A polynomial is an expression consisting of the sum of monomials. The latter are the product of a constant (number) and the root (or roots) of the expression to the power k. In this case, one speaks of a polynomial of degree k. The decomposition of a polynomial involves the transformation of the expression, in which the terms are replaced by factors. Let us consider the main ways of carrying out this kind of transformation.

Method for expanding a polynomial by extracting a common factor

This method is based on the laws of the distribution law. So, mn + mk = m * (n + k).

• Example: expand 7y 2 + 2uy and 2m 3 – 12m 2 + 4lm.

7y 2 + 2uy = y * (7y + 2u),

2m 3 - 12m 2 + 4lm = 2m(m 2 - 6m + 2l).

However, a factor that is necessarily present in each polynomial may not always be found, therefore this way is not universal.

Polynomial expansion method based on abbreviated multiplication formulas

Abbreviated multiplication formulas are valid for a polynomial of any degree. AT general view the conversion expression looks like this:

u k – l k = (u – l)(u k-1 + u k-2 * l + u k-3 *l 2 + … u * l k-2 + l k-1), where k is a representative of natural numbers .

Most often in practice, formulas for polynomials of the second and third orders are used:

u 2 - l 2 \u003d (u - l) (u + l),

u 3 - l 3 \u003d (u - l) (u 2 + ul + l 2),

u 3 + l 3 = (u + l)(u 2 - ul + l 2).

• Example: expand 25p 2 - 144b 2 and 64m 3 - 8l 3 .

25p 2 - 144b 2 \u003d (5p - 12b) (5p + 12b),

64m 3 - 8l 3 = (4m) 3 - (2l) 3 = (4m - 2l)((4m) 2 + 4m * 2l + (2l) 2) = (4m - 2l)(16m 2 + 8ml + 4l 2 ).

Polynomial decomposition method - grouping terms of an expression

This method is somewhat similar to the derivation technique. common multiplier, but has some differences. In particular, before isolating the common factor, one should group the monomials. Grouping is based on the rules of associative and commutative laws.

All monomials presented in the expression are divided into groups, in each of which general meaning such that the second factor will be the same in all groups. In general, such a decomposition method can be represented as an expression:

pl + ks + kl + ps = (pl + ps) + (ks + kl) ⇒ pl + ks + kl + ps = p(l + s) + k(l + s),

pl + ks + kl + ps = (p + k)(l + s).

• Example: expand 14mn + 16ln - 49m - 56l.

14mn + 16ln - 49m - 56l = (14mn - 49m) + (16ln - 56l) = 7m * (2n - 7) + 8l * (2n - 7) = (7m + 8l)(2n - 7).

Polynomial Decomposition Method - Full Square Formation

This method is one of the most efficient in the course of polynomial decomposition. At the initial stage, it is necessary to determine the monomials that can be “folded” into the square of the difference or sum. For this, one of the following relations is used:

(p - b) 2 \u003d p 2 - 2pb + b 2,

• Example: expand the expression u 4 + 4u 2 – 1.

We single out among its monomials the terms that form full square: u 4 + 4u 2 - 1 = u 4 + 2 * 2u 2 + 4 - 4 - 1 =

\u003d (u 4 + 2 * 2u 2 + 4) - 4 - 1 \u003d (u 4 + 2 * 2u 2 + 4) - 5.

Complete the transformation using the rules of abbreviated multiplication: (u 2 + 2) 2 - 5 = (u 2 + 2 - √5) (u 2 + 2 + √5).

That. u 4 + 4u 2 - 1 = (u 2 + 2 - √5)(u 2 + 2 + √5).

Very often, the numerator and denominator of a fraction are algebraic expressions that must first be decomposed into factors, and then, finding the same among them, divide both the numerator and denominator into them, that is, reduce the fraction. A whole chapter of a textbook on algebra in the 7th grade is devoted to tasks to factorize a polynomial. Factoring can be done 3 ways, as well as a combination of these methods.

1. Application of abbreviated multiplication formulas

As is known to multiply a polynomial by a polynomial, you need to multiply each term of one polynomial by each term of the other polynomial and add the resulting products. There are at least 7 (seven) common cases of multiplication of polynomials that are included in the concept. For example,

Table 1. Factorization in the 1st way

2. Taking the common factor out of the bracket

This method is based on the application of the distributive law of multiplication. For example,

We divide each term of the original expression by the factor that we take out, and at the same time we get the expression in brackets (that is, the result of dividing what was by what we take out remains in brackets). First of all, you need correctly determine the multiplier, which must be bracketed.

The polynomial in brackets can also be a common factor:

When performing the “factorize” task, one must be especially careful with the signs when taking the common factor out of brackets. To change the sign of each term in a parenthesis (b - a), we take out the common factor -1 , while each term in the bracket is divided by -1: (b - a) = - (a - b) .

In the event that the expression in brackets is squared (or to any even power), then numbers inside brackets can be swapped completely free, since the minuses taken out of brackets will still turn into a plus when multiplied: (b - a) 2 = (a - b) 2, (b - a) 4 = (a - b) 4 and so on…

3. Grouping method

Sometimes not all terms in the expression have a common factor, but only some. Then you can try group terms in brackets so that some factor can be taken out of each. Grouping method is double bracketing of common factors.

4. Using several methods at once

Sometimes you need to apply not one, but several ways to factorize a polynomial into factors at once.

This is a synopsis on the topic. "Factorization". Choose next steps:

• Go to the next abstract:

The concepts of "polynomial" and "factorization of a polynomial" in algebra are very common, because you need to know them in order to easily perform calculations with large multi-valued numbers. This article will describe several decomposition methods. All of them are quite simple to use, you just need to choose the right one in each case.

The concept of a polynomial

A polynomial is the sum of monomials, that is, expressions containing only the multiplication operation.

For example, 2 * x * y is a monomial, but 2 * x * y + 25 is a polynomial, which consists of 2 monomials: 2 * x * y and 25. Such polynomials are called binomials.

Sometimes, for the convenience of solving examples with multivalued values, the expression must be transformed, for example, decomposed into a certain number of factors, that is, numbers or expressions between which the multiplication operation is performed. There are a number of ways to factorize a polynomial. It is worth considering them starting from the most primitive, which is used even in primary classes.

Grouping (general entry)

The formula for factoring a polynomial into factors by the grouping method in general looks like this:

ac + bd + bc + ad = (ac + bc) + (ad + bd)

It is necessary to group the monomials so that a common factor appears in each group. In the first parenthesis, this is the factor c, and in the second - d. This must be done in order to then take it out of the bracket, thereby simplifying the calculations.

Decomposition algorithm on a specific example

The simplest example of factoring a polynomial into factors using the grouping method is given below:

10ac + 14bc - 25a - 35b = (10ac - 25a) + (14bc - 35b)

In the first bracket, you need to take the terms with the factor a, which will be common, and in the second - with the factor b. Pay attention to the + and - signs in the finished expression. We put before the monomial the sign that was in the initial expression. That is, you need to work not with the expression 25a, but with the expression -25. The minus sign, as it were, is “glued” to the expression behind it and always take it into account in calculations.

At the next step, you need to take out the factor, which is common, out of the bracket. That's what grouping is for. To take it out of the bracket means to write out before the bracket (omitting the multiplication sign) all those factors that are repeated exactly in all the terms that are in the bracket. If there are not 2, but 3 or more terms in the bracket, the common factor must be contained in each of them, otherwise it cannot be taken out of the bracket.

In our case, only 2 terms in brackets. The overall multiplier is immediately visible. The first parenthesis is a, the second is b. Here you need to pay attention to the digital coefficients. In the first bracket, both coefficients (10 and 25) are multiples of 5. This means that not only a, but also 5a can be bracketed. Before the bracket, write out 5a, and then divide each of the terms in brackets by the common factor that was taken out, and also write down the quotient in brackets, not forgetting the + and - signs. Do the same with the second bracket, take out 7b, since 14 and 35 multiple of 7.

10ac + 14bc - 25a - 35b = (10ac - 25a) + (14bc - 35b) = 5a (2c - 5) + 7b (2c - 5).

It turned out 2 terms: 5a (2c - 5) and 7b (2c - 5). Each of them contains a common factor (the whole expression in brackets here is the same, which means it is a common factor): 2c - 5. It also needs to be taken out of the bracket, that is, the terms 5a and 7b remain in the second bracket:

5a(2c - 5) + 7b(2c - 5) = (2c - 5)*(5a + 7b).

So the full expression is:

10ac + 14bc - 25a - 35b \u003d (10ac - 25a) + (14bc - 35b) \u003d 5a (2c - 5) + 7b (2c - 5) \u003d (2c - 5) * (5a + 7b).

Thus, the polynomial 10ac + 14bc - 25a - 35b is decomposed into 2 factors: (2c - 5) and (5a + 7b). The multiplication sign between them can be omitted when writing

Sometimes there are expressions of this type: 5a 2 + 50a 3, here you can bracket not only a or 5a, but even 5a 2. You should always try to take the largest possible common factor out of the bracket. In our case, if we divide each term by a common factor, we get:

5a 2 / 5a 2 = 1; 50a 3 / 5a 2 = 10a(when calculating the quotient of several powers with equal bases, the base is preserved, and the exponent is subtracted). Thus, one remains in the bracket (in no case do not forget to write one if you take out one of the terms entirely from the bracket) and the quotient of division: 10a. It turns out that:

5a 2 + 50a 3 = 5a 2 (1 + 10a)

Square formulas

For the convenience of calculations, several formulas have been derived. They are called reduced multiplication formulas and are used quite often. These formulas help factorize polynomials containing powers. It's another one effective way factorizations. So here they are:

• a 2 + 2ab + b 2 = (a + b) 2 - the formula, called the "square of the sum", since as a result of the expansion into a square, the sum of the numbers enclosed in brackets is taken, that is, the value of this sum is multiplied by itself 2 times, which means it is a multiplier.
• a 2 + 2ab - b 2 = (a - b) 2 - the formula of the square of the difference, it is similar to the previous one. The result is a difference enclosed in brackets, contained in a square power.
• a 2 - b 2 \u003d (a + b) (a - b)- this is the formula for the difference of squares, since initially the polynomial consists of 2 squares of numbers or expressions between which subtraction is performed. It is perhaps the most commonly used of the three.

Examples for calculating by formulas of squares

Calculations on them are made quite simply. For example:

1. 25x2 + 20xy + 4y 2 - use the formula "square of the sum".
2. 25x 2 is the square of 5x. 20xy is twice the product of 2*(5x*2y), and 4y 2 is the square of 2y.
3. So 25x 2 + 20xy + 4y 2 = (5x + 2y) 2 = (5x + 2y)(5x + 2y). This polynomial is decomposed into 2 factors (the factors are the same, therefore it is written as an expression with a square power).

Operations according to the formula of the square of the difference are performed similarly to these. What remains is the difference of squares formula. Examples for this formula are very easy to identify and find among other expressions. For example:

• 25a 2 - 400 \u003d (5a - 20) (5a + 20). Since 25a 2 \u003d (5a) 2, and 400 \u003d 20 2
• 36x 2 - 25y 2 \u003d (6x - 5y) (6x + 5y). Since 36x 2 \u003d (6x) 2, and 25y 2 \u003d (5y 2)
• c 2 - 169b 2 \u003d (c - 13b) (c + 13b). Since 169b 2 = (13b) 2

It is important that each of the terms is the square of some expression. Then this polynomial is to be factored by the difference of squares formula. For this, it is not necessary that the second power is above the number. There are polynomials containing large powers, but still suitable for these formulas.

a 8 +10a 4 +25 = (a 4) 2 + 2*a 4 *5 + 5 2 = (a 4 +5) 2

AT this example and 8 can be represented as (a 4) 2 , that is, the square of a certain expression. 25 is 5 2 and 10a is 4 - this is the double product of the terms 2*a 4 *5. That is, this expression, despite the presence of degrees with large exponents, can be decomposed into 2 factors in order to work with them later.

Cube formulas

The same formulas exist for factoring polynomials containing cubes. They are a little more complicated than those with squares:

• a 3 + b 3 \u003d (a + b) (a 2 - ab + b 2)- this formula is called the sum of cubes, since in its initial form the polynomial is the sum of two expressions or numbers enclosed in a cube.
• a 3 - b 3 \u003d (a - b) (a 2 + ab + b 2) - a formula identical to the previous one is denoted as the difference of cubes.
• a 3 + 3a 2 b + 3ab 2 + b 3 = (a + b) 3 - sum cube, as a result of calculations, the sum of numbers or expressions is obtained, enclosed in brackets and multiplied by itself 3 times, that is, located in the cube
• a 3 - 3a 2 b + 3ab 2 - b 3 = (a - b) 3 - the formula, compiled by analogy with the previous one with a change in only some signs of mathematical operations (plus and minus), is called the "difference cube".

The last two formulas are practically not used for the purpose of factoring a polynomial, since they are complex, and it is quite rare to find polynomials that completely correspond to just such a structure so that they can be decomposed according to these formulas. But you still need to know them, since they will be required for actions in the opposite direction - when opening brackets.

Examples for cube formulas

Consider an example: 64a 3 − 8b 3 = (4a) 3 − (2b) 3 = (4a − 2b)((4a) 2 + 4a*2b + (2b) 2) = (4a−2b)(16a 2 + 8ab + 4b 2 ).

We've taken fairly prime numbers here, so you can immediately see that 64a 3 is (4a) 3 and 8b 3 is (2b) 3 . Thus, this polynomial is expanded by the formula difference of cubes into 2 factors. Actions on the formula of the sum of cubes are performed by analogy.

It is important to understand that not all polynomials can be decomposed in at least one of the ways. But there are such expressions that contain larger powers than a square or a cube, but they can also be expanded into abbreviated multiplication forms. For example: x 12 + 125y 3 =(x 4) 3 +(5y) 3 =(x 4 +5y)*((x 4) 2 − x 4 *5y+(5y) 2)=(x 4 + 5y)( x 8 − 5x 4 y + 25y 2).

This example contains as many as 12 degrees. But even it can be factored using the sum of cubes formula. To do this, you need to represent x 12 as (x 4) 3, that is, as a cube of some expression. Now, instead of a, you need to substitute it in the formula. Well, the expression 125y 3 is the cube of 5y. The next step is to write the formula and do the calculations.

At first, or when in doubt, you can always check by inverse multiplication. You only need to open the brackets in the resulting expression and perform actions with similar terms. This method applies to all listed ways abbreviations: both to work with a common factor and grouping, and to operations on the formulas of cubes and square powers.

We already know how to partially use the factorization of the difference of degrees - when studying the topic “Difference of Squares” and “Difference of Cubes”, we learned to represent as a product the difference of expressions that can be represented as squares or as cubes of some expressions or numbers.

Abbreviated multiplication formulas

According to the formulas of abbreviated multiplication:

the difference of squares can be represented as the product of the difference of two numbers or expressions by their sum

The difference of cubes can be represented as the product of the difference of two numbers by the incomplete square of the sum

Transition to the difference of expressions in 4 powers

Based on the difference of squares formula, let's try to factorize the expression $a^4-b^4$

Recall how a power is raised to a power - for this, the base remains the same, and the exponents are multiplied, i.e. $((a^n))^m=a^(n*m)$

Then you can imagine:

$a^4=(((a)^2))^2$

$b^4=(((b)^2))^2$

So our expression can be represented as $a^4-b^4=(((a)^2))^2$-$(((b)^2))^2$

Now in the first bracket we again got the difference of numbers, which means we can again factorize as the product of the difference of two numbers or expressions by their sum: $a^2-b^2=\left(a-b\right)(a+b)$.

Now we calculate the product of the second and third brackets using the rule for the product of polynomials - we multiply each term of the first polynomial by each term of the second polynomial and add the result. To do this, we first multiply the first term of the first polynomial - $a$ - by the first and second terms of the second one (by $a^2$ and $b^2$), i.e. we get $a\cdot a^2+a\cdot b^2$, then we multiply the second term of the first polynomial -$b$- by the first and second terms of the second polynomial (by $a^2$ and $b^2$), those. get $b\cdot a^2 + b\cdot b^2$ and sum the resulting expressions

$\left(a+b\right)\left(a^2+b^2\right)=a\cdot a^2+a\cdot b^2+ b \cdot a^2 + b\cdot b^ 2 = a^3+ab^2+a^2b+b^3$

We write the difference of monomials of the 4th degree, taking into account the calculated product:

$a^4-b^4=(((a)^2))^2$-$(((b)^2))^2=((a)^2-b^2)(a^2 +b^2)$=$\ \left(a-b\right)(a+b)(a^2+b^2)\ $=

Transition to the difference of expressions in the 6th power

Based on the difference of squares formula, let's try to factorize the expression $a^6-b^6$

Recall how a power is raised to a power - for this, the base remains the same, and the exponents are multiplied, i.e. $((a^n))^m=a^(n\cdot m)$

Then you can imagine:

$a^6=(((a)^3))^2$

$b^6=(((b)^3))^2$

So our expression can be represented as $a^6-b^6=(((a)^3))^2-(((b)^3))^2$

In the first bracket we got the difference of cubes of monomials, in the second the sum of cubes of monomials, now we can again factorize the difference of cubes of monomials as the product of the difference of two numbers by the incomplete square of the sum $a^3-b^3=\left(a-b\right)( a^2+ab+b^2)$

The original expression takes the form

$a^6-b^6=((a)^3-b^3)\left(a^3+b^3\right)=\left(a-b\right)(a^2+ab+b^ 2)(a^3+b^3)$

We calculate the product of the second and third brackets using the rule for the product of polynomials - we multiply each term of the first polynomial by each term of the second polynomial and add the result.

$(a^2+ab+b^2)(a^3+b^3)=a^5+a^4b+a^3b^2+a^2b^3+ab^4+b^5$

We write the difference of monomials of the 6th degree, taking into account the calculated product:

$a^6-b^6=((a)^3-b^3)\left(a^3+b^3\right)=\left(a-b\right)(a^2+ab+b^ 2)(a^3+b^3)=(a-b)(a^5+a^4b+a^3b^2+a^2b^3+ab^4+b^5)$

Factoring the power difference

Let us analyze the formulas for the difference of cubes, the difference of $4$ degrees, the difference of $6$ degrees

We see that in each of these expansions there is some analogy, generalizing which we get:

Example 1

Factorize $(32x)^(10)-(243y)^(15)$

Solution: First, we represent each monomial as some monomial to the power of 5:

\[(32x)^(10)=((2x^2))^5\]\[(243y)^(15)=((3y^3))^5\]

We use the power difference formula

Picture 1.

The factorization of polynomials is an identical transformation, as a result of which a polynomial is transformed into a product of several factors - polynomials or monomials.

There are several ways to factorize polynomials.

Method 1. Bracketing the common factor.

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

Let us factorize the polynomial 28x 3 - 35x 4.

Solution.

1. We find a common divisor for 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 7x3.

2. We represent each of the elements as a product of factors, one of which
7x 3: 28x 3 - 35x 4 \u003d 7x 3 ∙ 4 - 7x 3 ∙ 5x.

3. Bracketing the common factor
7x 3: 28x 3 - 35x 4 \u003d 7x 3 ∙ 4 - 7x 3 ∙ 5x \u003d 7x 3 (4 - 5x).

Method 2. Using abbreviated multiplication formulas. The "mastery" of mastering this method is to notice in the expression one of the formulas for abbreviated multiplication.

Let us factorize the polynomial x 6 - 1.

Solution.

1. We can apply the difference of squares formula to this expression. To do this, we represent x 6 as (x 3) 2, and 1 as 1 2, i.e. 1. The expression will take the form:
(x 3) 2 - 1 \u003d (x 3 + 1) ∙ (x 3 - 1).

2. To the resulting expression, we can apply the formula for the sum and difference of cubes:
(x 3 + 1) ∙ (x 3 - 1) \u003d (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 consists in combining the components of a polynomial in such a way that it is easy to perform operations on them (addition, subtraction, taking out a common factor).

We factorize the polynomial x 3 - 3x 2 + 5x - 15.

Solution.

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

3. We take out the common factor x - 3 and get:
x 2 (x - 3) + 5 (x - 3) \u003d (x - 3) (x 2 + 5).

So,
x 3 - 3x 2 + 5x - 15 \u003d (x 3 - 3x 2) + (5x - 15) \u003d x 2 (x - 3) + 5 (x - 3) \u003d (x - 3) ∙ (x 2 + 5 ).

Let's fix the material.

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

Solution.

1. We 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. Group the components of the polynomial in this way: the 1st with the 2nd and the 3rd with the 4th. We get:
(a 2 - 3ab) - (4ab - 12b 2).

3. Let's take out the common factors:
(a 2 - 3ab) - (4ab - 12b 2) \u003d a (a - 3b) - 4b (a - 3b).

4. Let's take out the common factor (a - 3b):
a(a – 3b) – 4b(a – 3b) = (a – 3b) ∙ (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) =
= (а – 3 b) ∙ (а – 4b).

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