Introduction

1. Theoretical part

1.1 Basic concepts and definitions

1.3 Cardano formula

2. Problem solving

Conclusion

Introduction

Equations. It can be said for sure that there is not a single person who would not be familiar with them. From an early age, children begin to solve “problems with X”. Further more. True, for many, acquaintance with equations ends with school affairs. The famous German mathematician Courant wrote: “For more than two thousand years, the possession of some, not too superficial, knowledge in the field of mathematics was a necessary integral part in the intellectual inventory of each educated person". And among this knowledge was the ability to solve equations.

Already in ancient times, people realized how important it is to learn how to solve algebraic equations of the form

a0xn + a1xn ​​- 1 + ... + an = 0

after all, very many and very diverse questions of practice and natural science are reduced to them (of course, here we can immediately assume that a0 ¹ 0, since otherwise the degree of the equation is actually not n, but less). Many, of course, came up with the tempting idea to find formulas for any power of n that would express the roots of the equation in terms of its coefficients, i.e., would solve the equation in radicals. However, the "gloomy Middle Ages" turned out to be as gloomy as possible in relation to the problem under discussion - for seven whole centuries no one found the required formulas! Only in the 16th century, Italian mathematicians managed to move further - to find formulas for n \u003d 3 and 4. The history of their discoveries and even the authorship of the found formulas are rather obscure to this day, and we will not find out here complicated relationship between Ferro, Cardano, Tartaglia and Ferrari, but let's put it better mathematical essence affairs.

The purpose of the work is to explore various methods for solving equations of the third degree.

To achieve this goal, it is necessary to perform a number of tasks:

-Analysis scientific literature;

-Analysis of school textbooks;

-Selection of examples for solution;

-Solution of equations by various methods.

The work consists of two parts. The first deals with various methods for solving equations. The second part is devoted to solving equations different ways.

1. Theoretical part

1 Basic concepts and definitions

A cubic equation is an equation of the third degree of the form:

The number x that turns the equation into an identity is called the root or solution of the equation. It is also the root of a polynomial of the third degree, which is on the left side of the canonical notation.

Over the field of complex numbers, according to the fundamental theorem of algebra, a cubic equation always has 3 roots (taking into account the multiplicity).

Since every real polynomial is not even degree has at least one real root, all possible cases of the composition of the roots of a cubic equation are exhausted by the three described below. These cases are easily distinguished using the discriminant

So there are only three possible cases:

If a? > 0, then the equation has three different real roots.

If a?< 0, то уравнение имеет один вещественный и пару комплексно сопряжённых корней.

If a? = 0, then at least two roots coincide. This can be when the equation has a double real root and another real root different from them; or, all three roots coincide, forming a root of multiplicity 3. The resultant of the cubic equation and its second derivative helps to separate these two cases: the polynomial has a root of multiplicity 3 if and only if the indicated resultant is also zero.

The roots of a cubic equation are related to the coefficients as follows:

1.2 Methods for solving cubic equations

The most common method for solving cubic equations is the enumeration method.

First, by enumeration, we find one of the roots of the equation. The fact is that cubic equations always have at least one real root, and the integer root of the cubic equation with integer coefficients is a divisor of the free term d. The coefficients of these equations are usually chosen so that the desired root lies among small integers, such as: 0, ± 1, ± 2, ± 3. Therefore, we will look for the root among these numbers and check it by substituting it into the equation. The success rate with this approach is very high. Let's assume this root.

The second stage of the solution is the division of the polynomial by the binomial x - x1. According to Bezout's theorem, this division without a remainder is possible, and as a result we get a polynomial of the second degree, which must be equated to zero. By solving the resulting quadratic equation, we will find (or not) the remaining two roots.

Solution of a two-term cubic equation

The two-term cubic equation has the form (2)

This equation is reduced to the form by dividing by a non-zero coefficient A. Next, the formula for the abbreviated multiplication of the sum of cubes is applied:

From the first bracket we find, and the square trinomial has only complex roots.

Recurrent cubic equations

The reciprocal cubic equation has the form and B-coefficients.

Let's group:

Obviously, x=-1 is the root of such an equation, and the roots of the resulting square trinomial are easily found through the discriminant.

1.3 Cardano formula

AT general case, the roots of the cubic equation are found by the Cardano formula.

For the cubic equation (1), the values ​​are found using the substitution: x= (2), and the equation is reduced to the form:

an incomplete cubic equation in which there will be no term containing the second degree.

We assume that the equation has coefficients complex numbers. This equation will always have complex roots.

Let's denote one of these roots: . We introduce an auxiliary unknown u and consider the polynomial f(u)=.

Let's denote the roots of this polynomial through? and?, according to the Viette theorem (see p. 8):

Substitute in equation (3), expression (4), we obtain:

From the other side of (5): (7)

It follows from here, i.e. from formulas (6), (7), that the numbers are the roots of the equation:

From the last equation:

The other two roots are found by the formula:

1.4 trigonometric formula Vieta

This formula finds solutions to the reduced cubic equation, that is, an equation of the form

Obviously, any cubic equation can be reduced to an equation of the form (4) by simply dividing it by the coefficient a. So, the algorithm for applying this formula:

Calculate

2. Calculate

3. a) If, then compute

And our equation has 3 roots (real):

b) If, then replace trigonometric functions hyperbolic.

Calculate

Then the only root (real):

Imaginary roots:

C) If, then the equation has less than three various solutions:

2. Problem solving

Example 1. Find the real roots of a cubic equation

We apply the formula for abbreviated multiplication of the difference of cubes:

From the first bracket we find that the square trinomial in the second bracket has no real roots, since the discriminant is negative.

Example 2. Solve the equation

This equation is reciprocal. Let's group:

is the root of the equation. Finding the roots of a square trinomial

Example 3. Find the roots of a cubic equation

Let's transform the equation to the reduced one: multiply by both parts and make a change of variable.

The free term is 36. Let's write down all its divisors:

We substitute them in turn into equality until we get the identity:

Thus, is the root. It matches

Divide by using Horner's scheme.

Polynomial coefficients2-11129-0.52-11+2*(-0.5)=-1212-12*(-0.5)=189+18*(-0.5)=0

We get

Let's find the roots of the square trinomial:

Obviously, that is, its multiple root is.

Example 4. Find the real roots of the equation

is the root of the equation. Find the roots of a square trinomial.

Since the discriminant less than zero, then the trinomial has no real roots.

Example 5. Find the roots of the cubic equation 2.

Hence,

We substitute into the Cardano formula:

takes three values. Let's write them down.

When we have

When we have

When we have

Let's break these values ​​into pairs, which in the product give

The first pair of values ​​and

The second pair of values ​​and

The third pair of values ​​and

Back to the Cardano formula

Thus,

Conclusion

cubic trinomial equation

As a result of the execution term paper various methods for solving equations of the third degree were investigated, such as the enumeration method, Carano's formula, Vieta's formula, methods for solving reciprocal, two-term equations.

List of sources used

1)Bronstein I.N., Semendyaev K.A. "Handbook of mathematics for engineers and students of technical universities", M., 1986.

2)Kolmogorov A.N. Algebra and the beginnings of analysis. Study guide for 9th grade high school, 1977.

)Omelchenko V.P. Mathematics: tutorial/ V.P. Omelchenko, E.V. Kurbatova. - Rostov n / a.: Phoenix, 2005.- 380s.

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Learn how to solve cubic equations. The case when one root is known is considered. Methods for finding integers and rational roots. Application of the Cardano and Vieta formulas to solve any cubic equation.

Here we consider the solution of cubic equations of the form
(1) .
Further, we assume that this is real numbers.

(2) ,
then dividing it by , we obtain an equation of the form (1) with coefficients
.

Equation (1) has three roots: , and . One of the roots is always real. We denote the real root as . The roots and can be either real or complex conjugate. Real roots can be multiple. For example, if , then and are double roots (or roots of multiplicity 2), and is a simple root.

If only one root is known

Let us know one root of the cubic equation (1). Denote known root as . Then dividing equation (1) by , we obtain a quadratic equation. Solving the quadratic equation, we find two more roots and .

For the proof, we use the fact that the cubic polynomial can be represented as:
.
Then, dividing (1) by , we obtain a quadratic equation.

Examples of division of polynomials are presented on the page
“Division and multiplication of a polynomial by a polynomial by a corner and a column”.
The solution of quadratic equations is considered on the page
"The roots of a quadratic equation".

If one of the roots is

If the original equation is:
(2) ,
and its coefficients , , , are integers, then you can try to find an integer root. If this equation has an integer root, then it is a divisor of the coefficient. The method of searching for integer roots is that we find all the divisors of a number and check if equation (2) holds for them. If equation (2) is satisfied, then we have found its root. Let's denote it as . Next, we divide equation (2) by . We get a quadratic equation. Solving it, we find two more roots.

Examples of defining integer roots are given on the page
Examples of factorization of polynomials > > > .

Finding Rational Roots

If in equation (2) , , , are integers, and , and there are no integer roots, then you can try to find rational roots, that is, roots of the form , where and are integers.

To do this, we multiply equation (2) by and make the substitution:
;
(3) .
Next, we look for integer roots of equation (3) among the divisors of the free term.

If we have found an integer root of equation (3), then, returning to the variable , we obtain rational root equations (2):
.

Cardano and Vieta formulas for solving a cubic equation

If we do not know a single root, and there are no integer roots, then we can find the roots of a cubic equation using Cardano's formulas.

Consider the cubic equation:
(1) .
Let's make a substitution:
.
After that, the equation is reduced to an incomplete or reduced form:
(4) ,
where
(5) ; .

References:
I.N. Bronstein, K.A. Semendyaev, Handbook of Mathematics for Engineers and Students of Higher Educational Institutions, Lan, 2009.
G. Korn, Handbook of Mathematics for scientists and engineers, 2012.

Cubic Equation - algebraic equation third degree. General view of the cubic equation: ax3 + bx2 + cx + d = 0, a ≠ 0

Replacing x in this equation with a new unknown y associated with x by the equality x \u003d y - (b / 3a), the cubic equation can be reduced to a simpler (canonical) form: y3 + pу + q \u003d 0, where p \u003d - b2 + c , q = 2b – bc + d

3a2 a 27a3 3a2 a the solution of this equation can be obtained using the Cardano formula.

1.1 History of cubic equations

The term "cubic equation" was introduced by R. Descartes (1619) and W. Outred (1631).

The first attempts to find solutions to problems reducing to cubic equations were made by ancient mathematicians (for example, the problems of doubling a cube and trisecting an angle).

The mathematicians of the Middle Ages of the East created quite developed theory(in geometric shape) cubic equations; it is described in most detail in the treatise on the proofs of the problems of algebra and almukabala "Omar Haya" (circa 1070), where the question of finding positive roots 14 types of cubic equations containing only terms with positive coefficients in both parts.

For the first time in Europe trigonometric form a solution to one case of a cubic equation was given by Viet (1953).

The first solution in radicals of one of the types of cubic equations was found by S. Ferro (circa 1515), but it was not published. The discovery was independently repeated by Tartaglia (1535), indicating a rule for solving two other types of cubic equations. These discoveries were published in 1545 by G. Cardano, who mentioned the authorship of N. Tartaglia.

At the end of the XV century. Professor of Mathematics at the Universities of Rome and Milan Luca Pacioli in his famous textbook "The sum of knowledge in arithmetic, geometry, relations and proportionality" the problem of finding general method for solving cubic equations, he put it on a par with the problem of squaring a circle. And yet, through the efforts of Italian algebraists, such a method was soon found.

Let's start with simplification

If the cubic equation general view ax3 + bx2 + cx + d = 0, where a ≠ 0, divided by a, then the coefficient at x3 becomes equal to 1. Therefore, in the future we will proceed from the equation x3 + Px2 + Qx + R = 0. (1)

Same as at the heart of the solution quadratic equation lies the formula for the square of the sum, the solution of the cubic equation is based on the formula for the cube of the sum:

(a + b)3 = a3 + 3a2b + 3ab2 + b3.

In order not to get confused in the coefficients, here we replace a by x and rearrange the terms:

(x + b)3 = x3 + 3bx2 + 3b2x + b3. (2)

We see that in a proper way b, namely, by taking b = P/3, we can achieve that right part of this formula will differ from the left side of the equation x3 + Px2 + Qx + R = 0 only by the coefficient at x and the free term. We add the equation x3 + Px2 + Qx + R = 0 and (x + b)3 = x3 + 3bx2 + 3b2x + b3 and give similar ones:

(x + b)3 + (Q - 3b2)x + R - b3 = 0.

If we make the change here y = x + b, we get a cubic equation for y without a term with y2: y3 + py + q = 0.

So, we have shown that in the cubic equation x3 + Px2 + Qx + R = 0, using an appropriate substitution, you can get rid of the term containing the square of the unknown. Therefore, now we will solve an equation of the form x3 + px + q = 0. (3)

1.2 History of the Cardano Formula

The Cardano formula is named after J. Cardano, who first published it in 1545.

The author of this formula is Niccolò Tartaglia. He created this solution in 1535 specifically for participation in a mathematical competition, in which, of course, he won. Tartaglia, giving the formula (in poetic form) Cardano, presented only that part of the solution of the cubic equation in which the root has one (real) value.

The results of Cardano in this formula refer to the consideration of the so-called irreducible case, in which the equation has three values ​​(real values, in those days there were no imaginary or even negative numbers, although there were attempts in this direction). However, contrary to the fact that Cardano indicated in his publication the authorship of Tartaglia, the formula is called by the name of Cardano.

1. 3 Cardano Formula

Now let's look at the sum cube formula again, but write it differently:

(a + b)3 = a3 + b3 + 3ab(a + b).

Compare this entry with the equation x3 + px + q = 0 and try to establish a connection between them. Substitute in our formula x = a + b: x3 = a3 + b3 + 3abx, or x3 - 3abx - (a3 + b3) = 0

Now it is already clear: in order to find the root of the equation x3 + px + q = 0, it is enough to solve the system of equations a3 + b3 = - q, a3 + b3 = - q, or

3аb \u003d - p, a3b3 \u003d - p 3,

3 and take as x the sum of a and b. By changing u = a3, v = b3 this system is reduced to a completely plain sight: and + v = - q, and v = - p 3.

Then you can act in different ways, but all the "roads" will lead to the same quadratic equation. For example, according to Vieta's theorem, the sum of the roots of the given quadratic equation is equal to the coefficient at x with a minus sign, and the product is equal to the free term. This implies that and and v are the roots of the equation t2 + qt – (p/3)3 = 0.

Let's write out these roots: t1,2 = - q ± q 2 + p 3.

Variables a and b are equal to the cube roots from t1 and t2, and the desired solution of the cubic equation x3 + px + q = 0 is the sum of these roots: x = 3 - q + q 2 + p 3+ 3 - q - q 2 + p 3 .

This formula is known as the Cardano formula.

Solving Equations

Before looking at the Cardano formula in the work, let us explain how to find its other roots, if any, from one root of the cubic equation x3 + px + q = 0.

Let it be known that our equation has a root h. Then its left-hand side can be decomposed into linear and square multipliers. This is done very simply. We substitute the expression of the free term through the root q \u003d - h3 - ph into the equation and use the formula for the difference of cubes:

0 \u003d x3 - h3 + px - ph \u003d (x - h) (x2 + hx + h2) + p (x - h) \u003d (x - h) (x2 + hx + h2 + p).

Now you can solve the quadratic equation x2 + hx + h2 + p = 0 and find the rest of the roots of this cubic equation.

So, we are fully armed and, it would seem, we can cope with any cubic equation. Let's try our hand.

1. Let's start with the equation x3 + 6x - 2 = 0

We substitute p = 6 and q = -2 into the Cardano formula and after simple reductions we get the answer: x = 3√4 - 3√2. Well, the formula is quite nice. Only the prospect of taking the factor x - (3√4 - 3√2) from the left side of the equation and solving the remaining quadratic equation with "terrible" coefficients to calculate other roots is not very inspiring. However, looking at the equation more closely, we can calm down: the function on the left side is strictly increasing and therefore can vanish only once. This means that the number found is the only real root of the equation.

y y \u003d x3 + 6x - 2

3√4 – 3√2 x

Rice. 1 The graph of the function y \u003d x3 + 6x - 2 crosses the x-axis at one point - 3√4 - 3√2.

2. Next example- the equation x3 + 3x - 4 = 0.

Cardano's formula gives x = 3 2 + √5 + 3 2 - √5.

As in the previous example, we see that this root is unique. But you don't have to be super insightful to look at the equation and guess its root: x = 1. We have to admit that the formula gave out the usual unit in such a bizarre form. By the way, to simplify this cumbersome but not without elegance expression algebraic transformations fails - the cubic irrationality in it is unavoidable.

3. Well, now let's take an equation that obviously has three real roots. It is easy to compose it - just multiply three brackets of the form x - b. You just need to take care that the sum of the roots is equal to zero, because, according to general theorem Vieta, it differs from the coefficient at x2 only in sign. The simplest set of such roots is 0, 1, and -1.

Let's apply the Cardano formula to the equation x (x - 1) (x + 1) = 0, or x3 - x = 0.

Assuming p = -1 and q = 0 in it, we get x = 3 √ - 1/27 + 3 - √ - 1/27.

y y \u003d x (x - 1) (x + 1)

Rice. 2 The equation x (x - 1) (x + 1) \u003d 0 has three real roots: -1, 0 and 1. Accordingly, the graph of the function y \u003d x (x - 1) (x + 1) intersects the x-axis at three points.

appeared under the sign of the square root a negative number. This also happens when solving quadratic equations. But the quadratic equation in this case does not have real roots, while the cubic one has three of them!

A closer analysis shows that we did not fall into this trap by accident. The equation x3 + px + q = 0 has three real roots if and only if the expression Δ = (q/2)2 + (p/3)3 under square root in the Cardano formula is negative. If Δ > 0, then there is one real root (Fig. 3b), and if Δ = 0, then there are two of them (one of them is double), except for the case p = q = 0, when all three roots merge.

y Δ 0 y \u003d -px - q y \u003d x3

0 x 0 x y \u003d -px - q y \u003d x3 a) b)

Rice. 3 The cubic equation x3 + px + q = 0 can be represented as x3 = -px - q. This shows that the roots of the equation will correspond to the abscissas of the intersection points of the two graphs: y \u003d x3 and y \u003d -px - q. If Δ 0 is one.

1.4 Vieta's theorem

Vieta's theorem. If an integer rational equation degree n reduced to standard form, has n distinct real roots x1, x2,. xn, then they satisfy the equalities: x1 + x2 + + xn = - a1, a0 x1x2 + x1x3 + + xn-1xn = a2 a0 x1 x2 xn = (-1)nаn.

For the roots of the equation of the third degree a0x3 + a1x2 + a2x + a3 = 0, where a0 ≠ 0, the equalities x1 + x2 + x3 = - a1, a0 x1x2 + x1x3 + x2x3 = a2, a0 x1x2x3 = - a3 are valid.

1. 5 Bezout's theorem. Horner's scheme

The solution of equations is closely related to the factorization of polynomials. Therefore, when solving equations, everything that is connected with the selection in the polynomial is important linear factors, i.e., with the division of the polynomial A(x) by the binomial x - α. The basis of much knowledge about the division of the polynomial A(x) by the binomial x - α is a theorem belonging to French mathematician Etienne Bez (1730-1783) and bearing his name.

Bezout's theorem. The remainder of the division of the polynomial A (x) by the binomial x - α is equal to A (α) (i.e., the value of the polynomial A (x) at x = α).

Find the remainder after dividing the polynomial A(x) = x4 - 6x3 + 8 by x + 2.

Decision. According to the Bezout theorem, the remainder of the division by x + 2 is A (-2) \u003d (-2) 4 - 6 (-2) 3 + 8 \u003d 72.

A convenient way to find the values ​​of a polynomial when set value The variable x was introduced by the English mathematician Williams George Horner (1786-1837). This method was later called Horner's scheme. It consists in filling in some table of two lines. For example, to calculate A(-2) in the previous example, in the top line of the table we list the coefficients given polynomial, written in the standard form x4 - 6x3 + 8 = x4 + (-6)x3 + 0 x2 + 0 x + 8.

We duplicate the coefficient at the highest degree in the bottom line, and before it we write the value of the variable x = -2, at which the value of the polynomial is calculated. This results in the following table:

Empty cells of the table are filled according to the following rule: the rightmost number of the bottom row is multiplied by -2 and added to the number above the empty cell. According to this rule, the first empty cell contains the number (-2) 1 + (-6) = -8, the second cell contains the number (-2) (-8) + 0 = 16, the third cell contains the number (- 2) 16 + 0 = - 32, in last cage- number (-2) (-32) + 8 \u003d 72. The table completely filled out according to Horner's scheme looks like this:

2 1 -8 16 -32 72

The number in the last cell is the remainder of dividing the polynomial by x + 2, A(-2) = 72.

In fact, from the resulting table, filled in according to Horner's scheme, one can write down not only the remainder, but also the incomplete quotient

Q(x) \u003d x3 - 8x2 + 16x - 32, since the number on the second line (not counting from the last one) is the coefficients of the polynomial Q (x) - the incomplete quotient of division by x + 2.

Solve the equation x3 - 2x2 - 5x + 6 = 0

We write out all the divisors of the free term of the equation: ± 1, ± 2, ± 3, ± 6.

x=1, x=-2, x=3

Answer: x = 1, x = -2, x = 3

2. CONCLUSION

I will formulate the main conclusions about the work done.

In the process of work, I got acquainted with the history of the development of the problem of solving an equation of the third degree. The theoretical significance of the results obtained lies in the fact that it deliberately takes the place of the Cardano formula in solving some equations of the third degree. I made sure that the formula for solving the equation of the third degree exists, but because of its cumbersomeness it is not popular and not very reliable, since it does not always reach the final result.

In the future, we can consider such questions: how to find out in advance what roots an equation of the third degree has; can a cubic equation be solved graphically if possible, how; how to estimate approximately the roots of a cubic equation?

Lesson goals.

1. To deepen students' knowledge on the topic “Solving equations of higher degrees” and summarize the educational material.
2. To introduce students to the methods of solving equations of higher degrees.
3. To teach students to apply the theory of divisibility when solving equations of higher degrees.
4. To teach students how to divide a polynomial into a polynomial by the “corner”.
5. Develop skills and abilities to work with equations of higher degrees.

Developing:

1. Development of student attention.
2. Development of the ability to achieve results of work.
3. Development of interest in learning algebra and independent work skills.

Nurturing:

1. Raising a sense of collectivism.
2. Formation of a sense of responsibility for the result of work.
3. Formation in students adequate self-esteem when choosing a mark for work in the lesson.

Equipment: computer, projector.

During the classes

1 stage of work. Organizing time.

2 stage of work. Motivation and problem solving

Equation one of the most important concepts mathematics. The development of methods for solving equations, starting from the birth of mathematics as a science, long time was the main subject of study of algebra.

AT school course the study of mathematics a lot of attention is paid to solving various types of equations. Until the ninth grade, we could only solve linear and quadratic equations. Equations of the third, fourth, etc. degrees are called equations of higher degrees. In the ninth grade, we got acquainted with two basic methods for solving some equations of the third and fourth degrees: factoring a polynomial into factors and using a change of variable.

Is it possible to solve equations of higher degrees? We will try to find an answer to this question today.

3 stage of work. Review previously learned material. Introduce the concept of an equation of higher degrees.

1) Solution of a linear equation.

Linear is an equation of the form , where by definition. This equation has only one root.

2) Solution of a quadratic equation.

An equation of the form , where . The number of roots and the roots themselves are determined by the discriminant of the equation. For the equation has no roots, for has one root (two identical roots)

, for has two different roots .

From the considered linear and quadratic equations, we see that the number of roots of the equation is not more than its degree. In the course of higher algebra, it is proved that the equation of the -th degree has no more than n roots. As for the roots themselves, the situation is much more complicated. For equations of the third and fourth degrees, formulas are known for finding roots. However, these formulas are very complex and cumbersome and practical application Dont Have. For equations of the fifth and higher degrees general formulas does not exist and cannot exist (as was proved in the 19th century by N. Abel and E. Galois).

We will call the equations the third, fourth, etc. degrees by equations of higher degrees. Some Equations high degrees can be solved using two basic techniques: factoring a polynomial into factors or using a change of variable.

3) Solution of the cubic equation.

Let's solve the cubic equation

We group the terms of the polynomial on the left side of the equation and factor it. We get:

The product of factors is equal to zero if one of the factors is equal to zero. We get three linear equations:

So, this cubic equation has three roots: ; ;.

4) Solution of the biquadratic equation.

Biquadratic equations are very common, which have the form (i.e., equations that are quadratic with respect to ). To solve them, a new variable is introduced.

We will decide biquadratic equation.

Let's introduce a new variable and get a quadratic equation , whose roots are the numbers and 4.

Let's go back to the old variable and get two simple quadratic equations:

(roots and ) (roots and )

So, this biquadratic equation has four roots:

; ;.

Let's try to solve the equation using the above methods.

FAIL!!!

4 stage of work. Give some statements about the roots of a polynomial of the form , where polynomial nth degree

Here are some statements about the roots of a polynomial of the form :

1) A polynomial of the th degree has at most roots (taking into account their multiplicities). For example, a third degree polynomial cannot have four roots.

2) A polynomial of odd degree has at least one root. For example, polynomials of the first, third, fifth, etc. degrees have at least one root. Polynomials of even degree may or may not have roots.

3) If at the ends of the segment the values ​​of the polynomial have different signs (i.e., ), then the interval contains at least one root. This statement is widely used for the approximate calculation of the roots of a polynomial.

4) If the number is the root of a polynomial of the form , then this polynomial can be represented as a product , where the polynomial (-th degree. In other words, the polynomial of the form can be divided without a remainder by the binomial. This allows the equation of the th degree to be reduced to the equation (-th degree (reduce the degree of the equation).

5) If an equation with all integer coefficients (moreover, the free term) has an integer root, then this root is a divisor of the free term. Such a statement allows you to choose the whole root of the polynomial (if it exists).

5 stage of work. Show how divisibility theory is applied to solve equations of higher degrees. Consider examples of solving equations of higher degrees, in which the left-hand side is factorized using the method of dividing a polynomial by a polynomial by a “corner”.

Example 1. Solve the equation .

If this equation has an integer root, then it is a divisor of the free term (-1), i.e. equals one of the numbers: . The check shows that the root of the equation is the number -1. Hence, the polynomial can be represented as a product, i.e. a polynomial can be divided into a binomial without remainder. Let's perform the following division by "corner":

Thus, we have actually decomposed the left side of the equation into factors:

The product of factors is equal to zero if one of the factors is equal to zero. We get two equations.

Cubic equations have the form ax 3 + bx 2 + cx + d= 0). A method for solving such equations has been known for several centuries (it was discovered in the 16th century by Italian mathematicians). Solving some cubic equations is quite difficult, but with the right approach (and good level theoretical knowledge) you will be able to solve even the most complex cubic equations.

Steps

Solution using a formula for solving a quadratic equation

As noted above, the cubic equations have the form a x 3 + b x 2 + c x + d = 0 (\displaystyle ax^(3)+bx^(2)+cx+d=0), where the coefficients c (\displaystyle c) and d (\displaystyle d) may be equal 0 (\displaystyle 0), that is, a cubic equation can consist of only one term (with a variable in the third degree). First, check if the cubic equation given to you has an intercept, that is, d (\displaystyle d). If there is no free term, you can solve this cubic equation using the formula for solving a quadratic equation.

• If there is an intercept, use a different solution method (see the following sections).

1. Since in given equation there is no free term, then all terms of this equation contain a variable x (\displaystyle x), which can be bracketed: x (a x 2 + b x + c) (\displaystyle x(ax^(2)+bx+c)).

   • Example. 3 x 3 + − 2 x 2 + 14 x = 0 (\displaystyle 3x^(3)+-2x^(2)+14x=0). If you endure x (\displaystyle x) brackets, you get x (3 x 2 + − 2 x + 14) = 0 (\displaystyle x(3x^(2)+-2x+14)=0).

2. Note that the equation in brackets is a quadratic equation of the form ( a x 2 + b x + c (\displaystyle ax^(2)+bx+c)), which can be solved using the formula ((- b +/-√ (). Solve a quadratic equation and you will solve a cubic equation.

   • In our example, substitute the values ​​of the coefficients a (\displaystyle a), b (\displaystyle b), c (\displaystyle c) (3 (\displaystyle 3), − 2 (\displaystyle -2), 14 (\displaystyle 14)) into the formula: − b ± b 2 − 4 a c 2 a (\displaystyle (\frac (-b\pm (\sqrt (b^(2)-4ac)))(2a))) − (− 2) ± ((− 2) 2 − 4 (3) (14) 2 (3) (\displaystyle (\frac (-(-2)\pm (\sqrt (((-2)^(2 )-4(3)(14))))(2(3)))) 2 ± 4 − (12) (14) 6 (\displaystyle (\frac (2\pm (\sqrt (4-(12)(14))))(6))) 2 ± (4 − 168 6 (\displaystyle (\frac (2\pm (\sqrt ((4-168)))(6))) 2 ± − 164 6 (\displaystyle (\frac (2\pm (\sqrt (-164)))(6)))
   • Solution 1: 2 + − 164 6 (\displaystyle (\frac (2+(\sqrt (-164)))(6))) 2 + 12.8 i 6 (\displaystyle (\frac (2+12.8i)(6)))
   • Solution 2: 2 − 12.8 i 6 (\displaystyle (\frac (2-12.8i)(6)))

3. Remember that quadratic equations have two solutions, while cubic equations have three solutions. You have found two solutions to a quadratic, and therefore a cubic equation. In cases where you put "x" out of brackets, the third solution is always 0 (\displaystyle 0).

   • This is true because any number or expression multiplied by 0 (\displaystyle 0), equals 0 (\displaystyle 0). Since you endured x (\displaystyle x) out of brackets, then you have decomposed the cubic equation into two factors ( x (\displaystyle x) and a quadratic equation), one of which must be equal to 0 (\displaystyle 0) so that the whole equation is equal to 0 (\displaystyle 0).

   Finding entire solutions using factorization

   1. Check if the cubic equation given to you has an intercept. The method described in the previous section is not suitable for solving cubic equations in which there is a free term. In this case, you will have to use the method described in this or the next section.

      • Example. 2 x 3 + 9 x 2 + 13 x = − 6 (\displaystyle 2x^(3)+9x^(2)+13x=-6). Here, move a loose dick d = − 6 (\displaystyle d=-6) to the left side of the equation so that right side get 0 (\displaystyle 0): 2 x 3 + 9 x 2 + 13 x + 6 = 0 (\displaystyle 2x^(3)+9x^(2)+13x+6=0).

   2. Find coefficient multipliers a (\displaystyle a)(coefficient at x 3 (\displaystyle x^(3))) and free member d (\displaystyle d). Factors of a number are numbers that, when multiplied, give original number. For example, the factors of the number 6 (\displaystyle 6) are the numbers 1 (\displaystyle 1), 2 (\displaystyle 2), 3 (\displaystyle 3), 6 (\displaystyle 6) (6×1 (\displaystyle 6\times 1) and 2 × 3 (\displaystyle 2\times 3)).

      • In our example a = 2 (\displaystyle a=2) and d = 6 (\displaystyle d=6). Multipliers 2 (\displaystyle 2) are numbers 1 (\displaystyle 1) and 2 (\displaystyle 2). Multipliers 6 (\displaystyle 6) are numbers 1 (\displaystyle 1), 2 (\displaystyle 2), 3 (\displaystyle 3), and 6 (\displaystyle 6).

   3. Divide coefficient multipliers a (\displaystyle a) by factors of the free term d (\displaystyle d). You will get fractions and whole numbers. The integer solution of the cubic equation given to you will be either one of these integers, or the negative value of one of these integers.

      • In our example, divide the factors a (\displaystyle a) (1 (\displaystyle 1), 2 (\displaystyle 2)) by factors d (\displaystyle d) (1 (\displaystyle 1), 2 (\displaystyle 2), 3 (\displaystyle 3), 6 (\displaystyle 6)) and get: 1 (\displaystyle 1), , , , 2 (\displaystyle 2) and . Now add to this row of numbers their negative values: 1 (\displaystyle 1), − 1 (\displaystyle -1), 1 2 (\displaystyle (\frac (1)(2))), − 1 2 (\displaystyle -(\frac (1)(2))), 1 3 (\displaystyle (\frac (1)(3))), − 1 3 (\displaystyle -(\frac (1)(3))), 1 6 (\displaystyle (\frac (1)(6))), − 1 6 (\displaystyle -(\frac (1)(6))), 2 (\displaystyle 2), − 2 (\displaystyle -2), 2 3 (\displaystyle (\frac (2)(3))) and − 2 3 (\displaystyle -(\frac (2)(3))). The integer solutions of the cubic equation given to you are in this series of numbers.

   4. Now you can find integer solutions to your cubic equation by substituting integers from the found series of numbers into it. But if you do not want to waste time on this, use. This scheme involves the division of integers into values a (\displaystyle a), b (\displaystyle b), c (\displaystyle c), d (\displaystyle d) given cubic equation. If the remainder is 0 (\displaystyle 0), the integer is one of the solutions to the cubic equation.

      • Horner's division is not an easy topic; to receive additional information follow the link given above. Here's an example of how to find one of the solutions to a cubic equation given to you using Horner's division: -1 | 2 9 13 6 __| -2-7-6 __| 2 7 6 0 Since the remainder 0 (\displaystyle 0), then one of the solutions to the equation is an integer − 1 (\displaystyle -1).

   Using the discriminant

   1. In this method, you will work with coefficient values a (\displaystyle a), b (\displaystyle b), c (\displaystyle c), d (\displaystyle d). Therefore, it is better to write out the values ​​of these coefficients in advance.

      • Example. math>x^3-3x^2+3x-1. Here a = 1 (\displaystyle a=1), b = − 3 (\displaystyle b=-3), c = 3 (\displaystyle c=3), d = − 1 (\displaystyle d=-1). Don't forget that when x (\displaystyle x) there is no coefficient, this means that the coefficient is equal to 1 (\displaystyle 1).

   2. Calculate △ = b 2 − 3 a c (\displaystyle \triangle _(0)=b^(2)-3ac). This method will require some complex calculations, but if you understand it, you will be able to solve the most complex cubic equations. To start, calculate △ 0 (\displaystyle \triangle _(0)), one of several important quantities that we will need by substituting the appropriate values ​​into the formula.

      • In our example: b 2 − 3 a c (\displaystyle b^(2)-3ac) (− 3) 2 − 3 (1) (3) (\displaystyle (-3)^(2)-3(1)(3)) 9 − 3 (1) (3) (\displaystyle 9-3(1)(3)) 9 − 9 = 0 = △ 0 (\displaystyle 9-9=0=\triangle _(0)) 2 (− 27) − 9 (− 9) + 27 (− 1) (\displaystyle 2(-27)-9(-9)+27(-1)) − 54 + 81 − 27 (\displaystyle -54+81-27) 81 − 81 = 0 = △ 1 (\displaystyle 81-81=0=\triangle _(1))

   3. Calculate Δ = Δ1 2 - 4Δ0 3) ÷ -27 a 2 . Now calculate the discriminant of the equation using the found values ​​of Δ0 and Δ1. The discriminant is a number that gives you information about the roots of a polynomial (you may already know that the discriminant of a quadratic equation is b 2 - 4ac). In the case of a cubic equation, if the discriminant is positive, then the equation has three solutions; if the discriminant is zero, then the equation has one or two solutions; if the discriminant is negative, then the equation has only one solution. A cubic equation always has at least one solution because the graph of such an equation intersects the x-axis in at least one point.

      • If you substitute the appropriate values ​​​​of the quantities into this formula, you get possible solutions the cubic equation given to you. Substitute them into the original equation and if equality is met, then the solutions are correct. For example, if you plug the values ​​into the formula and get 1, plug the 1 into x 3 - 3x 2 + 3x- 1 and get 0. That is, equality is observed, and 1 is one of the solutions to the cubic equation given to you.