the Gauss plane.
Img
3
2
Re
Example 39 Solve the equation 𝑥 5 = 2 − 2𝑖 .
Solution: When finding a radius for the number 𝑧 = 𝑎 ± 𝑏𝑖 , there applies |𝑧| = √𝑎 2 + 𝑏 2 , 𝑛 𝑟 = √√𝑎 2 + 𝑏 2
. Therefore, in our example there applies |𝑧| = √4 + 4 = √8 = 2√2, 𝑟 =
5
√2√2
.
The number 2 − 2𝑖 lies in the fourth quadrant, in which case the angle 𝛼 can be determined as 𝛼 = 2𝜋 − 𝛼
0
, where 𝛼
0
can be computed from the following right triangle:
43
Img a
0
2
2
Re 𝛼
0
= 𝜋
4
, 𝛼 = 2𝜋 − 𝑥 𝑘 𝜋
4
=
7𝜋
. There are five roots for which there applies
4
5
= √2√2 (cos
7𝜋
4 + 2𝑘𝜋
5
+ 𝑖 sin
7𝜋
4 + 2𝑘𝜋
) , 𝑘 = 0,1,2,3,4.
5
Example 40 Solve the following binomial equations individually:
1) 𝑥 3 − 27 = 0
2) 𝑥 3 = 4√3 + 4𝑖
3) 𝑥 4 + 16 = 0
Results:
1) 𝑥
0
= 3, 𝑥
1
= 3 (cos
2𝜋
3
+ 𝑖 sin
2𝜋
) , 𝑥
3
2
= 3 (cos
4𝜋
3
+ 𝑖 sin
4𝜋
)
3
2) 𝑥
0 𝜋
= 2 (cos
18 𝑖 sin
25𝜋
)
18
+ 𝑖 sin 𝜋
18
) , 𝑥
1
= 2 (cos
13𝜋
18
+ 𝑖 sin
13𝜋
18
) , 𝑥
2
= 2 (cos
25𝜋
18
+
3) 𝑥
0
= 2 (cos 𝜋
4 𝑖 sin
5𝜋
) , 𝑥
4
3
+ 𝑖 sin 𝜋
4
) , 𝑥
1
= 2 (cos
7𝜋
4
= 2 (cos
+ 𝑖 sin
7𝜋
)
4
3𝜋
4
+ 𝑖 sin
3𝜋
4
) , 𝑥
2
= 2 (cos
5𝜋
4
+
Reciprocal Equation
A reciprocal equation is an equation of the form 𝑎 𝑛 𝑥 𝑛 + 𝑎 𝑛−1 𝑥 𝑛−1 … 𝑎
1 𝑥 + 𝑎
0
= 0, where 𝑎 𝑛−𝑖
= 𝑎 𝑖
for 𝑖 = 1,2, … , 𝑛 , or 𝑎 𝑛−𝑖
= −𝑎 𝑖
for 𝑖 = 1,2, … , 𝑛 .
It is called reciprocal on the account of the observation that when 𝑐 is one of its roots then 𝑐
1
(the reciprocal value) is also its root.
The reciprocal equation of the even degree for an even 𝑛 can be distinguished from the one of the odd degree for an odd 𝑛 , just as the equation of the 1 st
class when 𝑎 𝑛−𝑖
𝑎 𝑖
, from the equation of 2 nd class when 𝑎 𝑛−𝑖
= −𝑎 𝑖
.
=
The procedure for solving a reciprocal equation is as follows:
Each reciprocal equation of the 2 nd class has the root 𝑐 = 1 . Upon dividing it by the binomial (𝑥 − 1) , we get a reciprocal equation of the 1 st class.
Each reciprocal equation of the 1 st class of an odd degree has the root 𝑐 = −1 .
When divided by the binomial (𝑥 + 1) , the result is a 1 st
class reciprocal equation of an even degree.
A 1 st class reciprocal equation of an even degree can be transformed into an algebraic equation of the half degree by dividing it by the expression 𝑥 𝑛
⁄
2 , and performing the substitution 𝑦 = 𝑥 +
1 𝑥
.
Example 41 Find roots of the equation 2𝑥 7 + 5𝑥 6 − 3𝑥 4 + 3𝑥 3 − 5𝑥 − 2 = 0 .
Solution: It is a 2 nd class reciprocal equation of an odd degree. We will proceed as follows:
1)
We employ the Horner’s method to verify the roots −1; 1 . Number 1 will be certainly one of the roots of the equation.
2 5 0 -3 3 0 -5 -2
1 2 7 7 4 7 7 2 0
-1 2
-1 2
5
3
2 2
-1 3
5
2
2
0
0
We can see that 1 is a simple root and −1 is a double root of the equation. If these numbers were to be tested further on, we would not get zero in the last column of the Horner schema anymore, as you can check by yourselves. Hence we can rewrite the equation into the form
(𝑥 − 1)(𝑥 + 1) 2 (2𝑥 4 + 3𝑥 3 − 𝑥 2 + 3𝑥 + 2) = 0 and we can notice the expression in the third parentheses form a reciprocal equation again, this time of the 1 st
class and of an even degree.
2) We rearrange the equation 2𝑥 4 + 3𝑥 3 − 𝑥 2 + 3𝑥 + 2 = 0 by dividing it by 𝑥 2
, writing the terms with the same coefficients together:
3) Before we perform the substitution 𝑥 +
1 𝑥 express 𝑥 2 + 𝑥
1
2
:
= 𝑡 , there needs to be determined how to
(𝑥 +
1 𝑥
)
2
2 (𝑥 2 +
1 𝑥 2
) + 3 (𝑥 +
1 𝑥
) − 1 = 0
= 𝑡 2 𝑥 2 + 2 +
1 𝑥 2
= 𝑡 2 𝑥 2 +
1 𝑥 2
= 𝑡 2 − 2
45
When substituting the obtained expressions into the equation 2 (𝑥 2 +
1 𝑥 2
) +
3 (𝑥 +
1 𝑥
) − 1 = 0 , we get
2(𝑡 2 − 2) + 3𝑡 − 1 = 0 2𝑡 2 + 3𝑡 − 5 = 0 𝑡
1
= −
5
2
, 𝑡
2
= 1.
Let us return to the original variable. When = −
5
2
, then 𝑥 +
1 𝑥
= − this equation with 2𝑥 (assuming that 𝑥 ≠ 0) and get the equation 2𝑥
5
2
2
, we multiply
+ 5𝑥 + 2 =
0 , whose solutions are the numbers 𝑥
1
= −2, 𝑥
2
= −
1
2
.
When 𝑡 = 1 , then 𝑥 +
1 𝑥
= 1 and with a rearrangement similar to the previous step we get the equation 𝑥 2 − 1 + 𝑥 = 0 , whose solutions are 𝑥
3,4
=
1±𝑖√3
.
2
The complete solution is represented by the set
{−2, −1, −
1
2
, 1,
1 − 𝑖√3
,
2
1 + 𝑖√3
}.
2
Example 42 Solve individually the reciprocal equation 3𝑥 8 𝑥 3 − 4𝑥 2 + 𝑥 + 3 = 0 .
+ 𝑥 7 − 4𝑥 6 − 𝑥 5 + 2𝑥 4 −
Result: 𝑥 ∈ {−1,1,
1 ± 𝑖√3
2
,
−2 ± 𝑖√5
}.
3
Exercises for Check
1.
Define algebraic equation and its roots.
2.
State and prove Bézout's theorem.
3.
Is the field of real numbers algebraically closed? Explain.
4.
Write down Vieta’s formulas for a polynomial of the 5 th
degree.
8𝑥 7 + 4𝑥 6 − 2𝑥 5 + 3𝑥 4 − 6𝑥 3 − 2𝑥 2
1
5.
Find out multiplicity of the roots 𝑐
1
= , 𝑐
2
= −2 for the polynomial 𝑓(𝑥) =
2
+ 4𝑥 − 1 .
6.
Apply a derivative to find out multiple roots of the polynomial 𝑓(𝑥) = 4𝑥 5
20𝑥 4 + 25𝑥 3 − 10𝑥 2 − 20𝑥 + 8 ∈ 𝑄[𝑥] .
+
7.
Given the quadratic equation 𝑥 the value of the expression (𝑐
1
2 − 5𝑥 + 3 = 0 , where 𝑐
1
, 𝑐
2
are its roots, determine
+ 𝑐
2
) 2 − 3𝑐
1 𝑐
2
.
8.
On the example of the polynomial 𝑓(𝑥) = 4𝑥 4 − 4𝑥 3 + 13𝑥 2 − 12𝑥 + 2 explain how rational roots can be found for a polynomial with integer coefficients. Find all rational roots.
9.
Solve the inequality 4𝑥 4 − 4𝑥 3 + 13𝑥 2 − 12𝑥 + 2 < 0 on the set of real numbers.
10.
On the set of complex numbers solve the equality 5𝑥 4
5 = 0 .
− 26𝑥 3 + 10𝑥 2 − 26𝑥 +
Results:
5.
𝑐
1
is a triple root, 𝑐
2
is not a root.
6.
𝑐
1
=
1
2
is a triple root, 𝑐
2
= −2 is a double root.
7.
16
8.
𝑐 =
1
2
.
9.
No solution.
10.
𝑥 ∈ {𝑖; −𝑖; 5;
1
5
} .
47
4 Multivariate Polynomials
Chapter Objectives
Having studied this chapter you will be able to:
define multivariate polynomial
determine the degree of a multivariate polynomial
lexicographically order a multivariate polynomial
define symmetric and homogeneous polynomial
define elementary symmetric polynomials.
Time to Study
Circa 2 hours
Study Guide
Definition 18
A polynomial of 𝑛 variables 𝑥
1
, … , 𝑥 𝑛
stands for a sum of a finite number of terms 𝑓 = 𝐴
1
+ 𝐴
2
+ ⋯ + 𝐴 𝑟
, where 𝐴 𝑖
= 𝑎 𝑖 𝑥 𝑘
1
1 𝑥 𝑘
2
2
… 𝑥 𝑘 𝑛 𝑛 , 𝑎 𝑖
∈ 𝑇 is the coefficient of the term 𝐴 𝑖
, and 𝑘
1
, … , 𝑘 𝑛
are non-negative integers. A polynomial of 𝑛 variables over a field 𝑇 is labelled with the symbol 𝑓 ∈ 𝑇[𝑥
1
, . . . , 𝑥 𝑛
] . The degree of a term 𝑥 𝑘
1
1 𝑥 𝑘
2
2 𝑘
2
+ ⋯ + 𝑘 𝑛
.
… 𝑥 𝑘 𝑛 𝑛
stands for the number 𝑘
1
+
Example 43 The degree of the term 5𝑥 3 𝑦 5
is 8.
Note:
In the following text the notion of term height will also come handy. Let 𝑓(𝑥
1
, … , 𝑥 𝑛
) be a polynomial of 𝑛 variables and 𝑎 𝑖 𝑥 𝑘
1
1 𝑥 𝑘
2
2
… 𝑥 𝑘 𝑛 𝑛
be its arbitrary term. Then the sorted 𝑛 -tuple of non-negative integers (𝑘
1
, … , 𝑘 𝑛
) is called a height of the term.
Definition 19
The degree of a non-zero polynomial 𝑓 ∈ 𝑇[𝑥
1
, . . . , 𝑥 𝑛
] is equal to the highest of the degrees of its terms with non-zero coefficients. A polynomial whose all terms have the same degree 𝑠 is called a homogenous polynomial (of the degree 𝑠 ).
Example 44 The polynomial 𝑥 + 𝑦 is a homogenous polynomial of the degree 1, the polynomial 𝑥 2 − 2𝑥𝑦 is a homogenous polynomial of the degree 2, while the polynomial 𝑥 2 𝑦 + 3𝑦 3
is a homogenous polynomial of the degree 3.
Definition 20
Terms 𝐴 = 𝑎𝑥 𝑘
1
1 and only if 𝑘
1
= 𝑙
1
, 𝑘
2
= 𝑙
2
, … , 𝑘 𝑥 𝑛 𝑘
2
2
… 𝑥
= 𝑙 𝑛
. 𝑘 𝑛 𝑛
, 𝐵 = 𝑏𝑥 𝑙
1
1 𝑥 𝑙
2
2 … 𝑥 𝑙 𝑛 𝑛
are called similar terms if
Note:
When possible we sum up all similar terms outright. There applies 𝐴 + 𝐵 = (𝑎 + 𝑏) 𝑥 𝑘
1
1 𝑥 𝑘
2
2
… 𝑥 𝑘 𝑛 𝑛
.
Definition 21
Considering operations +, ∙ , the set 𝑇[𝑥
1
, … , 𝑥 𝑛
] is an integral domain
(operations are defined generally, (𝑇[𝑥
1
, … , 𝑥 𝑛
], +) is a group and (𝑇[𝑥
1
, … , 𝑥 𝑛
],∙) is a commutative semigroup).
Theorem 26
Every polynomial 𝑓 ∈ 𝑇[𝑥
1
, . . . , 𝑥 𝑛
] can be written in the form of a sum of homogenous polynomials, each of them of a different degree. This expression is unambiguous (except for their order).
Note:
In case of univariate polynomials the terms can be ordered in the ascending or descending manner with respect to the variable exponent. Therefore some other procedure must be defined.
Theorem 27
Let 𝐴 = 𝑥 𝑘
1
1 𝑥 𝑘
2
2
… 𝑥 𝑘 𝑛 𝑛
, 𝐵 = 𝑥 𝑙
1
1 𝑥 𝑙
2
2 … 𝑥 𝑙 𝑛 𝑛
be two terms of 𝑛 variables. Let us state that the term 𝐴 comes before the term 𝐵 if there exists an index 𝑖, 1 𝑖 𝑛 , for which there applies 𝑘
1
= 𝑙
1
, . . . , 𝑘 𝑖
1
= 𝑙 𝑖
1
, 𝑘 𝑖
𝑙 𝑖
.
The condition that the term 𝐴 comes before the term 𝐵 , or 𝐴 = 𝐵 , is denoted as 𝐴 ≫ 𝐵 .
Theorem 28
The relation ≫ is a relation of a linear ordering over the set of all terms containing 𝑛 variables (the relation is irreflexive, antisymmetric and transitive).
Definition 22
The relation ≫ is called a relation of lexicographical order of terms of 𝑛 variables. When the terms of a polynomial 𝑓 𝑻[𝑥
1
, . . . , 𝑥 𝑛
] are ordered by this relation, we can state that the terms of the polynomial f are lexicographically ordered. The term that comes before all other terms of the polynomial 𝑓 is called a leading term of the polynomial 𝑓 .
Example 45 Determine a lexicographical order of the polynomial 𝑓(𝑥, 𝑦, 𝑧) =
(𝑥 − 2𝑦 2 )(𝑥 2 𝑧 + 𝑧) and its leading term.
49
Solution: Upon factorization we get 𝑓(𝑥, 𝑦, 𝑧) = 𝑥 3 𝑧 + 𝑥𝑧 − 2𝑥 2 𝑦 2 𝑧 − 2𝑦 2 𝑧 = 𝑥 3 𝑧 −
2𝑥 2 𝑦 2 𝑧 + 𝑥𝑧 − 2𝑦 2 𝑧 , which is the lexicographical order of the polynomial. We can verify it using the term heights: (3,0,1), (2,2,1), (1,0,1), (0,2,1). The leading term of the polynomial is 𝑥 3 𝑧 .
Theorem 29
Let 𝑓, 𝑔 ∈ 𝑇[𝑥
1
, . . . , 𝑥 𝑛
] be two arbitrary real non-zero polynomials of n variables. Then the product of the leading terms of the polynomials 𝑓 and 𝑔 is the leading term of the product 𝑓 . 𝑔 .
Definition 23
A polynomial 𝑓(𝑥
1
, . . . , 𝑥 𝑛
) ∈ 𝑇[𝑥
1
, . . . , 𝑥 𝑛
] is called to be symmetric , when it does not change after any permutation of variables, i.e. for any arbitrary permutation ( a
1
, . . . , a 𝑛
) of indices 1, 2, . . . , 𝑛 there applies: 𝑓(𝑥 𝛼
1
, … , 𝑥 𝛼 𝑛
) = 𝑓(𝑥
1
, . . . , 𝑥 𝑛
)
The set of all symmetric polynomials of 𝑛 variables over the field of real numbers is denoted as 𝑅 𝑠
[𝑥
1
, . . . , 𝑥 𝑛
] .
Note:
A symmetric polynomial of two variables fulfils the condition 𝑓(𝑥, 𝑦) = 𝑓(𝑦, 𝑥) , whereas a symmetric polynomial of three variables fulfils the condition 𝑓(𝑥, 𝑦, 𝑧) = 𝑓(𝑥, 𝑧, 𝑦) = 𝑓(𝑦, 𝑥, 𝑧) = 𝑓(𝑦, 𝑧, 𝑥) = 𝑓(𝑧, 𝑥, 𝑦) = 𝑓(𝑧, 𝑦, 𝑥) . The polynomial 𝑥 + 𝑦 is a symmetric polynomial of the degree 1, the polynomial 𝑥 2 + 𝑥 + 𝑦 + 𝑦 2
is a symmetric polynomial of the degree 2.
Theorem 30
Let 𝐴 = 𝑥 𝑘
1
1 𝑓(𝑥
1
, . . . , 𝑥 𝑛 𝑥 𝑘
2
2
… 𝑥 𝑘 𝑛 𝑛
be the leading term of a symmetric polynomial
) .
Then there applies 𝑘
1
𝑘
2
. . . 𝑘 𝑛
.
Example 46 Find out a lexicographical order of the symmetric polynomial 𝑓(𝑥, 𝑦, 𝑧) =
(𝑥 + 𝑦)(𝑥 + 𝑧)(𝑦 + 𝑧) and its leading term.
Solution: We proceed by factorization, getting 𝑓(𝑥, 𝑦, 𝑧) = 𝑥 𝑥𝑧 2
2 𝑦 + 𝑥 2 𝑧 + 𝑥𝑦 2 + 2𝑥𝑦𝑧 +
+ 𝑦 2 𝑧 + 𝑦𝑧 2
, which is the lexicographic order of the polynomial. The leading term of the polynomial is 𝑥 2 𝑦 .
Theorem 31
Let 𝐴 = 𝑥 𝑘
1
1 𝑥 𝑘
2
2
… 𝑥 𝑘 𝑛 𝑛
be a term consisting of 𝑛 variables. Then there exists only a finite number of leading terms of symmetrical polynomials of 𝑛 variables that come after the term 𝐴 .
Example 47 Given the polynomial 𝑓(𝑥, 𝑦, 𝑧) = 𝑥 3 𝑦 + 2𝑥 2 𝑦 2 𝑥 2 𝑦 2 𝑧 2 + 𝑥𝑦 3 + 𝑥𝑧 3 + 2𝑦 2 𝑧 2 + 𝑦 3 𝑧 + 𝑦𝑧 3
+ 𝑥
, determine all its leading terms.
3 𝑧 + 2𝑥 2 𝑧 2 −
Solution: The heights of the terms of the polynomial are
(3,1,0), (2,2,0), (3,0,1), (2,0,2), (2,2,2), (1,3,0), (1,0,3), (0,2,2), (0,3,1), (0,1,3) respectively. When we order these heights by the relation > we get (3,1,0) > (3,0,1) >
(2,2,2) > (2,2,0) > (2,0,2) > (1,3,0) > (1,0,3) > (0,3,1) > (0,2,2) > (0,1,3) . The leading term of 𝑓 is thus the term 𝑥 3 𝑦 . However, the inequalities 𝑘
1
𝑘
2
. . . 𝑘 𝑛
from
Theorem 30 apply also for heights (2,2,2) and (2,2,0) , and hence there are other leading terms of the polynomial 𝑥 2 𝑦 2 𝑧 2
and 2𝑥 2 𝑦 2
.
We could conclude that the given polynomial is the sum of three elementary symmetric polynomials, whose leading terms are 𝑥 3 𝑦 , 𝑥 2 𝑦 2 𝑧 2
, 2𝑥 2 𝑦 2
. The procedure for the decomposition of symmetric polynomials into a sum of elementary symmetric polynomials will be presented in the following chapter.
4.1
Elementary Symmetric Polynomials
Definition 24
Polynomials from 𝑅[𝑥
1
, . . . , 𝑥 𝑛
] in the form: 𝜎
1
(𝑥
1
, … , 𝑥 𝑛
) = 𝑥
1
+ 𝑥
2
+ ⋯ + 𝑥 𝑛 𝜎
2
(𝑥
1
, … , 𝑥 𝑛
) = 𝑥
1 𝑥
2
+ 𝑥
1 𝑥
3
+ ⋯ + 𝑥 𝑛−1 𝑥 𝑛
⋮ 𝜎 𝑛−1
(𝑥
1
, … , 𝑥 𝑛
) = 𝑥
1 𝑥
2
… 𝑥 𝑛−1
+ 𝑥
1 𝑥
2
… 𝑥 𝑛−2 𝑥 𝑛
+ ⋯ + 𝑥
2 𝑥
3
… 𝑥 𝑛−1 𝑥 𝑛 𝜎 𝑛
(𝑥
1
, … , 𝑥 𝑛
) = 𝑥
1 𝑥
2 𝑥
3
… 𝑥 𝑛−1 𝑥 𝑛 are called elementary symmetric polynomials of 𝑛 variables.
Note:
The elementary polynomial
1 is therefore the sum of all variables 𝑥
1 polynomial
2 and so forth. Finally, the polynomial
𝑛 is the product of all variables
, . . . , 𝑥 𝑛
, the is the sum of products of all pairs, the polynomial
3 is the sum of triples, 𝑥
1
, . . . , 𝑥 𝑛
.
Considering the case 𝑛 = 3 and the familiar variables notation 𝑥, 𝑦, 𝑧 then:
1
= 𝑥 + 𝑦 + 𝑧
2
= 𝑥𝑦 + 𝑥𝑧 + 𝑦𝑧
3
= 𝑥𝑦𝑧
Symmetric polynomials appear in Vieta’s formulas and in addition they can be used to solve some systems of equations which it is not possible to solve by methods known from elementary and high schools, as will be demonstrated.
The following is the fundamental theorem of symmetric polynomials which deals with the existence and unambiguity of an arbitrary symmetric polynomial expression, using elementary symmetric polynomials.
Theorem 32
Every symmetric polynomial 𝑓(𝑥
1
, . . . , 𝑥 𝑛
) ∈ 𝑅 expressed as a polynomial of 𝑛 variables
1
,
2
, . . ., 𝑛 over 𝑅 , i.e.
𝑆
[𝑥
1
, . . . , 𝑥 𝑛
] can be
51
𝑓(𝑥
1
, . . . , 𝑥 𝑛
) = 𝜑(
1
,
2
, . . . ,
𝑛
) and this expression is unambiguous.
Example 48 Concerning power sums 𝑠 𝑘 theorem
(𝑎 + 𝑏) 𝑛 = ( 𝑛
0
) 𝑎 𝑛
= 𝑥 𝑘
1
+ 𝑥 𝑘
2
upon application of the binomial
+ ( 𝑛
1
) 𝑎 𝑛−1 𝑛 𝑏 + ⋯ + ( 𝑛 − 1
) 𝑎𝑏 𝑛−1 + ( 𝑛 𝑛
) 𝑏 𝑛 we get 𝑠 𝑠 𝑠
1
2
3
= 𝑥
1
= 𝑥 2
1
= 𝑥 3
1
+ 𝑥
2
+ 𝑥 2
2
+ 𝑥 3
2
= 𝜎
1
,
= (𝑥
= (𝑥
1
1
+ 𝑥
2
) 2
+ 𝑥
2
) 3
− 2𝑥
1 𝑥
2
− 3𝑥
1 𝑥
2
= 𝜎
(𝑥
1
1
2 − 2𝜎
2
,
+ 𝑥
2
) = 𝜎
1
3 − 3𝜎
1 𝜎
2
, and so on. The following algorithm can be generically applied to the transformation of a symmetric polynomial into the sum of elementary symmetric polynomials.
Algorithm for Determining All Leading Terms of Symmetric Polynomials of the
Degree
( 𝑘
1
𝑘
2 𝑘
1
+ 𝑘
. . .
2
+ ⋯ + 𝑘
𝑘 𝑛
): 𝑛
, That Come After The Leading Term 𝑎𝑥 𝑘
1
1 𝑥 𝑘
2
2
… 𝑥 𝑛 𝑘 𝑛
1.
Write down the height of the leading term of the polynomial, i.e. the sequence of exponents 𝑘
1
, … , 𝑘 𝑛
.
2.
Find in this sequence the last number greater than 1. Decrement this number by 1, leave the fore-going numbers as they are and select the numbers that go “after” as small as possible for it to apply 𝑙
1
𝑙
2
. . .
𝑙 𝑠
and 𝑙
1
+ 𝑙
2
+ ⋯ + 𝑙 𝑠
= 𝑘
1
+ 𝑘
2
+ ⋯ + 𝑘 𝑛
.
3.
Repeat the procedure until obtaining the sequence 1, 1, …, 1.
Then the obtained sequence of the term heights describes all leading terms of all symmetric polynomials of the given order that come after the given leading term.
When seeking the symmetric polynomial 𝑓(𝑥
1
, . . . , 𝑥 𝑛
) in the form 𝜑(
1
,
2
, . . . ,
𝑛
) , to the first leading term 𝑎𝑥 𝑘
1
1 corresponding term 𝜑
1
= 𝜎
1 𝑘
1
−𝑘
2 ∙ 𝜎
2 𝑘
2
−𝑘
3 ∙ … ∙ 𝜎 𝑘 𝑛 𝑛 𝑥 𝑘
2
2
… 𝑥 𝑘 𝑛 𝑛
we will create the
. The similar process is then used for other leading terms.
The polynomial 𝑓 can be written in the form 𝑓 = 𝑎𝜑
1
+ 𝐴𝜑
2
+ 𝐵𝜑
3
+ ⋯ + 𝑊𝜑 𝑡
, where 𝑎 is the coefficient of the first leading term and the constants 𝐴, 𝐵, … are undetermined coefficients that can be found by substitution of concrete values for 𝑥
1
, … , 𝑥 𝑛
.
Example 49 Transform the polynomial 𝑓 = 𝑥 2
1 𝑥 2
2 form 𝜑(
1
,
2
, . . . , 𝑛
) .
+ 𝑥 2
1 𝑥 2
3
+ 𝑥 2
1 𝑥 2
4
+ 𝑥 2
2 𝑥 2
3
+ ⋯ into the
Solution: Firstly, we set up a sequence of exponents and create corresponding leading terms of a general symmetric polynomial whose leading term is 𝑥 2
1 𝑥 2
2
. Such polynomial
should be of the form: 𝑥 2
1 𝑥 2
2
+ 𝑥 2
1 𝑥 2
3
+ 𝑥 2
1 𝑥 2
4
+ ⋯ + 𝑥 2
1 𝑥
2 𝑥
3
For each leading term we then create a corresponding term 𝜑 𝑘
:
+ 𝑥 2
1 𝑥
2 𝑥
4
+ ⋯ + 𝑥
1 𝑥
2 𝑥
3 𝑥
4
.
2 2 𝜑
1
= 𝜎
1
2−2 𝜎 2
2
= 𝜎
2
2
2
1
1
1
1
1 1 𝜑
2
= 𝜎 2−1
1 𝜎 1−1
2 𝜎 1
3
= 𝜎
1 𝜎
3 𝜑
3
= 𝜎 1−1
1 𝜎 1−1
2 𝜎 1−1
3 𝜎 1
4
= 𝜎
4
Therefore, for the polynomial 𝑓 there applies 𝑓 = 𝑎𝜎 2
2
+ 𝐴𝜎
1 𝜎
3
+ 𝐵𝜎
4
, where 𝑎 = 1 and the coefficients 𝐴, 𝐵 will be computed by two choices of substitution for 𝑥
1
, 𝑥
2
, 𝑥
3
, 𝑥
4
.
Case 1: (𝑥
1
, 𝑥
2
, 𝑥
3
, 𝑥
4
) = (1,1,1,0) , 𝑓 = 𝑥 2
1 𝑥 2
2
+ ⋯ = 1 + 1 + 1 = 3 , 𝜎
1
= 𝑥
1
+ 𝑥
2 𝑥 𝑥
1
3 𝑥
= 1 + 1 + 1 = 3 , 𝜎
2
2 𝑥
3
= 1 , 𝜎
4
= 𝑥
1 𝑥
2
+ 𝑥
1 𝑥
3
+ 𝑥
2 𝑥
3
= 1 + 1 + 1 = 3 , 𝜎
3
= 0 . After substitution into the equation 𝑓 = 𝜎 2
2
+ 𝐴𝜎
1 𝜎
3
𝐵𝜎
4
we get
3 = 9 + 3𝐴
𝐴 = −2.
+
=
+
Case 2: (𝑥
1
, 𝑥
2
, 𝑥
3
, 𝑥
4
) = (1,1,1,1) , you can perform the substitution by yourselves to check that 𝑓 = 6, 𝜎
1
= 4, 𝜎
2
= 6, 𝜎
3
= 4, 𝜎
4
= 1 . Hence there applies
6 = 36 + 16𝐴 + 𝐵
𝐵 = 2.
This leads us to the sought form of the polynomial 𝑓 = 𝜎 2
2
− 2𝜎
1 𝜎
3
+ 2𝜎
4
.
Example 50 Transform the symmetric polynomial 𝑓 = (𝑥 − 𝑦) 2 (𝑥 − 𝑧) 2 (𝑦 − 𝑧) 2
into the variables
1
,
2
, . . . , 𝑛
.
Solution: Upon factorization we would get 𝑓 = 𝑥 4 𝑦 2 − 2𝑥 4 𝑦𝑧 + 𝑥 4 𝑧 2 − 2𝑥 3 𝑦 3 + ⋯
Thus we can write down the sequence of exponents of leading terms, striking out the rows where the sequence has more than 3 terms, since the polynomial 𝑓 has only 3 variables.
4
4
3
2
1 1
3 𝜑
1
= 𝜎 2
1 𝜎 2
2 𝜑
2
= 𝜎
1
3 𝜎
3 𝜑
3
= 𝜎 3
2
3
3
2
2
2 1
1 1 1
2 2
2 1 1 𝜑
4
= 𝜎
1 𝜎
2 𝜎
3
𝜑
6
= 𝜎 2
3
53
2
1
1 1 1 1
1 1 1 1 1
𝑓 = 𝜎 2
1 𝜎
2
2 + 𝐴𝜎 3
1 𝜎
3
+ 𝐵𝜎 3
2
+ 𝐶𝜎
1 𝜎
2 𝜎
3
+ 𝐷𝜎 2
3
.
We have 4 undetermined coefficients, therefore 4 different choices will have to be performed (notice that for instance the cases (1,1,0) a (0,1,1) are equivalent with respect to the result, which is caused by the symmetrisation of the polynomial):
Case 1: (𝑥, 𝑦, 𝑧) = (1,1,0) , 𝑓 = (𝑥 − 𝑦) 2 (𝑥 − 𝑧) 2 (𝑦 − 𝑧) 2 = 0, 𝜎
1
= 𝑥 + 𝑦 + 𝑧 =
2, 𝜎
2
= 𝑥𝑦 + 𝑥𝑧 + 𝑦𝑧 = 1, 𝜎
3
= 𝑥𝑦𝑧 = 0
0 = 4 + 𝐵 𝐵 = −4
Case 2: (𝑥, 𝑦, 𝑧) = (1,1,1) , 𝑓 = 0, 𝜎
1
= 3, 𝜎
2
= 3, 𝜎
3
= 1
0 = 81 + 27𝐴 − 27.4 + 9𝐶 + 𝐷 27 = 27𝐴 + 9𝐶 + 𝐷
Case 3: (𝑥, 𝑦, 𝑧) = (1,1, −2) , 𝑓 = 0, 𝜎
1
−2
= 0, 𝜎
2
= 1.1 + 1. (−2) + 1. (−2) = −3, 𝜎
0 = 27.4 + 4𝐷 𝐷 = −27
3
=
Case 4: (𝑥, 𝑦, 𝑧) = (1,1, −1) , 𝑓 = 0, 𝜎
1
−1
= 1, 𝜎
2
= 1.1 + 1. (−1) + 1. (−1) = −1, 𝜎
0 = 1 − 𝐴 + 4 + 𝐶 − 27 22 = −𝐴 + 𝐶
3
=
Now we will solve the following system of equations for the remaining undetermined coefficients:
27 = 27𝐴 + 9𝐶 − 27
22 = −𝐴 + 𝐶
The solution of this system is 𝐴 = −4, 𝐶 = 18 . To summarise 𝑓 = 𝜎 2
1 𝜎 2
2
− 4𝜎 3
1 𝜎
3
− 4𝜎 3
2
+ 18𝜎
1 𝜎
2 𝜎
3
− 27𝜎 2
3
.
Example 51 Solve individually: Transform the polynomial 𝑓 = 𝑥 3 variables
1
,
2
, . . . ,
𝑛
.
+ 𝑦 3 + 𝑧 3
into the
Result: 𝑓 = 𝜎
1
3 − 3𝜎
1 𝜎
2
+ 3𝜎
3
Example 52 Solve individually: Transform the polynomial 𝑓 = (𝑥 2 𝑧 2 )(𝑦 2 + 𝑧 2 ) into the variables
1
,
2
, . . . , 𝑛
.
+ 𝑦 2 )(𝑥 2 +
Result: The leading term 𝑥 4 𝑦 2
, 𝑓 = 𝜎 2
1 𝜎 2
2
− 2𝜎
1
3 𝜎
3
− 2𝜎
2
3 + 4𝜎
1 𝜎
2 𝜎
3
− 𝜎 2
3
Symmetric polynomials have their use in application tasks. However, only two simple examples will be shown here to illustrate the usefulness of symmetric polynomials in various types of tasks.
Example 53 Solve the following system of equations in the domain of real numbers 𝑥 3 + 𝑦 3 = 35 𝑥𝑦 = 6 .
Solution: We can compute and check that 𝑥 3 + 𝑦 3 = 𝜎
1
3 − 3𝜎
1 𝜎
2
and 𝑥𝑦 = 𝜎
2
. Thus we get a new system of equations 𝜎 3
1
− 3𝜎
1 𝜎
2
= 35 𝜎
2
= 6
After substitution of the second equation into the first one, we get the equation 𝜎
1
3 −
18𝜎
1
− 35 = 0 , by estimating and substituting into the Horner scheme stepwise we eventually find one real root 𝜎
1
= 5 . As 𝜎
1
= 𝑥 + 𝑦 , we will now be solving the system 𝑥 + 𝑦 = 5 𝑥𝑦 = 6, whose solution is the set of two ordered pairs {[3,2], [2,3]} .
We can perform the check by substituting into the original assignment and having done this in both cases, we come to the identity 35 = 35 6 = 6 .
Example 54 A cuboid has edges of lengths 𝑎, 𝑏, 𝑐 . There applies 𝑎 + 𝑏 + 𝑐 = 6 and 𝑎𝑏 + 𝑎𝑐 + 𝑏𝑐 = 9 . Determine the volume of the cube that has the body diagonal of the same length as the given cuboid.
Solution: Let us mark the cube edge length with 𝑘 . The body diagonal of the cuboid is of the length √𝑎 2 + 𝑏 2 + 𝑐 2
while the length of the cube body diagonal is √3𝑘 . This leads to the equation
√𝑎 2 + 𝑏 2 + 𝑐 2 = √3𝑘, which after exponentiation turns into 𝑎 2 + 𝑏 2 + 𝑐 2 = 3𝑘 2 .
The polynomial 𝑎 2 𝜎 2
1
− 2𝜎
2
, hence
+ 𝑏 2 + 𝑐 2
is a symmetric polynomial and there applies 𝑎 2 + 𝑏 2 + 𝑐 2 = 𝜎
1
2 − 2𝜎
2
= 3𝑘 2 .
At the same time we know from the assignment that 𝜎
1 𝑎𝑐 + 𝑏𝑐 = 9 . Thus
= 𝑎 + 𝑏 + 𝑐 = 6 and 𝜎
2
= 𝑎𝑏 +
6 2 − 2.9 = 3𝑘 2 ,
After rearrangement 𝑘 2 = 6 , 𝑘 = √6 . The sought volume of the cube is then
𝑉 = 𝑘 3 = 6√6.
Example 55 Solve the following system of equations individually: 𝑥 3 + 𝑦 3 = 28
55
𝑥 + 𝑦 = 4
Solution: {[3,1], [1,3]} .
Exercises for Check
1.
Define polynomial of 𝑛 variables and its degrees.
2.
Define homogeneous and symmetric polynomial.
3.
Determine the degrees of the following polynomials. Are they homogeneous or symmetric? a.
𝑥 − 𝑦 , b.
3𝑥𝑦 + 2𝑥 2
, c.
4𝑥𝑦 2 + 3𝑥 + 3𝑦 + 4𝑥 d.
5𝑥 + 𝑦 2 − 7 ,
2 𝑦 , e.
𝑥𝑦 f.
𝑥𝑦
2
2
+ 𝑦𝑧 2
+ 𝑦𝑧 2
+ 𝑧𝑥
+ 𝑧𝑥
2
2
,
+ 𝑥 2 𝑦 + 𝑦 2 𝑧 + 𝑧 2 𝑥 .
4.
Apply the binomial theorem to rewrite the sum of powers 𝑠
4 elementary symmetric polynomials.
= 𝑥 4
1
+ 𝑥 4
2
by
5.
Express the polynomial (𝑥 + 𝑦)(𝑦 + 𝑧)(𝑧 + 𝑥) using elementary symmetric polynomials.
Results:
3.
a. 1, homogeneous, b. 2, homogeneous, c. 3, symmetric, d. 2, neither homogeneous
4.
nor symmetric,, e. 3, homogeneous, f. 3, homogeneous and symmetric 𝑠
4
= 𝜎 4
1
5.
𝜎
1 𝜎
2
− 𝜎
− 4𝜎
3
.
2
1 𝜎
2
+ 2𝜎 2
2
57
References
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J.: Polynomials. Springer Verlag New York Inc., 1989
B
IRKHOFF
, G., M
AC
L
ANE
, S.: Prehľad modernej algebry. Prague: SNTL, 1979
B LAŽEK , J., K
OMAN
, M., V OJTÁŠKOVÁ , B.:
Algebra a teoretická aritmetika.
Prague: SPN,
1985
H ERMAN , J., K UČERA , R.,
Š IMŠA , J.:
Metody řešení matematických úloh I.
Prague: PřF MU,
1996
H ORÁK , P.: Polynomy . 1. vydání. Brno: UJEP, 1978 .
