Properties of factorization. Complex cases of factoring polynomials

Any algebraic polynomial of degree n can be represented as a product of n-linear factors of the form and a constant number, which is the coefficients of the polynomial at the highest degree x, i.e.

Where - are the roots of the polynomial.

The root of a polynomial is the number (real or complex) that makes the polynomial vanish. The roots of a polynomial can be either real roots or complex conjugate roots, then the polynomial can be represented in the following form:

Let's consider methods for decomposing polynomials of degree “n” into the product of factors of the first and second degrees.

Method No. 1.Method of undetermined coefficients.

The coefficients of such a transformed expression are determined by the method of indefinite coefficients. The essence of the method is that the type of factors into which a given polynomial is decomposed is known in advance. When using the method of uncertain coefficients, the following statements are true:

P.1. Two polynomials are identically equal if their coefficients are equal for the same powers of x.

P.2. Any polynomial of the third degree is decomposed into the product of linear and quadratic factors.

P.3. Any fourth-degree polynomial can be decomposed into the product of two second-degree polynomials.

Example 1.1. It is necessary to factorize the cubic expression:

P.1. In accordance with the accepted statements, the identical equality holds for the cubic expression:

P.2. The right side of the expression can be represented as terms as follows:

P.3. We compose a system of equations from the condition of equality of coefficients at the corresponding powers of the cubic expression.

This system of equations can be solved by selecting coefficients (if it is a simple academic problem) or methods for solving nonlinear systems of equations can be used. Solving this system of equations, we find that the uncertain coefficients are determined as follows:

Thus, the original expression is factorized in the following form:

This method can be used both in analytical calculations and in computer programming to automate the process of finding the root of an equation.

Method No. 2.Vieta formulas

Vieta's formulas are formulas connecting the coefficients of algebraic equations of degree n and its roots. These formulas were implicitly presented in the works of the French mathematician François Vieta (1540 - 1603). Due to the fact that Vieth considered only positive real roots, he therefore did not have the opportunity to write these formulas in a general explicit form.

For any algebraic polynomial of degree n that has n-real roots,

The following relations are valid that connect the roots of a polynomial with its coefficients:

Vieta's formulas are convenient to use to check the correctness of finding the roots of a polynomial, as well as to construct a polynomial from given roots.

Example 2.1. Let us consider how the roots of a polynomial are related to its coefficients using the example of a cubic equation

In accordance with Vieta’s formulas, the relationship between the roots of a polynomial and its coefficients has the following form:

Similar relations can be made for any polynomial of degree n.

Method No. 3. Factoring a quadratic equation with rational roots

From Vieta's last formula it follows that the roots of a polynomial are divisors of its free term and leading coefficient. In this regard, if the problem statement specifies a polynomial of degree n with integer coefficients

then this polynomial has a rational root (irreducible fraction), where p is the divisor of the free term, and q is the divisor of the leading coefficient. In this case, a polynomial of degree n can be represented as (Bezout’s theorem):

A polynomial, the degree of which is 1 less than the degree of the initial polynomial, is determined by dividing a polynomial of degree n binomial, for example, using Horner's scheme or in the simplest way - "column".

Example 3.1. It is necessary to factor the polynomial

P.1. Due to the fact that the coefficient of the highest term is equal to one, the rational roots of this polynomial are divisors of the free term of the expression, i.e. can be integers . We substitute each of the presented numbers into the original expression and find that the root of the presented polynomial is equal to .

Let's divide the original polynomial by a binomial:

Let's use Horner's scheme

The coefficients of the original polynomial are set in the top line, while the first cell of the top line remains empty.

In the first cell of the second line, the found root is written (in the example under consideration, the number “2” is written), and the following values ​​in the cells are calculated in a certain way and they are the coefficients of the polynomial, which is obtained by dividing the polynomial by the binomial. The unknown coefficients are determined as follows:

The value from the corresponding cell of the first row is transferred to the second cell of the second row (in the example under consideration, the number “1” is written).

The third cell of the second row contains the value of the product of the first cell and the second cell of the second row plus the value from the third cell of the first row (in the example under consideration 2 ∙1 -5 = -3).

The fourth cell of the second row contains the value of the product of the first cell and the third cell of the second row plus the value from the fourth cell of the first row (in the example under consideration, 2 ∙ (-3) +7 = 1).

Thus, the original polynomial is factorized:

Method number 4.Using abbreviated multiplication formulas

Abbreviated multiplication formulas are used to simplify calculations, as well as factoring polynomials. Abbreviated multiplication formulas allow you to simplify the solution of individual problems.

Formulas used to factorize

Factoring polynomials is an identity transformation, as a result of which a polynomial is transformed into the product of several factors - polynomials or monomials.

There are several ways to factor polynomials.

Method 1. Taking the common factor out of brackets.

This transformation is based on the distributive law of multiplication: ac + bc = c(a + b). The essence of the transformation is to isolate the common factor in the two components under consideration and “take” it out of brackets.

Let us factor the polynomial 28x 3 – 35x 4.

Solution.

1. Find a common divisor for the elements 28x3 and 35x4. For 28 and 35 it will be 7; for x 3 and x 4 – x 3. In other words, our common factor is 7x 3.

2. We represent each of the elements as a product of factors, one of which
7x 3: 28x 3 – 35x 4 = 7x 3 ∙ 4 – 7x 3 ∙ 5x.

3. We take the common factor out of brackets
7x 3: 28x 3 – 35x 4 = 7x 3 ∙ 4 – 7x 3 ∙ 5x = 7x 3 (4 – 5x).

Method 2. Using abbreviated multiplication formulas. The “mastery” of using this method is to notice one of the abbreviated multiplication formulas in the expression.

Let us factor the polynomial x 6 – 1.

Solution.

1. We can apply the difference of squares formula to this expression. To do this, imagine x 6 as (x 3) 2, and 1 as 1 2, i.e. 1. The expression will take the form:
(x 3) 2 – 1 = (x 3 + 1) ∙ (x 3 – 1).

2. We can apply the formula for the sum and difference of cubes to the resulting expression:
(x 3 + 1) ∙ (x 3 – 1) = (x + 1) ∙ (x 2 – x + 1) ∙ (x – 1) ∙ (x 2 + x + 1).

So,
x 6 – 1 = (x 3) 2 – 1 = (x 3 + 1) ∙ (x 3 – 1) = (x + 1) ∙ (x 2 – x + 1) ∙ (x – 1) ∙ (x 2 + x + 1).

Method 3. Grouping. The grouping method is to combine the components of a polynomial in such a way that it is easy to perform operations on them (addition, subtraction, subtraction of a common factor).

Let's factor the polynomial x 3 – 3x 2 + 5x – 15.

Solution.

1. Let's group the components in this way: the 1st with the 2nd, and the 3rd with the 4th
(x 3 – 3x 2) + (5x – 15).

2. In the resulting expression, we take the common factors out of brackets: x 2 in the first case and 5 in the second.
(x 3 – 3x 2) + (5x – 15) = x 2 (x – 3) + 5(x – 3).

3. We take the common factor x – 3 out of brackets and get:
x 2 (x – 3) + 5(x – 3) = (x – 3)(x 2 + 5).

So,
x 3 – 3x 2 + 5x – 15 = (x 3 – 3x 2) + (5x – 15) = x 2 (x – 3) + 5(x – 3) = (x – 3) ∙ (x 2 + 5 ).

Let's secure the material.

Factor the polynomial a 2 – 7ab + 12b 2.

Solution.

1. Let us represent the monomial 7ab as the sum 3ab + 4ab. The expression will take the form:
a 2 – (3ab + 4ab) + 12b 2.

Let's open the brackets and get:
a 2 – 3ab – 4ab + 12b 2.

2. Let's group the components of the polynomial in this way: 1st with 2nd and 3rd with 4th. We get:
(a 2 – 3ab) – (4ab – 12b 2).

3. Let’s take the common factors out of brackets:
(a 2 – 3ab) – (4ab – 12b 2) = a(a – 3b) – 4b(a – 3b).

4. Let’s take the common factor (a – 3b) out of brackets:
a(a – 3b) – 4b(a – 3b) = (a – 3 b) ∙ (a – 4b).

So,
a 2 – 7ab + 12b 2 =
= a 2 – (3ab + 4ab) + 12b 2 =
= a 2 – 3ab – 4ab + 12b 2 =
= (a 2 – 3ab) – (4ab – 12b 2) =
= a(a – 3b) – 4b(a – 3b) =
= (a – 3 b) ∙ (a – 4b).

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Factoring polynomials is an identity transformation, as a result of which a polynomial is transformed into the product of several factors - polynomials or monomials.

There are several ways to factor polynomials.

Method 1. Taking the common factor out of brackets.

This transformation is based on the distributive law of multiplication: ac + bc = c(a + b). The essence of the transformation is to isolate the common factor in the two components under consideration and “take” it out of brackets.

Let us factor the polynomial 28x 3 – 35x 4.

Solution.

1. Find a common divisor for the elements 28x3 and 35x4. For 28 and 35 it will be 7; for x 3 and x 4 – x 3. In other words, our common factor is 7x 3.

2. We represent each of the elements as a product of factors, one of which
7x 3: 28x 3 – 35x 4 = 7x 3 ∙ 4 – 7x 3 ∙ 5x.

3. We take the common factor out of brackets
7x 3: 28x 3 – 35x 4 = 7x 3 ∙ 4 – 7x 3 ∙ 5x = 7x 3 (4 – 5x).

Method 2. Using abbreviated multiplication formulas. The “mastery” of using this method is to notice one of the abbreviated multiplication formulas in the expression.

Let us factor the polynomial x 6 – 1.

Solution.

1. We can apply the difference of squares formula to this expression. To do this, imagine x 6 as (x 3) 2, and 1 as 1 2, i.e. 1. The expression will take the form:
(x 3) 2 – 1 = (x 3 + 1) ∙ (x 3 – 1).

2. We can apply the formula for the sum and difference of cubes to the resulting expression:
(x 3 + 1) ∙ (x 3 – 1) = (x + 1) ∙ (x 2 – x + 1) ∙ (x – 1) ∙ (x 2 + x + 1).

So,
x 6 – 1 = (x 3) 2 – 1 = (x 3 + 1) ∙ (x 3 – 1) = (x + 1) ∙ (x 2 – x + 1) ∙ (x – 1) ∙ (x 2 + x + 1).

Method 3. Grouping. The grouping method is to combine the components of a polynomial in such a way that it is easy to perform operations on them (addition, subtraction, subtraction of a common factor).

Let's factor the polynomial x 3 – 3x 2 + 5x – 15.

Solution.

1. Let's group the components in this way: the 1st with the 2nd, and the 3rd with the 4th
(x 3 – 3x 2) + (5x – 15).

2. In the resulting expression, we take the common factors out of brackets: x 2 in the first case and 5 in the second.
(x 3 – 3x 2) + (5x – 15) = x 2 (x – 3) + 5(x – 3).

3. We take the common factor x – 3 out of brackets and get:
x 2 (x – 3) + 5(x – 3) = (x – 3)(x 2 + 5).

So,
x 3 – 3x 2 + 5x – 15 = (x 3 – 3x 2) + (5x – 15) = x 2 (x – 3) + 5(x – 3) = (x – 3) ∙ (x 2 + 5 ).

Let's secure the material.

Factor the polynomial a 2 – 7ab + 12b 2.

Solution.

1. Let us represent the monomial 7ab as the sum 3ab + 4ab. The expression will take the form:
a 2 – (3ab + 4ab) + 12b 2.

Let's open the brackets and get:
a 2 – 3ab – 4ab + 12b 2.

2. Let's group the components of the polynomial in this way: 1st with 2nd and 3rd with 4th. We get:
(a 2 – 3ab) – (4ab – 12b 2).

3. Let’s take the common factors out of brackets:
(a 2 – 3ab) – (4ab – 12b 2) = a(a – 3b) – 4b(a – 3b).

4. Let’s take the common factor (a – 3b) out of brackets:
a(a – 3b) – 4b(a – 3b) = (a – 3 b) ∙ (a – 4b).

So,
a 2 – 7ab + 12b 2 =
= a 2 – (3ab + 4ab) + 12b 2 =
= a 2 – 3ab – 4ab + 12b 2 =
= (a 2 – 3ab) – (4ab – 12b 2) =
= a(a – 3b) – 4b(a – 3b) =
= (a – 3 b) ∙ (a – 4b).

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Let's look at specific examples of how to factor a polynomial.

We will expand the polynomials in accordance with .

Factor polynomials:

Let's check if there is a common factor. yes, it is equal to 7cd. Let's take it out of brackets:

The expression in parentheses consists of two terms. There is no longer a common factor, the expression is not a formula for the sum of cubes, which means the decomposition is complete.

Let's check if there is a common factor. No. The polynomial consists of three terms, so we check to see if there is a formula for a complete square. Two terms are the squares of the expressions: 25x²=(5x)², 9y²=(3y)², the third term is equal to the double product of these expressions: 2∙5x∙3y=30xy. This means that this polynomial is a perfect square. Since the double product has a minus sign, it is:

We check whether it is possible to take the common factor out of brackets. There is a common factor, it is equal to a. Let's take it out of brackets:

There are two terms in brackets. We check if there is a formula for difference of squares or difference of cubes. a² is the square of a, 1=1². This means that the expression in brackets can be written using the difference of squares formula:

There is a common factor, it is equal to 5. Let’s take it out of brackets:

in brackets there are three terms. We check whether the expression is a perfect square. Two terms are squares: 16=4² and a² - the square of a, the third term is equal to the double product of 4 and a: 2∙4∙a=8a. Therefore, it is a perfect square. Since all terms have a “+” sign, the expression in parentheses is the perfect square of the sum:

We take the general multiplier -2x out of brackets:

In parentheses is the sum of two terms. We check whether this expression is a sum of cubes. 64=4³, x³- cube x. This means that the binomial can be expanded using the formula:

There is a common multiplier. But, since the polynomial consists of 4 terms, we will first, and only then, take the common factor out of brackets. Let’s group the first term with the fourth, and the second with the third:

From the first brackets we take out the common factor 4a, from the second - 8b:

There is no common multiplier yet. To get it, we take out the “-“ from the second brackets, and each sign in the brackets changes to the opposite:

Now let’s take the common factor (1-3a) out of brackets:

In the second brackets there is a common factor 4 (this is the same factor that we did not put out of brackets at the beginning of the example):

Since the polynomial consists of four terms, we perform grouping. Let’s group the first term with the second, the third with the fourth:

In the first brackets there is no common factor, but there is a formula for the difference of squares, in the second brackets the common factor is -5:

A common multiplier has appeared (4m-3n). Let's take it out of the equation.

In general, this task requires a creative approach, since there is no universal method for solving it. But let's try to give a few tips.

In the overwhelming majority of cases, the factorization of a polynomial is based on a corollary of Bezout’s theorem, that is, the root is found or selected and the degree of the polynomial is reduced by one by dividing by . The root of the resulting polynomial is sought and the process is repeated until complete expansion.

If the root cannot be found, then specific expansion methods are used: from grouping to introducing additional mutually exclusive terms.

Further presentation is based on the skills of solving equations of higher degrees with integer coefficients.

Bracketing out the common factor.

Let's start with the simplest case, when the free term is equal to zero, that is, the polynomial has the form .

Obviously, the root of such a polynomial is , that is, we can represent the polynomial in the form .

This method is nothing more than putting the common factor out of brackets.

Example.

Factor a third degree polynomial.

Solution.

Obviously, what is the root of the polynomial, that is X can be taken out of brackets:

Let's find the roots of the quadratic trinomial

Thus,

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Factoring a polynomial with rational roots.

First, let's consider a method for expanding a polynomial with integer coefficients of the form , the coefficient of the highest degree is equal to one.

In this case, if a polynomial has integer roots, then they are divisors of the free term.

Example.

Solution.

Let's check if there are intact roots. To do this, write down the divisors of the number -18 : . That is, if a polynomial has integer roots, then they are among the written numbers. Let's check these numbers sequentially using Horner's scheme. Its convenience also lies in the fact that in the end we obtain the expansion coefficients of the polynomial:

That is, x=2 And x=-3 are the roots of the original polynomial and we can represent it as a product:

It remains to expand the quadratic trinomial.

The discriminant of this trinomial is negative, therefore it has no real roots.

Answer:

Comment:

Instead of Horner's scheme, one could use the selection of the root and subsequent division of the polynomial by a polynomial.

Now consider the expansion of a polynomial with integer coefficients of the form , and the coefficient of the highest degree is not equal to one.

In this case, the polynomial can have fractionally rational roots.

Example.

Factor the expression.

Solution.

By performing a variable change y=2x, let's move on to a polynomial with a coefficient equal to one at the highest degree. To do this, first multiply the expression by 4 .

If the resulting function has integer roots, then they are among the divisors of the free term. Let's write them down:

Let us sequentially calculate the values ​​of the function g(y) at these points until zero is reached.