Solving typical problems. “Factoring a fifth-degree polynomial into quadratic factors using the Lagrange interpolation polynomial Factoring a quadratic trinomial
Keywords: equations, Polynomial, Roots of the equation
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Lesson type: A lesson in mastering and consolidating primary knowledge.
The purpose of the lesson:
- Introduce students to the concept of roots of a polynomial and teach them how to find them. Improve skills in using Horner's scheme for expanding a polynomial by powers and dividing a polynomial by a binomial.
- Learn to find the roots of an equation using Horner's scheme.
- Develop abstract thinking.
- Foster a computing culture.
- Development of interdisciplinary connections.
During the classes
1. Organizational moment.
Inform the topic of the lesson, formulate goals.
2. Checking homework.
3. Studying new material.
Let Fn(x) = a n x n +a n-1 x n-1 +...+ a 1 x +a 0 - a polynomial for x of degree n, where a 0 , a 1 ,...,a n are given numbers, and a 0 is not equal to 0. If the polynomial F n (x) is divided with the remainder by the binomial x-a, then the quotient (incomplete quotient) is polynomial Q n-1 (x) of degree n-1, the remainder R is a number, and the equality is true F n (x)=(x-a) Q n-1 (x) +R. The polynomial F n (x) is divisible by the binomial (x-a) only in the case of R=0.
Bezout's theorem: The remainder R from dividing the polynomial F n (x) by the binomial (x-a) is equal to the value of the polynomial F n (x) at x=a, i.e. R=Pn(a).
A little history. Bezout's theorem, despite its apparent simplicity and obviousness, is one of the fundamental theorems of the theory of polynomials. This theorem relates the algebraic properties of polynomials (which allow polynomials to be treated as integers) with their functional properties (which allow polynomials to be treated as functions). One way to solve higher degree equations is to factor the polynomial on the left side of the equation. The calculation of the coefficients of the polynomial and the remainder is written in the form of a table called the Horner scheme.
Horner's scheme is an algorithm for dividing polynomials, written for the special case when the quotient is equal to a binomial x–a.
Horner William George (1786 - 1837), English mathematician. The main research concerns the theory of algebraic equations. Developed a method for approximate solution of equations of any degree. In 1819 he introduced an important method for algebra of dividing a polynomial by a binomial x - a (Horner's scheme).
Derivation of the general formula for Horner's scheme.
Dividing a polynomial f(x) with a remainder by a binomial (x-c) means finding a polynomial q(x) and a number r such that f(x)=(x-c)q(x)+r
Let us write this equality in detail:
f 0 x n + f 1 x n-1 + f 2 x n-2 + ...+f n-1 x + f n =(x-c) (q 0 x n-1 + q 1 x n-2 + q 2 x n-3 +...+ q n-2 x + q n-1)+r
Let us equate the coefficients at the same degrees:
xn: f 0 = q 0 => q 0 = f 0 xn-1: f 1 = q 1 - c q 0 => q 1 = f 1 + c q 0 xn-2: f 2 = q 2 - c q 1 => q 2 = f 2 + c q 1 ... ... x0: f n = q n - c q n-1 => q n = f n + c q n-1.
Demonstration of Horner's circuit using an example.
Exercise 1. Using Horner's scheme, we divide the polynomial f(x) = x 3 - 5x 2 + 8 with remainder by the binomial x-2.
| 1 | -5 | 0 | 8 | |
| 2 | 1 | 2*1+(-5)=-3 | 2*(-3)+0=-6 | 2*(-6)+8=-4 |
f(x) = x 3 - 5x 2 + 8 =(x-2)(x 2 -3x-6)-4, where g(x)= (x 2 -3x-6), r = -4 remainder.
Expansion of a polynomial in powers of a binomial.
Using Horner's scheme, we expand the polynomial f(x)=x 3 +3x 2 -2x+4 in powers of the binomial (x+2).
As a result, we should obtain the expansion f(x) = x 3 +3x 2 -2x+4 = (x+2)(x 2 +x-4)+12 = (x+2)((x-1)(x+ 2)-2)+12 = (((1*(x+2)-3)(x+2)-2)(x+2))+12 = (x+2) 3 -3(x+2 ) 2 -2(x+2)+12
Horner's scheme is often used when solving equations of the third, fourth and higher degrees, when it is convenient to expand the polynomial into a binomial x-a. Number a called root of the polynomial F n (x) = f 0 x n + f 1 x n-1 + f 2 x n-2 + ...+f n-1 x + f n, if at x=a the value of the polynomial F n (x) is equal to zero: F n (a)=0, i.e. if the polynomial is divisible by the binomial x-a.
For example, the number 2 is the root of the polynomial F 3 (x)=3x 3 -2x-20, since F 3 (2)=0. it means. That the factorization of this polynomial contains a factor x-2.
F 3 (x)=3x 3 -2x-20=(x-2)(3x 2 +6x+10).
Any polynomial F n(x) of degree n 1 can have no more n real roots.
Any integer root of an equation with integer coefficients is a divisor of its free term.
If the leading coefficient of an equation is 1, then all rational roots of the equation, if they exist, are integers.
Consolidation of the studied material.
To consolidate the new material, students are invited to complete numbers from the textbook 2.41 and 2.42 (p. 65).
(2 students solve at the board, and the rest, having decided, check the assignments in the notebook with the answers on the board).
Summarizing.
Having understood the structure and principle of operation of the Horner scheme, it can also be used in computer science lessons, when the issue of converting integers from the decimal number system to the binary system and vice versa is considered. The basis for transferring from one number system to another is the following general theorem
Theorem. To convert a whole number Ap from p-ary number system to base number system d necessary Ap sequentially divide with remainder by number d, written in the same p-ary system until the resulting quotient becomes equal to zero. The remainders from the division will be d-numerical digits Ad, starting from the youngest category to the most senior. All actions must be carried out in p-ary number system. For a person, this rule is convenient only when p= 10, i.e. when translating from decimal system. As for the computer, on the contrary, it is “more convenient” for it to perform calculations in the binary system. Therefore, to convert “2 to 10”, sequential division by ten in the binary system is used, and “10 to 2” is the addition of powers of ten. To optimize the calculations of the “10 in 2” procedure, the computer uses Horner's economical computing scheme.
Homework. It is proposed to complete two tasks.
1st. Using Horner's scheme, divide the polynomial f(x)=2x 5 -x 4 -3x 3 +x-3 by the binomial (x-3).
2nd. Find the integer roots of the polynomial f(x)=x 4 -2x 3 +2x 2 -x-6. (considering that any integer root of an equation with integer coefficients is a divisor of its free term)
Literature.
- Kurosh A.G. “Course of Higher Algebra.”
- Nikolsky S.M., Potapov M.K. and others. Grade 10 “Algebra and the beginnings of mathematical analysis.”
- http://inf.1september.ru/article.php?ID=200600907.
In order to factorize, it is necessary to simplify the expressions. This is necessary so that it can be further reduced. The expansion of a polynomial makes sense when its degree is not lower than two. A polynomial with the first degree is called linear.
The article will cover all the concepts of decomposition, theoretical foundations and methods of factoring a polynomial.
Theory
Theorem 1When any polynomial with degree n, having the form P n x = a n x n + a n - 1 x n - 1 + . . . + a 1 x + a 0, are represented as a product with a constant factor with the highest degree a n and n linear factors (x - x i), i = 1, 2, ..., n, then P n (x) = a n (x - x n) (x - x n - 1) · . . . · (x - x 1) , where x i, i = 1, 2, …, n are the roots of the polynomial.
The theorem is intended for roots of complex type x i, i = 1, 2, …, n and for complex coefficients a k, k = 0, 1, 2, …, n. This is the basis of any decomposition.
When coefficients of the form a k, k = 0, 1, 2, …, n are real numbers, then the complex roots will occur in conjugate pairs. For example, roots x 1 and x 2 related to a polynomial of the form P n x = a n x n + a n - 1 x n - 1 + . . . + a 1 x + a 0 are considered complex conjugate, then the other roots are real, from which we obtain that the polynomial takes the form P n (x) = a n (x - x n) (x - x n - 1) · . . . · (x - x 3) x 2 + p x + q, where x 2 + p x + q = (x - x 1) (x - x 2) .
Comment
The roots of a polynomial can be repeated. Let's consider the proof of the algebra theorem, a consequence of Bezout's theorem.
Fundamental theorem of algebra
Theorem 2Any polynomial with degree n has at least one root.
Bezout's theorem
After dividing a polynomial of the form P n x = a n x n + a n - 1 x n - 1 + . . . + a 1 x + a 0 on (x - s), then we get the remainder, which is equal to the polynomial at point s, then we get
P n x = a n x n + a n - 1 x n - 1 + . . . + a 1 x + a 0 = (x - s) · Q n - 1 (x) + P n (s) , where Q n - 1 (x) is a polynomial with degree n - 1.
Corollary to Bezout's theorem
When the root of the polynomial P n (x) is considered to be s, then P n x = a n x n + a n - 1 x n - 1 + . . . + a 1 x + a 0 = (x - s) · Q n - 1 (x) . This corollary is sufficient when used to describe the solution.
Factoring a quadratic trinomial
A square trinomial of the form a x 2 + b x + c can be factorized into linear factors. then we get that a x 2 + b x + c = a (x - x 1) (x - x 2) , where x 1 and x 2 are roots (complex or real).
This shows that the expansion itself reduces to solving the quadratic equation subsequently.
Example 1
Factor the quadratic trinomial.
Solution
It is necessary to find the roots of the equation 4 x 2 - 5 x + 1 = 0. To do this, you need to find the value of the discriminant using the formula, then we get D = (- 5) 2 - 4 · 4 · 1 = 9. From here we have that
x 1 = 5 - 9 2 4 = 1 4 x 2 = 5 + 9 2 4 = 1
From this we get that 4 x 2 - 5 x + 1 = 4 x - 1 4 x - 1.
To perform the check, you need to open the parentheses. Then we get an expression of the form:
4 x - 1 4 x - 1 = 4 x 2 - x - 1 4 x + 1 4 = 4 x 2 - 5 x + 1
After checking, we arrive at the original expression. That is, we can conclude that the decomposition was performed correctly.
Example 2
Factor the quadratic trinomial of the form 3 x 2 - 7 x - 11 .
Solution
We find that it is necessary to calculate the resulting quadratic equation of the form 3 x 2 - 7 x - 11 = 0.
To find the roots, you need to determine the value of the discriminant. We get that
3 x 2 - 7 x - 11 = 0 D = (- 7) 2 - 4 3 (- 11) = 181 x 1 = 7 + D 2 3 = 7 + 181 6 x 2 = 7 - D 2 3 = 7 - 181 6
From this we get that 3 x 2 - 7 x - 11 = 3 x - 7 + 181 6 x - 7 - 181 6.
Example 3
Factor the polynomial 2 x 2 + 1.
Solution
Now we need to solve the quadratic equation 2 x 2 + 1 = 0 and find its roots. We get that
2 x 2 + 1 = 0 x 2 = - 1 2 x 1 = - 1 2 = 1 2 i x 2 = - 1 2 = - 1 2 i
These roots are called complex conjugate, which means the expansion itself can be depicted as 2 x 2 + 1 = 2 x - 1 2 · i x + 1 2 · i.
Example 4
Decompose the quadratic trinomial x 2 + 1 3 x + 1 .
Solution
First you need to solve a quadratic equation of the form x 2 + 1 3 x + 1 = 0 and find its roots.
x 2 + 1 3 x + 1 = 0 D = 1 3 2 - 4 1 1 = - 35 9 x 1 = - 1 3 + D 2 1 = - 1 3 + 35 3 i 2 = - 1 + 35 · i 6 = - 1 6 + 35 6 · i x 2 = - 1 3 - D 2 · 1 = - 1 3 - 35 3 · i 2 = - 1 - 35 · i 6 = - 1 6 - 35 6 · i
Having obtained the roots, we write
x 2 + 1 3 x + 1 = x - - 1 6 + 35 6 i x - - 1 6 - 35 6 i = = x + 1 6 - 35 6 i x + 1 6 + 35 6 i
Comment
If the discriminant value is negative, then the polynomials will remain second-order polynomials. It follows from this that we will not expand them into linear factors.
Methods for factoring a polynomial of degree higher than two
When decomposing, a universal method is assumed. Most of all cases are based on a corollary of Bezout's theorem. To do this, you need to select the value of the root x 1 and reduce its degree by dividing by a polynomial by 1 by dividing by (x - x 1). The resulting polynomial needs to find the root x 2, and the search process is cyclical until we obtain a complete expansion.
If the root is not found, then other methods of factorization are used: grouping, additional terms. This topic involves solving equations with higher powers and integer coefficients.
Taking the common factor out of brackets
Consider the case when the free term is equal to zero, then the form of the polynomial becomes P n (x) = a n x n + a n - 1 x n - 1 + . . . + a 1 x .
It can be seen that the root of such a polynomial will be equal to x 1 = 0, then the polynomial can be represented as the expression P n (x) = a n x n + a n - 1 x n - 1 +. . . + a 1 x = = x (a n x n - 1 + a n - 1 x n - 2 + . . . + a 1)
This method is considered to be taking the common factor out of brackets.
Example 5
Factor the third degree polynomial 4 x 3 + 8 x 2 - x.
Solution
We see that x 1 = 0 is the root of the given polynomial, then we can remove x from the brackets of the entire expression. We get:
4 x 3 + 8 x 2 - x = x (4 x 2 + 8 x - 1)
Let's move on to finding the roots of the square trinomial 4 x 2 + 8 x - 1. Let's find the discriminant and roots:
D = 8 2 - 4 4 (- 1) = 80 x 1 = - 8 + D 2 4 = - 1 + 5 2 x 2 = - 8 - D 2 4 = - 1 - 5 2
Then it follows that
4 x 3 + 8 x 2 - x = x 4 x 2 + 8 x - 1 = = 4 x x - - 1 + 5 2 x - - 1 - 5 2 = = 4 x x + 1 - 5 2 x + 1 + 5 2
To begin with, let us take into consideration a decomposition method containing integer coefficients of the form P n (x) = x n + a n - 1 x n - 1 + . . . + a 1 x + a 0, where the coefficient of the highest degree is 1.
When a polynomial has integer roots, then they are considered divisors of the free term.
Example 6
Decompose the expression f (x) = x 4 + 3 x 3 - x 2 - 9 x - 18.
Solution
Let's consider whether there are complete roots. It is necessary to write down the divisors of the number - 18. We get that ±1, ±2, ±3, ±6, ±9, ±18. It follows that this polynomial has integer roots. You can check using Horner's scheme. It is very convenient and allows you to quickly obtain the expansion coefficients of a polynomial:
It follows that x = 2 and x = - 3 are the roots of the original polynomial, which can be represented as a product of the form:
f (x) = x 4 + 3 x 3 - x 2 - 9 x - 18 = (x - 2) (x 3 + 5 x 2 + 9 x + 9) = = (x - 2) (x + 3) (x 2 + 2 x + 3)
We proceed to the expansion of a quadratic trinomial of the form x 2 + 2 x + 3.
Since the discriminant is negative, it means there are no real roots.
Answer: f (x) = x 4 + 3 x 3 - x 2 - 9 x - 18 = (x - 2) (x + 3) (x 2 + 2 x + 3)
Comment
It is allowed to use root selection and division of a polynomial by a polynomial instead of Horner's scheme. Let's move on to considering the expansion of a polynomial containing integer coefficients of the form P n (x) = x n + a n - 1 x n - 1 + . . . + a 1 x + a 0 , the highest of which is equal to one.
This case occurs for rational fractions.
Example 7
Factorize f (x) = 2 x 3 + 19 x 2 + 41 x + 15 .
Solution
It is necessary to replace the variable y = 2 x, you should move on to a polynomial with coefficients equal to 1 at the highest degree. You need to start by multiplying the expression by 4. We get that
4 f (x) = 2 3 x 3 + 19 2 2 x 2 + 82 2 x + 60 = = y 3 + 19 y 2 + 82 y + 60 = g (y)
When the resulting function of the form g (y) = y 3 + 19 y 2 + 82 y + 60 has integer roots, then their location is among the divisors of the free term. The entry will look like:
±1, ±2, ±3, ±4, ±5, ±6, ±10, ±12, ±15, ±20, ±30, ±60
Let's move on to calculating the function g (y) at these points in order to get zero as a result. We get that
g (1) = 1 3 + 19 1 2 + 82 1 + 60 = 162 g (- 1) = (- 1) 3 + 19 (- 1) 2 + 82 (- 1) + 60 = - 4 g (2) = 2 3 + 19 2 2 + 82 2 + 60 = 308 g (- 2) = (- 2) 3 + 19 (- 2) 2 + 82 (- 2) + 60 = - 36 g (3) = 3 3 + 19 3 2 + 82 3 + 60 = 504 g (- 3) = (- 3) 3 + 19 (- 3) 2 + 82 (- 3) + 60 = - 42 g (4) = 4 3 + 19 · 4 2 + 82 · 4 + 60 = 756 g (- 4) = (- 4) 3 + 19 · (- 4) 2 + 82 · (- 4) + 60 = - 28 g (5) = 5 3 + 19 5 2 + 82 5 + 60 = 1070 g (- 5) = (- 5) 3 + 19 (- 5) 2 + 82 (- 5) + 60
We find that y = - 5 is the root of an equation of the form y 3 + 19 y 2 + 82 y + 60, which means x = y 2 = - 5 2 is the root of the original function.
Example 8
It is necessary to divide with a column 2 x 3 + 19 x 2 + 41 x + 15 by x + 5 2.
Solution
Let's write it down and get:
2 x 3 + 19 x 2 + 41 x + 15 = x + 5 2 (2 x 2 + 14 x + 6) = = 2 x + 5 2 (x 2 + 7 x + 3)
Checking the divisors will take a lot of time, so it is more profitable to factorize the resulting quadratic trinomial of the form x 2 + 7 x + 3. By equating to zero we find the discriminant.
x 2 + 7 x + 3 = 0 D = 7 2 - 4 1 3 = 37 x 1 = - 7 + 37 2 x 2 = - 7 - 37 2 ⇒ x 2 + 7 x + 3 = x + 7 2 - 37 2 x + 7 2 + 37 2
It follows that
2 x 3 + 19 x 2 + 41 x + 15 = 2 x + 5 2 x 2 + 7 x + 3 = = 2 x + 5 2 x + 7 2 - 37 2 x + 7 2 + 37 2
Artificial techniques for factoring a polynomial
Rational roots are not inherent in all polynomials. To do this, you need to use special methods to find factors. But not all polynomials can be expanded or represented as a product.
Grouping method
There are cases when you can group the terms of a polynomial to find a common factor and put it out of brackets.
Example 9
Factor the polynomial x 4 + 4 x 3 - x 2 - 8 x - 2.
Solution
Because the coefficients are integers, then the roots can presumably also be integers. To check, take the values 1, - 1, 2 and - 2 in order to calculate the value of the polynomial at these points. We get that
1 4 + 4 1 3 - 1 2 - 8 1 - 2 = - 6 ≠ 0 (- 1) 4 + 4 (- 1) 3 - (- 1) 2 - 8 (- 1) - 2 = 2 ≠ 0 2 4 + 4 2 3 - 2 2 - 8 2 - 2 = 26 ≠ 0 (- 2) 4 + 4 (- 2) 3 - (- 2) 2 - 8 (- 2) - 2 = - 6 ≠ 0
This shows that there are no roots; it is necessary to use another method of expansion and solution.
It is necessary to group:
x 4 + 4 x 3 - x 2 - 8 x - 2 = x 4 + 4 x 3 - 2 x 2 + x 2 - 8 x - 2 = = (x 4 - 2 x 2) + (4 x 3 - 8 x) + x 2 - 2 = = x 2 (x 2 - 2) + 4 x (x 2 - 2) + x 2 - 2 = = (x 2 - 2) (x 2 + 4 x + 1)
After grouping the original polynomial, you need to represent it as the product of two square trinomials. To do this we need to factorize. we get that
x 2 - 2 = 0 x 2 = 2 x 1 = 2 x 2 = - 2 ⇒ x 2 - 2 = x - 2 x + 2 x 2 + 4 x + 1 = 0 D = 4 2 - 4 1 1 = 12 x 1 = - 4 - D 2 1 = - 2 - 3 x 2 = - 4 - D 2 1 = - 2 - 3 ⇒ x 2 + 4 x + 1 = x + 2 - 3 x + 2 + 3
x 4 + 4 x 3 - x 2 - 8 x - 2 = x 2 - 2 x 2 + 4 x + 1 = = x - 2 x + 2 x + 2 - 3 x + 2 + 3
Comment
The simplicity of grouping does not mean that choosing terms is easy enough. There is no specific solution method, so it is necessary to use special theorems and rules.
Example 10
Factor the polynomial x 4 + 3 x 3 - x 2 - 4 x + 2 .
Solution
The given polynomial has no integer roots. The terms should be grouped. We get that
x 4 + 3 x 3 - x 2 - 4 x + 2 = = (x 4 + x 3) + (2 x 3 + 2 x 2) + (- 2 x 2 - 2 x) - x 2 - 2 x + 2 = = x 2 (x 2 + x) + 2 x (x 2 + x) - 2 (x 2 + x) - (x 2 + 2 x - 2) = = (x 2 + x) (x 2 + 2 x - 2) - (x 2 + 2 x - 2) = (x 2 + x - 1) (x 2 + 2 x - 2)
After factorization we get that
x 4 + 3 x 3 - x 2 - 4 x + 2 = x 2 + x - 1 x 2 + 2 x - 2 = = x + 1 + 3 x + 1 - 3 x + 1 2 + 5 2 x + 1 2 - 5 2
Using abbreviated multiplication formulas and Newton's binomial to factor a polynomial
Appearance often does not always make it clear which method should be used during decomposition. After the transformations have been made, you can build a line consisting of Pascal’s triangle, otherwise they are called Newton’s binomial.
Example 11
Factor the polynomial x 4 + 4 x 3 + 6 x 2 + 4 x - 2.
Solution
It is necessary to convert the expression to the form
x 4 + 4 x 3 + 6 x 2 + 4 x - 2 = x 4 + 4 x 3 + 6 x 2 + 4 x + 1 - 3
The sequence of coefficients of the sum in parentheses is indicated by the expression x + 1 4 .
This means we have x 4 + 4 x 3 + 6 x 2 + 4 x - 2 = x 4 + 4 x 3 + 6 x 2 + 4 x + 1 - 3 = x + 1 4 - 3.
After applying the difference of squares, we get
x 4 + 4 x 3 + 6 x 2 + 4 x - 2 = x 4 + 4 x 3 + 6 x 2 + 4 x + 1 - 3 = x + 1 4 - 3 = = x + 1 4 - 3 = x + 1 2 - 3 x + 1 2 + 3
Consider the expression that is in the second bracket. It is clear that there are no knights there, so we should apply the difference of squares formula again. We get an expression of the form
x 4 + 4 x 3 + 6 x 2 + 4 x - 2 = x 4 + 4 x 3 + 6 x 2 + 4 x + 1 - 3 = x + 1 4 - 3 = = x + 1 4 - 3 = x + 1 2 - 3 x + 1 2 + 3 = = x + 1 - 3 4 x + 1 + 3 4 x 2 + 2 x + 1 + 3
Example 12
Factorize x 3 + 6 x 2 + 12 x + 6 .
Solution
Let's start transforming the expression. We get that
x 3 + 6 x 2 + 12 x + 6 = x 3 + 3 2 x 2 + 3 2 2 x + 2 3 - 2 = (x + 2) 3 - 2
It is necessary to apply the formula for abbreviated multiplication of the difference of cubes. We get:
x 3 + 6 x 2 + 12 x + 6 = = (x + 2) 3 - 2 = = x + 2 - 2 3 x + 2 2 + 2 3 x + 2 + 4 3 = = x + 2 - 2 3 x 2 + x 2 + 2 3 + 4 + 2 2 3 + 4 3
A method for replacing a variable when factoring a polynomial
When replacing a variable, the degree is reduced and the polynomial is factored.
Example 13
Factor the polynomial of the form x 6 + 5 x 3 + 6 .
Solution
According to the condition, it is clear that it is necessary to make the replacement y = x 3. We get:
x 6 + 5 x 3 + 6 = y = x 3 = y 2 + 5 y + 6
The roots of the resulting quadratic equation are y = - 2 and y = - 3, then
x 6 + 5 x 3 + 6 = y = x 3 = y 2 + 5 y + 6 = = y + 2 y + 3 = x 3 + 2 x 3 + 3
It is necessary to apply the formula for abbreviated multiplication of the sum of cubes. We get expressions of the form:
x 6 + 5 x 3 + 6 = y = x 3 = y 2 + 5 y + 6 = = y + 2 y + 3 = x 3 + 2 x 3 + 3 = = x + 2 3 x 2 - 2 3 x + 4 3 x + 3 3 x 2 - 3 3 x + 9 3
That is, we obtained the desired decomposition.
The cases discussed above will help in considering and factoring a polynomial in different ways.
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Factoring a polynomial of the fifth degree into quadratic factors using the Lagrange interpolation polynomial 1. Definition of the Lagrange interpolation polynomial of the fifth degree. To factorize the reduced polynomial of the fifth degree, it is necessary to fulfill the equality: f(x)=φ(x)·g(x). In this case, the degree of the polynomials φ(x) and g(x) should not be higher than five. To determine an integer polynomial of no higher than the fifth degree with a given table of values, there is a formula for the Lagrange interpolation polynomial (IPL): 6 Ak k=1 F"(xk)(x−xk) , where F(x)=(x-x1)·( x-x2)·(x-x3)·(x-x4)·(x- φ(x) = F(x)· ∑ x5)(x-x6), Fʹ(xk) values of the derivative of the function F(x) at points xk. Where it is necessary to set the coordinates of six points on the plane. To determine the factors φ(x) and g(x), we choose arbitrarily six integer values x = x1; x2; x3; x4; x5; x6 and substitute them into the equality f (x)= φ(x) g(x) We get: f(x1)= φ(x1) g(x1); f(x2)= φ(x2) g(x2); f(x3) = φ(x3) g(x3); f(x4)= φ(x4) g(x4); f(x5)=φ(x5) g(x5); f(x6)= φ(x6) · g(x6). These equalities show that each value φ(xk) of the desired factor φ(x) is a divisor of the number f(xk). To construct the factor φ(x), we will use the IML and substitute arbitrary values as f(xk) integers Аk, and choose the values xk in the form of successive integers close to zero, i.e. x1= -3; x2= -2; x3= -1; x4=0; x5=1; x6=2. In expanded IML form φ(x) looks like this:
F(x) φ(x) A4 + A2 A3 + A1 A5 F"(1)(x−1) + +A6 F"(−3)(x+3) F"(−2)(x+2) + + F"(0)x F"(−1)(x+1) F"(2)(x−2)) , ·(where F(x)=(x+3)·(x+2) ·(x+1)·x·(x-1)·(x-2) (2).To construct the factor φ(x) using IML, you need to specify the numbers A1; A2; A3; A4; A5; A6. Definition: numbers A1; A2; A3; A4; A5; A6 taken from the IML formula written in a series are called Lagrange series. 2. Decomposition of a polynomial into linear factors using IML. Theorem 1 (Generalization of Horner’s scheme) The polynomial φ(x) is linear , if the numbers A1; A2; A3; A4; A5; A6 form an increasing sequence of integers. Proof: we reduce the polynomial (2) to the lowest common denominator, i.e. to 120· F(x), we write the resulting numerator as a polynomial fifth degree whose coefficients contain the numbers A1; A2; A3; A4; A5; A6. In order for the polynomial (2) to be linear, it is necessary to equate to zero the coefficients at “x” of the fifth, fourth, third and second degree, and the coefficient at “x” of the first degree is equated to 120. As a result, we obtain the following system of five equations with six variables: -A1+5 A2-10 A3+10 A4-5 A5+A6=0 5 A2-20 A3 +30 A4-20 A5+5 A6=0 5 A1-35 A2+70 A3-50 A4+5 A5+5 A6=0 -5 A2+80 A3-150 А4+80·А4-5·А6=0 -4·А1+30·А2-120·А3+40·А4+60·А5-6·А6=120. If we fix the number A6, then all the rest will be expressed by the following formulas: A1=A6-5; A2=A6-4; A3=A6- 3; A4=A6-2; A5=A6-1.
We have obtained an increasing sequence of integers. It follows from the theorem that the linear factor has the following form: φ(x)=x+A4 (3). Definition: sequence of numbers given by these relationships A1=A6-5; A2=A6-4; A3=A6-3; A4=A6-2; A5=A6-1; A6 is called a linear Lagrangian series. Definition: a linear Lagrange series is called a “candidate” if all its numbers Аk are divisors of the corresponding values of the function f(xk), where k=1;2;3;4;5;6. For all candidates, we construct a linear factor φ(x) using formula (3) and check it for divisibility with f(x). It follows from the theorem that the linear factor has the following form φ(x)=x+A4, where A4 is the divisor of the free term, i.e. Similar to the reduced polynomial according to Horner's scheme. Example: f(x)= x5-8x4+2x3-16x2+x-8. Using Horner's scheme we find the value of the polynomial at x = -3; -2; -1; 0;1;2. To do this, let's compile table 1: -8 -11 -10 -9 -8 -7 -6 -3 -2 -1 0 1 2 We will rewrite the last column of table 1 with the first row of table 2. Select in this row the number that has the smallest number of divisors. In our example, this number is -8. Let's write down all its divisors in a column. For each divisor of the number -8, we write a linear Lagrangian series in a line. From the resulting Lagrangian series we will select “candidates”. Let us construct a polynomial φ(x) in f(0) using “candidates”. linear multiplier -8 -1100 -250 -36 -8 -28 -150 is determined by 1 1 1 1 1 1 1 2 35 22 11 2 -5 -10 -16 -121 -60 -27 -16 -21 -36 1 364 121 28 1 -20 -71
36 A3 0 -2 1 -3 3 -5 7 -9 -8 A4 1 -1 2 -2 4 -4 8 -8 -28 A5 2 0 3 -1 5 -3 9 -7 -150 A6 3 1 4 0 6 -2 10 -6 formula (3) and check them for divisibility with the given polynomial f(x)= x5-8x4+2x3-16x2+x-8. Table 2: -250 -1100 A2 A1 -2 -1 -3 -4 0 -1 -5 -4 2 1 -6 -7 5 6 -11 “candide -10 at” In the above table 2, rectangles are shaded in gray, which contain numbers that are not divisors of the corresponding values of the function f(x). This table contains a row or Lagrangian series of all numbers, which are divisors of the corresponding values of the function f(x). This series is the only candidate. In this series A4 = -8, substituting φ(x)=x- A4 into the formula, we find φ(x)=x- 8. We highlight the actual candidate in black. 3. Expansion of polynomial factors using IML. Check:x5-8x4+2x3-16x2+x-8=(x-8)·(x4+2x2+1). into quadratic ones Theorem 2. The factor φ(x) is quadratic if the numbers A1; A2; A3; A4; A5; A6 are interconnected by the following relationships: A1=5·(A5+4)-4·A6 A2=4·(A5+3)-3·A6 A3=3·(A5+2)-2·A6 A4=2· (A5+1)-1 A6
Proof: Proof: let us reduce the polynomial (1) to the lowest common denominator, i.e. to 120· F(x), we write the resulting numerator in the form of a fifth-degree polynomial whose coefficients contain numbers A1; A2; A3; A4; A5; A6. In order for the polynomial (1) to be quadratic, it is necessary to equate the coefficients of “x” of the fifth, fourth and third degrees to zero, and the coefficient of “x” of the second degree to 120. As a result, we obtain the following system of four equations with six variables: -A1+5 A2-10 A3+10 A4-5 A5+A6=0 5 A2-20 A3+30 A4-20 A5+5 A6=0 5 A1-35 A2 +70 A3-50 A4+5 A5+5 A6=0 -5 A2+80 A3-150 A4+80 A5-5 A6=120. If we fix two numbers A5 and A6, then all the rest will be expressed by the following formulas: A1=5·(A5+4)-4·A6; A2=4·(A5+3)-3·A6; A3=3·(A5+2)-2·A6; A4=2·(A5+1)-1·A6. It follows from the theorem that the quadratic factor will be expressed by the formula φ(x)=x2+(A6-A5-3) x+ A4. (4) Definition: A sequence of integers given by the following relations; A3=3·(A5+2)-2·A6; A4=2·(A5+1)-1·A6 is called a quadratic Lagrangian series Definition: a quadratic Lagrangian series is called a “candidate” if all its numbers Ak are divisors of the corresponding values of the function f(xk), k=1;2;3;4 ;5;6. For all candidates, we construct the quadratic factor φ(x) using formula (4) and check it for divisibility with f(x). A1=5·(A5+4)-4·A6; A2=4·(A5+3)-3·A6
A3 A4+ d+4 A4 A5+ d+2 A5 A5 4. Simplified form of quadratic Lagrangian series. The formulas for the quadratic Lagrangian series can be simplified. To do this, the letter “d” will denote the difference A5-A6, then the numbers of the quadratic Lagrange series will look like simpler formulas and convenient for their construction: A1 A2 A2+ d+8 A3+ d+6 Example: A5=7; A6=10 compose a quadratic Lagrangian series. Let's find d=7-10=-3, then using the formulas in the table we will find the numbers of this series: A1 A2+ d+8 10+(- 3)+8 15 Answer: 15; 10; 7; 6; 7; 10. Consider an example of factoring the reduced polynomial of the fifth degree: f(x)=x5-5x4+13x3-22x2+27x- 20. A5 A2 A3+ d+6 A5 7+(-3)+6 6+(-3) +4 7+(-3)+2 7 7 10 A4 A5+ d+2 A3 A4+ d+4 7 6 A6 A6 A6 A6 10 10 1) Using Horner’s scheme, we find the values of the function at x=-3; -2;-1; 0;1;2. To do this, let's make a table: 1 1 1 1 1 1 1 -5 -8 -7 -6 -5 -4 -3 13 37 27 19 13 9 7 -22 -133 -76 -41 -22 -13 -8 -3 - 2 -1 0 1 2 2) Determine whether this polynomial has linear factors. To do this, we write down the resulting function values in table line No. 3. From these we choose the number that has the smallest number of divisors. In our example, this is the number “2”. Let's write all its integer divisors in a column. For each divisor of the number "2" in -20 -1298 -378 -88 -20 -6 2 27 426 179 68 27 14 11
line we write linear Lagrangian series. We will select candidates from them and check for divisibility with the given polynomial f(x). Table No. 3: -1298 A1 -378 A2 -88 A3 -20 A4 -3 0 -4 -5 -6 A5 0 -2 1 -3 2 A6 1 -1 2 -2 In this table No. 3, cells are marked in gray which contain numbers that are not divisors of the corresponding values of the function f(x). There is no need to fill in empty cells, since the constructed quadratic Lagrangian series with a number in a gray cell is certainly not a “candidate”. From this table No. 3 it is clear that there are no “candidates”. This means that this polynomial f(x)=x5-5x4+13x3- 22x2+27x-20 cannot be expanded into linear factors. 3) Determine whether this polynomial has quadratic factors. To do this, we write down the resulting function values in table line No. 4. From these we choose two numbers that have the smallest number of divisors. In our example, these are the numbers “2” and “-6”; we will write their divisors in columns. For each pair of divisors of the numbers “2” and “-6”, we write quadratic Lagrangian series in a line. We will select candidates from them and check them for divisibility with the given polynomial f(x). Table No. 4: -1298 A1 A2+ d+8 -378 A2 A3+ d+6 5 -88 A3 A4+ d+4 1 10 -5 -20 A4 A5+ d+2 3 -1 5 -3 7 -5 -6 A5 A5 1 -1 2 -2 3 -3 2 A6 A6 1 1 1 1 1 1 d d= A5- A6 d=0 d=-2 d=1 d=-3 d=2 d=-4
19 7 2 14 -2 14 7 22 2 13 6 11 5 2 5 -1 8 -4 7 19 1 13 -11 5 1 7 -1 9 -3 15 -9 2 -2 4 -4 6 -6 12 -12 6 2 8 0 10 -2 16 -8 6 -6 1 -1 2 -2 3 -3 6 -6 1 -1 2 -2 3 -3 6 -6 1 -1 2 -2 3 -3 6 -6 1 1 -1 -1 -1 -1 -1 -1 -1 -1 2 2 2 2 2 2 2 2 -2 -2 -2 -2 -2 -2 -2 -2 d=5 d=-7 d= 2 d=0 d=3 d=-1 d=4 d=-2 d=7 d=-5 d=-1 d=-3 d=0 d=-4 d=1 d=-5 d=4 d=-8 d=3 d=1 d=4 d=0 d=5 d=-1 d= 8 d=-4 “cand.” "cand." In this table No. 4 we see two “candidates”. With their help, using the formula φ(x)=x2+(A6- A5-3) x+ A4 we find the square factors: φ1(x)=x2-3x+ 4; φ2(x)=x2+x-4. The check shows that one of the two factors is true, this is φ1(x)=x2-3x+ 4, and the other factor turned out to be extraneous. Answer: x5-5x4+13x3-22x2+27x-20=(x2-3x+ 4)·(x3-2x2+3x-5). In this table No. 4 we obtained 32 quadratic Lagrangian series. This number is determined by the number of different pairs of divisors, both positive and negative, which are located in two adjacent columns. two function values,
5. Reducing the number of quadratic Lagrange series. By definition, If the values of the function, the number of divisors, which are minimal, are not located nearby, then you can use the following theorem: Theorem 3 Let A4 and A6 be known, then A5=(A4+ A6 ·1):2-1 Let A3 and A6 be known, then A5= (A3+ A6 ·2):3-2 Let A2 and A6 be known, then A5=(A2+ A6 ·3):4-3 Let A1 and A6 be known, then A5=(A1+ A6 ·4):5-4. Proof: let’s prove the last equality A5=(A1+A6·4):5-4. quadratic Lagrangian numbers, A1=5·(A5+4)-4·A6, we substitute this number into the original equality and obtain A5=(5·(A5+4)-4·A6+A6·4):5-4=(5 ·A5+20):5-4=A5+4-4=A5, which is what needed to be proven. Other equalities can be proved in a similar way. This theorem allows us to reduce the number of quadratic Lagrangian series. Let's consider the example we have already solved f(x)=x5-5x4+13x3-22x2+27x-20 and solve it for the case when we consider quadratic Lagrangian series constructed using divisors A4 and A6. Table No. 5: -1298 -378 A2 A1 A2+ A3+ d+6 d+8 d d = A5- A6 -88 A3 A4+ d+4 -20 A4 A5+ d+2 1 -1 5 -5 1 -1 -6 A5 ( A4+ A6 ·1):2-1 0 -1 2 -3 -1 -2 2 A 6 A 6 1 1 d =-2 1 d =1 1 d =-4 - d =0 1 d =-1 - 1 5 7 1 10 -5 5 2 14
19 11 7 22 2 2 14 -2 13 6 5 -1 8 -4 7 1 19 5 -5 2 -2 4 -4 10 -10 20 -20 2 -2 4 -4 10 -10 20 -20 1 -4 1 -1 2 -2 5 -5 10 -10 -1 -3 0 -4 3 -7 8 -12 “cand.” "cand." d =2 - 1 - 1 2 d =-1 2 d =-3 2 d =0 2 d =-4 2 2 2 2 - 2 - 2 - 2 - 2 - 2 - 2 - 2 - 2 d =1 d =-1 d =5 In this table No. 5 we received 24 quadratic Lagrange series. Since in the formula the sum of A4 and A6 must be divided by 2, therefore the divisors of A4 and A6 must be either both even or both odd. Due to this, the number of quadratic Lagrange series has decreased. If we use this Theorem 3 to write quadratic Lagrange series constructed using A1 and A6, then the number of series will be reduced to 12. Table No. 6: -378 -1298 A1 A2 2 A6 d -88 A3 -20 A4 -6 A5
"cand." A3+d+ 6 5 d=-4 d=0 “cand.” "cand." A5+d+ 2 -5 -1 A4+d+ 4 -5 1 (4A1+A6): 5-4 -3 -1 -15 -5 -7 7 -2 2 -26 -6 -10 12 A6 d=A5- A6 d=-4 1 1 d=-2 1 -1 -1 -1 2 2 2 -2 d=-4 -2 -2 A2+d+ 8 1 11 -59 -1 -11 -59 2 22 -118 - 2 -22 118 In table No. 6, the number of quadratic Lagrange series has been reduced to 12, since A5 is found according to the formula (4A1 + A6): 5-4 and A5 as an integer must be less than or equal to -6. In all tables, the black highlighted row is the "valid candidate". The remaining candidates are “imaginary”. For a polynomial of the sixth degree, it can be proven that the quadratic factor can be found using the formula: φ(x)=x2+ (A7 - A6 - 5) x+ A4, where the numbers are A1; A2; A3; A4; A5; A6; A7 form a quadratic Lagrangian series. 6. Conclusions: 1. This decomposition method using IML -2 14 -4 8 -4 4 -8 is a generalization of the “Horner scheme”. 2. Using this method, you can determine quadratic factors for polynomials above the fifth degree. 3. Using this method, you can study the properties of Lagrangian numbers for determining cubic polynomials in the expansion of polynomials of the fifth and higher degrees. 7. Literature: 1. A. N. Chebotarev “Fundamentals of Galois theory”, OMTI GTTI, 1934, 1 hour.
2. “Numbers and polynomials”, compiled by A.A. Egorov - M.: Quantum Bureau, 2000 / supplement to the magazine “Quantum” No. 6, 2000.
Task 1. Find gcd of polynomials
f(x)=x 4 –2x 3 –x+2, g(x)=x 4 –x 3 +x–1, h(x)=x 4 –4x 2 –x+2.
Solution. The GCD of polynomials can be found uniquely only up to a constant factor (constant non-zero factors do not affect the divisibility of polynomials). Therefore, we can agree to take as the GCD of polynomials the one whose leading coefficient is equal to 1.
By applying the Euclidean algorithm to polynomials with integer coefficients, we can, in order to avoid fractional coefficients, multiply the dividend or divisor by any non-zero number, not only starting with any of the successive divisions, but also during this division itself. This will, of course, lead to a distortion of the quotient, but the remainders of interest to us will acquire only a certain factor of zero degree.
To find the GCD of three polynomials, we first use the Euclidean algorithm to find the GCD of any two polynomials, for example d(x)=(f(x),h(x)), and then find the gcd d(x) And g(x).
Euclid's algorithm consists of sequential division of polynomials with a remainder. Let's divide first f(x) on h(x), then h(x) by the remainder obtained by division r(X) (the first remainder), then the first remainder by the second remainder, etc., until we get zero in the remainder. GCD of polynomials f(x) And h(x) will be the last non-zero remainder. The division process will be carried out using an “angle”.
| _ | x 4 -2x 3 -x+2 | x 4 -4x 2 -x+2 | _ | x 4 -4x 2 -x+2 | x 3 -2x 2 | ||||
| x 4 -4x 2 -x+2 | 1 | x 4 -2x 3 | x+2 | ||||||
| -2x 3 +4x 2 | _ | 2x 3 -4x 2 -x+2 | |||||||
| x 3 -2x 2 | 2x 3 -4x 2 | ||||||||
| _ | -x+2 | ||||||||
| x-2 | |||||||||
| 0 | |||||||||
| _ | x 3 -2x 2 | x-2 | ||
| x 3 -2x 2 | x 2 | |||
| 0 | ||||
This means the gcd of polynomials f(x) And h(x) is equal to binomial x–2.
d(x)=(f(x), h(x))=x–2.
Similarly, we find the gcd of polynomials d(x) And g(x), it will be equal to 1. Thus, ( f(x), g(x), h(x))=(g(x), (f(x), h(x)))=1.
Note . The "=" or "!!" sign means that during division, multiplication was performed by some number other than zero.
Task 2.Using the Euclidean algorithm to find polynomials u(x) And v(x), satisfying the equality f(x)u(x)+g(x)v(x)=d(x), Where d(x) – gcd of polynomials f(x) And g(x): f(x)=4x 4 –2x 3 –16x 2 +5x+9, g(x)=2x 3 –x 2 –5x+4.
Solution. Apply to polynomials f(x) And g(x) Euclidean algorithm. It must be remembered that here the arbitrariness that consists in multiplying polynomials by constant factors, which is possible when finding GCD, cannot be allowed, since here we will also use quotients, which can be distorted with the indicated arbitrariness.
As a result of division we get:
f(x)=g(x)q 1 (x)+r 1 (x),
Where q 1 (x)=2x, r 1 (x)= –6x 2 –3x+9,
g(x)=r 1 (x)q 2 (x)+r 2 (x),
Where q 2 (x)= –x/3+1/3, r 2 (x)= –x+1,
r 1 (x)=r 2 (x)q 3 (x)+r 3 (x),
Where q 3 (x)=6x+9, r 3 (x)=0.
Thus, the Euclidean algorithm is written here in three lines, and the greatest common divisor is equal to - r 2 (x)=x–1=d(x). To express d(x) through polynomials f(x) And g(x), we will find r 2 (x) from the second line of the Euclidean algorithm:
r 2 (x)=g(x)–r 1 (x)q 2 (x).
Substituting into this equality instead r 1 (x) its expression, found from the first line of the Euclidean algorithm, we obtain:
r 2 (x)=f(x)[–q 2 (x)]+g(x),
to get equality f(x)u(x)+g(x)v(x)=d(x), you need to multiply the previous equality by (–1), we get:
–r 2 (x)=f(x)q 2 (x) +g(x)[–1–q 1 (x)q 2 (x)]=d(x),
Where u(x)=q 2 (x), v(x)= –1–q 1 (x)q 2 (x).
After substituting polynomials into this equality q 1 (x), q 2 (x) we get:
u(x)= , v(x)= .
Task 3. Using the method of indefinite coefficients to select polynomials u(x) And v(x) so that f(x)u(x)+g(x)v(x)=1, (1) for polynomials f(x)=x 2 –2x–1, g(x)=2x 4 –3x 3 –6x 2 +2x+2.
Solution. Let's use the theorem: if d(x) is the gcd of polynomials f(x) And g(x), then we can find such polynomials u(x) And v(x), What
f(x)u(x)+g(x)v(x)=d(x).
In this case, we can assume that the degrees of the polynomials f(x) And g(x) is greater than zero, which is the degree u(x) less than degree g(x), and the degree v(x) less than degree f(x).
Polynomials f(x) And g(x) satisfy equality (1) if ( f(x),g(x))=1. In our case f(x) And g(x) are relatively prime polynomials, which means we can find the polynomial u(x)=ax 3 +bx 2 +cx+d and polynomial v(x)=ex+f.
Substituting into equality (1) instead f(x), g(x), u(x), v(x) their expressions, we get:
(x 2 – 2x– 1)(ax 3 +bx 2 +cx+d)+(2x 4 – 3x 3 – 6x 2 + 2x+ 2)(ex+f)=1
(a+ 2e)x 5 + (b– 2a+ 2f– 3e)x 4 + (c– 2b–a– 3f– 6e)x 3 + (d– 2c–b– 6f+ 2e)x 2 +(–2d–c+ 2f+ 2e)x––d+ 2f= 1.
Thus, we have the equality of two polynomials: on the left side a polynomial of degree five with undetermined coefficients, and on the right side a polynomial of degree zero. Two polynomials are equal if their coefficients are equal for the same powers of the unknown.
Equating the coefficients for the same degrees of the unknown, we obtain a system of six linear equations with unknowns a, b, c, d, e, f:
Solving it, we get: d= 3, e=–1, f= 2, c=–4, b=–3, a= 2.
Thus, the required polynomials u(x) And v(x) will be:
u(x)=2x 3 –3x 2 –4x+3, v(x)= –x+2.
Task 4. Using Horner’s scheme, calculate f(A) and expand the polynomial f(x) by degrees x–A, Where f(x)=x 4 +2x 3 –7x 2 +3X–1, A=2.
Solution. According to Bezout's theorem, the remainder of a polynomial is f(x) to a linear binomial x–A equal to the value f(A) polynomial at x=A.
Division by “angle” can be written more simply: if f(x)=a 0 x n+a 1 x n –1 +a 2 xn– 2 + …+a n –1 x+a n, then the coefficients of the quotient q(x)=b 0 x n–1 + b 1 x n –2 + b 2 x n –3 + …+b n–1 and remainder r from division f(x) on x–a can be found using Horner's scheme:
f(2)=9=r 1, and the quotient of division f(x) on x–2 yes q 1 (x)=x 3 +4x 2 +x+5, i.e. f(x)=
=(x–2)q 1 (x)+r 1
Then, according to Horner’s scheme, we divide q 1 (x) on x–2, we get the quotient q 2 (x) and the remainder r 2, further q 2 (x) divide by x–2, we get q 3 (x) And r 3, etc.
For a polynomial f(x) we get:
f(x)=(x–2)q 1 (x)+r 1 =(x–2)[(x–2)q 2 (x)+r 2 ]+r 1 =(x–2) 2 q 2 (x)+r 2 (x–2)+r 1 =
=(x––2) 2 [(x–2)q 3 (x)+r 3 ]+r 2 (x–2)+r 1 =(x–2) 3 q 3 (x)+r 3 (x–2) 2 +r 2 (x–2)+r 1 =
=(x–2) 3 [(x––2)q 4 (x)+r 4 ]+r 3 (x–2) 2 +r 2 (x–2)+r 1 =(x–2) 4 q 4 (x)+r 4 (x–2) 3 +r 3 (x–2) 2 +r 2 (x–2)+ +r 1 = r 5 (x–2) 4 +r 4 (x–2) 3 +r 3 (x–2) 2 +r 2 (x–2)+r 1.
Thus, the coefficients in the expansion of the polynomial f(x) by degrees x–2 are equal, respectively, to the remainders from the division of polynomials f(x), q 1 (x), q 2 (x), q 3 (x), q 4 (x) on x–2.
The entire solution can be written in a table:
| –7 | –1 | ||||
From the table it is clear that r 5 =1, r 4 =10, r 3 =29, r 2 =31, r 1 =9 and
f(x)= (x–2) 4 +10(x–2) 3 +29(x–2) 2 +31(x–2)+9.
Task 5.Prove that .
Solution. Let's consider a polynomial. Number X= –1 is the root of the polynomial f(x) and by Bezout's theorem f(x) is completely divisible by X+1, i.e. f(x)=(x+1)g(x), Where g(x) is a polynomial with integer coefficients, therefore X 11 +1 is divided by X+1 for any integer X. Let's put X=3 5 . We get, i.e. , and because , we conclude that .
Comment. From the rules for “dividing by an angle” a polynomial f(x) to a polynomial g(x) it is immediately clear that if the polynomials f(x) And g(x) with integer coefficients, and g(x) reduced, then the quotient and remainder are polynomials with integer coefficients.
Task 6. Residues from division of a polynomial f(x) into binomials X+5 and X-3 equals –9 and 7 respectively. Find remainders when dividing this polynomial by a polynomial g(x)=(x+5)(x-3).
Solution. By Bezout's theorem f(–5)= –9, f(3)=7. When dividing a polynomial f(x) to a polynomial g(x)=x 2 +2x–15 we get some quotient q(x) and the remainder p(x)=ax+b, i.e. f(x)=(x 2 +2x–15)q(x)+(ax+b) .
Substituting into the last equality instead of X values –5 and 3 we obtain a system of two equations with two unknowns a And b:
Having solved it, we find a=2, b=1. Then the required remainder of the division of the polynomial f(x) to a polynomial g(x) will be equal to 2 X+1.
Task 7. Given a polynomial f(x) with integer coefficients and . Prove that .
Solution. Consider the expansion of the polynomial f(x) by degrees ( x–10):
due to the fact that it is divisible by 21, i.e. is divisible by 7. Similarly, it is divisible by 3. Due to the relative simplicity of 3 and 7, the number f(10)=a n divisible by 21.
Task 8. Expand the polynomial x 7 +3 into the product of polynomials of no higher than the second degree with real coefficients.
Solution. Let's find the roots of the polynomial x 7 +3, they will be
Giving k values 0, 1, …, 6, we get seven roots of the polynomial x 7 +3;
x 0 = ; x 1 = ; x 2 = ;
x 3 = = – ; x 4 = = ;
x 5 = = ;
x 6 = = .
Among them, only one is valid - this is x 3 = – , the rest are complex, and pairwise conjugate: x 6 = , x 5 = , x 4 = . In general
X k = , x k= .
Let's look at the work
(x–x k)(x– )=(x 2 –(x k+ )x+x k)=x 2 – x+ , where k=0, 1, 2.
We have a quadratic trinomial with real coefficients. Polynomial x 7 +3 can be decomposed into a product of 7 linear factors (a consequence of the fundamental theorem of algebra). By multiplying the factors that correspond to the conjugate roots, we obtain the desired expansion:
x 7 +3=(x–x 0)(x–x 1)(x–x 2)(x–x 3)(x–x 4)(x–x 5)(x–x 6)=(x–x 3)(x–x 0)(x–x 6)(x–x 1)
(x–x 5)(x–x 2)(x––x 4)=(x–x 3)(x–x 0)(x– )(x–x 1)(x– )(x–x 2)(x– )=(x+ )
(x 2 –(2· ) x+ )(x 2 –(2· ) x+ ) (x 2 ––(2· ) x+ ).
Task 9. Present the polynomial as the sum of the squares of two polynomials.
Solution. Any polynomial f(x) with real coefficients, positive for any, is represented as the sum of the squares of two polynomials. To do this, let's find the roots of the polynomial f(x): , decompose into linear factors, then multiply and , we obtain the required representation:
Let's denote , , we get f(x)=p 2 (x)+q 2 (x).
Task 10. Determine the multiplicity of the root of the polynomial. Find a polynomial of the greatest degree with simple roots, each root of which is the roots of a polynomial f(x).
1) Let's check whether the polynomial is a root f(x).
2) Let's check whether the first derivative of the polynomial is the root f(x)
. f¢(–1)=0, therefore – root
polynomial f(x), multiplicity not less than 2.
3), therefore the root of the multiplicity is not less than 3.
4) , root of the polynomial f(x) multiplicity 3, i.e. . To find a polynomial of the greatest degree with simple roots, each root of which is a root f(x), needed in the polynomial f(x) get rid of multiple roots. To do this, we divide the polynomial f(x) by the greatest common divisor of polynomials f(x) And f¢( x): . Therefore, the required polynomial will be , where , X=2 – simple roots of the polynomial.
Note: The multiplicity of the root could be checked using Horner's scheme.
Task 11. Separate multiples of a polynomial
Solution. By the theorem on multiple factors: if some irreducible polynomial over the field P g(x) is k- multiple of the polynomial f(x) with coefficients from the field P, then g(x) is ( k–1) – multiple factor of the derivative f(x). Thus, when moving from f(x) To f′( x) the multiplicity of all factors is reduced by 1. However, for the polynomial f′( x) there may be factors that do not exist f(x). To get rid of them we will find a gcd f(x) and f′( x). It will include only those factors that are included in f(x), however with a factor of 1 less.
Applying the Euclidean algorithm, we obtain
Since there is a polynomial of the third degree, the decomposition of which into factors is generally difficult, but which in turn can have multiple factors, we will apply to it a similar process of reducing the multiplicity of factors. We'll get it. So the multiplier X–1 is included in with a multiplicity of 1, and therefore, it is included in with a multiplicity of 2. Divide by ( X–1) 2 , let's find . Hence we have: multiplier ( X–1) included in f(x) with a multiplicity of 3, and X+3 with a multiple of 2. Dividing f(x) to the polynomial , we get
Task 12. Prove that the number is irrational.
Solution. This number is the root of the reduced integer polynomial, which has no rational roots, because all its rational roots are integers and must be divisors of the number 5.
Task 13. Find rational roots of the polynomial
f(x)=6x 4 +19x 3 –7x 2 –26x+12.
Solution. If a rational irreducible fraction that is the root of a polynomial f(x)=A 0 x n +a 1 xn– 1 +a 2 xn– 2 +…+a n– 1 x+a n with integer coefficients, then:
1. k there is a divisor A 0 ;
2. p there is a divisor a n ;
3. p–mk there is a divisor f(m) for any integer m.
In our case: k can take values: ±1, ±2, ±3, ±6, and p– ±1,±2, ±3, ±4, ±6, ±12. Now it would be possible to check each of these numbers of the form by substitution into a polynomial or using Horner’s scheme. However, many of these numbers can be "weeded out" in a simpler way. Let us find the boundaries of the real roots of this polynomial VG x =1+, NG x = –(1+), where A is the largest of the absolute values of the coefficients, and A 0 – coefficient at x n or VG x =1+, where k– index of the first negative coefficient of the polynomial f(x), A B– the largest of the absolute values of its negative coefficients (this method is applicable when A 0 >0). In our example k=2, B=26, A 0 =6. VG x =1+< 4.
To find the lower boundary using this method, it is enough to f(x) instead of x substitute (– x) and use the following rule: the lower bound of the real roots of the polynomial f(x) is equal to the upper bound of the real roots of the polynomial f(–x), taken with the opposite sign. In our case
f(–x)=6x 4 –19x 3 –7x 2 +26x+12, and 0 =6, k=1, B=19. VG x =1+<5, значит, нижняя граница – НГ х = –5. Итак, корни многочлена заключены в интервале (–5,4). Более точные границы можно было найти по методу Ньютона. Воспользуемся еще тем, что если – корень f(x), then integer. We'll find f(1)=4,
f(–1)=13, then – integer, – integer, if – root f(x).
We check all kinds of fractions, taking into account the boundaries of the roots.
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| ts | d | ts | d | d | d | ts | d | ts |
During this check, rational numbers 2, –3, , appeared - “candidate roots”, we check them according to Horner’s scheme, making sure that f(2)≠0, , f(–3)=0, . For a fourth degree polynomial we found two roots, which means f(x) multiple ( x+3) or f(x)=(6x 2 +4x–8)(x+3) . Roots of a polynomial g(x)=6x 2 +4x–8 we find directly x= are non-rational numbers.
Task 14. Prove that this equation does not have non-zero integer solutions.
Solution. The left side of the equality is a homogeneous polynomial of the fourth degree. Let's divide both sides of the equality by X 4 . We get
Let's put it then. A given equation has a nonzero integer solution if and only if the polynomial has rational roots. A reduced polynomial is integer, all its rational roots are: firstly, integers; secondly, divisors of the free term 9, i.e. must belong to the set (±1, ±3, ±9). By direct verification, you can make sure that not a single element of this set is the root of a polynomial, i.e. this polynomial has no rational roots, which means that the given equation has non-zero integer roots.
Task 15. For what natural n will the number be prime?
Solution. Let's show that. Indeed, if A is an arbitrary root of the polynomial, then A will be the root of the polynomial, i.e. A 3 =1 and A 2 +A+1=0.
Let's consider, i.e. A– root of a polynomial. Because A is an arbitrary root of a polynomial, then each root of a polynomial is a root of a polynomial, therefore, where P(x) is a polynomial with integer coefficients.
Suppose then, i.e. .
Let's consider the cases and .
2. When is a prime number.
A natural number is represented as the product of two natural numbers. From this we can see that it can be simple, if or , we discard it.
When , and is presented as the product of two natural numbers greater than 1, which means that this number is composite.
Task 16. Solve equations in the field of complex numbers:
1)x 3 +6x+2=0; 2) x 3 –9x 2 +18x–28=0; 3) x 4 -2x 3 +4x 2 -2x+ 3=0.
1. Solve the equation x 3 +6x+2=0.
For the roots of a cubic equation x 3 +ax+b=0 there is the so-called Cardano formula: x i =u i +v i (i=0, 1, 2), where u 0 , u 1 , u 2 – radical value
u= and v i= . In our case, A=6, b=2,
u= = = = = (cos + i sin), where l=0, 1, 2. Substituting instead l values 0, 1, 2, we get: u 0 = , u 1 =
= (cos + i sin )= (– + i), u 2 = (cos + i sin )= (– – i ),
v 0 = = = = ,
v 1 = = = = ( +i ),
v 2 = = = = ( –i ),
x 0 = u 0 +v 0 = – , x 1 =u 1 +v 1 = , x 2 = u 2 +v 2 =
Answer: - ; .
2. Solve the equation x 3 –9x 2 +18x–28=0.
Let us reduce our equation to an equation of the form y 3 +ay+b=0, making the substitution x=y– =y+3, (a 0 , a 1 – coefficients for x 3 and x 2). We get:
y 3 –9y–28=0. Its solutions are found using the Cardano formula: y i =u i+v i, (i=0, 1,…2),
Where u 0 =3, u 1 = , u 2 = , v 0 =1 , v 1 = , v 2= ,
y 0 =4, y 1 = , y 2 = , x 0 =7, x 1 = , x 2 = .
Answer: 7; .
3. Solve the equation x 4 -2x 3 +4x 2 -2x+ 3=0.
Let's use the Ferrari method. Let us leave terms with on the left side of the equation X 4 and X 3 and add it to a complete square:
Now let’s add terms with a new unknown to both sides y so that the left side becomes a square again (regardless of the value y)
Here are the coefficients before the powers x on the right side depend on an uncertain quantity y. Let's choose the value of y so that the right side becomes a square. To do this, it is necessary that the discriminant of the square (relative to x) of the trinomial on the right side was equal to zero. Equating this discriminant to zero we get:
from here y=4 and .
Substituting y=4 into equation (*), we get: or . Taking the square root from both sides of the resulting equation, we obtain two quadratic equations: and or and . Having solved them, we find the 4 roots of our equation: , .
Answer: , .
Task 17. Given polynomials
f(x)=x 3 –3x 2 +2x–5, g(x)=x 3 +3x 2 –1.
1) Determine the number of real roots of each;
2) Using Sturm’s theorem, find the interval ( a, b), Where b–a=1 containing the largest root x 0 polynomial g(x);
3) Calculate the root with an accuracy of 0.0001 x 0 using the linear interpolation method and Newton's method;
1. If the odds a And b equations x 3 +ax+b=0 are real, then the number of real roots of this equation is completely determined by the sign of the number D = – 4a 3 – 27b 2, called the discriminant of the polynomial x 3 +ax+b, in the following way:
a) for D=0, all three roots are real, two of them are equal;
b) for D>0 – all three roots are valid;
c) at D<0 – один корень действительный, два мнимых.
In our case: f(x)=x 3 –3x 2 +2x–5 or putting x=y+1, y 3 –y–5=0, i.e. D=4–27·25<0, поэтому многочлен f(x) has one real root.
2. For a polynomial g(x) we determine the number of real roots by establishing the number of sign changes in the Sturm system of the polynomial g(x) when moving from –∞ to +∞. We will also find the entire boundaries between which each of these roots is located, and we will not build a graph of this function in advance.
Any polynomial g(x) with real coefficients and no multiple roots, has the Sturm system. If a polynomial has multiple roots, then you need to get rid of them by dividing the polynomial g(x) on the gcd of polynomials g(x) And g"(x). Sturm polynomial system g(x) can be constructed as follows: put g 1 (x)=g"(x), then divide g(x) on g 1 (x) and the remainder of this division, taken with the opposite sign, is taken as g 2 (x), i.e. g(x)=g 1 (x)h 1 (x)–g 2 (x). In general, if polynomials g k–1 ( x) And g To ( x) have already been found, then g k+1 ( x) will be the remainder of the division g k–1 ( x) on g To ( x), taken with the opposite sign:
g k–1 ( x)=g To ( x)q To ( x)– g k+1 ( x).
Let's find the Sturm system for g(x), using the specified method. Moreover, in the process of division, we will, unlike the Euclidean algorithm, multiply and reduce only by arbitrary positive numbers, because the signs of the remainders play an important role in Sturm's method. We will get such a system
g(x)=x 3 +3x 2 –1,
g 1 (x)=3x 2 +6x,
g 2 (x)=2x+1,
g 3 (x)=1.
Let us determine the signs of the polynomials of this system at x=–∞ and x= +∞, for which we look only at the signs of the leading coefficients and at the degrees of these polynomials. At +∞, the signs of all polynomials of the Sturm system will coincide with the signs of their highest terms, and at –∞ the signs of the polynomials of the Sturm system coincide with the signs of their highest coefficients for polynomials of even degree and are opposite to the signs of the highest polynomials of odd degree.
Thus, upon transition x from –∞ to +∞ the Sturm system loses three sign changes, so the polynomial g(x) has exactly three real roots (Sturm's theorem).
Let us continue the study of signs in the Sturm system, considering the intervals (0,1), (1,2), (2,3), etc., (0,–1), (–1,–2), (–2 ,–3), etc. Thus, we define the intervals ( A, b), Where a–b=1 containing three real roots and find the interval for x 0 .
Thus, the Sturm system of the polynomial g(x) loses one change of signs during the transition x-3 to -2, -1 to 0 and 0 to 1. Roots x 1 , x 2 , x 3 of this polynomial therefore satisfy the inequalities:
–3<x 1 <–2, –1<x 2 <0, 0<x 3 <1, т.е. наибольший корень x 0 (0,1).
3. Let us construct a schematic graph of the polynomial in the interval (0, 1) g(x), calculating the following values of the polynomials:
g(0)=–1, g(1)=3, g"(0)=0, g"(1)=9 (the function increases on the interval under consideration), g""(0)>0g""(1)>0 (function is convex).
A schematic graph of the function is presented in Fig. 1.
First, using the chord method on the segment (0,1), the curve y=g(x) is replaced by the chord AB and the abscissa is taken as the first approximate value of the root x=from the point of intersection of this chord with the axis x. Triangle KBC is similar to triangle CAE, therefore , or , or . In general .
Then, using Newton’s method, we draw a tangent y to schedule g(x) at point A(1, g(1)) (we draw a tangent at the point x=1, because g(1) and g""(1) of the same sign) and take the abscissa as another approximate value of the root x=R the point of intersection of this tangent with the Ox axis.
Let's write down the equation of the tangent passing through point A
y–g(1)=g"(1)(x–1).
Since this tangent passes through the point ( p, 0), then substituting these values into the tangent equation, we obtain
0–g(1)=g"(1)(p–1) or p=1– =1– .
In general p=b– .
More precise value of the desired root x 0 can now be searched in the new
interval ( A 1 , b 1), putting A 1 =0,3, b 1 =0.7. By repeating the chord method and Newton's method in the interval ( A 1 , b 1) we have: g(A 1)=–0,703; g(b 1)=0,813; g"(b 1)=5,67.
Because g(A 1) and g(b 1) different signs, then x 0 (A 1 ,b 1)
p 1 =0,7– .
Let's consider a new interval ( A 2 , b 2), putting A 2 =0,5, b 2 =0,55, g(A 2)=–0,125, g(b 2)=0,073875, g"(b 2)=4.2075, because g(A 2) and g(b 2) – different signs, then x 0 (A 2 ,b 2),
, p 2 =0,55– .
And finally, considering the interval ( A 3 , b 3), where A 3 =0,531, b 3 =0.532, let's find it more accurately x 0 .
Task 18.The following rational fraction, where
f(x)= 2x 4 –10x 3 +7x 2 +4x+3, g(x)=x 5 –2x 3 +2x 2 –3x+2,
expand into the sum of simple fractions in the field of rational numbers.
Solution. Every proper rational fraction has a unique decomposition into the sum of simple fractions. In our case, the degree f(x) less than degree g(x), so the fraction is correct.
