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) = 0 is the degree of the polynomial P(x), i.e. the largest of the powers of its terms with a non-zero coefficient.

So, for example, the equation (x 3 - 1) 2 + x 5 \u003d x 6 - 2 has a fifth degree, because after the operations of opening brackets and bringing similar ones, we obtain an 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 a polynomial and its divisors:

1. Polynomial nth degree has a number of roots not exceeding the number n, and the roots of multiplicity m occur exactly m times.

2. A polynomial of odd degree has at least one real root.

3. If α is the root of Р(х), then Р n (х) = (х – α) · Q n – 1 (x), where Q n – 1 (x) is a polynomial of degree (n – 1).

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

6. For a third degree polynomial

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

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

7. Any polynomial of the fourth degree expands into the product of two square trinomials.

8. A polynomial f(x) is divisible by a polynomial g(x) without remainder if there exists a polynomial q(x) such that f(x) = g(x) q(x). To divide polynomials, the rule of "division by a corner" is applied.

9. For the polynomial P(x) to be divisible by the binomial (x – c), it is necessary and sufficient that the number c 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) = 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 of examples

Example 1

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

Solution.

According to 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." Let's find P(1/3) = 0. Therefore, the remainder is 0 and the number 1/3 is the root of the polynomial.

Answer: R = 0.

Example 2

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

Solution:

2x 3 + 3x 2 – 2x + 3| x + 2

2x 3 + 4x 2 2x 2 – x

X 2 – 2 x

Answer: R = 3; quotient: 2x 2 - x.

Basic methods for solving equations higher degrees

1. Introduction of a new variable

The method of introducing a new variable is already familiar from 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 through t, obtaining a new equation r (t). Then solving the equation r(t), find the roots:

(t 1 , t 2 , …, t n). After that, a set of n equations q(x) = t 1 , q(x) = t 2 , ... , q(x) = 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 = 0.

Solution:

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

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

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

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

2. Factorization by the method of grouping and abbreviated multiplication formulas

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

Example 1

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

Solution.

Imagine - 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 \u003d 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 \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 decomposed into factors with unknown coefficients. Using the property that polynomials are equal if their coefficients are equal at the same powers, the unknown expansion coefficients are found.

Example 1

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

Solution.

A polynomial of the 3rd degree 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 = 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.

Solving the system:

(b – a = 4,
(c – ab = 5,
(-ac=2,

(a = -1,
(b=3,
(c = 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) = 0 are easy to find.

Answer: -1; -2.

4. The method of selecting the root by the highest and free coefficient

The method is based on the application of theorems:

1) Any 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 natural) to be the root of an equation with integer coefficients, it is necessary that the number p is an integer divisor of the free term a 0, and q is a natural divisor of the highest coefficient.

Example 1

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

Solution:

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

Hence p/q = ±1, ±2, ±1/2, ±1/3, ±2/3, ±1/6.

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

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

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In general, an equation that has a degree higher than 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 a degree of no more than 4. The solution of such equations is based on the decomposition of the polynomial into factors, so we advise you to review this topic before studying this article.

Most often, one has 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 so that we can 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 degree equations with integer coefficients

All equations of the form a n x n + a n - 1 x n - 1 + . . . + a 1 x + a 0 = 0 , we can reduce to an equation of the same degree by multiplying both sides by a n n - 1 and changing the variable of the form 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

The resulting coefficients will also be integers. Thus, we will need to solve the reduced equation of the nth degree with integer coefficients, which has the form x n + a n x n - 1 + ... + a 1 x + a 0 = 0.

We calculate the integer 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. Let's write them down and substitute them into the original equality one by one, checking the result. Once we have obtained an identity and found one of the roots of the equation, we can write it in the form x - x 1 · P n - 1 (x) = 0 . Here x 1 is the root of the equation, and P n - 1 (x) is the quotient of x n + a n x n - 1 + ... + a 1 x + a 0 divided by x - x 1 .

Substitute the remaining divisors in P n - 1 (x) = 0 , starting with x 1 , since the roots can be repeated. After obtaining the identity, the root x 2 is considered found, and the equation can be written as (x - x 1) (x - x 2) P n - 2 (x) \u003d 0. Here P n - 2 (x) will be quotient from dividing P n - 1 (x) by x - x 2 .

We continue to sort through the divisors. Find all integer roots and denote their number as m. After that, the original equation can be represented as x - x 1 x - x 2 · … · x - x m · P n - m (x) = 0 . Here P n - m (x) is a polynomial of n - m -th degree. For calculation it is convenient to use Horner's scheme.

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

As a result, we got the equation P n - m (x) = 0, the roots of which can be found by any convenient way. They can be irrational or complex.

Let's show on specific example how such a solution scheme is applied.

Example 1

Condition: find the solution of the equation x 4 + x 3 + 2 x 2 - x - 3 = 0 .

Solution

Let's start with finding integer roots.

We have an intercept equal to minus three. It has divisors equal to 1 , - 1 , 3 and - 3 . Let's substitute them into the original equation and see which of them will give identities as a result.

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

Now let's divide the polynomial x 4 + x 3 + 2 x 2 - x - 3 by (x - 1) into a column:

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

We divide the polynomial x 3 + 2 x 2 + 4 x + 3 by (x + 1) in a column:

We get that

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)

We substitute the next divisor into the equation x 2 + x + 3 = 0, starting from - 1:

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

The resulting equalities will be incorrect, which means that the equation no longer has integer 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 from this that this square trinomial does not have real roots, but there are complex conjugate ones: x = - 1 2 ± i 11 2 .

Let us clarify that instead of dividing into a column, Horner's scheme can be used. This is done like this: after we have determined the first root of the equation, we fill in the table.

In the table of coefficients, we can immediately see the coefficients of the quotient from the division of polynomials, which means x 4 + x 3 + 2 x 2 - x - 3 = 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 = 0.

Solution

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

Let's check them 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

So x = 2 will be the root of the equation. Divide x 4 - x 3 - 5 x 2 + 12 by x - 2 using Horner's scheme:

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

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

So 2 will again be a root. Divide x 3 + x 2 - 3 x - 6 = 0 by x - 2:

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

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

Let's solve the quadratic equation:

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

We get a complex conjugate pair of roots: x = - 3 2 ± i 3 2 .

Answer: x = - 3 2 ± i 3 2 .

Example 3

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

Solution

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

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

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

We replace the variables 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 got a standard equation of the 4th degree, which can be solved according to the standard scheme. Let's check the divisors, divide and in the end we get that it has 2 real roots y \u003d - 2, y \u003d 3 and two complex ones. We will not present the entire solution here. By virtue of the replacement, the real roots of this equation will be x = y 2 = - 2 2 = - 1 and x = y 2 = 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) = 0 is the degree of the polynomial P(x), i.e. the largest of the powers of its terms with a non-zero coefficient.

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

Statements about the roots of a polynomial and its divisors:

1. Polynomial nth degree has a number of roots not exceeding the number n, and the roots of multiplicity m occur exactly m times.

2. A polynomial of odd degree has at least one real root.

3. If α is the root of Р(х), then Р n (х) = (х – α) · Q n – 1 (x), where Q n – 1 (x) is a polynomial of degree (n – 1).

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

6. For a third degree polynomial

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

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

7. Any polynomial of the fourth degree expands into the 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 c 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) = 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 of examples

Example 1

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

Solution.

Answer: R = 0.

Example 2

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

Solution:

2x 3 + 3x 2 – 2x + 3| x + 2

2x 3 + 4x 2 2x 2 – x

X 2 – 2 x

Answer: R = 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) = t 1 , q(x) = t 2 , ... , q(x) = 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 = 0.

Solution:

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

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

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

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

2. Factorization by the method of grouping and abbreviated multiplication formulas

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

Example 1

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

Solution.

Imagine - 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 \u003d 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 \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 = 0.

Solution.

A polynomial of the 3rd degree 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 = 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.

Solving the system:

(b – a = 4,
(c – ab = 5,
(-ac=2,

(a = -1,
(b=3,
(c = 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) = 0 are easy to find.

Answer: -1; -2.

4. The method of selecting the root by the highest and free coefficient

The method is based on the application of theorems:

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

Example 1

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

Solution:

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

Hence p/q = ±1, ±2, ±1/2, ±1/3, ±2/3, ±1/6.

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

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

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To get help from a tutor -.
The first lesson is free!

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The use of equations is widespread in our lives. They are used in many calculations, construction of structures and even sports. Equations have been used by man since ancient times and since then their use has only increased. In mathematics, equations of higher degrees with integer coefficients are quite common. To solve this kind of equation, you need:

Determine the rational roots of the equation;

Factor out 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:

Let's find all its real roots. Multiply the left and right sides of the equation by \

Let's change the variables \

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

Where can I solve an equation of higher powers online with a solver?

You can solve the equation on our website https: // site. Free online solver will allow you to solve an online equation of any complexity in seconds. All you have to do is just enter your data into the solver. You can also watch the video instruction and learn how to solve the equation on our website. And if you have any questions, 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, it is often necessary to factorize a polynomial. To factor a polynomial is to represent it as a product of two or more polynomials. We use some methods of expanding polynomials quite often: taking out a common factor, using abbreviated multiplication formulas, highlighting the full square, grouping. Let's look at 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 you somehow choose the root of a 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 by the binomial “column”, or by the corresponding grouping of the terms of the polynomial and extracting a factor from them, or by the method of indefinite coefficients.

Example. Factorize a polynomial

Solution. Since the coefficient at x4 is equal to 1, then the rational roots of this polynomial exist, are divisors of the number 6, that is, they can be integers ±1, ±2, ±3, ±6. We denote this polynomial by P4(x). Since Р Р4 (1) = 4 and Р4 (-4) = 23, the numbers 1 and -1 are not roots of the polynomial PA (x). Since P4(2) = 0, then x = 2 is the root of the polynomial P4(x), and, therefore, this polynomial is divisible 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

Therefore, P4(x) = (x - 2)(x3 - 3x2 + x - 3). Since xz - Zx2 + 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. The essence of this method will be explained by the following example.

Example. x3 - (√3 + 1) x2 + 3.

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

Since the roots of this trinomial square with respect to a are a1 = x and a2 = x2 - x, then the equality a2 - ax2 + (xs - x2) = (a - x) (a - x2 + x) is true. Therefore, the polynomial x3 - (√3 + 1)x2 + 3 decomposes into factors √3 - x and √3 - x2 + x, i.e.

x3 - (√3+1)x2+3=(x-√3)(x2-x-√3).

Method for introducing a new unknown

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

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

Solution. Let's transform this polynomial as follows: x(x+1)(x+2)(x+3)-15= [x (x + 3)][(x + 1)(x + 2)] - 15 = (x2 + 3x) (x2 + 3x + 2) - 15.

Denote x2 + 3x by y. Then we have 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).

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

Example. Factorize the polynomial (x-4)4+(x+2)4

Solution. Denote x - 4 + x + 2 = x - 1 by 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 factoring a polynomial, one has to apply successively several of the methods discussed above.

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

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

Applying the method of selecting a full square to the first bracket, we have x4 - 3x3 + 4x - 3 = (x4 - 2 1 x2 + 12) - (x2 -4x + 4).

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

Finally, applying the difference of squares formula, 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, a ≠ 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 = 0. (2)

Solution. Equation (2) is a symmetrical equation of the third degree.

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

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

Answer: x = -1.

2. Symmetric equations of the fourth degree

Type equation

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

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

Let us rewrite equation (4) in the form:

In this equation, we make a replacement, then we get a 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 у0, 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

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

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

Assuming, we obtain an equation that has two roots y1 = 2 and y2 = 3. Therefore, the original equation is equivalent to the set of equations

The solution of the first equation of this set is x1 = 1, and the solution of the second is u.

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

Answer: x1=1.

§3. Algebraic equations

1. Reducing the degree of the equation

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

Example 1. Solve the equation

Solution. Denote by, then equation (1) can be rewritten as The last equation has roots and Therefore, equation (1) is equivalent to the set of equations and. The solution of the first equation of this set is and The solution of the second equation is

The solutions of equation (1) are

Example 2. Solve the equation

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

We obtain the equation We rewrite this equation in the form

(3) and denoting through we rewrite equation (3) in the form The last equation has roots and Therefore, we obtain that equation (3) is equivalent to the set of two equations and 4)

The solutions of set (4) are and, 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 using the replacement of the unknown, i.e., the replacement

Example 3. Solve the equation

Solution. Let's denote by, i.e. we make a change of variables or Then equation (6) can be rewritten in the form or, using the formula, in the form

Because the roots quadratic equation is and then the solutions of equation (7) are the solutions of the set of equations u. 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

Solution. Let's make a change of unknowns, i.e. y=x+3 or x = y – 3. Then equation (9) can be rewritten as

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

(y2-4)(y2-1)=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 = 0, therefore, dividing equation (11) by x2, we obtain an equivalent equation

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

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

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

This collection has two roots: x1= -1 and x2 = -2.

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

Comment. type equation,

Which can always be reduced to the form (11) and, moreover, considering α > 0 and λ > 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 = γδ ≠ 0, can be rewritten by multiplying the first bracket with the second, and the third with the fourth, in the form i.e. equation (13) is 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

Solution. Equation (14) has the form (13) , so we rewrite it as

Since x = 0 is not a solution to this equation, dividing both sides of it by x2, we obtain an equivalent original equation. Making a change of variables, we obtain a quadratic equation whose solution is and. Therefore, the original equation (14) is equivalent to the set of equations u.

The solution to the first equation of this set 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, has no root x = 0, therefore, dividing equation (15) by x2. we obtain an equation equivalent to it, which, after the replacement of the unknown, will be rewritten in the form of a quadratic equation, the solution of which is not difficult.

Example 7. Equation solution

Solution. 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 change of the unknown, we can rewrite equation (17) in the form

Quadratic equation (18) has 2 roots: y1 = 1 and y2 = -1. Therefore, equation (17) is equivalent to the set of equations and (19)

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

They will be the roots of equation (16).

§4. Rational Equations

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

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

Consider some methods for solving an equation of the form = 0.

1. Equations of the form

The equation

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

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

Summing up the terms in each bracket, we rewrite Eq. (3) as

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

, (5) equivalent to equation (4). Let us make a change of the unknown, then equation (5) will 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

There are also solutions to the equation. None of these numbers set the denominator to zero. 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 of 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 can be solved as follows: it is necessary to select the integer part in each of the fractions of the equation, i.e., replace equation (8) with the equation

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

Example. solve the equation

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

Summing up the terms in parentheses, we rewrite Eq. (10) as

Making the change of 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, of course, this is possible) each of the fractions on the left side of equation (14) into the sum of simple fractions

Reduce equation (14) to the form (1), then, having carried out a convenient rearrangement of the terms of the resulting equation, solve it by the method described in paragraph 1).

Example. solve the equation

Solution. 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 the set of equations and

To solve the second equation of the set (18), we will make a change 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, equation (20) does not have them either.

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

4. Equations of the form

The equation

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

Example. solve the equation

Solution. Let us 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 set equation (24) can be rewritten as

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 of the original equation.

5. Equations of the form.

Equation of the form (25)

Under certain conditions on numbers, by replacing the unknown, one can reduce to an equation of the form

Example. solve the equation

Solution. 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

Having made a change of variables, we rewrite equation (27) in the form

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