What can be cooked from squid: quick and tasty
Definition. If each of two polynomials is divisible by the third without remainder, then it is called the common divisor of the first two.
Greatest common divisor (GCD) two polynomials is called their highest common divisor.
GCD can be found using the factorization into irreducible factors or using the Euclidean algorithm.
Example 40 Find gcd polynomials and 
.
Solution. Let us factor out both polynomials:
The decomposition shows that the required GCD is the polynomial ( NS– 1).
Example 41 Find the gcd of polynomials 
and 
.
Solution. Let us factor out both polynomials.
For polynomial 
NSNS- 1) according to Horner's scheme.
For polynomial 
possible rational roots are the numbers 1, 2, 3 and 6. Using the substitution, we make sure that NS= 1 is the root. Divide the polynomial by ( NS- 1) according to Horner's scheme.
Therefore, where the decomposition of the square trinomial 
was produced by Vieta's theorem.
Comparing the factorization of polynomials, we find that the required GCD is the polynomial ( NS– 1)(NS– 2).
Similarly, you can find the GCD for several polynomials.
However, the method of finding the GCD by factoring is not always available. The way to find GCD for all cases is called Euclid's algorithm.
The scheme of Euclid's algorithm is as follows. One of the two polynomials is divided by the other, the degree of which is not higher than the degree of the first. Further, for the dividend, each time the polynomial that served as the divisor in the previous operation is taken, and for the divisor, the remainder obtained in the same operation is taken. This process stops as soon as the remainder is equal to zero. Let's show this algorithm by examples.
Consider the polynomials used in the previous two examples.
Example 42 Find the gcd of polynomials 
and 
.
Solution. Divide 
on 
"Corner":
x
Now let's divide the divisor 
for the remainder NS– 1:
x+ 1
Since the last division occurred without a remainder, the GCD will be NS- 1, that is, the polynomial used as the divisor in this division.
Example 43 Find the gcd of polynomials 
and 
.
Solution... To find the GCD, we will use the Euclidean algorithm. Divide 
on 
"Corner":
1
Let's make the second division. To do this, you would have to divide the previous divisor 
for the remainder 
, but since 
=
, for convenience we will divide the polynomial 
not on 
and on 
... This replacement does not change the solution of the problem, since the GCD of a pair of polynomials is determined up to a constant factor. We have:
The remainder turned out to be equal to zero, which means that the last divisor, i.e., the polynomial
and will be the desired GCD.
Fractional rational functions
Definitions and statements for 2.5 can be found in.
A fractional rational function with real coefficients is an expression of the form 
, where 
and 
Are polynomials.
A fractional rational function (in what follows we will call it "fraction") is called correct if the degree of the polynomial in the numerator is strictly less than the degree of the polynomial in the denominator. Otherwise, it is called wrong.
The algorithm for reducing an incorrect fraction to a correct one is called "separating the whole part".
Example 44 Select whole fraction: 
.
Solution. In order to select the whole part of a fraction, it is necessary to divide the numerator of the fraction by its denominator. Divide the numerator of this fraction by its denominator with a "corner":
Since the degree of the resulting polynomial is less than the degree of the divisor, the division process is over. Eventually:
=
... The resulting fraction 
is correct.
Fraction of the species 
is called the simplest if φ ( x
)
Is an irreducible polynomial, and the degree 
less than the degree φ ( x
).
Comment. Note that the degrees of the numerator and the irreducible polynomial in the denominator are compared (excluding the degree of α).
For fractions with real coefficients, there are 4 types of simple fractions:
Any correct fraction 
can be represented as a sum of simple fractions, the denominators of which are all possible divisors 
.
Algorithm for decomposition of a fraction into the simplest:
If the fraction is incorrect, then we select the whole part, and decompose the resulting correct fraction into the simplest ones.
Factor the denominator of the correct fraction.
We write a regular fraction as a sum of the simplest fractions with undefined coefficients.
Bring the sum of the fractions on the right side to a common denominator.
We find undefined coefficients:
Either by equating the coefficients at the same degrees of the left and right given numerators;
Or by substituting specific (usually the roots of their common denominator) values x.
We write down the answer taking into account the integer part of the fraction.
Example 45 Decompose into protozoa 
.
Solution. Since this fractional-rational function is incorrect, we select the whole part:
1
= 1 +
.
Expand the resulting fraction 
into the simplest. First, factor the denominator. To do this, find its roots using the standard formula:
Let us write the decomposition of a fractional rational function into elementary ones using indefinite coefficients:
Let us bring the right-hand side of the equality to a common denominator:
We compose a system by equating the coefficients at the same degrees in the numerators of the left and right fractions:
Answer: 
.
Example 46 Decompose into protozoa 
.
Solution. Since this fraction is correct (that is, the degree of the numerator is less than the degree of the denominator), it is not necessary to select the whole part. Let us factor out the denominator of the fraction:
Let us write the decomposition of this fraction into elementary ones using undefined coefficients:
According to the statement, the denominators of the simplest fractions should be all kinds of divisors of the denominator of the fraction:
... (2.2) It would be possible to compose a system of equations by equating the numerators of the left and right fractions, but in this example the calculations will be too cumbersome. The following technique will help to simplify them: substitute the roots of the denominator in the numerators in turn.
At x = 1:
At NS= ‑1:
Now to determine the remaining coefficients A and WITH it will be enough to equate the coefficients at the highest degree and the free terms. They can be found without expanding the parentheses:
There is 0 on the left side of the first equation, since the numerator of the left fraction in (2.2) does not contain a term with 
, and in the right fraction of the term with 
coefficient A +
C... The left side of the second equation contains 0, since the free term in the numerator of the left fraction in (2.2) is equal to zero, and the free term in the numerator of the right fraction in (2.2) is (- A +
B +
C
+
D). We have:
Answer: 
.
DIVISION OF POLYMONIES. ALGORITHM Euclidean
§1. Division of polynomials
When dividing, polynomials are represented in canonical form and are arranged in decreasing degrees of any letter, relative to which the degree of the dividend and the divisor is determined. The degree of the dividend must be greater than or equal to the degree of the divisor.
The result of division is the only pair of polynomials - quotient and remainder, which must satisfy equality:
< делимое > = < делитель > ´ < частное > + < остаток > .
If a polynomial of degree n Pn (x ) is divisible,
Degree polynomial m Rk (x ) is a divisor of ( n ³ m),
The polynomial Qn - m (x ) - private. The degree of this polynomial is equal to the difference between the degrees of the dividend and the divisor,
A polynomial of degree k Rk (x ) is the remainder ( k< m ).
That equality
Pn (x) = Fm (x) × Qn - m (x) + Rk (x) (1.1)
must be fulfilled identically, that is, remain valid for any real values of x.
Note again that the degree of the remainder k must be less than the divisor power m ... The purpose of the remainder is to complete the product of polynomials Fm (x) and Qn - m (x ) to a polynomial equal to the dividend.
If the product of polynomials Fm (x) × Qn - m (x ) gives a polynomial equal to the dividend, then the remainder R = 0. In this case, they say that division is performed without a remainder.
Let us consider the algorithm for dividing polynomials using a specific example.
Let it be required to divide the polynomial (5x5 + x3 + 1) by the polynomial (x3 + 2).
1. Divide the senior term of the dividend 5x5 by the senior term of the divisor x3:
It will be shown below that this is the first member of the quotient.
2. The divisor is multiplied by the next (initially first) term of the quotient and this product is subtracted from the dividend:
5x5 + x3 + 1 - 5x2 (x3 + 2) = x3 - 10x2 + 1.
3. The dividend can be represented as
5x5 + x3 + 1 = 5x2 (x3 + 2) + (x3 - 10x2 +
If in action (2) the degree of the difference turns out to be greater than or equal to the degree of the divisor (as in the example under consideration), then with this difference the actions indicated above are repeated. Wherein
1. The senior term of the difference x3 is divided by the senior term of the divisor x3:
It will be shown below that in this way the second term in the quotient is found.
2. The divisor is multiplied by the next (now, second) term of the quotient and this product is subtracted from the last difference
X3 - 10x2 + 1 - 1 × (x3 + 2) = - 10x2 - 1.
3. Then, the last difference can be represented as
X3 - 10x2 + 1 = 1 × (x3 + 2) + (–10x2 +
If the degree of the next difference turns out to be less than the degree of the divisor (as when repeating in action (2)), then the division is completed with a remainder equal to the last difference.
To confirm that the quotient is the sum (5x2 + 1), we substitute in equality (1.2) the result of the transformation of the polynomial x3 - 10x2 + 1 (see (1.3)): 5x5 + x3 + 1 = 5x2 (x3 + 2) + 1× (x3 + 2) + (- 10x2 - 1). Then, after taking the common factor (x3 + 2) outside the brackets, we finally get
5x5 + x3 + 1 = (x3 + 2) (5x2 + 1) + (- 10x2 - 1).
That, in accordance with equality (1.1), should be considered as the result of dividing the polynomial (5x5 + x3 + 1) by the polynomial (x3 + 2) with the quotient (5x2 + 1) and remainder (- 10x2 - 1).
It is customary to formalize these actions in the form of a scheme called "dividing by a corner". In this case, in the record of the dividend and subsequent differences, it is desirable to produce the terms of the sum over all decreasing powers of the argument without a gap.
| |
5x5 + 10x2 5x2 + 1
| |
X3 + 2
–10x2 + 0x - 1
position: relative; z-index: 1 ">We see that the division of polynomials is reduced to a sequential repetition of actions:
1) at the beginning of the algorithm, the senior term of the dividend, subsequently, the senior term of the next difference is divided by the senior term of the divisor;
2) the result of division gives the next term in the quotient, by which the divisor is multiplied. The resulting work is written under the dividend or the next difference;
3) the lower polynomial is subtracted from the upper polynomial and, if the degree of the resulting difference is greater than or equal to the degree of the divisor, then actions 1, 2, 3 are repeated with it.
If the degree of the obtained difference is less than the degree of the divisor, then the division is complete. In this case, the last difference is the remainder.
Example # 1
position: absolute; z-index: 9; left: 0px; margin-left: 190px; margin-top: 0px; width: 2px; height: 27px ">
4x2 + 0x - 24x2 ± 2x ± 2
Thus, 6x3 + x2 - 3x - 2 = (2x2 - x - 1) (3x + 2) + 2x.
Example No. 2
A3b2 + b5
A3b2 a2b3
- a2b3 + b5
± a2b3 ± ab4
Ab4 + b5
- ab4 b5
Thus , a5 + b5 = (a + b) (a4 –a3b + a2b2 - ab3 + b4).
Example №3
| |
X5 x4y x4 + x3y + x2y2 + xy3 + y4
X3y2 - y5
X3y2 ± x2y3
Hu 4 - y 5
Hu 4 - y 5
Thus, x5 - y5 = (x - y) (x4 + x3y + x2y2 + xy3 + y4).
A generalization of the results obtained in examples 2 and 3 are two formulas for abbreviated multiplication:
(x + a) (x2 n - x2 n –1 a + x2 n –2 a 2 -… + a2n) = x 2n + 1 + a2n + 1;
(x - a) (x 2n + x 2n – 1 a + x 2n – 2 a2 +… + a2n) = x 2n + 1 - a2n + 1, where n Î N.
Exercises
Perform actions
1. (- 2x5 + x4 + 2x3 - 4x2 + 2x + 4): (x3 + 2).
Answer: - 2x2 + x +2 - quotient, 0 - remainder.
2. (x4 - 3x2 + 3x + 2): (x - 1).
Answer: x3 + x2 - 2x + 1 - quotient, 3 - remainder.
3. (x2 + x5 + x3 + 1): (1 + x + x2).
Answer: x3 - x2 + x + 1 - quotient, 2x - remainder.
4. (x4 + x2y2 + y4): (x2 + xy + y2).
Answer: x2 - xy + y2 - quotient, 0 - remainder.
5. (a 3 + b 3 + c 3 - 3 abc): (a + b + c).
Answer: a 2 - (b + c) a + (b 2 - bc + c 2 ) - quotient, 0 - remainder.
§2. Finding the Greatest Common Divisor of Two Polynomials
1. Euclid's algorithm
If each of two polynomials is divisible by a third without a remainder, then this third polynomial is called a common divisor of the first two.
The greatest common divisor (GCD) of two polynomials is called their greatest common divisor.
Note that any number that is not equal to zero is a common divisor of any two polynomials. Therefore, any number that is not equal to zero is called the trivial common divisor of these polynomials.
Euclid's algorithm proposes a sequence of actions that either leads to finding the GCD of two given polynomials, or shows that such a divisor in the form of a polynomial of the first or greater degree does not exist.
Euclid's algorithm is implemented as a sequence of divisions. In the first division, the polynomial is considered more as a dividend, and to a lesser extent as a divisor. If the polynomials for which the GCD is found have the same degrees, then the dividend and the divisor are chosen arbitrarily.
If, during the next division, the remainder of the polynomial has degree greater than or equal to 1, then the divisor becomes divisible, and the remainder becomes a divisor.
If at the next division of the polynomials the remainder is equal to zero, then the GCD of these polynomials is found. It is the divisor at the last division.
If, at the next division of polynomials, the remainder turns out to be a non-zero number, then for these polynomials there is no GCD other than trivial ones.
Example # 1
Reduce fraction 
.
Solution
Find the GCD of these polynomials using the Euclidean algorithm
| |
X3 + 7x2 + 14x + 8 1
- x2 - 3x - 2
| |
X3 + 3x2 + 2x - x - 4
3x2 + 9x + 6
3x2 + 9x + 6
| |
position: absolute; z-index: 49; left: 0px; margin-left: 209px; margin-top: 6px; width: 112px; height: 20px "> 
font-size: 14.0pt; line-height: 150% "> Answer:
font-size: 14.0pt; line-height: 150% "> 2. Possibilities of simplifying the calculations of GCD in the Euclidean algorithm
Theorem
When multiplying the dividend by a number that is not zero, the quotient and the remainder are multiplied by the same number.
Proof
Let P be the dividend, F be the divisor, Q be the quotient, R - remainder. Then,
P = F × Q + R.
Multiplying this identity by the number a ¹ 0, we get
a P = F × (a Q) + a R,
where the polynomial a P can be viewed as a dividend, and the polynomials a Q and a R - as a quotient and remainder obtained by dividing a polynomial a P by a polynomial F ... Thus, when multiplying the dividend by the number a ¹ 0, quotient and remainder are also multiplied by a, h. d.
Consequence
Multiplying a divisor by a number a ¹ 0 can be thought of as multiplying the dividend by a number.
Therefore, when multiplying the divisor by the number a ¹ 0 is the quotient and the remainder is multiplied by.
Example No. 2
Find the quotient Q and remainder R when dividing polynomials
Font-size: 14.0pt; line-height: 150% "> Solution
To pass in the dividend and divisor to integer coefficients, we multiply the dividend by 6, which will lead to multiplication by 6 of the desired quotient Q and remainder R ... After that, we multiply the divisor by 5, which will lead to the multiplication of the quotient 6 Q and remainder 6 R on . As a result, the quotient and remainder obtained by dividing polynomials with integer coefficients will differ by times from the sought values of the quotient Q and remainder R obtained by dividing these polynomials.
| |
| |
12y4 ± 18x3 30x2y2 6y2 - 2xy - 9x2 =
- 4х3 - 12х2у2 - 11х3у + 3х4
± 4х3 6х2у2 ± 10х3у
- 18x2y2 - x3y + 3x4
± 18x2y2 27x3y ± 45x4
- 28x3y + 48x4 = font-size: 14.0pt; line-height: 150% "> Therefore;
Answer: 
, 
.
Note that if the greatest common divisor of these polynomials is found, then multiplying it by any number that is not equal to zero, we also get the greatest divisor of these polynomials. This circumstance makes it possible to simplify calculations in the Euclidean algorithm. Namely, before the next division, the dividend or divisor can be multiplied by numbers selected in a special way so that the coefficient of the first term in the quotient is an integer. As shown above, the multiplication of the dividend and the divisor will lead to a corresponding change in the partial remainder, but such that, as a result, the GCD of these polynomials will be multiplied by some number equal to zero, which is acceptable.
Example No. 3
Reduce fraction 
.
Solution
Applying Euclid's algorithm, we get
| |
X4 x3 ± 3x2 font-size: 14.0pt; line-height: 150% "> 4 1
2x3 + 6x2 + 3x - 2
font-size: 14.0pt; line-height: 150% "> 2) 2 (x4 + x3 - 3x2 + 4) = 2x4 + 2x3 - 6x2 + 8 2x3 + 6x2 + 3x - 2
2x4 6x3 3x2 ± 2x x - 2
- 4x3 - 9x2 + 2x + 8
± 4x3 ± 12x2 ± 6x font-size: 14.0pt; line-height: 150% "> 4
3x2 + 8x + 4
| |
| |
6x3 font-size: 14.0pt "> 16x2 font-size: 14.0pt"> 8x 2x +
Let there be given nonzero polynomials f (x) and φ (x). If the remainder of dividing f (x) by φ (x) is zero, then the polynomial φ (x) is called the divisor of the polynomial f (x). The following statement holds: the polynomial φ (x) if and only if will be a divisor of the polynomial f (x) when there is a polynomial ψ (x) satisfying the equality f (x) = φ (x) ψ (x). A polynomial φ (x) is called a common divisor of arbitrary polynomials f (x) and g (x) if it is a divisor of each of these polynomials. According to the divisibility properties, all polynomials of degree zero belong to the number of common divisors of the polynomials f (x) and g (x). If these polynomials have no other common divisors, then they are called coprime and are written (f (x), g (x)) = 1. In the general case, the polynomials f (x) and g (x) can have common divisors depending on x.
As for integers, the concept of their greatest common divisor is introduced for polynomials. The greatest common divisor of nonzero polynomials f (x) and g (x) is their common divisor d (x), which is divisible by any common divisor of these polynomials. The greatest common divisor of the polynomials f (x) and g (x) is denoted by the symbols gcd, d (x), (f (x), g (x)). Note that this definition of GCD also holds for integers, although more often a different one is used, which is known to all students.
This definition raises a number of questions:
1. Is there a GCD for arbitrary nonzero polynomials f (x) and g (x)?
2. How to find the gcd of the polynomials f (x) and g (x)?
3. How many greatest common factors do the polynomials f (x) and g (x) have? And how do you find them?
There is a way to find the GCD of integers called the sequential division algorithm or Euclid's algorithm. It is applicable to polynomials and is as follows.
Euclid's Algorithm. Suppose given polynomials f (x) and g (x), degree f (x) ≥ degree g (x). We divide f (x) by g (x), we get the remainder r 1 (x). Divide g (x) by r 1 (x), we get the remainder r 2 (x). Divide r 1 (x) by r 2 (x). So we continue the division until the division is complete. The remainder r k (x) by which the previous remainder r k -1 (x) is completely divisible, and will be the greatest common divisor of the polynomials f (x) and g (x).
Let's make the following remark, which is useful in solving examples. Applying the Euclidean algorithm to polynomials to find the GCD, we can, in order to avoid fractional coefficients, multiply the dividend or cancel the divisor by any non-zero number, and not only starting any of the successive divisions, but also in the process of this division itself. This will lead to a distortion of the quotient, but the remainders of interest to us will acquire only a certain factor of zero degree, which, as we know, is allowed when looking for divisors.
Example 1. Find the GCD of the polynomials f (x) = x 3 –x 2 –5x – 3,
g (x) = x 2 + x – 12. Divide f (x) by g (x):
The first remainder r 1 (x) after reduction by 9 will be x – 3. We divide g (x) by r 1 (x):
.
The division happened entirely. Therefore, r 1 (x) = x – 3 is the GCD of the polynomials x 3 –x 2 –5x – 3 and x 2 + x – 12.
Example 2. Find the GCD of the polynomials f (x) = 3x 3 + 2x 2 –4x – 1,
g (x) = 5x 3 –3x 2 + 2x – 4. Multiply f (x) by 5 and divide 5f (x) by g (x):
The first remainder r 1 (x) will be 19x 2 –26x + 7. We divide g (x) by the first remainder, having previously multiplied g (x) by 19:
Multiply by 19 and continue dividing:
Reduce by 1955 and get the second remainder r 2 (x) = x-1. We divide r 1 (x) by r 2 (x):
.
The division was made entirely, therefore, r 2 (x) = x-1 is the GCD of the polynomials f (x) and g (x).
Example 3. Find the GCD of the polynomials f (x) = 3x 3 –x 2 + 2x – 4,
g (x) = x 3 –2x 2 +1.
. 
.
.
Answer:(f (x), g (x)) = x – 1.
This method of finding the GCD shows that if the polynomials f (x) and g (x) have both rational or real coefficients, then the coefficients of their greatest common divisor will also be rational or, respectively, real.
The polynomials f (x), g (x), and d (x) are related by the following relation, which is often used in various questions and is described by the theorem.
If d (x) is the greatest common divisor of the polynomials f (x) and g (x), then we can find polynomials u (x) and v (x) such that f (x) u (x) + g (x) v (x) = d (x). In this case, we can assume that if the degrees of the polynomials f (x) and g (x) are greater than zero, then the degree of u (x) is less than the degree of g (x), and the degree of v (x) is less than the degree of f (x).
Let us show by an example how to find the polynomials u (x) and v (x) for the given polynomials f (x) and g (x).
Example 4. Find the polynomials u (x) and v (x) so that f (x) u (x) + g (x) v (x) = d (x) if
A) f (x) = x 4 -3x 3 +1, g (x) = x 3 -3x 2 +1;
B) f (x) = x 4 -x 3 + 3x 2 -5x + 2, g (x) = x 3 + x-2.
A. We find the GCD of the polynomials f (x) and g (x), using the Euclidean algorithm, only now in the process of division it is impossible to reduce and multiply by suitable numbers, as we did in examples 1, 2, 3.
(1) 
(2)
Thus, the common divisor of the polynomials f (x) and g (x) is –1.
According to the division made, we write down the equalities:
f (x) = g (x) x + (- x + 1) (1 *)
g (x) = (- x + 1) (- x 2 + 2x + 2) –1. (2 *)
From equality (2 *) we express d (x) = –1 = g (x) - (- x + 1) (- x 2 + 2x + 2). From equality (1 *) we find –x + 1 = f (x) –g (x) x and substitute its value into equality (2 *): d (x) = –1 = g (x) - (f (x ) –G (x) x) (- x 2 + 2x + 2).
Now we group the terms on the right-hand side with respect to f (x) and g (x):
d (x) = –1 = g (x) –f (x) (- x 2 + 2x + 2) + g (x) x (–x 2 + 2x + 2) = f (x) (x 2 - 2x – 2) + g (x) (1 – x 3 + 2x 2 + 2x) = f (x) (x 2 –2x – 2) + g (x) (- x 3 + 2x 2 + 2x + 1) ...
Therefore, u (x) = x 2 –2x – 2, v (x) = –x 3 + 2x 2 + 2x + 1.
The greatest common divisor of the polynomials f (x) and g (x) is the polynomial 2x-2. We express it using equalities (1) and (2):
Answer:
LABORATORY OPTIONS
Option 1
1. Find the GCD of polynomials:
a) x 4 -2x 3 -x 2 -4x-6, 2x 4 -5x 3 + 8x 2 -10x + 8.
b) (x – 1) 3 (x + 2) 2 (2x + 3), (x – 1) 4 (x + 2) x.
f (x) = x 6 -4x 5 + 11x 4 -27x 3 + 37x 2 -35x + 35,
g (x) = x 5 -3x 4 + 7x 3 -20x 2 + 10x-25.
Option 2
1. Find the GCD of polynomials:
a) x 4 -3x 3 -3x 2 + 11x-6, x 4 -5x 3 + 6x 2 + x-3.
b) (2x + 3) 3 (x-2) 2 (x + 1) and its derivative.
2. Find the polynomials u (x) and v (x) so that f (x) u (x) + g (x) v (x) = d (x), d (x) = (f (x), g (x)) if
f (x) = 3x 7 + 6x 6 -3x 5 + 4x 4 + 14x 3 -6x 2 -4x + 4, g (x) = 3x 6 -3x 4 + 7x 3 -6x + 2.
Option 3
1. Find the GCD of polynomials:
a) 2x 4 + x 3 + 4x 2 -4x-3, 4x 4 -6x 3 -4x 2 + 2x + 1.
b) (x + 1) 2 (2x + 4) 3 (x + 5) 5, (x-2) 2 (x + 2) 4 (x-1).
2. Find the polynomials u (x) and v (x) so that f (x) u (x) + g (x) v (x) = d (x), d (x) = (f (x), g (x)) if
f (x) = 3x 3 -2x 2 + 2x + 2, g (x) = x 2 -x + 1.
Option 4
1. Find the GCD of polynomials:
a) 3x 4 -8 3 + 7x 2 -5x + 2, 3x 4 -2x 3 -3x 2 + 17x-10.
b) (x + 7) 2 (x-3) 3 (2x + 1) and its derivative.
2. Find the polynomials u (x) and v (x) so that f (x) u (x) + g (x) v (x) = d (x), d (x) = (f (x), g (x)) if
f (x) = x 4 -x 3 -4x 2 + 4x + 1, g (x) = x 2 -x-1.
Option 5
1. Find the GCD of polynomials:
a) 2x 4 -3x 3 -x 2 + 3x-1, x 4 + x 3 -x-1.
b) x 4 (x-1) 2 (x + 1) 3, x 3 (x-1) 3 (x + 3).
2. Find the polynomials u (x) and v (x) so that f (x) u (x) + g (x) v (x) = d (x), d (x) = (f (x), g (x)) if
f (x) = 3x 5 + 5x 4 -16x 3 -6x 2 -5x-6, g (x) = 3x 4 -4x 3 -x 2 -x-2.
Option 6
1. Find the GCD of polynomials:
a) x 4 -2x 3 + 4x 2 -2x + 3, x 4 + 5x 3 + 8x 2 + 5x + 7.
b) x 3 (x + 1) 2 (x-1) and its derivative.
2. Find the polynomials u (x) and v (x) so that f (x) u (x) + g (x) v (x) = d (x), d (x) = (f (x), g (x)) if
f (x) = x 5 -5x 4 -2x 3 + 12x 2 -2x + 12, g (x) = x 3 -5x 2 -3x + 17.
Option 7
1. Find the GCD of polynomials:
a) x 4 + 3x 3 -3x 2 + 3x-4, x 4 + 5x 3 + 5x 2 + 5x + 4.
b) (2x + 1) (x-8) (x + 1), (x 3 +1) (x-1) 2 x 3.
2. Find the polynomials u (x) and v (x) so that f (x) u (x) + g (x) v (x) = d (x), d (x) = (f (x), g (x)) if
f (x) = 4x 4 -2x 3 -16x 2 + 5x + 9, g (x) = 2x 3 -x 2 -5x + 4.
Option 8
1. Find the GCD of polynomials:
a) x 4 -3x 3 -2x 2 + 4x + 6, 2x 4 -6x 3 + 2x 2 -7x + 3.
b) (x 3 -1) (x 2 -1) (x 2 +1), (x 3 +1) (x-1) (x 2 +2).
2. Find the polynomials u (x) and v (x) so that f (x) u (x) + g (x) v (x) = d (x), d (x) = (f (x), g (x)) if
f (x) = 2x 4 + 3x 3 -3x 2 -5x + 2, g (x) = 2x 3 + x 2 -x-1.
Option 9
1. Find the GCD of polynomials:
a) 2x 4 + x 3 -5x 2 + 3x + 2, 3x 4 + 8x 3 + 3x 2 -3x-2.
b) (x 3 +1) (x + 1) 2 (2x + 3) and its derivative.
2. Find the polynomials u (x) and v (x) so that f (x) u (x) + g (x) v (x) = d (x), d (x) = (f (x), g (x)) if
f (x) = 3x 4 -5x 3 + 4x 2 –2x + 1, g (x) = 3x 3 -2x 2 + x-1.
Option 10
1. Find the GCD of polynomials:
a) x 4 -5x 3 + 7x 2 -3x + 2, 2x 4 -x 3 -7x 2 + 3x-2.
b) (x + 1) (x 2 -1) (x 3 +1), (x 3 -1) (x 2 + x) x.
2. Find the polynomials u (x) and v (x) so that f (x) u (x) + g (x) v (x) = d (x), d (x) = (f (x), g (x)) if
f (x) = x 5 + 5x 4 + 9x 3 + 7x 2 + 5x + 3, g (x) = x 4 + 2x 3 + 2x 2 + x + 1.
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BASIC INFORMATION FROM THE THEORY
Definition 4.1.
A polynomial j (x) from P [x] is called common divisor polynomials g (x) and f (x) from P [x] if f (x) and g (x) are divisible by j (x) without remainder.
Example 4.1. Two polynomials are given: (x) g (x)= x 4 - 3x 3 - 4x 2 + 2x + 2 Î R [x]. Common divisors of these polynomials: j 1 (x) = x 3 - 4x 2 + 2 = Î R [x], j 2 (x) =(x 2 - 2x - 2) Î R [x], j 3 (x) =(x - 1) Î R [x], j 4 (x) = 1 Î R [x]. (Check it out!)
Definition 4.2.
Greatest common divisornonzero polynomials f (x) and g (x) from P [x] is a polynomial d (x) from P [x], which is their common divisor and is itself divisible by any other common divisor of these polynomials.
Example 4.2. For the polynomials from Example 4.1. f (x)= x 4 - 4x 3 + 3x 2 + 2x - 6 Î R [x], g (x)= x 4 - 3x 3 - 4x 2 + 2x + 2 Î R [x] the greatest common factor is the polynomial d (x) = j 1 (x) = x 3 - 4x 2 + 2 Î R [x], since this o polynomial d (x) is divisible by all their other common factors j 2 (x), j 3 (x),j 4 (x).
The greatest common denominator (GCD) is indicated by the symbol:
d (x) = (f (x), g (x)).
The greatest common divisor exists for any two polynomials f (x), g (x) Î P [x] (g (x)¹ 0). Its existence determines Euclid's algorithm, which is as follows.
Divide f (x) on g (x)... The remainder and the quotient obtained by division are denoted by r 1 (x) and q 1 (x). Then if r 1 (x)¹ 0, divide g (x) on r 1 (x), we get the remainder r 2 (x) and private q 2 (x) etc. Degrees of Residue r 1 (x), r 2 (x),... will decrease. But the sequence of non-negative integers is bounded from below by the number 0. Therefore, the division process will be finite, and we come to the remainder r k (x), into which the previous remainder will be completely divided r k - 1 (x). The whole process of division can be written as follows:
f (x)= g (x) × q 1 (x) + r 1 (x), deg r 1 (x)< deg g (x);
g (x)= r 1 (x)× q 2 (x) + r 2 (x), deg r 2 (x) < deg r 1 (x);
. . . . . . . . . . . . . . . . . . . . . . . .
r k - 2 (x)= r k - 1 (x)× q k (x) + r k (x), deg r k (x)< deg r k - 1 (x);
r k - 1 (x) = r k (x) × q k +1 (x).(*)
Let us prove that r k (x) will be the greatest common divisor of polynomials f (x) and g (x).
1) Let us show that r k (x) is an common divisor given polynomials.
Let's turn to the penultimate equality:
r k –-2 (x)= r k –-1 (x)× q k (x) + r k (x), or r k –-2 (x)= r k (x) × q k +1 (x) × q k (x) + r k (x).
Its right side is divided into r k (x). Therefore, the left side is also divisible by r k (x), those. r k –-2 (x) divided by r k (x).
r k –- 3 (x)= r k –- 2 (x)× q k - 1 (x) + r k –- 1 (x).
Here r k –- 1 (x) and r k –- 2 (x) are divided into r k (x), whence it follows that the sum on the right-hand side of the equality is divisible by r k (x). This means that the left side of the equality is divisible by r k (x), those. r k –- 3 (x) divided by r k (x). Moving upwards in this manner, we obtain that the polynomials f (x) and g (x) are divided into r k (x). Thus, we have shown that r k (x) is an common divisor given polynomials (definition 4.1.).
2) Let us show that r k (x) divided by any other common divisor j (x) polynomials f (x) and g (x), that is, is greatest common factor these polynomials .
Let's turn to the first equality: f (x)=g (x) × q 1 (x) + r 1 (x).
Let be d (x)- some common divisor f (x) and g (x)... Then, by the properties of divisibility, the difference f (x)–g (x) × q 1 (x) also divided into d (x), that is, the left side of the equality f (x)–g (x) × q 1 (x)= r 1 (x) divided by d (x). Then and r 1 (x) will be divided into d (x). Continuing the reasoning in a similar way, successively descending downward by equalities, we obtain that r k (x) divided by d (x). Then, according to Definition 4.2.r k (x) will be greatest common factor polynomials f (x) and g (x): d (x) = (f (x), g (x)) = r k (x).
Greatest common divisor of polynomials f (x) and g (x) is unique up to a factor - a polynomial of degree zero, or, one might say, up to association(definition 2.2.).
Thus, we have proved the following theorem:
Theorem 4.1. / Euclid's algorithm /.
If for polynomials f (x), g (x) Î P [x] (g (x)¹ 0) the system of equalities and inequalities holds(*), then the last non-zero remainder will be the greatest common divisor of these polynomials.
Example 4.3. Find the Greatest Common Divisor of Polynomials
f (x)= x 4 + x 3 + 2x 2 + x + 1 and g (x)= x 3 –2x 2 + x –2.
Solution.
Step 1 Step 2
| x 4 + x 3 + 2x 2 + x + 1 | x 3 –2x 2 + x –2 | x 3 –2x 2 + x –2 | 7x 2 + 7 | |
| –(x 4 –2x 3 + x 2 - 2x) | x + 3 = q 1 (x) | –(x 3 + x) | 1 / 7x. – 2/7 = q 2 (x) | |
| 3x 3 + x 2 + 3x + 1 - ( 3x 3 –6x 2 + 3x –6) | –2x 2 –2 - ( –2x 2 –2) | |||
| 7x 2 + 7 = r 1 (x) | 0 = r 2 (x) |
Let us write the division steps in the form of a system of equalities and inequalities, as in (*) :
f (x)= g (x) × q 1 (x) + r 1 (x), deg r 1 (x)< deg g (x);
g (x)= r 1 (x)× q 2 (x).
According to Theorem 4.1./ Euclid's algorithm / the last non-zero remainder r 1 (x) = 7x 2 + 7 will be the greatest common divisor d (x) these polynomials :
(f (x), g (x)) = 7x 2 + 7.
Since divisibility in the ring of polynomials is defined up to associate ( Property 2.11.), then as GCD you can take not 7x 2 + 7, but ( 7x 2 + 7) = x 2 + 1.
Definition 4.3.
The greatest common divisor with the highest coefficient 1 will be called normalized greatest common factor.
Example 4.4. Example 4.2. the greatest common factor was found d (x) = (f (x), g (x)) = 7x 2 + 7 polynomials f (x)= x 4 + x 3 + 2x 2 + x + 1 and g (x)= x 3 –2x 2 + x –2. Replacing it with the polynomial associated with it d 1 (x)= x 2 + 1, we obtain the normalized greatest common divisor of these polynomials ( f (x), g (x)) = x 2 + 1.
Comment. Applying Euclid's algorithm to find the greatest common divisor of two polynomials, we can draw the following conclusion. Greatest common divisor of polynomials f (x) and g (x) does not depend on whether we consider f (x) and g (x) over the field P or above its extension P '.
Definition 4.4.
Greatest common divisorpolynomials f 1 (x), f 2 (x), f 3 (x), ... f n (x) Î P [x] is called such a polynomial d (x)Î P [x], which is their common divisor and is itself divisible by any other common divisor of these polynomials.
Since Euclid's algorithm is suitable only for finding the greatest common divisor of two polynomials, then in order to find the greatest common divisor of n polynomials, one needs to prove the following theorem.
Euclid's algorithm for polynomials. Euclid's algorithm allows you to find the greatest common divisor of two polynomials, i.e. polynomial of greatest degree, by which both given polynomials are divisible without remainder.
The algorithm is based on the fact that for any two polynomials in one variable, f(x) and g(x), there are such polynomials q(x) and r(x), called the quotient and remainder, respectively, that
f(x) = g(x)∙q(x) + r(x), (*)
in this case, the degree of the remainder is less than the degree of the divisor, the polynomial g(x), and, in addition, according to the given polynomials f(x) and g(x) the quotient and the remainder are found unambiguously. If in equality (*) the remainder r(x) is equal to the zero polynomial (zero), then they say that the polynomial f(x) divided by g(x) without a remainder.
The algorithm consists of sequential division with the remainder first of the first given polynomial, f(x), On the second, g(x):
f(x) = g(x)∙q 1 (x) + r 1 (x), (1)
then if r 1 (x) ≠ 0, is the second given polynomial, g(x), by the first remainder - by the polynomial r 1 (x):
g(x) = r 1 (x)∙q 2 (x) + r 2 (x), (2)
r 1 (x) = r 2 (x)∙q 3 (x) + r 3 (x), (3)
then if r 3 (x) ≠ 0, - the second remainder for the third:
r 2 (x) = r 3 (x)∙q 4 (x) + r 4 (x), (4)
etc. Since at each stage the degree of the next remainder decreases, the process cannot continue indefinitely, so at some stage we will definitely come to a situation where the next, n+ 1st remainder r n+ 1 equals zero:
| r n–2 (x) = r n–1 (x)∙q n (x) + r n (x), | (n) |
| r n–1 (x) = r n (x)∙q n+1 (x) + r n+1 (x), | (n+1) |
| r n+1 (x) = 0. | (n+2) |
Then the last nonzero remainder r n
and will be the greatest common divisor of the original pair of polynomials f(x) and g(x).
Indeed, if by virtue of the equality ( n+ 2) substitute 0 instead of r n + 1 (x) into equality ( n+ 1), then the resulting equality r n – 1 (x) = r n
(x)∙q n + 1 (x) instead of r n – 1
(x) - into equality ( n), it turns out that r n – 2 (x) = r n
(x)∙q n + 1 (x) q n
(x) + r n
(x), i.e. r n – 2 (x) = r n
(x)(q n + 1 (x) q n
(x) + 1), etc. In equality (2), after substitution, we obtain g(x) = r n
(x)∙Q(x), and, finally, from equality (1) - that f(x) = r n
(x)∙S(x), where Q and S- some polynomials. Thus, r n
(x) Is the common divisor of the two original polynomials, and the fact that it is the largest (i.e., the greatest possible degree) follows from the procedure of the algorithm.
If the greatest common divisor of two polynomials does not contain a variable (i.e. is a number), the original polynomials f(x) and g(x) are called relatively simple.