Equations with large powers. Equations of higher degrees Methods for solving equations n

Consider solving equations with one variable of degree higher than the second.

The degree of the equation P (x) \u003d 0 is the degree of the polynomial P (x), i.e. the largest of the degrees of its members with a coefficient not equal to zero.

So, for example, the equation (x 3 - 1) 2 + x 5 \u003d x 6 - 2 has a fifth power, because after the operations of opening the brackets and bringing down similar ones, we obtain the equivalent equation x 5 - 2x 3 + 3 \u003d 0 of the fifth degree.

Recall the rules that will be needed to solve equations of degree higher than the second.

Statements about the roots of the polynomial and its divisors:

1. A polynomial of degree n has a number of roots that does not exceed number n, and roots of multiplicity m occur exactly m times.

2. An odd degree polynomial has at least one real root.

3. If α is the root of P (x), then P n (x) \u003d (x - α) · Q n - 1 (x), where Q n - 1 (x) is a polynomial of degree (n - 1).

5. The reduced polynomial with integer coefficients cannot have fractional rational roots.

6. For a polynomial of the third degree

P 3 (x) \u003d ax 3 + bx 2 + cx + d one of the two is possible: either it decomposes into the product of three binomials

P 3 (x) \u003d a (x - α) (x - β) (x - γ), or decomposes into the product of the binomial and the square trinomial P 3 (x) \u003d a (x - α) (x 2 + βx + γ )

7. Any fourth-degree polynomial decomposes into a product of two square trinomials.

8. The polynomial f (x) is divisible by the polynomial g (x) without remainder if there exists a polynomial q (x) such that f (x) \u003d g (x) · q (x). For division of polynomials, the “division by corner” rule applies.

9. For the polynomial P (x) to be divisible by the binomial (x - c), it is necessary and sufficient that the number with be the root of P (x) (Corollary to Bezout's theorem).

10. Vieta's theorem: If x 1, x 2, ..., x n are the real roots of the polynomial

P (x) \u003d a 0 x n + a 1 x n - 1 + ... + a n, then the following equalities hold:

x 1 + x 2 + ... + x n \u003d -a 1 / a 0,

x 1 · x 2 + x 1 · x 3 + ... + x n - 1 · x n \u003d a 2 / a 0,

x 1 · x 2 · x 3 + ... + x n - 2 · x n - 1 · x n \u003d -a 3 / a 0,

x 1 · x 2 · x 3 · x n \u003d (-1) n a n / a 0.

Solution Examples

Example 1

Find the remainder of the division of P (x) \u003d x 3 + 2/3 x 2 - 1/9 by (x - 1/3).

Decision.

By the corollary of Bezout’s theorem: “The remainder of dividing a polynomial by a binomial (x - c) is equal to the value of the polynomial in c”. We find P (1/3) \u003d 0. Therefore, the remainder is 0 and the number 1/3 is the root of the polynomial.

Answer: R \u003d 0.

Example 2

Divide the "corner" 2x 3 + 3x 2 - 2x + 3 by (x + 2). Find the remainder and the partial quotient.

Decision:

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

2x 3 + 4 x 2 2x 2 - x

X 2 - 2 x

Answer: R \u003d 3; quotient: 2x 2 - x.

Basic methods for solving equations of higher degrees

1. Introduction of a new variable

The method of introducing a new variable is already familiar with the example of biquadratic equations. It consists in the fact that to solve the equation f (x) \u003d 0, a new variable (substitution) t \u003d x n or t \u003d g (x) is introduced and f (x) is expressed in terms of t, obtaining a new equation r (t). Solving then the equation r (t), find the roots:

(t 1, t 2, ..., t n). After that, a set of n equations q (x) \u003d t 1, q (x) \u003d t 2, ..., q (x) \u003d t n is obtained, from which the roots of the original equation are found.

Example 1

(x 2 + x + 1) 2 - 3x 2 - 3x - 1 \u003d 0.

Decision:

(x 2 + x + 1) 2 - 3 (x 2 + x) - 1 \u003d 0.

(x 2 + x + 1) 2 - 3 (x 2 + x + 1) + 3 - 1 \u003d 0.

Replacement (x 2 + x + 1) \u003d t.

t 2 - 3t + 2 \u003d 0.

t 1 \u003d 2, t 2 \u003d 1. Reverse replacement:

x 2 + x + 1 \u003d 2 or x 2 + x + 1 \u003d 1;

x 2 + x - 1 \u003d 0 or x 2 + x \u003d 0;

Answer: From the first equation: x 1, 2 \u003d (-1 ± √5) / 2, from the second: 0 and -1.

2. Factorization by grouping and shorthand multiplication formulas

The basis of this method is also not new and consists in grouping the terms in such a way that each group contains a common factor. To do this, sometimes you have to apply some artificial tricks.

Example 1

x 4 - 3x 2 + 4x - 3 \u003d 0.

Decision.

Imagine - 3x 2 \u003d -2x 2 - x 2 and group:

(x 4 - 2x 2) - (x 2 - 4x + 3) \u003d 0.

(x 4 - 2x 2 +1 - 1) - (x 2 - 4x + 3 + 1 - 1) \u003d 0.

(x 2 - 1) 2 - 1 - (x - 2) 2 + 1 \u003d 0.

(x 2 - 1) 2 - (x - 2) 2 \u003d 0.

(x 2 - 1 - x + 2) (x 2 - 1 + x - 2) \u003d 0.

(x 2 - x + 1) (x 2 + x - 3) \u003d 0.

x 2 - x + 1 \u003d 0 or x 2 + x - 3 \u003d 0.

Answer: There are no roots in the first equation, from the second: x 1, 2 \u003d (-1 ± √13) / 2.

3. Factorization by the method of indefinite coefficients

The essence of the method is that the original polynomial is factorized with unknown coefficients. Using the property that polynomials are equal if their coefficients are equal at the same degrees, unknown expansion coefficients are found.

Example 1

x 3 + 4x 2 + 5x + 2 \u003d 0.

Decision.

A polynomial of degree 3 can be decomposed into a product of linear and square factors.

x 3 + 4x 2 + 5x + 2 \u003d (x - a) (x 2 + bx + c),

x 3 + 4x 2 + 5x + 2 \u003d x 3 + bx 2 + cx - ax 2 - abx - ac,

x 3 + 4x 2 + 5x + 2 \u003d x 3 + (b - a) x 2 + (cx - ab) x - ac.

Having solved the system:

(b - a \u003d 4,
(c - ab \u003d 5,
(-ac \u003d 2,

(a \u003d -1,
(b \u003d 3,
(c \u003d 2, i.e.

x 3 + 4x 2 + 5x + 2 \u003d (x + 1) (x 2 + 3x + 2).

The roots of the equation (x + 1) (x 2 + 3x + 2) \u003d 0 are easily found.

Answer: -1; -2.

4. Method of root selection by senior and free coefficient

The method relies on the application of theorems:

1) Every integer root of a polynomial with integer coefficients is a divisor of the free term.

2) In order for the irreducible fraction p / q (p is an integer, q is a positive integer) to be the root of the equation with integer coefficients, it is necessary that the number p be the integer divisor of the free term and 0, and q is the natural divisor of the highest coefficient.

Example 1

6x 3 + 7x 2 - 9x + 2 \u003d 0.

Decision:

6: q \u003d 1, 2, 3, 6.

Therefore, p / q \u003d ± 1, ± 2, ± 1/2, ± 1/3, ± 2/3, ± 1/6.

Having found one root, for example - 2, we will find other roots using angle division, the method of indefinite coefficients or Horner's scheme.

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

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In the general case, an equation having a degree above 4 cannot be solved in radicals. But sometimes we can still find the roots of the polynomial on the left in the equation of the highest degree, if we represent it as a product of polynomials in the degree of no more than 4. The solution of such equations is based on factorization of the polynomial, so we recommend that you repeat this topic before studying this article.

Most often you have to deal with equations of higher degrees with integer coefficients. In these cases, we can try to find rational roots, and then factor the polynomial, then convert it to an equation of a lower degree, which will be easy to solve. In the framework of this material, we will consider just such examples.

Yandex.RTB R-A-339285-1

Higher equations with integer coefficients

All equations of the form a n x n + a n - 1 x n - 1 +. . . + a 1 x + a 0 \u003d 0, we can lead to an equation of the same degree by multiplying both parts by a n n - 1 and replacing a variable of the form y \u003d a n x:

a n x n + a n - 1 x n - 1 +. . . + a 1 x + a 0 \u003d 0 ann · xn + an - 1 · ann - 1 · xn - 1 + ... + a 1 · (an) n - 1 · x + a 0 · (an) n - 1 \u003d 0 y \u003d anx ⇒ yn + bn - 1 yn - 1 + ... + b 1 y + b 0 \u003d 0

Those coefficients that turned out in the end will also be integer. Thus, we will need to solve the reduced equation of degree n with integer coefficients, having the form x n + a n x n - 1 + ... + a 1 x + a 0 \u003d 0.

We calculate the whole roots of the equation. If the equation has integer roots, you need to look for them among the divisors of the free term a 0. We write them out and substitute them into the original equality in turn, checking the result. As soon as we get the identity and find one of the roots of the equation, we can write it in the form x - x 1 · P n - 1 (x) \u003d 0. Here x 1 is the root of the equation, and P n - 1 (x) is the quotient of dividing x n + a n x n - 1 + ... + a 1 x + a 0 by x - x 1.

We substitute the remaining written divisors into P n - 1 (x) \u003d 0, starting with x 1, since the roots can repeat. After obtaining the identity, the root x 2 is considered to be found, and the equation can be written in the form (x - x 1) (x - x 2) · P n - 2 (x) \u003d 0. Here P n - 2 (x) will be the quotient of dividing P n - 1 (x) by x - x 2.

We continue to sort through the dividers. Find all integer roots and denote their number as m. After that, the initial equation can be represented as x - x 1 x - x 2 · ... · x - x m · P n - m (x) \u003d 0. Here P n - m (x) is a polynomial of the n - mth degree. For calculation it is convenient to use the Horner scheme.

If our original equation has integer coefficients, we cannot end up with fractional roots.

We ended up with the equation P n - m (x) \u003d 0, the roots of which can be found in any convenient way. They can be irrational or complex.

Let us show on a concrete example how this solution scheme is applied.

Example 1

Condition: find the solution to the equation x 4 + x 3 + 2 x 2 - x - 3 \u003d 0.

Decision

Let's start with finding the whole roots.

We have a free term equal to minus three. It has divisors equal to 1, - 1, 3 and - 3. We substitute them into the original equation and see which of them will give the identity.

For x equal to one, we get 1 4 + 1 3 + 2 · 1 2 - 1 - 3 \u003d 0, which means that the unit will be the root of this equation.

Now let us divide the polynomial x 4 + x 3 + 2 x 2 - x - 3 by (x - 1) in the column:

Therefore, x 4 + x 3 + 2 x 2 - x - 3 \u003d x - 1 x 3 + 2 x 2 + 4 x + 3.

1 3 + 2 · 1 2 + 4 · 1 + 3 \u003d 10 ≠ 0 (- 1) 3 + 2 · (- 1) 2 + 4 · - 1 + 3 \u003d 0

We got the identity, which means we found another root of the equation, equal to - 1.

Divide the polynomial x 3 + 2 x 2 + 4 x + 3 by (x + 1) in the column:

We get that

x 4 + x 3 + 2 x 2 - x - 3 \u003d (x - 1) (x 3 + 2 x 2 + 4 x + 3) \u003d (x - 1) (x + 1) (x 2 + x + 3)

Substitute the next divisor into the equality x 2 + x + 3 \u003d 0, starting with - 1:

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

The equalities obtained as a result will be incorrect, which means that the equation no longer has whole roots.

The remaining roots will be the roots of the expression x 2 + x + 3.

D \u003d 1 2 - 4 · 1 · 3 \u003d - 11< 0

It follows that this square trinomial has no real roots, but there are complex conjugates: x \u003d - 1 2 ± i 11 2.

We clarify that instead of dividing in a column, you can apply the Horner scheme. This is done as follows: after we have determined the first root of the equation, fill out the table.

In the coefficient table, we can immediately see the coefficients of the quotient of the division of polynomials, which means that x 4 + x 3 + 2 x 2 - x - 3 \u003d x - 1 x 3 + 2 x 2 + 4 x + 3.

After finding the next root equal to - 1, we get the following:

Answer: x \u003d - 1, x \u003d 1, x \u003d - 1 2 ± i 11 2.

Example 2

Condition: solve the equation x 4 - x 3 - 5 x 2 + 12 \u003d 0.

Decision

A free member has divisors 1, - 1, 2, - 2, 3, - 3, 4, - 4, 6, - 6, 12, - 12.

Check them in order:

1 4 - 1 3 - 5 · 1 2 + 12 \u003d 7 ≠ 0 (- 1) 4 - (- 1) 3 - 5 · (- 1) 2 + 12 \u003d 9 ≠ 0 2 4 · 2 3 - 5 · 2 2 + 12 \u003d 0

Therefore, x \u003d 2 will be the root of the equation. Divide x 4 - x 3 - 5 x 2 + 12 by x - 2 using the Horner scheme:

As a result, we get x - 2 (x 3 + x 2 - 3 x - 6) \u003d 0.

2 3 + 2 2 - 3 · 2 - 6 \u003d 0

So 2 again will be the root. Divide x 3 + x 2 - 3 x - 6 \u003d 0 by x - 2:

As a result, we obtain (x - 2) 2 · (x 2 + 3 x + 3) \u003d 0.

Checking the remaining divisors does not make sense, since the equality x 2 + 3 x + 3 \u003d 0 is faster and more convenient to solve using the discriminant.

Solve the quadratic equation:

x 2 + 3 x + 3 \u003d 0 D \u003d 3 2 - 4 · 1 · 3 \u003d - 3< 0

We obtain a complex conjugate pair of roots: x \u003d - 3 2 ± i 3 2.

Answer: x \u003d - 3 2 ± i 3 2.

Example 3

Condition: find for the equation x 4 + 1 2 x 3 - 5 2 x - 3 \u003d 0 the real roots.

Decision

x 4 + 1 2 x 3 - 5 2 x - 3 \u003d 0 2 x 4 + x 3 - 5 x - 6 \u003d 0

We perform a 2 3 multiplication of both parts of the equation:

2 x 4 + x 3 - 5 x - 6 \u003d 0 2 4 · x 4 + 2 3 x 3 - 20 · 2 · x - 48 \u003d 0

Replace the variables y \u003d 2 x:

2 4 · x 4 + 2 3 x 3 - 20 · 2 · x - 48 \u003d 0 y 4 + y 3 - 20 y - 48 \u003d 0

As a result, we got a standard equation of the 4th degree, which can be solved according to the standard scheme. We check the divisors, divide, and get in the end that it has 2 real roots y \u003d - 2, y \u003d 3 and two complex ones. The entire solution will not be given here. By virtue of replacing the real roots of this equation are x \u003d y 2 \u003d - 2 2 \u003d - 1 and x \u003d y 2 \u003d 3 2.

Answer: x 1 \u003d - 1, x 2 \u003d 3 2

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Consider solving equations with one variable of degree higher than the second.

The degree of the equation P (x) \u003d 0 is the degree of the polynomial P (x), i.e. the largest of the degrees of its members with a coefficient not equal to zero.

Recall the rules that will be needed to solve equations of degree higher than the second.

Statements about the roots of the polynomial and its divisors:

1. A polynomial of degree n has a number of roots that does not exceed number n, and roots of multiplicity m occur exactly m times.

2. An odd degree polynomial has at least one real root.

3. If α is the root of P (x), then P n (x) \u003d (x - α) · Q n - 1 (x), where Q n - 1 (x) is a polynomial of degree (n - 1).

5. The reduced polynomial with integer coefficients cannot have fractional rational roots.

6. For a polynomial of the third degree

P 3 (x) \u003d ax 3 + bx 2 + cx + d one of the two is possible: either it decomposes into the product of three binomials

P 3 (x) \u003d a (x - α) (x - β) (x - γ), or decomposes into the product of the binomial and the square trinomial P 3 (x) \u003d a (x - α) (x 2 + βx + γ )

7. Any fourth-degree polynomial decomposes into a product of two square trinomials.

9. For the polynomial P (x) to be divisible by the binomial (x - c), it is necessary and sufficient that the number with be the root of P (x) (Corollary to Bezout's theorem).

10. Vieta's theorem: If x 1, x 2, ..., x n are the real roots of the polynomial

P (x) \u003d a 0 x n + a 1 x n - 1 + ... + a n, then the following equalities hold:

x 1 + x 2 + ... + x n \u003d -a 1 / a 0,

x 1 · x 2 + x 1 · x 3 + ... + x n - 1 · x n \u003d a 2 / a 0,

x 1 · x 2 · x 3 + ... + x n - 2 · x n - 1 · x n \u003d -a 3 / a 0,

x 1 · x 2 · x 3 · x n \u003d (-1) n a n / a 0.

Solution Examples

Example 1

Find the remainder of the division of P (x) \u003d x 3 + 2/3 x 2 - 1/9 by (x - 1/3).

Decision.

Answer: R \u003d 0.

Example 2

Divide the "corner" 2x 3 + 3x 2 - 2x + 3 by (x + 2). Find the remainder and the partial quotient.

Decision:

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

2x 3 + 4 x 2 2x 2 - x

X 2 - 2 x

Answer: R \u003d 3; quotient: 2x 2 - x.

Basic methods for solving equations of higher degrees

1. Introduction of a new variable

(t 1, t 2, ..., t n). After that, a set of n equations q (x) \u003d t 1, q (x) \u003d t 2, ..., q (x) \u003d t n is obtained, from which the roots of the original equation are found.

Example 1

(x 2 + x + 1) 2 - 3x 2 - 3x - 1 \u003d 0.

Decision:

(x 2 + x + 1) 2 - 3 (x 2 + x) - 1 \u003d 0.

(x 2 + x + 1) 2 - 3 (x 2 + x + 1) + 3 - 1 \u003d 0.

Replacement (x 2 + x + 1) \u003d t.

t 2 - 3t + 2 \u003d 0.

t 1 \u003d 2, t 2 \u003d 1. Reverse replacement:

x 2 + x + 1 \u003d 2 or x 2 + x + 1 \u003d 1;

x 2 + x - 1 \u003d 0 or x 2 + x \u003d 0;

Answer: From the first equation: x 1, 2 \u003d (-1 ± √5) / 2, from the second: 0 and -1.

2. Factorization by grouping and shorthand multiplication formulas

The basis of this method is also not new and consists in grouping the terms in such a way that each group contains a common factor. To do this, sometimes you have to apply some artificial tricks.

Example 1

x 4 - 3x 2 + 4x - 3 \u003d 0.

Decision.

Imagine - 3x 2 \u003d -2x 2 - x 2 and group:

(x 4 - 2x 2) - (x 2 - 4x + 3) \u003d 0.

(x 4 - 2x 2 +1 - 1) - (x 2 - 4x + 3 + 1 - 1) \u003d 0.

(x 2 - 1) 2 - 1 - (x - 2) 2 + 1 \u003d 0.

(x 2 - 1) 2 - (x - 2) 2 \u003d 0.

(x 2 - 1 - x + 2) (x 2 - 1 + x - 2) \u003d 0.

(x 2 - x + 1) (x 2 + x - 3) \u003d 0.

x 2 - x + 1 \u003d 0 or x 2 + x - 3 \u003d 0.

Answer: There are no roots in the first equation, from the second: x 1, 2 \u003d (-1 ± √13) / 2.

3. Factorization by the method of indefinite coefficients

Example 1

x 3 + 4x 2 + 5x + 2 \u003d 0.

Decision.

A polynomial of degree 3 can be decomposed into a product of linear and square factors.

x 3 + 4x 2 + 5x + 2 \u003d (x - a) (x 2 + bx + c),

x 3 + 4x 2 + 5x + 2 \u003d x 3 + bx 2 + cx - ax 2 - abx - ac,

x 3 + 4x 2 + 5x + 2 \u003d x 3 + (b - a) x 2 + (cx - ab) x - ac.

Having solved the system:

(b - a \u003d 4,
(c - ab \u003d 5,
(-ac \u003d 2,

(a \u003d -1,
(b \u003d 3,
(c \u003d 2, i.e.

x 3 + 4x 2 + 5x + 2 \u003d (x + 1) (x 2 + 3x + 2).

The roots of the equation (x + 1) (x 2 + 3x + 2) \u003d 0 are easily found.

Answer: -1; -2.

4. Method of root selection by senior and free coefficient

The method relies on the application of theorems:

1) Every integer root of a polynomial with integer coefficients is a divisor of the free term.

Example 1

6x 3 + 7x 2 - 9x + 2 \u003d 0.

Decision:

6: q \u003d 1, 2, 3, 6.

Therefore, p / q \u003d ± 1, ± 2, ± 1/2, ± 1/3, ± 2/3, ± 1/6.

Having found one root, for example - 2, we will find other roots using angle division, the method of indefinite coefficients or Horner's scheme.

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

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The application of equations is widespread in our lives. They are used in many calculations, construction of structures and even sports. Man used the equations in ancient times and since then their application has only increased. In mathematics, equations of higher degrees with integer coefficients are quite often found. To solve this kind of equation it is necessary:

Determine the rational roots of the equation;

Factor the polynomial that is on the left side of the equation;

Find the roots of the equation.

Suppose we are given an equation of the following form:

Find all its real roots. Multiply the left and right sides of the equation by \\

Let's change the variables \\

Thus, we have obtained the fourth-degree reduced equation, which is solved by the standard algorithm: we check the divisors, divide, and as a result, we find out that the equation has two real roots \\ and two complex roots. We get the following answer to our fourth degree equation:

Where can I solve the equation of higher degrees online solver?

You can solve the equation on our website https: // site. A free online solver allows you to solve an equation online of any complexity in seconds. All you have to do is just enter your data in the solver. You can also watch a video instruction and learn how to solve the equation on our website. And if you still have questions, then you can ask them in our Vkontakte group http://vk.com/pocketteacher. Join our group, we are always happy to help you.

When solving algebraic equations, one often has to factor the polynomial. Factorizing a polynomial means representing it as a product of two or more polynomials. We use some methods of decomposing polynomials quite often: taking out a common factor, applying reduced multiplication formulas, extracting a full square, grouping. Let's consider some more methods.

Sometimes, when factoring a polynomial, the following statements are useful:

1) if a polynomial with integer coefficients has a rational root (where is an irreducible fraction, then is the divisor of the free term and the divisor of the highest coefficient:

2) If in some way we choose the root of the polynomial of degree, then the polynomial can be represented in the form where the polynomial of degree

The polynomial can be found either by dividing the polynomial into a binomial “column”, or by correspondingly grouping the terms of the polynomial and extracting a factor from them, or by the method of indefinite coefficients.

Example. Factor polynomial

Decision. Since the coefficient at x4 is 1, the rational roots of this polynomial exist, are divisors of 6, that is, they can be integers ± 1, ± 2, ± 3, ± 6. Denote this polynomial by P4 (x). Since P P4 (1) \u003d 4 and P4 (-4) \u003d 23, the numbers 1 and -1 are not the roots of the polynomial PA (x). Since P4 (2) \u003d 0, then x \u003d 2 is the root of the polynomial P4 (x), and, therefore, this polynomial is divided by the binomial x - 2. Therefore, x4 -5x3 + 7x2 -5x +6 x-2 x4 -2x3 x3 -3x2 + x-3

3x3 + 7x2 -5x +6

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

Consequently, P4 (x) \u003d (x - 2) (x3 - 3x2 + x - 3). Since xz - 3x2 + x - 3 \u003d x2 (x - 3) + (x - 3) \u003d (x - 3) (x2 + 1), then x4 - 5x3 + 7x2 - 5x + 6 \u003d (x - 2) (x - 3) (x2 + 1).

Parameter Input Method

Sometimes, when factoring a polynomial, the method of introducing a parameter helps. We explain the essence of this method in the following example.

Example. x3 - (√3 + 1) x2 + 3.

Decision. Consider a polynomial with the parameter a: x3 - (a + 1) x2 + a2, which for a \u003d √3 turns into a given polynomial. We write this polynomial as a quadratic trinomial with respect to a: ar - ax2 + (x3 - x2).

Since the roots of this quadratic trinomial with respect to a are a1 \u003d x and a2 \u003d x2 - x, then the equality a2 - ax2 + (xs - x2) \u003d (a - x) (a - x2 + x) is valid. Therefore, the polynomial x3 - (√3 + 1) x2 + 3 is factorized as √3 - x and √3 - x2 + x, i.e.

x3 - (√3 + 1) x2 + 3 \u003d (x-√3) (x2-x-√3).

Method for introducing a new unknown

In some cases, by replacing the expression f (x) in the polynomial Pn (x), we can obtain a polynomial with respect to y through y, which can be easily factorized. Then, after changing y to f (x), we obtain the factorization of the polynomial Pn (x).

Example. Factor the polynomial x (x + 1) (x + 2) (x + 3) -15.

Decision. We transform this polynomial as follows: x (x + 1) (x + 2) (x + 3) -15 \u003d [x (x + 3)] [(x + 1) (x + 2)] - 15 \u003d (x2 + 3x) (x2 + 3x + 2) - 15.

Denote x2 + 3x by y. Then we have y (y + 2) - 15 \u003d y2 + 2y - 15 \u003d y2 + 2y + 1 - 16 \u003d (y + 1) 2 - 16 \u003d (y + 1 + 4) (y + 1 - 4) \u003d ( y + 5) (y - 3).

Therefore, x (x + 1) (x + 2) (x + 3) - 15 \u003d (x2 + 3x + 5) (x2 + 3x - 3).

Example. Factor polynomial (x-4) 4+ (x + 2) 4

Decision. Denote x - 4 + x + 2 \u003d x - 1 by y.

(x - 4) 4 + (x + 2) 2 \u003d (y - 3) 4 + (y + 3) 4 \u003d y4 - 12y3 + 54y3 - 108y + 81 + y4 + 12y3 + 54y2 + 108y + 81 \u003d

2y4 + 108y2 + 162 \u003d 2 (y4 + 54y2 + 81) \u003d 2 [(yy + 27) 2 - 648] \u003d 2 (y2 + 27 - √b48) (y2 + 27 + √b48) \u003d

2 ((x-1) 2 + 27-√b48) ((x-1) 2 + 27 + √b48) \u003d 2 (x2-2x + 28--18√2) (x2-2x + 28 + 18√ 2 )

Combination of different methods

Often, when factoring a polynomial, several of the methods discussed above must be applied sequentially.

Example. Factor the polynomial x4 - 3x2 + 4x-3.

Decision. Using the grouping, we rewrite the polynomial in the form x4 - 3x2 + 4x - 3 \u003d (x4 - 2x2) - (x2 -4x + 3).

Applying the full square selection method to the first bracket, we have x4 - 3x3 + 4x - 3 \u003d (x4 - 2 · 1 · x2 + 12) - (x2 -4x + 4).

Using the full square formula, we can now write that x4 - 3x2 + 4x - 3 \u003d (x2 -1) 2 - (x - 2) 2.

Finally, applying the formula of the difference of squares, we get that x4 - 3x2 + 4x - 3 \u003d (x2 - 1 + x - 2) (x2 - 1 - x + 2) \u003d (x2 + x-3) (x2-x + 1 )

§ 2. Symmetric equations

1. Symmetric equations of the third degree

Equations of the form ax3 + bx2 + bx + a \u003d 0, and ≠ 0 (1) are called symmetric equations of the third degree. Since ax3 + bx2 + bx + a \u003d a (x3 + 1) + bx (x + 1) \u003d (x + 1) (ax2 + (b-a) x + a), then equation (1) is equivalent to the set of equations x + 1 \u003d 0 and ax2 + (b-a) x + a \u003d 0, which is not difficult to solve.

Example 1. Solve the equation

3x3 + 4x2 + 4x + 3 \u003d 0. (2)

Decision. Equation (2) is a symmetric equation of the third degree.

Since 3x3 + 4xg + 4x + 3 \u003d 3 (x3 + 1) + 4x (x + 1) \u003d (x + 1) (3x2 - 3x + 3 + 4x) \u003d (x + 1) (3x2 + x + 3) , then equation (2) is equivalent to the set of equations x + 1 \u003d 0 and 3x3 + x + 3 \u003d 0.

The solution to the first of these equations is x \u003d -1, the second equation has no solutions.

Answer: x \u003d -1.

2. Symmetric equations of the fourth degree

Equation of the form

(3) is called a symmetric equation of the fourth degree.

Since x \u003d 0 is not the root of equation (3), then, dividing both sides of equation (3) by x2, we obtain an equation equivalent to the original (3):

We rewrite equation (4) in the form:

In this equation we make a replacement, then we obtain the quadratic equation

If equation (5) has 2 roots y1 and y2, then the original equation is equivalent to the set of equations

If equation (5) has one root y0, then the original equation is equivalent to the equation

Finally, if equation (5) has no roots, then the original equation also has no roots.

Example 2. Solve the equation

Decision. This equation is a fourth-degree symmetric equation. Since x \u003d 0 is not its root, then, dividing equation (6) by x2, we obtain the equation equivalent to it:

Having grouped the terms, we rewrite equation (7) in the form or in the form

Putting, we obtain an equation having two roots y1 \u003d 2 and y2 \u003d 3. Therefore, the original equation is equivalent to a set of equations

The solution to the first equation of this set is x1 \u003d 1, and the solution to the second is and.

Therefore, the original equation has three roots: x1, x2 and x3.

Answer: x1 \u003d 1 ,.

§3. Algebraic equations

1. The decrease in the degree of equation

Some algebraic equations by replacing a certain polynomial in them with one letter can be reduced to algebraic equations whose degree is less than the degree of the original equation and the solution of which is simpler.

Example 1. Solve the equation

Decision. We denote by, then equation (1) can be rewritten in the form The last equation has roots and Therefore, equation (1) is equivalent to the set of equations and. There is a solution to the first equation of this set, and there are solutions to the second equation

The solutions of equation (1) are

Example 2. Solve the equation

Decision. Multiplying both sides of the equation by 12 and denoting by,

We get the equation Rewrite this equation in the form

(3) and denoting by rewriting equation (3) in the form The last equation has roots, and therefore we find that equation (3) is equivalent to the combination of two equations and there are solutions to this set of equations, i.e., equation (2) is equivalent to the combination of equations and ( four)

The solutions of (4) are and, they are the solutions of equation (2).

2. Equations of the form

The equation

(5) where are given numbers, can be reduced to a biquadratic equation by replacing the unknown, i.e., replacing

Example 3. Solve the equation

Decision. Denote by, t. e. make a change of variables or Then equation (6) can be rewritten in the form or, using the formula, in the form

Since the roots of the quadratic equation are, and then the solutions of equation (7) are the solutions of the set of equations and. This set of equations has two solutions and, therefore, the solutions of equation (6) are and

3. Equations of the form

The equation

(8) where the numbers α, β, γ, δ, and Α are such that α

Example 4. Solve the equation

Decision. We replace the unknowns, i.e., y \u003d x + 3 or x \u003d y - 3. Then equation (9) can be rewritten in the form

(y-2) (y-1) (y + 1) (y + 2) \u003d 10, i.e., in the form

(y2-4) (y2-1) \u003d 10 (10)

Biquadratic equation (10) has two roots. Therefore, equation (9) also has two roots:

4. Equations of the form

Equation, (11)

Where, has no root x \u003d 0, therefore, dividing equation (11) by x2, we obtain the equation equivalent to it

Which, after replacing the unknown, will be rewritten in the form of a quadratic equation, the solution of which is not difficult.

Example 5. Solve the equation

Decision. Since h \u003d 0 is not the root of equation (12), dividing it by x2, we obtain the equation equivalent to it

Making the substitution unknown, we obtain the equation (y + 1) (y + 2) \u003d 2, which has two roots: y1 \u003d 0 and y1 \u003d -3. Therefore, the original equation (12) is equivalent to the set of equations

This set has two roots: x1 \u003d -1 and x2 \u003d -2.

Answer: x1 \u003d -1, x2 \u003d -2.

Comment. Equation of the form

In which, one can always bring to the form (11) and, moreover, considering α\u003e 0 and λ\u003e 0 to the form.

5. Equations of the form

The equation

, (13) where the numbers, α, β, γ, δ, and Α are such that αβ \u003d γδ ≠ 0, we can rewrite it by multiplying the first bracket with the second and the third with the fourth, in the form, i.e., equation (13) now written in the form (11), and its solution can be carried out in the same way as the solution of equation (11).

Example 6. Solve the equation

Decision. Equation (14) has the form (13), so we rewrite it in the form

Since x \u003d 0 is not a solution to this equation, dividing both its parts by x2, we obtain an equivalent initial equation. Making a change of variables, we obtain a quadratic equation, the solution of which is and. Therefore, the original equation (14) is equivalent to the set of equations and.

The solution to the first equation of this population is

The second equation of this set of solutions has no. So, the original equation has roots x1 and x2.

6. Equations of the form

The equation

(15) where the numbers a, b, c, q, A are such that, does not have a root x \u003d 0, therefore, dividing equation (15) by x2. we obtain an equation equivalent to it, which, after replacing the unknown, will be rewritten in the form of a quadratic equation, the solution of which is not difficult.

Example 7. The solution of the equation

Decision. Since x \u003d 0 is not the root of equation (16), then, dividing both its parts by x2, we obtain the equation

, (17) equivalent to equation (16). Having made the replacement unknown, we rewrite equation (17) in the form

Quadratic equation (18) has 2 roots: y1 \u003d 1 and y2 \u003d -1. Therefore, equation (17) is equivalent to a combination of equations and (19)

The set of equations (19) has 4 roots:,.

They will be the roots of equation (16).

§four. Rational equations

Equations of the form \u003d 0, where H (x) and Q (x) are polynomials, are called rational.

Having found the roots of the equation H (x) \u003d 0, then we need to check which of them are not the roots of the equation Q (x) \u003d 0. These roots and only they will be solutions to the equation.

Consider some methods for solving equations of the form \u003d 0.

1. Equations of the form

The equation

(1) under certain conditions on numbers, it can be solved as follows. Grouping the terms of equation (1) in two and summing each pair, you need to get polynomials of the first or zero degree in the numerator, differing only in numerical factors, and in the denominators - trinomials with the same two terms containing x, then after changing the variables, the equation will either have also the form (1), but with a smaller number of terms, either will be equivalent to a combination of two equations, one of which will be the first degree, and the second will be an equation of the form (1), but with a smaller number of terms.

Example. Solve the equation

Decision. Grouping on the left side of equation (2) the first term with the last, and the second with the penultimate, we rewrite equation (2) in the form

Summing up the terms in each bracket, we rewrite equation (3) in the form

Since there is no solution to equation (4), dividing this equation by, we obtain the equation

, (5) equivalent to equation (4). We make a replacement for the unknown, then equation (5) can be rewritten in the form

Thus, the solution of equation (2) with five terms on the left side is reduced to the solution of equation (6) of the same form, but with three terms on the left side. Summing up all the terms on the left side of equation (6), we rewrite it in the form

The solutions to the equation are and. None of these numbers vanish the denominator of the rational function on the left side of equation (7). Therefore, equation (7) has these two roots, and therefore, the original equation (2) is equivalent to the set of equations

The solutions to the first equation of this set are

The solutions of the second equation from this set are

Therefore, the original equation has roots

2. Equations of the form

The equation

(8) under certain conditions on numbers, one can solve this way: it is necessary to select the integer part in each of the fractions of the equation, i.e., replace equation (8) with equation

Reduce it to the form (1) and then solve it by the method described in the previous paragraph.

Example. Solve the equation

Decision. We write equation (9) in the form or in the form

Summing up the terms in brackets, we rewrite equation (10) in the form

Making a replacement for the unknown, we rewrite equation (11) in the form

Summing up the terms on the left side of equation (12), we rewrite it in the form

It is easy to see that equation (13) has two roots: and. Therefore, the original equation (9) has four roots:

3) Equations of the form.

An equation of the form (14) under certain conditions on numbers can be solved as follows: by expanding (if it is, of course, possible) each of the fractions on the left side of equation (14) into a sum of simple fractions

Reduce equation (14) to form (1), then, after conveniently rearranging the terms of the resulting equation, solve it by the method described in paragraph 1).

Example. Solve the equation

Decision. Since and, then, multiplying the numerator of each fraction in equation (15) by 2 and noting that equation (15) can be written as

Equation (16) has the form (7). Regrouping the terms in this equation, we rewrite it in the form or in the form

Equation (17) is equivalent to a combination of equations and

To solve the second equation of the set (18), we make the replacement of the unknown Then it will be rewritten in the form or in the form

Summing up all the terms on the left side of equation (19), rewrite it as

Since the equation has no roots, then equation (20) also does not have them.

The first equation of the population (18) has a single root Since this root is included in the ODZ of the second equation of the population (18), it is the only root of the population (18), and hence the original equation.

4. Equations of the form

The equation

(21) under certain conditions on the numbers and A, after presenting each term on the left side in the form, it can be reduced to the form (1).

Example. Solve the equation

Decision. We rewrite equation (22) in the form or in the form

Thus, equation (23) is reduced to the form (1). Now, grouping the first term with the last, and the second with the third, we rewrite equation (23) in the form

This equation is equivalent to the set of equations and. (24)

The last equation of the set (24) can be rewritten in the form

There are solutions to this equation, and since it is included in the ODZ of the second equation of the set (30), then the set (24) has three roots :. All of them are solutions to the original equation.

5. Equations of the form.

Equation of the form (25)

Under certain conditions, by replacing the unknown, the numbers can be reduced to an equation of the form

Example. Solve the equation

Decision. Since it is not a solution to equation (26), then dividing the numerator and denominator of each fraction on the left side by, we rewrite it in the form

After changing the variables, we rewrite equation (27) in the form

Solving equation (28) is and. Therefore, equation (27) is equivalent to the set of equations and. (29)