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Большой Мастер-класс профессора Васильевой

Большой Мастер-класс профессора Васильевой

Channel: ПРИКЛАДНАЯ КИНЕЗИОЛОГИЯ UAAK

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Gimp Tutorial : Burger House Poster | Food Poster Design

Channel: Zakey Design

Евангелие от Иоанна. Глава 16. Андрей Солодков. Новый Завет

Евангелие от Иоанна. Глава 16. Андрей Солодков. Новый Завет

Channel: БИБЛИЯ и толкования - Экзегет.ру

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Создание сайта | Урок #13 - Обратная связь

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El MEJOR método para editar LOG | DaVinci Resolve Tutorial

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الحل الوحيد مشكلة سلسلة الدراجة النارية بيجو #peagout #103

الحل الوحيد مشكلة سلسلة الدراجة النارية بيجو #peagout #103

Channel: المعلم للصيانة

Разделить многочлен на многочлен

Разложить многочлен на множители

Деление многочлена уголком

Разделить многочлен в столбик

Решить уравнение четвертой степени

Решить уравнение разложив на множители

00:00:02

to factor a polynomial of the fourth degree

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as the main

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expansion tool we will

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use the division of a polynomial by a

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polynomial with a corner and so first we will

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find the first root of the equation f of x

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equals zero x 4 minus x 3 minus 19 x

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square minus 11 x plus 30 is equal to zero,

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the main idea of finding the roots in such an

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equation is as follows: if

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the equation has integer roots, then they

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should be looked for among the divisors of the

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free term in our equation, the

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free term is equal to 30,

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we will write down several integer

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divisors of the number 30, these are the divisors plus

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minus 1 plus minus 2 plus -3 and so on, let's

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start checking each of

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these divisors in turn, start with x equal to one,

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substitute the values x equals one into

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our equation,

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we get 1 in 4 minus 1 in the third minus

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19 multiplied by 1 squared minus 11

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multiplied by 1 plus 30,

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count 1 minus 10 minus 19 -11 will be

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minus 30 plus 30 we get 0

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we deduce the polynomial x in 4 minus x 3

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minus 19 x square minus 11 x plus 30

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can be represented in the following form

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as the product of 2 parentheses the first of

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which is x minus 1 since x is equal

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unit is the equation

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that will be in the second bracket, we are

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now going to look for the

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expression in the second bracket,

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divide our polynomial

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by x minus 1 with the corner and see that here

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is x 4 and here x to the first power we

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need to multiply by x in 3

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I write mix 3 and now by x 3 we multiply

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each of the terms

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x 3 we multiply by x we get x 4 x 3

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we multiply by -1 we get minus x in 3

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we put a minus sign and subtract

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x 4 minus x 4 minus x 3 minus minus x 3

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we get along

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we write the next two terms from

00:02:50

our original polynomial

00:02:52

this is minus 19 x square minus

00:02:57

11 x

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again we compare the first terms in

00:03:03

our polynomials here minus 19 x

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square here x we need to take

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minus 19 x now we multiply minus 19 x

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we multiply by x we get minus 19 x

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square

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then minus 19 watts multiplied by -1

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we get plus 19 x

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hard line put a minus sign and

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subtract we see that each time we choose a

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value here so that after

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subtracting the expression with the greatest

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degree it is reduced, that is, minus 19 x

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square minus minus 19 x square this is 0

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then minus 11 x minus 19 x it turns out

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minus 30 x

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take the next term plus 30

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now we take as you guessed by

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-30

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-30 multiply by x will be minus 30 x

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minus 30 multiply by -1 we get 30 . 30 is a

00:04:12

fractional line again, we put a minus sign and

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we see that our expression is reduced to

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0, so we have

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expanded the original polynomial, we need or 1

00:04:23

bracket x minus 1 2 x in 3 minus 19 x

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minus 30 we will continue further factoring

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our polynomial now we will

00:04:33

consider the equation x 3 minus 19 x

00:04:36

minus 3 is equal to zero, the roots of this

00:04:39

equation will also be among the

00:04:41

divisors of the free term 30, we begin

00:04:44

to check x is equal to one, we substitute it

00:04:47

turns out one in 3 minus 19 multiplied by

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1 minus 30 is not equal to zero,

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then we substitute x equals minus 1

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we get -1 in 3 minus 19 multiplied by -1

00:05:04

minus 30 we see that this expression is also not equal to zero,

00:05:07

we substitute the next

00:05:10

divisor of the number 30 x is equal to 2

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we get two in the third minus 19 multiplied

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by 2 minus 30

00:05:21

we see that this expression is not equal to zero the

00:05:25

next root we check x is equal to minus

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2 minus two to the third power minus

00:05:32

19 multiplied by minus 2 minus 30 read

00:05:37

minus 8 plus thirty-eight minus 30 is

00:05:40

equal to zero means the polynomial xv 3 minus

00:05:44

19 x minus 30 can be

00:05:48

factorized 1 bracket will be x plus 2

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since the root x is equal to minus 2 us

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approached and the second bracket needs to be found

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in order to find the expression

00:06:00

in the second bracket we divide the polynomial x 3 minus

00:06:03

19 x minus 37 by express 2

00:06:07

with a corner when you divide one

00:06:10

polynomial by another with a corner, pay attention to

00:06:12

how for example in our case we don’t have

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enough add y you x square

00:06:18

I recommend not to skip this term but to

00:06:20

write it with a coefficient of 0, that is,

00:06:22

write x 3 plus 0 multiplied by x square

00:06:26

minus 19 x minus 30 when you have already mastered

00:06:31

division by a corner, you don’t have to do this step for

00:06:34

now, let’s write it in this way, we see

00:06:36

divide by x + 2

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we see that here there are faces in 3 here it’s

00:06:41

just x we need to let’s click on

00:06:44

x-square

00:06:45

multiply x square by each term

00:06:47

will be x 3

00:06:50

+ 2 x square

00:06:54

put a minus sign and

00:06:56

subtract

00:06:58

x 3 minus x

00:07:01

300 x square minus 2 x square we get

00:07:05

minus 2 x square minus

00:07:08

19 x

00:07:11

here we have minus 2x square and here x

00:07:14

that is, we need to take minus 2x each,

00:07:18

multiplying minus 2 x by x will be minus 2 x

00:07:22

square minus

00:07:23

2 x multiply by 2 we get minus 4x

00:07:28

put a minus sign and

00:07:31

count minus 2 x square minus minus 2x

00:07:35

square is 0 minus 19 x minus minus 4x

00:07:39

in other words minus 19x plus 4x

00:07:43

we get minus 15x and

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take the next term -30 take

00:07:51

-15 each multiply minus 15 by x we get

00:07:55

minus 15x -15 multiply by 2 we get

00:07:59

minus 30 put the sign minus

00:08:02

we make sure that we get 0, so

00:08:05

we expanded the polynomial xv 3 minus 19 x

00:08:10

minus 30 into two brackets x plus 2 and x

00:08:14

squared minus 2 x minus 15, then we

00:08:17

move on to the expansion of the following

00:08:19

expression x squared minus 2 x minus 15

00:08:22

we will look for the roots of this equation

00:08:25

among divisors -15 since the value of x

00:08:28

is 1 minus 1 we checked they are

00:08:31

no longer suitable for us so we check the next

00:08:34

divisor this value 3x is equal to three

00:08:37

we substitute 3 squared minus 2 times 3

00:08:40

minus 15 we read 9 minus 6

00:08:44

minus 15 is not equal to zero we substitute the

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following teaching x is equal to minus 3 I get

00:08:50

minus 3 squared minus 2 multiplied by

00:08:54

minus 3 minus 15 we read 9 + 6 minus 15 is

00:09:00

equal to zero, which means our equation x

00:09:03

squared minus 2 x minus 15 can be

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factorized x + 3 in 1 brackets

00:09:11

and the value that will be in

00:09:13

we will find the second bracket using

00:09:15

corner division of course now you can say

00:09:18

that quadratic equations and there are

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many ways to find the

00:09:23

second root but we are studying the

00:09:27

corner division method so we will also divide

00:09:29

our polynomial of the second degree x square

00:09:32

minus 2 x minus 15 by x + 3 with a corner,

00:09:38

since we are interested in the method of

00:09:40

dividing with a corner, we see that here x is a

00:09:42

square, here x we take x by x, we multiply x by

00:09:45

x and x square x by 3 and we get plus 3x

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we put a minus sign, I subtract zero here,

00:09:53

here minus 5x we write the following

00:09:57

terms we take by -5 it turns out minus 5

00:10:01

x minus 15 we put a minus sign we get 0

00:10:05

so we factored our polynomial x squared

00:10:09

minus 2 x minus 15 into

00:10:12

factors in the first brackets x plus 3 in the

00:10:14

second bracket x minus 5 so

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we return to our original

00:10:19

polynomial of the fourth degree and write it

00:10:22

in what

00:10:24

We factored this polynomial, we write x into 4 minus x into

00:10:29

3 minus 19 x squared minus 11 x plus 30

00:10:35

breaks down into the product of the following

00:10:37

brackets 1 bracket is x minus 1 2 bracket

00:10:40

is x plus 2 3 bracket is x plus 3 and 4

00:10:45

bracket is x minus 5, so the problem is solved,

00:10:48

subscribe to our channel,

00:10:50

like it, thanks for your attention

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Разложить многочлен на множители с помощью деления уголком. Решить уравнение путем разложения на множители. Разделить многочлен на многочлен в столбик Разбираем задачи наших подписчиков! ➖➖➖➖➖➖➖➖➖➖➖➖➖➖ Следующая задача может быть Ваша, для этого нужно: 1. Быть подписанным на наш канал (открыт доступ к информации о подписках). 2. Написать условие своей задачи или интересующую тему в комментариях. ➖➖➖➖➖➖➖➖ Подписаться на канал: https://www.youtube.com/channel/UCVCg09Zf6_LEcM_hlERRCIA?sub_confirmation=1 ➖➖➖➖➖➖➖➖

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