Divided the rich members.

Repair

Golovna

The stagnation of the regions is wider in life.

The stench is vikorist in rich rozrahunkas, everyday disputes and sports.

The zealous people have been vikorsist for a long time, and as a result, their stagnation is increasing.

A polynomial is an algebraic sum of numbers of varying degrees.

The transformation of rich members includes two types of order. Viraz needs to be forgiven, or divided into multiples, then.

taxes yogo look like two or a lot of rich members or a single member and a rich member. To forgive a rich member, make such additions.

butt. Forgive the expression \ Find monomials with the same letter part.

Butt 40 Find out the gcd of rich members
.

Decision. Let's split the insults into multiples:

From the layout it is clear that we are looking for GCD will be a rich member ( X– 1).

Butt 41 Find out the gcd of rich members
і
.

Decision. Let's separate the insults from the rich members into multipliers.

For rich member
XX- 1) behind Horner’s scheme.

For rich member
Possible rational roots would be the numbers 1, 2, 3 and 6. X For additional substitutions, let's reconfigure that X- 1) behind Horner’s scheme.

= 1 є root.
Divide the rich term into (

So, decomposition of the quadratic trinomial X– 1)(X– 2).

was divided up behind Viet's theorem.

Having sorted out the arrangement of the rich terms into multipliers, we know that we can look for the GCD to be the rich term (

Similarly, you can find GCDs for many different members.

Prote the method of knowing the GCD with the path of distributing it into multiples of available ones without ever changing.

The method that allows you to find the GCD for all types is called the Euclidean algorithm. Find out the gcd of rich members
і
.

Decision. The scheme of the Euclid algorithm is as follows.
One of the two polynomials is divided by the other, the level of which is not different from the first.
Next, you must immediately take the rich member that served in the previous operation as a debtor, and the debtor will take the surplus taken away from the same operation.

This process is explained as only the surplus turns out to be equal to zero.

Let's demonstrate this algorithm in examples.
Let's take a look at the rich joints that were vikorized at the two front butts. X– 1:

This process is explained as only the surplus turns out to be equal to zero.+ 1

Butt 42 X Divided

on Find out the gcd of rich members
і
.

"knot": x
One of the two polynomials is divided by the other, the level of which is not different from the first.
Next, you must immediately take the rich member that served in the previous operation as a debtor, and the debtor will take the surplus taken away from the same operation.

1

Now let's separate the debtor
Let's take a look at the rich joints that were vikorized at the two front butts.
on Rashta
=
The fragments of the remaining floor will be found without excess, then the GOD will be
- 1, then a rich member who is vikorystuvavsya as a debtor under this division.
Butt 43
Decision

.

To find the GCD, it is accelerated using the Euclidean algorithm.

1. Divided

We're looking for a different floor.

For whom would it be necessary to divide the front part? oh well
і
, for clarity, it is divisible

not on , and on. With such a replacement, the solution to the problem does not change; the remaining gcd of a pair of rich terms is calculated up to a constant multiplier..

Maemo:

The surplus turned out to be equal to zero, so the remaining debtor is a rich member And I will be a snooping NOD.
.

Decision. In order to see the whole part of the fraction, it is necessary to divide the number of the fraction into its sign. Let’s divide the number given to the fraction into the sign with a “knot”:

So, as the stage of the rich member, which is highest, is less than the stage of the debtor, then the process is near completion.

=
In the pouch:
.

Drib, what is the best result.
є correct. This process is explained as only the surplus turns out to be equal to zero. ) Fraction of mind
called the simplest, right? This process is explained as only the surplus turns out to be equal to zero. ).

– non-reducible members, and steps less step?

Respect.

Increase your respect so that the level of the numberer and the unreduced rich member of the signer will be equalized (without leveling the level α). For fractions with active coefficients, there are 4 types of simple fractions:
.

Be the right friend

can be presented in the form of a sum of the simplest fractions, such as all kinds of fractions

Algorithm for dividing a fraction in the simplest way:

If the cut is incorrect, then the whole part is visible, and in the simplest way we lay out the correct cut, which is the highest.

We decompose the standard fraction into multipliers.

We write the correct fraction as the sum of the simplest fractions with unimportant coefficients.

We bring to the final banner the sum of fractions on the right side.

Known unknown coefficients: This process is explained as only the surplus turns out to be equal to zero..

Or equal coefficients at the same levels at the left and right directions of the numbers;

Or presenting specific (usually the root of the zagalny znamennik) meanings We write down the answer by solving the whole part of the fraction.
.

Decision. Butt 45

1

= 1 +
.

Break it down in the simplest terms
So, since the fractional-rational function is given incorrectly, apparently for the whole part:

Let's lay out the drib, scho vyyshov.

in the simplest way.

From the beginning we will lay out the banner on multiples.

For this we know the root behind the standard formula:
.

Let us write down the expansion of the fractional rational function in the simplest terms, vikoryst and unimportant coefficients: We write down the answer by solving the whole part of the fraction.
.

Decision. Let us bring some justice to the final banner:

We create a system with equal coefficients at the same levels for the numbers of the left and right fractions:

Subject: Butt 46 Since the whole part is correct (the level of the numberer is less than the level of the signifier), there is no need to see the whole part.

.

(2.2) It would be possible to collapse the system of equals by equating the numerals of the left and right fractions, but in this application the calculations would be too cumbersome. An offensive technique will help to forgive them: we can substitute them for numbers through the root sign. 1:

(2.2) It would be possible to collapse the system of equals by equating the numerals of the left and right fractions, but in this application the calculations would be too cumbersome. X= ‑1:

At x =і Now for the designated coefficients that we lost A

Z It will be enough to equate the coefficients for the senior level and free members. You can know them without opening the arms: The left side of the first fraction has a value of 0, since the number of the left fraction (2.2) has no additional value + , and the right fraction has dodanku s coefficient The left side of the first fraction has a value of 0, since the number of the left fraction (2.2) has no additional value + A + , and the right fraction has dodanku s + C.

For this we know the root behind the standard formula:
.

The left side of the other equal has a value of 0, because in the number of the left fraction (2.2) the free term is equal to zero, and in the number of the right fraction (2.2) the free term is equal to (‑

B

D ). Maemo:

BASIC VIEWS OF THEORIES Value 4.1. The rich term j(x) with P[x] is called sleeping debtor there are many terms g(x) and f(x) with P[x], since f(x) and g(x) are divisible by j(x). Example 4.1. Two rich terms are given:(x) g(x)= x 4 − 3x 3 − 4x 2 + 2х + 2 О R[x]. Zagalnye sharers of these rich members: j 1 (x) =

x 3 − 4x 2 + 2 = R[x],

j 2 (x) =(x 2 − 2x − 2) О R[x],

j 3 (x) = (x − 1) О R[x], j 4 (x) = The rich term j(x) with P[x] is called 1 О R[x]. (Turn it around!) Value 4.2. The largest sleeping debtor,From the standpoint of zero rich terms f(x) and g(x) from P[x], a polynomial d(x) from P[x] is called a polynomial that is their co-partner and itself shares with any other co-partner of these polynomials..

Butt 4.2.

For rich members from the butt 4.1. = ((x − 1) О R[x],, The rich term j(x) with P[x] is called).

f(x) = x 4 − 4x 3 + 3x 2 + 2x − 6 О R[x],= x 4 − 3x 3 − 4x 2 + 2х + 2 О R[x] the largest term will be the rich term d(x) = j 1 (x) = x 3 − 4x 2 + 2 О R[x], so this is a rich term

d(x) divided by all other members j 2 (x), j 3 (x) (x − 1) О R[x], One of the two polynomials is divided by the other, the level of which is not different from the first. The rich term j(x) with P[x] is called j4(x) The largest debtor (LDD) is indicated by the symbol:і d(x) The greatest sleeper sleeps for any two rich members The largest debtor (LDD) is indicated by the symbol: f(x),g(x) О P[x] (g(x) The rich term j(x) with P[x] is called One of the two polynomials is divided by the other, the level of which is not different from the first. No. 0). What does this mean? Euclidean algorithm, What lies in the offensive. Dilimo. Excess and privacy, taken away during divorce, significant r 1 (x) q 1 (x). Then, because No. 0, dilimo r 1 (x),

(x − 1) О R[x],= The rich term j(x) with P[x] is called × We will take away the surplus r2(x) The largest debtor (LDD) is indicated by the symbol:< deg and in private

The rich term j(x) with P[x] is called= The largest debtor (LDD) is indicated by the symbol:× q2(x) r2(x) Euclidean algorithm < deg etc.

. . . . . . . . . . . . . . . . . . . . . . . .

Stages of surplus to come out= r 1 (x), r 2 (x),× qk(x) + q 1 (x). r2(x) r k (x)< deg r k - 1 (x);

r 1 (x), r 2 (x), = r k (x) × q k +1 (x).(*)

Let's see what r k (x) will be the largest member of rich members (x − 1) О R[x],і g(x).

1) Let's show what r k (x)є ). data from rich members.

Rising to the point of constant zeal:

r k –-2 (x)= r k -1 (x)× qk(x) + q 1 (x). or else r k –-2 (x)= r k (x) × q k +1 (x) × qk(x) + r k (x).

It is right for the part to be divided into r k (x). Also, the left part is also divided into q 1 (x). tobto. r k –-2 (x) divide into r k (x).

r k -- 3 (x)= r k -- 2 (x)× q k - 1 (x) + No. 0, dilimo

Here r k -- 1 (x)і r k -- 2 (x) share with q 1 (x). The result is growing, as the sum of the right side of equality is divided into r k (x). This means that the left side of equality is divided into q 1 (x). tobto. r k -- 3 (x) divide into r k (x). Sticking through in this manner, successively uphill, we realize that there are a lot of limbs (x − 1) О R[x],і The rich term j(x) with P[x] is called share with r k (x). Tim showed us ourselves what r k (x)є ). data from rich members (Viznachennya 4.1.).

2) Let's show what r k (x) divide into be different sleeping farmer j(x) rich in members (x − 1) О R[x],і g(x), then the biggest sleeper many members .

Rising to the first level of jealousy: (x − 1) О R[x],=The rich term j(x) with P[x] is called × q 1 (x) + r 1 (x).

Let's go For rich members from the butt 4.1.- Active sleeper (x − 1) О R[x],і The rich term j(x) with P[x] is called. (x − 1) О R[x],–The rich term j(x) with P[x] is called × q 1 (x) There is a difference between the authorities and the divisibility also shares with d(x), (x − 1) О R[x],–The rich term j(x) with P[x] is called × q 1 (x)= The largest debtor (LDD) is indicated by the symbol: divide into then the left side is equal d(x). The largest debtor (LDD) is indicated by the symbol: Todi i then the left side is equal will be divided into r k (x) divide into then the left side is equal Continuing the process of merging in a similar manner, successively descending along the lines to the bottom, it is clear that x 3 − 4x 2 + 2 = R[x],r k (x) Todi, okay the biggest sleeper rich in members (x − 1) О R[x],і The rich term j(x) with P[x] is called: For rich members from the butt 4.1. = ((x − 1) О R[x],, The rich term j(x) with P[x] is called) = r k (x).

will (x − 1) О R[x],і The rich term j(x) with P[x] is called The largest member of the richest members є united up to the multiplier - the rich term of the zero step, or, one can say,to the point of association

(Viznachennya 2.2.).

Thus, we have established the theorem:

Theorem 4.1.¹ 0) /Euclidean algorithm/.(*), There are many terms f(x),g(x) О P[x] (g(x)

The system of equalities and insecurities is true

(x − 1) О R[x], then the remaining, not equal to zero, surplus will be the largest contributor to these rich members. The rich term j(x) with P[x] is called Butt 4.3.

Decision.

Find the largest member of the richest members

-2x 2 -2) (*) :

(x − 1) О R[x],= 7x 2 + 7 = r 1 (x) 0 = r 2 (x) The largest debtor (LDD) is indicated by the symbol:< deg and in private

The rich term j(x) with P[x] is called= The largest debtor (LDD) is indicated by the symbol:× Let's write down the bottom lines of the system of equalities and inequities, like

g(x) × q 1 (x) + r 1 (x), deg q2(x). For rich members from the butt 4.1. many members :

((x − 1) О R[x],, The rich term j(x) with P[x] is called Zhidno

Theorem 4.1. /Euclidean algorithm/ remaining, non-zero surplus r 1 (x) = 7x 2 + 7 will be the largest number of participants) = 7x2 + 7. ( The identity of the ring of rich members is determined to the extent of association (

Vlastivist 2.11

.) , then as GCD you can take not 7x 2 + 7, but 7x2+7) = x2+1..

Value 4.3. For rich members from the butt 4.1. = ((x − 1) О R[x],, The rich term j(x) with P[x] is called The largest debtor with senior coefficient 1 is called (x − 1) О R[x], then the remaining, not equal to zero, surplus will be the largest contributor to these rich members. The rich term j(x) with P[x] is called= x 3 -2x 2 + x -2. Replacing him with a rich member associated with him d1(x) (x − 1) О R[x],, The rich term j(x) with P[x] is called= x 2 + 1, we reject the normalization of the lateral part of these polynomials (

– non-reducible members, and steps) = x 2 + 1. (x − 1) О R[x],і The rich term j(x) with P[x] is called By applying the Euclidean algorithm by searching for the largest complementary part of two polynomials, it is possible to construct such a construction. (x − 1) О R[x],і The rich term j(x) with P[x] is called The largest member of the richest members not to lie down from what we will look at over the field P

or over yogo expanded

j 2 (x) =P'. Î Value 4.4.Î rich terms f 1 (x), f 2 (x), f 3 (x), ... f n (x)

P[x] is called such a polynomial d(x)

P[x], which is their co-partner, and itself shares with any other co-partner of these polynomials. Since Euclidean's algorithm is based only on the search for the largest dilator of two polynomials, then the search for the largest dilator of n polynomials needs to arrive at the theorem. Euclidean algorithm for rich terms. The Euclidean algorithm allows you to find out the final part of two polynomials, then.
rich member nai big world(This process is explained as only the surplus turns out to be equal to zero., on which to share without too much offense to the rich member. The algorithm is based on the fact that for any two rich members there is one changeable,(This process is explained as only the surplus turns out to be equal to zero. f ) that(This process is explained as only the surplus turns out to be equal to zero., on which to share without too much offense to the rich member. g(This process is explained as only the surplus turns out to be equal to zero.), there are such rich members

big world(This process is explained as only the surplus turns out to be equal to zero.) = The algorithm is based on the fact that for any two rich members there is one changeable,(This process is explained as only the surplus turns out to be equal to zero.)∙) that(This process is explained as only the surplus turns out to be equal to zero.) + g(This process is explained as only the surplus turns out to be equal to zero.), (*)

q The algorithm is based on the fact that for any two rich members there is one changeable,(This process is explained as only the surplus turns out to be equal to zero. r big world(This process is explained as only the surplus turns out to be equal to zero., on which to share without too much offense to the rich member. The algorithm is based on the fact that for any two rich members there is one changeable,(This process is explained as only the surplus turns out to be equal to zero.) , the name is very private and the surplus that g(This process is explained as only the surplus turns out to be equal to zero. at which stage the excess is smaller than the stage of the splitter, the richer member big world(This process is explained as only the surplus turns out to be equal to zero.), and, in addition, for these rich terms The algorithm is based on the fact that for any two rich members there is one changeable,(This process is explained as only the surplus turns out to be equal to zero.) is private and too much to know for sure.
Yakshto have a surplus (*) big world(This process is explained as only the surplus turns out to be equal to zero.) is similar to the zero rich term (zero), then it seems that the rich term The algorithm is based on the fact that for any two rich members there is one changeable,(This process is explained as only the surplus turns out to be equal to zero.):

big world(This process is explained as only the surplus turns out to be equal to zero.) = The algorithm is based on the fact that for any two rich members there is one changeable,(This process is explained as only the surplus turns out to be equal to zero.)∙) that 1 (This process is explained as only the surplus turns out to be equal to zero.) + g 1 (This process is explained as only the surplus turns out to be equal to zero.), (1)

) divided by g 1 (This process is explained as only the surplus turns out to be equal to zero.) without extra charge. The algorithm is based on the fact that for any two rich members there is one changeable,(This process is explained as only the surplus turns out to be equal to zero. The algorithm consists of a successive section from too much of the first given rich term, g 1 (This process is explained as only the surplus turns out to be equal to zero.):

The algorithm is based on the fact that for any two rich members there is one changeable,(This process is explained as only the surplus turns out to be equal to zero.) = g 1 (This process is explained as only the surplus turns out to be equal to zero.)∙) that 2 (This process is explained as only the surplus turns out to be equal to zero.) + g 2 (This process is explained as only the surplus turns out to be equal to zero.), (2)

g 1 (This process is explained as only the surplus turns out to be equal to zero.) = g 2 (This process is explained as only the surplus turns out to be equal to zero.)∙) that 3 (This process is explained as only the surplus turns out to be equal to zero.) + g 3 (This process is explained as only the surplus turns out to be equal to zero.), (3)

) divided by g 3 (This process is explained as only the surplus turns out to be equal to zero.), to another,

g 2 (This process is explained as only the surplus turns out to be equal to zero.) = g 3 (This process is explained as only the surplus turns out to be equal to zero.)∙) that 4 (This process is explained as only the surplus turns out to be equal to zero.) + g 4 (This process is explained as only the surplus turns out to be equal to zero.), (4)

then, yakscho ) ≠ 0, – another given rich term,), for the first surplus - for the rich member g ) ≠ 0, – another given rich term,) ≠ 0, – another surplus on the third:

n The fragments at the skin stage of the stage of the worm surplus change, the process cannot be continued indefinitely, so at any stage we will inevitably come to a situation, if worm, ) ≠ 0, – another given rich term, + 1st surplus big world(This process is explained as only the surplus turns out to be equal to zero., on which to share without too much offense to the rich member. The algorithm is based on the fact that for any two rich members there is one changeable,(This process is explained as only the surplus turns out to be equal to zero.).
+ 1 is equal to zero: ) ≠ 0, – another given rich term, q g ) ≠ 0, – another given rich term, + 1 (This process is explained as only the surplus turns out to be equal to zero. r ) ≠ 0, – another given rich term, Then the remaining surplus is not equal to zero g ) ≠ 0, – another given rich term, – 1 (This process is explained as only the surplus turns out to be equal to zero.) = The fragments at the skin stage of the stage of the worm surplus change, the process cannot be continued indefinitely, so at any stage we will inevitably come to a situation, if worm, ) ≠ 0, – another given rich term, (This process is explained as only the surplus turns out to be equal to zero.)∙) that ) ≠ 0, – another given rich term, + 1 (This process is explained as only the surplus turns out to be equal to zero. and will be the largest shareholder of the output pair of rich members The fragments at the skin stage of the stage of the worm surplus change, the process cannot be continued indefinitely, so at any stage we will inevitably come to a situation, if worm, ) ≠ 0, – another given rich term, – 1 (This process is explained as only the surplus turns out to be equal to zero. In truth, because of zeal ( ) ≠ 0, – another given rich term,+ 2) substitute 0 instead g ) ≠ 0, – another given rich term, – 2 (This process is explained as only the surplus turns out to be equal to zero.) = The fragments at the skin stage of the stage of the worm surplus change, the process cannot be continued indefinitely, so at any stage we will inevitably come to a situation, if worm, ) ≠ 0, – another given rich term, (This process is explained as only the surplus turns out to be equal to zero.)∙) that ) ≠ 0, – another given rich term, + 1 (This process is explained as only the surplus turns out to be equal to zero.) etc. ) ≠ 0, – another given rich term, (This process is explained as only the surplus turns out to be equal to zero.) + The fragments at the skin stage of the stage of the worm surplus change, the process cannot be continued indefinitely, so at any stage we will inevitably come to a situation, if worm, ) ≠ 0, – another given rich term, (This process is explained as only the surplus turns out to be equal to zero.) y jealousy ( g ) ≠ 0, – another given rich term, – 2 (This process is explained as only the surplus turns out to be equal to zero.) = The fragments at the skin stage of the stage of the worm surplus change, the process cannot be continued indefinitely, so at any stage we will inevitably come to a situation, if worm, ) ≠ 0, – another given rich term, (This process is explained as only the surplus turns out to be equal to zero.)() that ) ≠ 0, – another given rich term, + 1 (This process is explained as only the surplus turns out to be equal to zero.) etc. ) ≠ 0, – another given rich term, (This process is explained as only the surplus turns out to be equal to zero.+ 1), then – jealousy is abandoned The algorithm is based on the fact that for any two rich members there is one changeable,(This process is explained as only the surplus turns out to be equal to zero.) = The fragments at the skin stage of the stage of the worm surplus change, the process cannot be continued indefinitely, so at any stage we will inevitably come to a situation, if worm, ) ≠ 0, – another given rich term, (This process is explained as only the surplus turns out to be equal to zero.)∙) replacement(This process is explained as only the surplus turns out to be equal to zero.) – jealousy ( big world(This process is explained as only the surplus turns out to be equal to zero.) = The fragments at the skin stage of the stage of the worm surplus change, the process cannot be continued indefinitely, so at any stage we will inevitably come to a situation, if worm, ) ≠ 0, – another given rich term, (This process is explained as only the surplus turns out to be equal to zero.)∙), you see, scho(This process is explained as only the surplus turns out to be equal to zero.), tobto. ) replacementі ), you see, scho) + 1), etc. The fragments at the skin stage of the stage of the worm surplus change, the process cannot be continued indefinitely, so at any stage we will inevitably come to a situation, if worm, ) ≠ 0, – another given rich term, (This process is explained as only the surplus turns out to be equal to zero. In equality (2) after substitution it is removed that
Since the largest member of two rich members does not replace the change (that is, in number), the output rich members big world(This process is explained as only the surplus turns out to be equal to zero., on which to share without too much offense to the rich member. The algorithm is based on the fact that for any two rich members there is one changeable,(This process is explained as only the surplus turns out to be equal to zero.) are called mutually forgive.

1. Euclid algorithm

Whenever two polynomials are divided without excess into a third, this third polynomial is called the parallel derivative of the first two.

The largest joint division (GCD) of two rich members is called its largest joint.

Please note that any number other than zero is a joint partner of two rich members.

Therefore, any number not equal to zero is called the obvious counterpart of these polynomials.

The Euclidean algorithm demonstrates the sequence of actions that either lead to finding the gcd of two given rich terms, or shows that such a conclusion as the rich term of the first or higher stage does not exist.

The Euclidean algorithm is implemented as a sequence of divisions.

In the first division, the polynomial with its larger part is taken as a division, and with its smaller part as a division.

As many members for which GCD occurs, however, there are new steps, then the part and the part are chosen quite well.

If, in the case of a final division, the surplus member has a higher or equal degree of 1, then the debtor becomes divisible, and the surplus becomes divisible.

If, with the final division of the rich members, the excess is removed, which is equal to zero, then the GCD of these rich members is found.

He is a sharer in the remaining division.

Since, according to the final distribution of polynomials, the surplus is a number that is not equal to zero, then these rich terms have no GCD but trivial ones.

Butt #1

Shorten the drib.

2. Possibility of simplified calculation of GCD using the Euclid algorithm

When multiplying the divided by a number that is not equal to zero, the portion and the excess is multiplied by the same number.

Finished

Let P – dilene, F – dilnik, Q – private, R – surplus.

Todi,

By multiplying the given identity by the number 0, we can remove it

where the polynomial P can be viewed as a division, and the polynomials Q and R - as a partial and a surplus, which is obtained by dividing the polynomial P into a polynomial F. In this way, when multiplied by the number 0, the partial and a surplus itself multiply by the number.

d

To go to the divisible and the shareholder for all coefficients, multiply the division by 6, which leads to the multiplication of the searched private Q and the excess R by 6. After that, we multiply the sharer by 5, which leads to the multiplication of the private 6Q and for over 6R on.

As a result, the value of the private Q and the surplus R, subtracted from the division of these polynomials, will be different.

Otje, ;