Factoring polynomials. Taking the common factor out of brackets. Taking the general factor out of brackets - Knowledge Hypermarket

Definition 1

First let's remember Rules for multiplying a monomial by a monomial:

To multiply a monomial by a monomial, you must first multiply the coefficients of the monomials, then, using the rule of multiplying powers with the same base, multiply the variables included in the monomials.

Example 1

Find the product of the monomials $(2x)^3y^2z$ and $(\frac(3)(4)x)^2y^4$

Solution:

First, let's calculate the product of the coefficients

$2\cdot\frac(3)(4) =\frac(2\cdot 3)(4)$ in this task we used the rule for multiplying a number by a fraction - to multiply a whole number by a fraction, you need to multiply the number by the numerator of the fraction, and the denominator put without changes

Now let's use the main property of a fraction - the numerator and denominator of a fraction can be divided by the same number, different from $0$. Let's divide the numerator and denominator of this fraction by $2$, that is, reduce this fraction by $2$ $2\cdot\frac(3)(4)$ =$\frac(2\cdot 3)(4)=\\frac(3 )(2)$

The resulting result turned out to be an improper fraction, that is, one in which the numerator is greater than the denominator.

Let's transform this fraction by isolating the whole part. Let us remember that to isolate an integer part, it is necessary to write down the incomplete quotient obtained by dividing the numerator by the denominator, as an integer part, the remainder of the division into the numerator of the fractional part, and the divisor into the denominator.

We found the coefficient of the future product.

Now we will sequentially multiply the variables $x^3\cdot x^2=x^5$,

$y^2\cdot y^4 =y^6$. Here we used the rule for multiplying powers with the same base: $a^m\cdot a^n=a^(m+n)$

Then the result of multiplying monomials will be:

$(2x)^3y^2z \cdot (\frac(3)(4)x)^2y^4=1\frac(1)(2)x^5y^6$.

Then based on of this rule you can perform the following task:

Example 2

Represent a given polynomial as the product of a polynomial and a monomial $(4x)^3y+8x^2$

Let us represent each of the monomials included in the polynomial as the product of two monomials in order to isolate a common monomial, which will be a factor in both the first and second monomials.

First, let's start with the first monomial $(4x)^3y$. Let's factorize its coefficient into simple factors: $4=2\cdot 2$. We will do the same with the coefficient of the second monomial $8=2\cdot 2 \cdot 2$. Note that two factors $2\cdot 2$ are included in both the first and second coefficients, which means $2\cdot 2=4$ - this number will be included in the general monomial as a coefficient

Now let us note that in the first monomial there is $x^3$, and in the second there is the same variable to the power of $2:x^2$. This means that it is convenient to represent the variable $x^3$ like this:

The variable $y$ is included in only one term of the polynomial, which means it cannot be included in the general monomial.

Let's imagine the first and second monomial included in the polynomial as a product:

$(4x)^3y=4x^2\cdot xy$

$8x^2=4x^2\cdot 2$

Note that the common monomial, which will be a factor in both the first and second monomial, is $4x^2$.

$(4x)^3y+8x^2=4x^2\cdot xy + 4x^2\cdot 2$

Now we apply the distributive law of multiplication, then the resulting expression can be represented as a product of two factors. One of the multipliers will be the total multiplier: $4x^2$ and the other will be the sum of the remaining multipliers: $xy + 2$. Means:

$(4x)^3y+8х^2 = 4x^2\cdot xy + 4x^2\cdot 2 = 4x^2(xy+2)$

This method is called factorization using a common factor.

Common factor in in this case the monomial $4x^2$ was used.

Algorithm

Note 1

Find the greatest common divisor of the coefficients of all monomials included in the polynomial - it will be the coefficient of the common factor-monomial, which we will put out of brackets

A monomial consisting of the coefficient found in paragraph 2 and the variables found in paragraph 3 will be a common factor. which can be taken out of brackets as a common factor.

Example 3

Take out the common factor $3a^3-(15a)^2b+4(5ab)^2$

Solution:

Let’s find the gcd of the coefficients; for this we will decompose the coefficients into simple factors

$45=3\cdot 3\cdot 5$

And we find the product of those included in the expansion of each:

Identify the variables that make up each monomial and select the variable with the smallest exponent

$a^3=a^2\cdot a$

The variable $b$ is included only in the second and third monomial, which means it will not be included in the common factor.

Let's compose a monomial consisting of the coefficient found in step 2, the variables found in step 3, we get: $3a$ - this will be the common factor. Then:

$3a^3-(15a)^2b+4(5ab)^2=3a(a^2-5ab+15b^2)$

Within the framework of the study of identity transformations, the topic of taking the common factor out of brackets is very important. In this article we will explain what exactly such a transformation is, derive the basic rule and analyze typical examples of problems.

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The concept of taking a factor out of brackets

To successfully apply this transformation, you need to know what expressions it is used for and what result should be obtained in the end. Let us clarify these points.

You can take the common factor out of brackets in expressions that represent sums in which each term is a product, and in each product there is one factor that is common (the same) for everyone. This is called the common factor. It is this that we will take out of the brackets. So, if we have works 5 3 And 5 4, then we can take the common factor 5 out of brackets.

What does this transformation consist of? During it, we represent the original expression as the product of a common factor and an expression in parentheses containing the sum of all original terms except the common factor.

Let's take the example given above. Let's add a common factor of 5 to 5 3 And 5 4 and we get 5 (3 + 4) . The final expression is the product of the common factor 5 by the expression in brackets, which is the sum of the original terms without 5.

This transformation is based on the distributive property of multiplication, which we have already studied before. In literal form it can be written as a (b + c) = a b + a c. By changing right side on the left, we will see a scheme for taking the common factor out of brackets.

The rule for taking the common factor out of brackets

Using everything said above, we derive the basic rule for such a transformation:

Definition 1

To remove the common factor from brackets, you need to write the original expression as the product of the common factor and brackets that include the original sum without the common factor.

Example 1

Let's take a simple example of rendering. We have a numeric expression 3 7 + 3 2 − 3 5, which is the sum of three terms 3 · 7, 3 · 2 and a common factor 3. Taking the rule we derived as a basis, we write the product as 3 (7 + 2 − 5). This is the result of our transformation. The entire solution looks like this: 3 7 + 3 2 − 3 5 = 3 (7 + 2 − 5).

We can put the factor out of brackets not only in numerical, but also in literal expressions. For example, in 3 x − 7 x + 2 you can take out the variable x and get 3 x − 7 x + 2 = x (3 − 7) + 2, in the expression (x 2 + y) x y − (x 2 + y) x 3– common factor (x2+y) and get in the end (x 2 + y) · (x · y − x 3).

It is not always possible to immediately determine which factor is common. Sometimes an expression must first be transformed by replacing numbers and expressions with identically equal products.

Example 2

So, for example, in the expression 6 x + 4 y it is possible to derive a common factor 2 that is not written down explicitly. To find it, we need to transform the original expression, representing six as 2 · 3 and four as 2 · 2. That is 6 x + 4 y = 2 3 x + 2 2 y = 2 (3 x + 2 y). Or in expression x 3 + x 2 + 3 x we can take out of brackets the common factor x, which is revealed after the replacement x 3 on x · x 2 . This transformation is possible due to the basic properties of the degree. As a result, we get the expression x (x 2 + x + 3).

Another case that should be discussed separately is the removal of a minus from brackets. Then we take out not the sign itself, but minus one. For example, let us transform the expression in this way − 5 − 12 x + 4 x y. Let's rewrite the expression as (− 1) 5 + (− 1) 12 x − (− 1) 4 x y, so that the overall multiplier is more clearly visible. Let's take it out of brackets and get − (5 + 12 · x − 4 · x · y) . This example shows that in brackets the same amount is obtained, but with opposite signs.

In conclusions, we note that transformation by placing the common factor out of brackets is very often used in practice, for example, to calculate the value of rational expressions. This method is also useful when you need to represent an expression as a product, for example, to factor a polynomial into individual factors.

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Math lesson in 7th grade

8. Lesson objectives

For the teacher:

educational

organize educational activities:

By mastering the algorithm for taking the common factor out of brackets and understanding the logic of its construction;

To develop the ability to apply the algorithm for taking the common factor out of brackets

developing

create conditions for the development of regulatory skills:

Determine your own goals educational activities;

Plan ways to achieve goals;

Correlate your actions with the planned results;

Monitor and evaluate educational activities based on results;

Organize educational cooperation and joint activities with the teacher and peers.

- educational

Create conditions for the formation of a responsible attitude towards learning;

Create conditions for the development of students’ independence in organizing and carrying out their educational activities.

Create conditions for patriotic education

Create conditions for environmental education

For students:

Master the algorithm for taking the common factor out of brackets and understanding the logic of its construction;

Develop the ability to apply the algorithm for taking the common factor out of brackets

9. UUDs used: regulatory (goal setting, activity planning, control and evaluation)

10.Lesson type: learning new material

11.Forms of student work: frontal, steam room, individual

12. Necessarytechnical equipment: computer, projector, lesson logo, mathematics textbooks, electronic presentation made in Power program Point, handout

Lesson structure and flow

Definition 1

First let's remember Rules for multiplying a monomial by a monomial:

To multiply a monomial by a monomial, you must first multiply the coefficients of the monomials, then, using the rule of multiplying powers with the same base, multiply the variables included in the monomials.

Example 1

Find the product of the monomials $(2x)^3y^2z$ and $(\frac(3)(4)x)^2y^4$

Solution:

First, let's calculate the product of the coefficients

$2\cdot\frac(3)(4) =\frac(2\cdot 3)(4)$ in this task we used the rule for multiplying a number by a fraction - to multiply a whole number by a fraction, you need to multiply the number by the numerator of the fraction, and the denominator put without changes

Now let's use the main property of a fraction - the numerator and denominator of a fraction can be divided by the same number, different from $0$. Let's divide the numerator and denominator of this fraction by $2$, that is, reduce this fraction by $2$ $2\cdot\frac(3)(4)$ =$\frac(2\cdot 3)(4)=\\frac(3 )(2)$

The resulting result turned out to be an improper fraction, that is, one in which the numerator is greater than the denominator.

Let's transform this fraction by isolating the whole part. Let us remember that to isolate an integer part, it is necessary to write down the incomplete quotient obtained by dividing the numerator by the denominator, as an integer part, the remainder of the division into the numerator of the fractional part, and the divisor into the denominator.

We found the coefficient of the future product.

Now we will sequentially multiply the variables $x^3\cdot x^2=x^5$,

$y^2\cdot y^4 =y^6$. Here we used the rule for multiplying powers with the same base: $a^m\cdot a^n=a^(m+n)$

Then the result of multiplying monomials will be:

$(2x)^3y^2z \cdot (\frac(3)(4)x)^2y^4=1\frac(1)(2)x^5y^6$.

Then, based on this rule, you can perform the following task:

Example 2

Represent a given polynomial as the product of a polynomial and a monomial $(4x)^3y+8x^2$

Let us represent each of the monomials included in the polynomial as the product of two monomials in order to isolate a common monomial, which will be a factor in both the first and second monomials.

First, let's start with the first monomial $(4x)^3y$. Let's factorize its coefficient into simple factors: $4=2\cdot 2$. We will do the same with the coefficient of the second monomial $8=2\cdot 2 \cdot 2$. Note that two factors $2\cdot 2$ are included in both the first and second coefficients, which means $2\cdot 2=4$ - this number will be included in the general monomial as a coefficient

Now let us note that in the first monomial there is $x^3$, and in the second there is the same variable to the power of $2:x^2$. This means that it is convenient to represent the variable $x^3$ like this:

The variable $y$ is included in only one term of the polynomial, which means it cannot be included in the general monomial.

Let's imagine the first and second monomial included in the polynomial as a product:

$(4x)^3y=4x^2\cdot xy$

$8x^2=4x^2\cdot 2$

Note that the common monomial, which will be a factor in both the first and second monomial, is $4x^2$.

$(4x)^3y+8x^2=4x^2\cdot xy + 4x^2\cdot 2$

Now we apply the distributive law of multiplication, then the resulting expression can be represented as a product of two factors. One of the multipliers will be the total multiplier: $4x^2$ and the other will be the sum of the remaining multipliers: $xy + 2$. Means:

$(4x)^3y+8х^2 = 4x^2\cdot xy + 4x^2\cdot 2 = 4x^2(xy+2)$

This method is called factorization using a common factor.

The common factor in this case was the monomial $4x^2$.

Algorithm

Note 1

Find the greatest common divisor of the coefficients of all monomials included in the polynomial - it will be the coefficient of the common factor-monomial, which we will put out of brackets

A monomial consisting of the coefficient found in paragraph 2 and the variables found in paragraph 3 will be a common factor. which can be taken out of brackets as a common factor.

Example 3

Take out the common factor $3a^3-(15a)^2b+4(5ab)^2$

Solution:

Let’s find the gcd of the coefficients; for this we will decompose the coefficients into simple factors

$45=3\cdot 3\cdot 5$

And we find the product of those included in the expansion of each:

Identify the variables that make up each monomial and select the variable with the smallest exponent

$a^3=a^2\cdot a$

The variable $b$ is included only in the second and third monomial, which means it will not be included in the common factor.

Let's compose a monomial consisting of the coefficient found in step 2, the variables found in step 3, we get: $3a$ - this will be the common factor. Then:

$3a^3-(15a)^2b+4(5ab)^2=3a(a^2-5ab+15b^2)$

>>Math: Taking the common factor out of brackets

Before starting to study this section, return to § 15. There we have already looked at an example in which it was required to present polynomial as the product of a polynomial and a monomial. We have established that this problem is not always correct. If, nevertheless, such a product was able to be compiled, then they usually say that the polynomial is factorized using general judgment common factor out of brackets. Let's look at a few examples.

Example 1. Factor the polynomial:

A) 2x + 6y, c) 4a 3 + 6a 2; e) 5a 4 - 10a 3 + 15a 8.
b) a 3 + a 2; d) 12ab 4 - 18a 2 b 3 c;

Solution.
a) 2x + 6y = 2 (x + 3). The common divisor of the coefficients of the polynomial terms has been taken out of brackets.

b) a 3 + a 2 = a 2 (a + 1). If the same variable is included in all terms of the polynomial, then it can be taken out of brackets to a degree equal to the smallest of the available ones (i.e., choose the smallest of the available exponents).

c) Here we use the same technique as when solving examples a) and b): for coefficients we find the common divisor (in this case the number 2), for variables - the smallest degree from those available (in this case a 2). We get:

4a 3 + 6a 2 = 2a 2 2a + 2a 2 3 = 2a 2 (2a + 3).

d) Usually for integer coefficients they try to find not just a common divisor, but a greatest common divisor. For coefficients 12 and 18, it will be the number 6. We note that the variable a is included in both terms of the polynomial, with the smallest exponent being 1. The variable b is also included in both terms of the polynomial, with the smallest exponent being 3. Finally, the variable c is included only in the second term of the polynomial is not included in the first term, which means that this variable cannot be taken out of brackets to any degree. As a result we have:

12ab 4 - 18a 2 b 3 c = 6ab 3 2b - 6ab 3 Zas = 6ab 3 (2b - Zas).

e) 5a 4 -10a 3 +15a 8 = 5a 3 (a-2 + For 2).

In fact, in this example we developed the following algorithm.

Comment . In some cases, it is useful to take out the fractional coefficient as a general factor.

For example:

Example 2. Factorize:

X 4 y 3 -2x 3 y 2 + 5x 2.

Solution. Let's use the formulated algorithm.

1) The greatest common divisor of the coefficients -1, -2 and 5 is 1.
2) The variable x is included in all terms of the polynomial with exponents 4, 3, 2, respectively; therefore, x 2 can be taken out of brackets.
3) The variable y is not included in all terms of the polynomial; This means that it cannot be taken out of the brackets.

Conclusion: x 2 can be taken out of brackets. True, in this case it makes more sense to put -x 2 out of brackets.

We get:
-x 4 y 3 -2x 3 y 2 + 5x 2 = - x 2 (x 2 y 3 + 2xy 2 - 5).

Example 3. Is it possible to divide the polynomial 5a 4 - 10a 3 + 15a 5 into the monomial 5a 3? If yes, then execute division.

Solution. In example 1d) we got that

5a 4 - 10a 3 + 15a 8 - 5a 3 (a - 2 + For 2).

This means that the given polynomial can be divided by 5a 3, and the quotient will be a - 2 + For 2.

Similar examples we discussed in § 18; Please look through them again, but this time from the point of view of taking the common factor out of brackets.

Factoring a polynomial by taking the common factor out of brackets is closely related to two operations that we studied in § 15 and 18 - multiplying a polynomial by a monomial and dividing a polynomial by monomial.

Now let’s somewhat expand our ideas about taking the common factor out of brackets. The thing is that sometimes algebraic expression is given in such a way that the common factor can be not a monomial, but the sum of several monomials.

Example 4. Factorize:

2x(x-2) + 5(x-2) 2 .

Solution. Let's introduce a new variable y = x - 2. Then we get:

2x (x - 2) + 5 (x - 2) 2 = 2xy + 5y 2.

We note that the variable y can be taken out of brackets:

2xy + 5y 2 - y (2x + 5y). Now let's return to the old notation:

y(2x + 5y) = (x- 2)(2x + 5(x - 2)) = (x - 2)(2x + 5x-10) = (x-2)(7x:-10).

In such cases, after gaining some experience, you can not introduce a new variable, but use the following

2x(x - 2) + 5(x - 2) 2 = (x - 2)(2x + 5(x - 2))= (x - 2)(2x + 5x~ 10) = (x - 2)( 7x - 10).

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