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Rіshennya rіvnyan n steps. Rishennya rivnyan vishnyh steps The winnings of a multiplier for the temples

When verifying algebraic equivalents, it is often possible to decompose a polynomial into factors. The division of a polynomial into multipliers is to represent it in the viewer with two or more decimal polynomials. Deyakі methods of arranging polynomials live often end up with: vinyinnya zagalny multiplier, storing fast multiplier formulas, seeing a popping square, ugrupovannya. The method is clear.

In some cases, when the polynomial is expanded into multipliers, they will come out on top of solidity:

1) as a polynomial, with a large number of functions, a racial root (de - non-short, then - a divisor of a vilny member and a dean of a senior officer:

2) If you are a rank, take the roots of the step polynomial, then the polynomial can be represented in the view of the step polynomial

A polynomial can be known either by the subdivision of a polynomial into a two-term "stovpchik", or by the fact that the addition of a polynomial and the appearance of a multiplier from them, or by the method of unvalued coefficients.

Butt. Factoring a polynomial

Decision. Oskilki efficiency for x4 is 1, then the rational root of this polynomial is equal to the number 6, ie E. It can be as many numbers ± 1, ± 2, ± 3, ± 6. Meaning the given polynomial in terms of P4 (x). So yak Р Р4 (1) = 4 і Р4 (-4) = 23, then the numbers 1 і -1 are not in the roots of the bagaton RA (x). Oskilki P4 (2) = 0, then x = 2 є the root of the polynomial P4 (x), і, the same, the polynomial extends to a dynomial x - 2. Tom x4 -5x3 + 7x2 -5x +6 x-2 x4 -2x3 x3 -3x2 + x-3

3x3 + 7x2 -5x +6

3x3 + 6x2 x2 - 5x + 6 X2 2x

Otzhe, P4 (x) = (x - 2) (x3 - Zx2 + x3). So yak xz - Zx2 + x - 3 = x2 (x - 3) + (x - 3) = (x - 3) (x2 + 1), then x4 - 5x3 + 7x2 - 5x + 6 = (x2) (x - 3) (x2 + 1).

Parameter input method

Inodi when expanding a polynomial into multipliers, an additional method of introducing a parameter. The essence of this method is understandable on such an application.

Butt. x3 - (√3 + 1) x2 + 3.

Decision. The polynomial with the parameter a can be easily seen: x3 - (a + 1) x2 + a2, which for a = √3 turns into a polynomial. We can write down the polynomial yak is the square trinomial of the trinomial schodo a: a - ax2 + (x3 - x2).

Since the root of the square root of the trinomial є a1 = x і a 2 = x2 - x, then the equality of A2 - ax2 + (xs - x2) = (a - x) (ax2 + x) is true. Also, the polynomial x3 - (√3 + 1) x2 + 3 is multiplied by √3 - x і √3 - x2 + x, i.e. E.

x3 - (√3 + 1) x2 + 3 = (x-√3) (x2-x-√3).

Method of introducing new

In some cases with a path, replace the rotation f (x), which is included in the polynomial Pn (x), through y it is possible to deduct the polynomial schodo y, which can be easily multiplied. If we change y by f (x), we can easily multiply the polynomial Pn (x).

Butt. Factoring the polynomial x (x + 1) (x + 2) (x + 3) -15.

Decision. Reconfigure the polynomial with the next rank: x (x + 1) (x + 2) (x + 3) -15 = [x (x + 3)] [(x + 1) (x + 2)] - 15 = (x2 +3 x) (x2 + 3x + 2) - 15.

Let x2 + 3x be denoted by y. Todi maєmo y (y + 2) - 15 = y2 + 2y - 15 = y2 + 2y + 1 - 16 = (y + 1) 2 - 16 = (y + 1 + 4) (y + 1 - 4) = ( y + 5) (y - 3).

Tom x (x + 1) (x + 2) (x + 3) - 15 = (x2 + 3x + 5) (x2 + 3x - 3).

Butt. Multiplied polynomial (x-4) 4+ (x + 2) 4

Decision. Let's say x- 4 + x + 2 = x - 1 through y.

(X - 4) 4 + (x + 2) 2 = (y - 3) 4 + (y + 3) 4 = y4 - 12y3 + 54y3 - 108y + 81 + y4 + 12y3 + 54y2 + 108y + 81 =

2y4 + 108y2 + 162 = 2 (y4 + 54y2 + 81) = 2 [(y2 + 27) 2 - 648] = 2 (y2 + 27 - √b48) (y2 + 27 + √b48) =

2 ((x-1) 2 + 27-√b48) ((x-1) 2 + 27 + √b48) = 2 (x2-2x + 28- 18√2) (x2- 2x + 28 + 18√2 ).

Combination of different methods

Often, when a polynomial is expanded into multipliers, it is possible to fix the last drop of the different methods.

Butt. Factoring the polynomial x4 - 3x2 + 4x-3.

Decision. Zastosovuchi ugrupovannya, rewritable polynomial in viglyadі x4 - 3x2 + 4x - 3 = (x4 - 2x2) - (x2 -4x + 3).

Zastosovyuchi to the first bow the method of seeing a regular square, maєmo x4 - 3x3 + 4x - 3 = (x4 - 2 · 1 · x2 + 12) - (x2 -4x + 4).

The zastosovoyuchi formula for the new square can now be written down, x4 - 3x2 + 4x - 3 = (x2 -1) 2 - (x2) 2.

Nareshtі, zasosovuchi formula of the difference squares, otrimaєmo, scho x4 - 3x2 + 4x - 3 = (x2 - 1 + x2) (x2 - 1 - x + 2) = (x2 + x-3) (x2 -x + 1).

§ 2. Symmetrical Rivnyannya

1. Symmetrical level of the third step

Equivalent to the form of ax3 + bx2 + bx + a = 0, and ≠ 0 (1) are called symmetric equals of the third step. Oskilki ax3 + bx2 + bx + a = a (x3 + 1) + bx (x + 1) = (x + 1) (ax2 + (b-a) x + a), then equal (1) is more powerful than rіvnyan x + 1 = 0 і ax2 + (b-a) x + a = 0, virіshiti yaku does not become difficult.

Butt 1. Virishiti Rivnyannya

3x3 + 4x2 + 4x + 3 = 0. (2)

Decision. Rivnyannya (2) є symmetrical rivnyannya of the third step.

Oskilki 3x3 + 4xg +4 x + 3 = 3 (x3 + 1) + 4x (x + 1) = (x + 1) (3x2 - Zx + 3 + 4x) = (x + 1) (3x2 + x + 3 ), then the equal (2) is more powerful than the equivalents x + 1 = 0 і 3х3 + x + 3 = 0.

The decision of the first one is from the qih of the rivnyan є x = -1, the other is not the answer.

View: x = -1.

2. Symmetrical level 4

equal to the mind

(3) be called the symmetrical level four.

Oskilki x = 0 is not a root of the rivnyannya (3), then, having divided the offense of a part of the rivnyannya (3) by x2, it is more equal to the rivnyannya (3):

Rewritable Rivnyannya (4) in viglyad:

At the same rivnyannya zrobimo zrobimu, todi otrimaєmo square rivnyannya

Yaksho rivnyannya (5) maє 2 roots b1 and b2, then vikhіdne іvnyannya is more powerful than sukupnostі іvnyan

If there is one root y0, then the last one is equal

Nareshty, if ryvnyannya (5) is not rooted, then it’s not rooted either.

Appendix 2. Virishiti Rivnyannya

Decision. Dane Rivnyannya є symmetrical Rivnyannya of the fourth step. So, since x = 0 is not yo root, then, having distributed the equal (6) to x2, obsessively equal to the equal of the equal:

Grouped dodankas, rewritable rіvnyаnnya (7) at the viglyad or at the viglyad

Poklavshi, otrimaєmo rivnyannya maє two roots b1 = 2 and y2 = 3. Otzhe, vikhidne rivnyannya is equal to sukupnosti rivnyan

The decision of the first x1 = 1, and the decision of the other є і.

Otzhe, vikhіdne іvnyannya has three roots: x1, x2 and x3.

View: x1 = 1,.

§3. algebraic rivnyannya

1. Decrease of the level of rivnyannya

Deyakі algebraic rіvnyannya by replacing the deyаkiy polynomial in one letter can be used to build algebraic rіvnyans, steps which are less than the level of ideological іvnyаnnya and virіshennya which are simpler.

Butt 1. Virishiti Rivnyannya

Decision. Apparently, through, the same rivnyannya (1) can be rewritten from the viglyadi Ostann rivnyannya maє root і Otzhe, rіvnyannya (1) more powerful than sukupnostі і. The decision of the first family і The decision of another family є

Rishennyi rivnyannya (1) є

Appendix 2. Virishiti Rivnyannya

Decision. Multiplying the offense of the part of the ryvnyannya by 12 and meaning through,

Otrimaєmo rivnyannya Rewrite the price of rіvnyannya at viglyadі

(3) і having recognized through the rewriting of the rivnyannya (3) at the viglyadі Ostann іvnyannya іvnyаnnya. 4)

Rishennyi sukupnosti (4) є і, stench і rіshennyi rіvnyannya (2).

2. Rivnyannya mind

rivnyannya

(5) de-Dany numbers, you can bring up to bicadrat equal for the additional substitution of the uncomfortable t. E. Zamini

Appendix 3. Rozvyazati rivnyannya

Decision. Significantly through, that is, I can easily replace the wicked ones, but Todi rivnyannya (6) can be rewritten from the viglyad, or, the stagnant formula, from the viglyad

Oskilki root of a square ryvnyannya є і current іvnyаnnya (7) є solution of sukupnostі і. Tse sukupnіst іvnyаnnya і two decisions і Otzhe, рішення івняння (6) є і

3. Rivnyannya mind

rivnyannya

(8) de numbers α, β, γ, δ, і Α such, where α

Appendix 4. Virishiti Rivnyannya

Decision. I will easily replace non-domiciled ones, i.e., Y = x + 3 or x = y - 3. Todi rivnyannya (9) can be rewritten in viglyadі

(Y-2) (y-1) (y + 1) (y + 2) = 10, i.e.

(Y2- 4) (y2-1) = 10 (10)

Bi-square root (10) can be two roots. Otzhe, rіvnyannya (9) so I have two roots:

4. Rivnyannya mind

Rivnyannya, (11)

De, not a root x = 0, to that, having distributed the equal (11) to x2, obsessively equal to the equal

If you want to replace the uncomfortable, you will rewrite at the viewer of a square ryvnyannya, the decision not to become difficult.

Appendix 5. Rozvyazati rivnyannya

Decision. So, since h = 0 is not є by the root of the equal

I will replace the rogues without a home, I will take away the rogues (y + 1) (y + 2) = 2, as I have two roots: y1 = 0 and y1 = -3. Otzhe, vikhіdne rivnyannya (12) is equally valid

Qia sukupnist has two roots: x1 = -1 і x2 = -2.

Like: x1 = -1, x2 = -2.

Respect. Rivnyannya mind,

For which, it is possible to bring to the form (11) і, moreover, it takes α> 0 and λ> 0 to form.

5. Rivnyannya mind

rivnyannya

, (13) de numbers, α, β, γ, δ, і Α such, but αβ = γδ ≠ 0, can be rewritten by multiplying the first bow by the other, and the third by the fourth, in the view of T. E. Rivnyannya (13) Now it is written in the viglyad (11), the first solution can be carried out in the same way as the solution (11).

Appendix 6. Rozvyazati rivnyannya

Decision. Rivnyannya (14) ma viglyad (13), to that rewritten yogo in viglyad

So, since x = 0 is not a solution to the same level, then, having distributed this offense to a part by x2, it is obsessively equal to the solution. I will replace the rogues for the winners, I will accept the square rіvnyаnnya, the solution of what є і. Otzhe, vikhіdne іvnyannya (14) is equal to the similarity of іvnyany і.

Rishennya of the first ryvnyannya tsієї sukupnosti є

Other rіvnyannya tsієї sukupnostі rіshen is not possible. Otzhe, vikhіdne іvnyannya maє korinnya x1 і x2.

6. Rivnyannya mind

rivnyannya

(15) De the numbers a, b, c, q, A are like, nho, not a root x = 0, therefore, having divided the equal (15) by x2. obsessively equal to your equal, as in order to replace the uncomfortable you will rewrite at the viglyad of a square equal, the decision not to become difficult.

Appendix 7. Rishennya Rivnyannya

Decision. So, since x = 0 is not є by the root of the rivnyannya (16), then, having divided the offense of the th part by x2,

, (17) is stronger than the equal (16). Having made a substitute for an unnamed person, a rivnyannya (17) is rewritten in a viglyad

Square equal (18) maє 2 roots: y1 = 1 і y2 = -1. To that іvnyannya (17) is equal to the sukupnostі іvnyany і (19)

Sukupnist rivnyans (19) maє 4 roots:,.

The stench will be the roots of the ryvnyannya (16).

§4. Rational Rivnyannya

Equivalent to the form = 0, de H (x) and Q (x) - polynomials, are called rational.

Know the root of the family H (x) = 0, because of the need to reconsider, since they are not the root of the family Q (x) = 0. The root and only the smell will be the root of the family.

Deyakі deyakі method of rіshennya rіvnyannya type = 0.

1. Rivnyannya mind

rivnyannya

(1) with good minds, the numbers can be viewed in such a way. Group members of the equal (1) two and one pair of skin pairs, it is required to reject the polynomials of the first or zero steps in the numeral, so that only numerical multipliers appear, and in the denominators - the three members are replaced by the same two members, For mothers are also of the form (1), with a smaller number of completions, for it would be equal to the number of two races, one of which will be the first step, and the other will be of the same type (1), but with a smaller number of completions.

Butt. Rozv'yazati rivnyannya

Decision. Having grouped in the left part of the ryvnyannya (2) the first term is from the left, and the other one is rewritten, the rewritten is from the viglyad

Pidsumovyuchi in the skin arches of the warehouses

But it’s not (4), then, having distributed the price of rivnyannya, obsessive

, (5) more than equal (4). Zrobimu replace the homeless person, todi the rivnyannya (5) rewrite at the viglyad

In such a rank, the solution of the warehouse (2) from five warehouse in the left part was brought up to the last one (6) of the same kind, but from three warehouses in the left part. All the members in the left part of the rivnyannya (6), rewritten in

Rіshennya rіvnyannya є i. The same number of numbers does not turn to zero the standard of the rational function in the left part of the rivnyannya (7). Otzhe, rivnyannya (7) there are two roots, and to that vikhidne rivnyannya (2) is more powerful than sukupnosti rivnyan

Rishennya of the first ryvnyannya tsієї sukupnosti є

Rishennya of another rivnyannya from the price of

To that vykhіdne іvnyannya maє kornnya

2. Rivnyannya mind

rivnyannya

(8) with good minds, one can think of numbers like this: you need to see the whole part in the skin from the fractions of the rivnyannya, i.e.

Bring him to the form (1) і by virtue of yogi in the way described in the foregoing paragraph.

Butt. Rozv'yazati rivnyannya

Decision. Writable іvnyannya (9) at the viglyad or at the viglyad

Warehouses in the arches

I will replace the rogue, the rewrite (11) at the viglyad

Pidsumovuchi members in the livi part of the rivnyannya (12), rewritten yogo in viglyadі

It is easy to bachiti, scho rivnyannya (13) I have two roots: i. Otzhe, vikhidne rivnyannya (9) machotiri root:

3) Rivnyannya mind.

Rivnyannya type (14) with good minds on numbers, you can check out as follows: open (yakscho tse, vividly, you can) skin with fractions in the left part of rivnyannya (14) in the sum of the simplest fractions

Bring the ryvnyannya (14) to the form (1), then, having performed the manual overloading of the members of the discarded ryvnyannya, by the method that was victorious in paragraph 1).

Butt. Rozv'yazati rivnyannya

Decision. Oskilki і, then, multiplying the number of skin fraction in the equal (15) by 2 and adding that the equal (15) can be written in the viglyadі

Rivnyannya (16) maє viglyad (7). Overgrouping warehouses in the whole house

Rivnyannya (17) is equally valid

For the new vernacular of the sukupnost (18), I’ll completely replace the unattended Todi, I’ll rewrite at the viglyad or at the viglyad.

All the members in the left part of the family (19), rewrite yogo in viglyad

So, since the ryvnyannya is not root, then the rivnyannya (20) is also not.

Perceiving the ryvnyannya sukupnosti (18), it is possible to enter the ODZ of another rivnyannya sukupnosti (18), then it is vin є dinimi root (18), and that means, і vichіdny rіvnyannya.

4. Rivnyannya mind

rivnyannya

(21) with good minds on the numbers і A, if a skin supplement is given in the left part of the eye, it can be brought up to the eye (1).

Butt. Rozv'yazati rivnyannya

Decision. Rewritable rivnyannya (22) for viglyad or for viglyad

With such a rank, rіvnyannya (23) is brought to the form (1). Now, the grouping first term is from the left, and the other from the third, rewritten the rivnyannya (23) in the viglyad

The price is equal to that of the family. (24)

The remaining rіvnyannya sukupnostі (24) can be rewritten at the viglyadі

The decision of the rivnyannya є i, so as to enter the ODZ of another rivnyannya sukupnost (30), then the sukupnist (24) has three roots:. All the stench is the solution of the wicked ryvnyannya.

5. Rivnyannya mind.

Rivnyannya vid (25)

With good minds on the numbers, the replacement of the homeless can be brought to the level of mind

Butt. Rozv'yazati rivnyannya

Decision. So, as not (26), then you have diluted the number and the denominator of the skin fraction in the left part on, rewritten from the viglyad

Having made a substitute for the winners, rewrite the ryvnyannya (27) at the viglyad

Virishuchi rivnyannya (28) і і. To that an equal (27), it is equally worthy of an equal to an equal. (29)

Horner's scheme

VIRISHENNI RIVNIAN WITH PARAMETERS
Z GROUP "C" PID HOUR OF PREPARATION UP TO ZNO

Kazantseva Lyudmila Viktorivna

teacher of mathematics MBOU "Uyarskaya ZOSH № 3"

On optional occupations, it is necessary to expand the amount of clear knowledge for the development of the advanced folding of the group "C".

This robot will be able to see a lot of busy people.

Completely introduce Horner's scheme for the variation of "Polynomial into polynomials". Tsei material allows the development of common orders in the way of grouping polynomials, and more rational way To save an hour.

The plan is to borrow.

Busy 1.

1. Explanation of theoretical material.

2. Decision of butts a B C D).

Busy 2.

1. Rishennya Rivnyan a B C D).

2. Knowledge of the rational roots of the polynomial

Zasosuvannya Horner's scheme with parameters.

Busy 3.

zavdannya a B C).

Busy 4.

1. Zavdannya d), e), f), g), h).

rіshennya rіvnyan other steps.

Horner's scheme.

theorem : Come on, a non-short drib є root of the rivnyannya

a o x n + a 1 x n-1 + ... + a n-1 x 1 + a n = 0

with a wide range of features. todі number Rє Senior Officer a about .

succession: Be-like the root of the family with all the functions and the partner of the first member.

succession: Yakshcho senior staff member of the family 1 Then all the rational roots, as the stench smells - the whole.

butt 1. 2x 3 - 7x 2 + 5x - 1 = 0

Nekhay not short drib є root of the rivnyannya, todR є number1: ± 1

q є Senior Member's Dealer: ± 1; ± 2

Rational Roots of Rivnyannya Requirements for the Middle Numbers:± 1; ±.

f (1) = 2 - 7 + 5 - 1 = - 1 ≠ 0

f (-1) = -2 - 7 - 5 - 1 ≠ 0

f () = – + – 1 = – + – = 0

Root є number .

rosodil polynomial P (x) = a about NS NS + a 1 x n -1 + … + a n two-term ( x - £) manually viconuvati for Horner's scheme.

Significantly different part P (x) on ( x - £) across Q (x ) = b o x n -1 + b 1 x n -2 + … b n -1 ,

and the surplus through b n

P (x) =Q (x ) (x – £) + b n That is the same thing

a about NS NS + a 1 x n-1 + ... + a n = (B o x n-1 + … + b n-1 ) (x - £) +b n

Q (x ) - polynomial, step on 1 lower than the step of the output polynomial. polynomial performance Q (x ) This is based on Horner's scheme.

but about

a 1

a 2

a n-1

a n

b o = a about

b 1 = a 1 + £· b o

b 2 = a 2 + £· b 1

b n-1 = a n-1 + £· b n-2

b n = a n + £· b n-1

In the first row of the table, write down the polynomial performance P (x).

As soon as the steps of the winter are on the way, then on the top of the class tables it is written 0.

Senior officer of the private door to the senior officer of the date ( a about = b o ). yaksho £ є by the root of a polynomial, then go to the left 0.

butt 2... Multipliers with a wide range of performance factors

P (x) = 2x 4 - 7x 3 - 3x 2 + 5x - 1

± 1.

go - 1.

dimo P (x) on (X + 1)

– 7

– 3

– 1

– 9

– 1

2x 4 - 7x 3 - 3x 2 + 5x - 1 = (x + 1) (2x 3 - 9x 2 + 6x - 1)

Shukaєmo tsіlі roots of the middle member: ± 1

So yak is a senior member of the 1, then roots can be fractions of numbers: - ; .

go .

– 9

– 1

– 8

2x 3 - 9x 2 + 6x - 1 = (x -) (2x 2 - 8x + 2) = (2x - 1) (x 2 - 4x + 1)

Trinomial NS 2 - 4x + 1 Multipliers with a wide range of performance are NOT available.

zavdannya:

1. Place on multipliers with a wide range of performance:

a) NS 3 - 2x 2 - 5x + 6

q: ± 1;

p: ± 1; ± 2; ± 3; ± 6

: ± 1; ± 2; ± 3; ± 6

We know the rational roots of the polynomial f (1) = 1 – 2 – 5 + 6 = 0

x = 1

– 2

– 5

– 1

– 6

x 3 - 2x 2 - 5x + 6 = (x - 1) (x 2x - 6) = (x - 1) (x - 3) (x + 2)

Visually, the roots of the square rivnyannya

x 2 - x - 6 = 0

x = 3; x = - 2

b) 2x 3 + 5x 2 + X - 2

p: ± 1; ± 2

q: ± 1; ± 2

: ± 1; ± 2; ±

We know the root of the third-degree polynomial

f (1) = 2 + 5 + 1 - 2 ≠ 0

f (-1) = - 2 + 5 - 1 - 2 = 0

One of the roots of the rivnyannya x = - 1

– 2

– 1

– 2

2x 3 + 5x 2 + x - 2 = (x + 1) (2x 2 + 3x - 2) = (x + 1) (x + 2) (2x - 1)

Expandable square trinomial 2x 2 +3 x - 2 on multipliers

2x 2 +3 x - 2 = 2 (x + 2) (x -)

D = 9 + 16 = 25

x 1 = - 2; x 2 =

v) NS 3 - 3x 2 + X + 1

p: ± 1

q: ± 1

: ± 1

f (1) = 1 - 3 + 1 - 1 = 0

One of the roots of the third-degree polynomial є x = 1

– 3

– 2

– 1

x 3 - 3x 2 + x + 1 = (x - 1) (x 2 - 2x - 1)

Know the root of the rivnyannya NS 2 - 2x - 1 = 0

D = 4 + 4 = 8

x 1 = 1 –

x 2 = 1 +

x 3 - 3x 2 + X + 1 = (x - 1) (x - 1 +
) (X - 1 -
)

G) NS 3 - 2x - 1

p: ± 1

q: ± 1

: ± 1

Visually, the roots of the polynomial

f (1) = 1 - 2 - 1 = - 2

f (-1) = - 1 + 2 - 1 = 0

first root x = - 1

– 2

– 1

x 3 - 2x - 1 = (x + 1) (x 2x - 1)

x 2 - x - 1 = 0

D = 1 + 4 = 5

x 1.2 =

x 3 - 2x - 1 = (x + 1) (x -
) (NS -
)

2. Virishiti Rivnyannya:

a) NS 3 - 5x + 4 = 0

Visually, the roots of the third-degree polynomial

: ± 1; ± 2; ± 4

f (1) = 1 - 5 + 4 = 0

One root є x = 1

– 5

– 4

x 3 - 5x + 4 = 0

(X - 1) (x 2 + x - 4) = 0

NS 2 + X - 4 = 0

D = 1 + 16 = 17

x 1 =
; NS 2 =

as follows: 1;
;

b) NS 3 - 8x 2 + 40 = 0

Visually, the roots of the third-degree polynomial.

: ± 1; ± 2; ± 4; ± 5; ± 8; ± 10; ± 20; ± 40

f (1) ≠ 0

f (-1) ≠ 0

f (-2) = - 8 - 32 + 40 = 0

One root є x = - 2

– 8

– 2

– 10

The third-order polynomial is decomposable into factors.

x 3 - 8x 2 + 40 = (x + 2) (x 2 - 10x + 20)

We know the root of the square ryvnyannya NS 2 - 10x + 20 = 0

D = 100 - 80 = 20

x 1 = 5 –
; NS 2 = 5 +

View: - 2; 5 –
; 5 +

v) NS 3 - 5x 2 +3 x + 1 = 0

Shukaєmo tsіlі roots of the middle of the long-term member: ± 1

f (-1) = - 1 - 5 - 3 + 1 ≠ 0

f (1) = 1 - 5 + 3 + 1 = 0

go x = 1

– 5

– 4

– 1

x 3 - 5x 2 + 3x + 1 = 0

(X - 1) (x 2 - 4x - 1) = 0

Viznachaєmo root of a square ryvnyannya NS 2 - 4x - 1 = 0

D = 20

x = 2 +
; x = 2 -

as follows: 2 –
; 1; 2 +

G) 2x 4 - 5x 3 + 5x 2 – 2 = 0

p: ± 1; ± 2

q: ± 1; ± 2

: ± 1; ± 2; ±

f (1) = 2 - 5 + 5 - 2 = 0

One of the roots of the rivnyannya x = 1

– 5

– 2

– 3

2x 4 - 5x 3 + 5x 2 - 2 = 0

(X - 1) (2x 3 - 3x 2 + 2x + 2) = 0

It is known for the same scheme of the root of the third step.

2x 3 - 3x 2 + 2x + 2 = 0

p: ± 1; ± 2

q: ± 1; ± 2

: ± 1; ± 2; ±

f (1) = 2 - 3 + 2 + 2 ≠ 0

f (-1) = - 2 - 3 - 2 + 2 ≠ 0

f (2) = 16 - 12 + 4 + 2 ≠ 0

f (-2) = - 16 - 12 - 4 + 2 ≠ 0

f() = – + 1 + 2 ≠ 0

f(–) = – – – 1 + 2 ≠ 0

The offensive root of the ryvnyannyax = -

– 3

– 4

2x 3 - 3x 2 + 2x + 2 = 0

(X +) (2x 2 - 4x + 4) = 0

Visually, the roots of the square rivnyannya 2x 2 - 4x + 4 = 0

x 2 - 2x + 2 = 0

D = - 4< 0

Otzhe, to the roots of the fourth step

1 i –

as follows: –; 1

3. Know the rational roots of the polynomial

a) NS 4 - 2x 3 - 8x 2 + 13x - 24

q: ± 1

: ± 1; ± 2; ± 3; ± 4; ± 6; ± 8; ± 12; ± 24

Pidberemo one of the roots of the fourth degree polynomial:

f (1) = 1 - 2 - 8 + 13 - 24 ≠ 0

f (-1) = 1 + 2 - 8 - 13 - 24 ≠ 0

f (2) = 16 - 16 - 32 + 26 - 24 ≠ 0

f (-2) = 16 + 16 - 72 - 24 ≠ 0

f (-3) = 81 + 54 - 72 - 39 - 24 = 0

One of the roots of the polynomial NS 0= – 3.

x 4 - 2x 3 - 8x 2 + 13x - 24 = (x + 3) (x 3 - 5x 2 + 7x + 8)

We know the rational roots of the polynomial

x 3 - 5x 2 + 7x + 8

p: ± 1; ± 2; ± 4; ± 8

q: ± 1

f (1) = 1 - 5 + 7 + 8 ≠ 0

f (-1) = - 1 - 5 - 7 - 8 ≠ 0

f (2) = 8 - 20 + 14 + 8 ≠ 0

f (-2) = - 8 - 20 - 14 + 8 ≠ 0

f (-4) = 64 - 90 - 28 + 8 ≠ 0

f (4) ≠ 0

f (-8) ≠ 0

f (8) ≠ 0

Krim numbers x 0 = – 3 іnshikh rational roots dumb.

b) NS 4 - 2x 3 - 13x 2 - 38x - 24

p: ± 1; ± 2; ± 3; ± 4; ± 6; ± 8; ± 12; ± 24

q: ± 1

f (1) = 1 + 2 - 13 - 38 - 24 ≠ 0

f (–1) = 1 – 2 – 13 + 38 – 24 = 39 – 39 = 0, tobto x = - 1 root of a polynomial

– 13

– 38

– 24

– 1

– 14

– 24

x 4 - 2x 3 - 13x 2 - 38x - 24 = (x + 1) (x 3 - x 2 - 14x - 24)

Visually, the roots of the third-degree polynomial NS 3 - NS 2 - 14x - 24

p: ± 1; ± 2; ± 3; ± 4; ± 6; ± 8; ± 12; ± 24

q: ± 1

f (1) = - 1 + 1 + 14 - 24 ≠ 0

f (-1) = 1 + 1 - 14 - 24 ≠ 0

f (2) = 8 + 4 - 28 - 24 ≠ 0

f (-2) = - 8 + 4 + 28 - 24 ≠ 0

This means that the other root of the polynomial x = - 2

– 14

– 24

– 2

– 1

– 12

x 4 - 2x 3 - 13x 2 - 38x - 24 = (x + 1) (x 2 + 2) (x 2x - 12) =

= (X + 1) (x + 2) (x + 3) (x - 4)

as follows: – 3; – 2; – 1; 4

Zasosuvannya Horner's scheme with the definition of a parameter.

Know the best value of the parameter a, at what level f (X) = 0 there are three different roots, one of them NS 0 .

a) f (X) = x 3 + 8x 2 + Ah +b , NS 0 = – 3

So one of the roots NS 0 = – 3 Then behind Horner's scheme there is:

– 3

– 15 + a

0 = - 3 (- 15 + a) + b

0 = 45 - 3a + b

b = 3а - 45

x 3 + 8x 2 + ax + b = (x + 3) (x 2 + 5x + (a - 15))

rivnyannya NS 2 + 5x + (a - 15) = 0 D > 0

a = 1; b = 5; s = (a - 15),

D = b 2 - 4ac = 25 - 4 (a - 15) = 25 + 60 - 4a> 0,

85 - 4a> 0;

4a< 85;

a< 21

The highest value of the parameter a, at what level

f (X) = 0 maє three roots, a = 21

as follows: 21.

b) f (x) = x 3 - 2x 2 + Ax + b, x 0 = – 1

So yak is one of the roots NS 0= – 1, then behind Horner's scheme

– 2

– 1

– 3

3 + a

x 3 - 2x 2 + ax + b = (x + 1) (x 2 - 3x + (3 + a))

rivnyannya x 2 – 3 x + (3 + a ) = 0 guilty of mother two roots. Tse vikonutsya only in the fall, if D > 0

a = 1; b = - 3; c = (3 + a),

D = b 2 - 4ac = 9 - 4 (3 + a) = 9 - 12 - 4a = - 3 - 4a> 0,

– 3 - 4a> 0;

– 4a< 3;

a < –

most significant a = - 1 a = 40

as follows: a = 40

G) f (x) = x 3 - 11x 2 + Ax + b, x 0 = 4

So yak is one of the roots NS 0 = 4 Then behind Horner's scheme

– 11

– 7

– 28 + a

x 3 - 11x 2 + ax + b = (x - 4) (x 2 - 7x + (a - 28))

f (x ) = 0, yaksho x = 4 abo x 2 – 7 x + (a – 28) = 0

D > 0, tobto

D = b 2 - 4ac = 49 - 4 (a - 28) = 49 + 112 - 4a = 161 - 4a> 0,

161 - 4a> 0;

– 4a< – 161; f x 0 = – 5 Then behind Horner's scheme

– 5

– 40 + a

x 3 + 13x 2 + ax + b = (x + 5) (x 2 + 8x + (a - 40))

f (x ) = 0, yaksho x = - 5 abo x 2 + 8 x + (a – 40) = 0

Rivnyannya maє two roots, yaksho D > 0

D = b 2 - 4ac = 64 - 4 (a - 40) = 64 + 1 60 - 4a = 224 - 4a> 0,

224- 4a> 0;

a< 56

rivnyannya f (x ) there are three roots with the greatest value a = 55

as follows: a = 55

g) f (x ) = x 3 + 19 x 2 + ax + b , x 0 = – 6

So yak is one of the roots – 6 Then behind Horner's scheme

– 6

a - 78

x 3 + 19x 2 + ax + b = (x +6) (x 2 + 13x + (a - 78)) = 0

f (x ) = 0, yaksho x = - 6 abo x 2 + 13 x + (a – 78) = 0

Another rivnyannya has two roots, yaksho

Zastosuvannya rivnyan is widely expanded in our life. The stench of vikoristovuyutsya in bagatokh rozrakhunkah, bud_vnstvі sporud and navіt sportі. Rivnyannya Lyudina has been victorious for a long time and from the quiet period of her growing up. In mathematics, it is often the case that there is a level of sophistication with a wide range of functions. The virginity of a given family must be:

Visnachiti of rational root of the family;

The multiplication of the polynomial, which is located in the left part of the polynomial;

Know the roots of the family.

Admittedly, we have been given the rivnyannya of such a viglyad:

We know all the ideas of the root. Multiplying the lіva і the right of the part of the family to \

Viconaєmo will replace the winners \

In such a rank, we have been guided to the level of the fourth step, as if we were going to follow the standard algorithm: the conversion of the names was carried out, and as a result, it was done, but there were two real roots \ and two complex ones. Otrimaєmo this is the example of our fourth degree ryvnyannya:

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Clear rіshennya іvnyаn with one wintry world and another.

The step of equal P (x) = 0 is called the step of the polynomial P (x), so that the most of the steps of its term with a coefficient, not equal to zero.

So, for example, equal (x 3 - 1) 2 + x 5 = x 6 - 2 maє n'yat steps, that is, for the operation of opening the bows and the given others are obsessively more equal than equal x 5 - 2x 3 + 3 = 0 the fifth step.

There are rules that are known to be used for the next step to the next step.

Confirmation about the root of the polynomial and its divisors:

1. polynomial n-th step I do not change the number of roots n, and the root of multiplicity m is created exactly m times.

2. A polynomial of an unpaired degree may have one valid root.

3. If α is a root of P (x), then P n (x) = (x - α) Q n - 1 (x), de Q n - 1 (x) is a polynomial of degree (n - 1).

5. A polynomial guidance with a number of performance factors cannot be used as a shot-type rational root.

6. For a third-degree polynomial

P 3 (x) = ax 3 + bx 2 + cx + d can be done in one of two: otherwise, it can be folded into three more two-membered

Р 3 (x) = а (х - α) (х - β) (х - γ), or fold into a double term і of the square trinomial Р 3 (x) = а (х - α) (х 2 + βх + γ ).

7. Be a polynomial of the fourth step to expand into two square trinomials.

8. The polynomial f (x) extends by the polynomial g (x) without excess, as well as the polynomial q (x), where f (x) = g (x) q (x). For a subset of polynomials, the rule is "subtle".

9. For the identity of P (x) to a two-term (x - c), it is necessary and sufficient, but the number is rooted P (x) (as a result of Bezout's theorems).

10. Vієta's theorem: Yakscho x 1, x 2, ..., x n are the action roots of the polynomial

P (x) = a 0 x n + a 1 x n - 1 + ... + a n, then maybe the same is true:

x 1 + x 2 + ... + x n = -a 1 / a 0,

x 1 x 2 + x 1 x 3 + ... + x n - 1 x n = a 2 / a 0,

x 1 x 2 x 3 + ... + x n - 2 x n - 1 x n = -a 3 / a 0,

x 1 x 2 x 3 x n = (-1) n a n / a 0.

solution of applications

Butt 1.

Know the surplus of the length P (x) = x 3 + 2/3 x 2 - 1/9 on (x - 1/3).

Decision.

Following Bezout's theorems: "The value of the polynomial length into a two-term (x - c) is the base value of the polynomial from with." We know P (1/3) = 0. Already, the surplus is 0 and the number 1/3 is the root of the polynomial.

Like: R = 0.

Butt 2.

Rozdіliti "kutoch" 2x 3 + 3x 2 - 2x + 3 on (x + 2). Know the surplus and not a small part.

Decision:

2x 3 + 3x 2 - 2x + 3 | x + 2

2x 3 + 4 x 2 2x 2x

X 2 - 2 x

View: R = 3; private: 2x 2x.

Basic methods of visualization of level steps

1. The introduction of new changes

The method of introducing new changes is already familiar on the butt of bicadrons. Win polyagaє in that, for the update f (x) = 0, introduce a new change (substitution) t = x n or t = g (x) and use f (x) through t, we can see the new one r (t). Virіshuyuchi poіm іvnyannya r (t), to know the root:

(T 1, t 2, ..., t n). For the sake of recognizing the root of the vernacular q (x) = t 1, q (x) = t 2, ..., q (x) = t n, which is the root of the vernacular.

Butt 1.

(X 2 + x + 1) 2 - 3x 2 - 3x - 1 = 0.

Decision:

(X 2 + x + 1) 2 - 3 (x 2 + x) - 1 = 0.

(X 2 + x + 1) 2 - 3 (x 2 + x + 1) + 3 - 1 = 0.

Replacement (x 2 + x + 1) = t.

t 2 - 3t + 2 = 0.

t 1 = 2, t 2 = 1. Zvorotn_y change:

x 2 + x + 1 = 2 or x 2 + x + 1 = 1;

x 2 + x - 1 = 0 or x 2 + x = 0;

Suggestion: From the first line: x 1, 2 = (-1 ± √5) / 2, from the other: 0 і -1.

2. Distribution into multipliers by the method of multiplying і speed multiplier formulas

The basis of the given method is also not new і polyaga in the presence of such a rank, the dermal group has replaced the zagalny multiplier. For a whole number of people it is necessary to make sure that the items are priyomy.

Butt 1.

x 4 - 3x 2 + 4x - 3 = 0.

Decision.

Clearly - 3x 2 = -2x 2 - x 2 and group:

(X 4 - 2x 2) - (x 2 - 4x + 3) = 0.

(X 4 - 2x 2 +1 - 1) - (x 2 - 4x + 3 + 1 - 1) = 0.

(X 2 - 1) 2 - 1 - (x - 2) 2 + 1 = 0.

(X 2 - 1) 2 - (x - 2) 2 = 0.

(X 2 - 1 - x + 2) (x 2 - 1 + x - 2) = 0.

(X 2 - x + 1) (x 2 + x - 3) = 0.

x 2 - x + 1 = 0 or x 2 + x - 3 = 0.

Suggestion: In the first place, there are no roots, from the other: x 1, 2 = (-1 ± √13) / 2.

3. Multiplier spreading by the method of non-valued coefficients

The essence of the method of polarity is that the output polynomial is multiplied with unavailable coefficients. Vikoristovuyu power, but the polynomials of the equal, as well as the efficiency at the same steps, there is no such thing as the distribution.

Butt 1.

x 3 + 4x 2 + 5x + 2 = 0.

Decision.

The polynomial of the 3rd step can be expanded into a linear and square multiplier.

x 3 + 4x 2 + 5x + 2 = (x - a) (x 2 + b x + c),

x 3 + 4x 2 + 5x + 2 = x 3 + bx 2 + cx - ax 2 - abx - ac,

x 3 + 4x 2 + 5x + 2 = x 3 + (b - a) x 2 + (cx - ab) x - ac.

Viruses the system:

(B - a = 4,
(C - ab = 5,
(-Ac = 2,

(A = -1,
(B = 3,
(C = 2, tobto

x 3 + 4x 2 + 5x + 2 = (x + 1) (x 2 + 3x + 2).

The root of the ryvnyannya (x + 1) (x 2 + 3x + 2) = 0 is easy to find.

View: -1; -2.

4. Method of choosing a root according to the oldest and most important factor

Method of spiraling into theorems:

1) Each root of a polynomial has a number of coefficients є the number of a member.

2) In order for a non-short dib p / q (p - whole, q - natural) to be the root of the equal number of factors, it is necessary that the number p will be the number of a member of a 0, and q is a natural partner of a senior member.

Butt 1.

6x 3 + 7x 2 - 9x + 2 = 0.

Decision:

6: q = 1, 2, 3, 6.

The same, p / q = ± 1, ± 2, ± 1/2, ± 1/3, ± 2/3, ± 1/6.

Knowing one root, for example - 2, the other roots known, vikoristovuchi rose in a cube, the method of unimportant coefficients or Horner's scheme.

View: -2; 1/2; 1/3.

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In a zalny vipadku ryvnyannya, scho there are 4 steps, cannot be allowed in radicals. If only one is one, one can still know the root of a polynomial, which is worth evil in a rational way, as if by looking at it there are more polynomials in steps of no more than 4 steps. The decision of such people is based on the decomposition of the polynomial into multipliers, so it is good for you to repeat the topic before the publication of this statute.

Most often, the mother is brought to the right with the level of the main steps with the most important functions. In chicks, we can try to know the rational root, and then by expanding the polynomial into multipliers, instead of converting it into a higher level, as if it were just virishity. Within the framework of the whole material, you can do it yourself.

Rivnyannya supervisingly with a wide range of functions

All ryvnyannya, which may look like a n x n + a n - 1 x n - 1 +. ... ... + A 1 x + a 0 = 0, we can reduce to equal the same world for an additional multiplicity of both parts by a n n - 1 and having changed the change y = a n x:

a n x n + a n - 1 x n - 1 +. ... ... + A 1 x + a 0 = 0 ann xn + an - 1 ann - 1 xn - 1 + ... + a 1 (an) n - 1 x + a 0 (an) n - 1 = 0 y = anx ⇒ yn + bn - 1 yn - 1 + ... + b 1 y + b 0 = 0

Those performances, which were the result, will also be tough. In such a rank, we will need to be sure to guide the n-th stage with a number of performance factors, which is a ma viglyad x n + a n x n - 1 + ... + a 1 x + a 0 = 0.

Numerous numbers of roots of the family. It’s just as good a root as it is, it’s necessary to tell the middle of a member a 0. We’ll see it and we will present it in return, according to the result, and will change the result. If only we rejected the sameness and knew one of the roots of the rivnyannya, then we can write it in the viglyad x - x 1 part of the list xn + anxn - 1 + ... + a 1 x + a 0 to x - x 1.

Predstavlyaєmo іnshі vypisanі podilniki in P n - 1 (x) = 0, after h x 1, fragments of the root can be repeated. If we reject the identity of the roots x 2, we will be familiar with the knowledge, and the equal can be written in the viglyad (x - x 1) (x - x 2) P n - 2 (x) = 0. Here P n - 2 (x) will be private. distribution P n - 1 (x) to x - x 2.

Prodovzhuєmo and give to sort out the files. We know all the roots and meaningfully the number of yak m. If you want to write, you can represent yak x - x 1 x - x 2 ... x - xm ... For p_drahunku manually vicoristovuvati Horner's scheme.

As long as we have a great deal of efficiency, we can’t be counted as a result of shot roots.

In our bag there was a ryvnyannya P n - m (x) = 0, the root of which can be known in a manual way. The stench can be irrational or complex.

It is shown on a specific application, as such is the scheme of solution.

butt 1

Umova: know the solution x 4 + x 3 + 2 x 2 - x - 3 = 0.

Decision

It’s almost impossible to know the roots.

We have a vilny member, an equal minus three. At the new one podilniki, pіvnі 1, - 1, 3 і - 3. Pіdstavami іkh wіchіdnе іvnyаnnya and wondered, as they will give in the result of the same.

For x, which is single, mi is otrimaєmo 1 4 + 1 3 + 2 · 1 2 - 1 - 3 = 0, which means that one will be the root of the given rivnyannya.

Now the viconamo of the division of the polynomial x 4 + x 3 + 2 x 2 - x - 3 on (x - 1) in the store:

Hence, x 4 + x 3 + 2 x 2 - x - 3 = x - 1 x 3 + 2 x 2 + 4 x + 3.

1 3 + 2 1 2 + 4 1 + 3 = 10 ≠ 0 (- 1) 3 + 2 (- 1) 2 + 4 - 1 + 3 = 0

We have the same identity, which means that we knew one more root of the ryvnyannya, the ryvny - 1.

Dilimo polynomial x 3 + 2 x 2 + 4 x + 3 on (x + 1) in the store:

Otrimuєmo, scho

x 4 + x 3 + 2 x 2 - x - 3 = (x - 1) (x 3 + 2 x 2 + 4 x + 3) = = (x - 1) (x + 1) (x 2 + x + 3)

Pidstavlyaєmo chergovy dilnik in parity x 2 + x + 3 = 0, repaired z - 1:

1 2 + (- 1) + 3 = 3 ≠ 0 3 2 + 3 + 3 = 15 ≠ 0 (- 3) 2 + (- 3) + 3 = 9 ≠ 0

Equality, rejected as a result, will be unsettled, even though the peasant has more dumb roots.

The root will be the root of the rotation x 2 + x + 3.

D = 1 2 - 4 1 3 = - 11< 0

For a given square trinomial, there are no functional roots, but a complex connection: x = - 1 | 2 ± i 11 2.

To clarify, the substitution of the pod in the stockpile can be made using Horner's scheme. The order is as follows: because of that, as we have found the first root of the ryvnyannya, the table is remembered.

In the tables of performance, you can immediately change the performance of the private type of the polynomial, meaning x 4 + x 3 + 2 x 2 - x - 3 = x - 1 x 3 + 2 x 2 + 4 x + 3.

Writing the znakhozhennya of the offensive root, rivnogo - 1, we will not recognize it:

as follows: x = - 1, x = 1, x = - 1 | 2 ± i 11 2.

butt 2

Umova: check the line x 4 - x 3 - 5 x 2 + 12 = 0.

Decision

In a vilny member є, the followers are 1, - 1, 2, - 2, 3, - 3, 4, - 4, 6, - 6, 12, - 12.

Reverse in order:

1 4 - 1 3 - 5 1 2 + 12 = 7 ≠ 0 (- 1) 4 - (- 1) 3 - 5 (- 1) 2 + 12 = 9 ≠ 0 2 4 2 3 - 5 2 2 + 12 = 0

Hence, x = 2 will be the root of the root. Rozdilimo x 4 - x 3 - 5 x 2 + 12 on x - 2, skimming with Horner's scheme:

In a small bag, we can accept x - 2 (x 3 + x 2 - 3 x - 6) = 0.

2 3 + 2 2 - 3 2 - 6 = 0

This means that 2 will be a root. Rozdilimo x 3 + x 2 - 3 x - 6 = 0 on x - 2:

As a result, we can make (x - 2) 2 (x 2 + 3 x + 3) = 0.

The reversal of the solution to the senses is not easy, the odds are x 2 + 3 x + 3 = 0 more and more quickly for the additional discriminant.

Virishimo is square with rivnyannya:

x 2 + 3 x + 3 = 0 D = 3 2 - 4 1 3 = - 3< 0

I will recognize a complex pair of roots: x = - 3 2 ± i 3 2.

vidpovid: X = - 3 2 ± i 3 2.

butt 3

Umova: know for the equal x 4 + 1 2 x 3 - 5 2 x - 3 = 0 valid roots.

Decision

x 4 + 1 2 x 3 - 5 2 x - 3 = 0 2 x 4 + x 3 - 5 x - 6 = 0

Viconuєmo multiplication 2 3 of both parts of the country:

2 x 4 + x 3 - 5 x - 6 = 0 2 4 x 4 + 2 3 x 3 - 20 2 x - 48 = 0

Change y = 2 x:

2 4 x 4 + 2 3 x 3 - 20 2 x - 48 = 0 y 4 + y 3 - 20 y - 48 = 0

As a result, we have seen the standard level of the 4th stage, as it is possible to follow the standard scheme. Converting the podilniki, distributed and dispensed into the bag, there are 2 root roots y = - 2, y = 3 and two complex roots. We will not be guided by the whole decision here. By virtue of replacing the real roots of this equal will be x = y 2 = - 2 2 = - 1 і x = y 2 = 3 2.

as follows: x 1 = - 1, x 2 = 3 2

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