Rules for the difference of numbers with bracketing. Taking the common factor out of brackets


\(5x+xy\) can be represented as \(x(5+y)\). These are indeed identical expressions, we can verify this if we open the brackets: \(x(5+y)=x \cdot 5+x \cdot y=5x+xy\). As you can see, as a result we get the original expression. This means that \(5x+xy\) is indeed equal to \(x(5+y)\). By the way, this reliable way to check the correctness of the common factors - open the resulting bracket and compare the result with the original expression.


The main rule for bracketing:

For example, in the expression \(3ab+5bc-abc\) only \(b\) can be taken out of the bracket, because it is the only one that is present in all three terms. The process of taking common factors out of brackets is shown in the diagram below:

Bracketing rules

    In mathematics, it is customary to take out all common factors at once.

    Example:\(3xy-3xz=3x(y-z)\)
    Please note that here we could expand like this: \(3(xy-xz)\) or like this: \(x(3y-3z)\). However, these would be incomplete decompositions. Both the C and the X must be taken out.

    Sometimes the common members are not immediately visible.

    Example:\(10x-15y=2·5·x-3·5·y=5(2x-3y)\)
    In this case, the common term (five) was hidden. However, having expanded \(10\) as \(2\) multiplied by \(5\), and \(15\) as \(3\) multiplied by \(5\) - we “pulled the five into the light of God”, after which they were easily able to take it out of the bracket.

    If a monomial is removed completely, one remains from it.

    Example: \(5xy+axy-x=x(5y+ay-1)\)
    We put \(x\) out of brackets, and the third monomial consists only of x. Why does one remain from him? Because if any expression is multiplied by one, it will not change. That is, this same \(x\) can be represented as \(1\cdot x\). Then we have the following chain of transformations:

    \(5xy+axy-\)\(x\) \(=5xy+axy-\)\(1 \cdot x\) \(=\)\(x\) \((5y+ay-\)\ (1\) \()\)

    Moreover, this is the only The right way removal, because if we do not leave one, then when we open the brackets we will not return to the original expression. Indeed, if we do the extraction like this \(5xy+axy-x=x(5y+ay)\), then when expanded we will get \(x(5y+ay)=5xy+axy\). The third member is missing. This means that such a statement is incorrect.

    You can place a minus sign outside the bracket, and the signs of the terms in the bracket are reversed.

    Example:\(x-y=-(-x+y)=-(y-x)\)
    Essentially, here we are putting out the “minus one”, which can be “selected” in front of any monomial, even if there was no minus in front of it. We use here the fact that one can be written as \((-1) \cdot (-1)\). Here is the same example, described in detail:

    \(x-y=\)
    \(=1·x+(-1)·y=\)
    \(=(-1)·(-1)·x+(-1)·y=\)
    \(=(-1)·((-1)·x+y)=\)
    \(=-(-x+y)=\)
    \(-(y-x)\)

    A parenthesis can also be a common factor.

    Example:\(3m(n-5)+2(n-5)=(n-5)(3m+2)\)
    We most often encounter this situation (removing brackets from brackets) when factoring using the grouping method or

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 basic 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 remainder of the division into the numerator of the fractional part, 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 by taking out 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 that are 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:

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

Chichaeva Darina 8th grade

In the work, an 8th grade student described the rule for factoring a polynomial by putting the common factor out of brackets with a detailed procedure for solving many examples on this topic. For each example discussed, 2 examples are offered for independent decision, to which there are answers. The work will help study this topic for those students who, for some reason, did not master it when passing the 7th grade program material and (or) when repeating the algebra course in 8th grade after the summer holidays.

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Municipal budgetary educational institution

secondary school No. 32

"UNESCO Associated School "Eureka Development"

Volzhsky, Volgograd region

Work completed:

8B class student

Chichaeva Darina

Volzhsky

2014

Taking the common factor out of brackets

  • - One way to factor a polynomial isputting the common factor out of brackets;
  • - When taking the general multiplier out of brackets, it is applieddistributive property;
  • - If all terms of a polynomial contain common factor then this factor can be taken out of brackets.

When solving equations, in calculations and a number of other problems, it can be useful to replace a polynomial with the product of several polynomials (which may include monomials). Representing a polynomial as a product of two or more polynomials is called factoring the polynomial.

Consider the polynomial 6a 2 b+15b 2 . Each of its terms can be replaced by the product of two factors, one of which is equal to 3b: →6a 2 b = 3b*2a 2 , + 15b 2 = 3b*5b →from this we get: 6a 2 b+15b 2 =3b*2a 2 +3b*5b.

The resulting expression based on the distribution property of multiplication can be represented as a product of two factors. One of them is the common multiplier 3b , and the other is the sum 2a 2 and 5b→ 3b*2a 2 +3b*5b=3b(2a 2 +5b) →Thus, we expanded the polynomial: 6a 2 b+15b 2 into factors, representing it as a product of a monomial 3b and the polynomial 2a 2 +5b. This method of factoring a polynomial is called taking the common factor out of brackets.

Examples:

Factor it out:

A) kx-px.

Multiplier x x we put it out of brackets.

kx:x=k; px:x=p.

We get: kx-px=x*(k-p).

b) 4a-4b.

Multiplier 4 exists in both the 1st term and the 2nd term. That's why 4 we put it out of brackets.

4a:4=a; 4b:4=b.

We get: 4a-4b=4*(a-b).

c) -9m-27n.

9m and -27n are divisible by -9 . Therefore, we take the numerical factor out of brackets-9.

9m: (-9)=m; -27n: (-9)=3n.

We have: -9m-27n=-9*(m+3n).

d) 5y 2 -15y.

5 and 15 are divisible by 5; y 2 and y are divided by y.

Therefore, we take the common factor out of brackets 5у.

5y 2 : 5y=y; -15y: 5y=-3.

So: 5y 2 -15y=5y*(y-3).

Comment: From two degrees with the same base, we take out the degree with the smaller exponent.

e) 16у 3 +12у 2.

16 and 12 are divisible by 4; y 3 and y 2 are divided by y 2.

So the common factor 4y 2 .

16y 3 : 4y 2 =4y; 12y 2 : 4y 2 =3.

As a result we get: 16y 3 +12y 2 =4y 2 *(4y+3).

f) Factor the polynomial 8b(7y+a)+n(7y+a).

IN this expression we see the same factor is present(7y+a) , which can be taken out of brackets. So, we get:8b(7y+a)+n(7y+a)=(8b+n)*(7y+a).

g) a(b-c)+d(c-b).

Expressions b-c and c-b are opposite. Therefore, to make them the same, before d change the “+” sign to “-”:

a(b-c)+d(c-b)=a(b-c)-d(b-c).

a(b-c)+d(c-b)=a(b-c)-d(b-c)=(b-c)*(a-d).

Examples for independent solutions:

  1. mx+my;
  2. ah+ay;
  3. 5x+5y ;
  4. 12x+48y;
  5. 7ax+7bx;
  6. 14x+21y;
  7. –ma-a;
  8. 8mn-4m2;
  9. -12y 4 -16y;
  10. 15y 3 -30y 2 ;
  11. 5c(y-2c)+y 2 (y-2c);
  12. 8m(a-3)+n(a-3);
  13. x(y-5)-y(5-y);
  14. 3a(2x-7)+5b(7-2x);

Answers.

1) m(x+y); 2) a(x+y); 3) 5(x+y); 4) 12(x+4y); 5) 7х(a+b); 6) 7(2x+3y); 7) -а(m+1); 8) 4m(2n-m);

9) -4y(3y 3 +4); 10) 15у 2 (у-2); 11) (y-2c)(5c+y 2); 12) (a-3)(8m+n); 13) (y-5)(x+y); 14) (2x-7)(3a-5b).

Algebra lesson in 7th grade.

Topic: “Putting the common factor out of brackets.”

Textbook Makarychev Yu.N., Mindyuk N.G. and etc.

Lesson objectives:

Educational –

    identify the level of students’ mastery of a complex of knowledge and skills in the use of multiplication and division skills;

    develop the ability to apply the factorization of a polynomial by placing the common factor out of brackets;

    apply the removal of the common factor from brackets when solving equations.

Developmental –

    promote the development of observation, the ability to analyze, compare, and draw conclusions;

    develop self-control skills when performing tasks.

Educational -

    fostering responsibility, activity, independence, objective self-esteem.

Lesson type: combined.

Key learning outcomes:

    be able to take the common factor out of brackets;

    be able to apply this method when solving exercises.

Movelesson.

1 module (30 min).

1. Organizing time.

    greetings;

    preparing students for work.

2. Examination homework.

    Checking availability (on duty), discussing issues that have arisen.

3 . Updating basic knowledge.

    N Find GCD (15,6), (30,60), (24,8), (4,3), (20,55), (16, 12).

    What is GCD?

How is division of powers with the same bases performed?

How is multiplication of powers with the same bases performed?

For these degrees (c 3) 7 ,b 45 ,c 5 , a 21 , a 11 b 7 ,d 5 Name the degree with the smallest exponent, the same bases, the same exponents

Let us repeat the distributive law of multiplication. Write it down in letter form

a (b + c) = ab + ac

* - multiplication sign

Complete oral tasks on the application of the distributive property. (Prepare on the board).

1) 2*(a + b) 4) (x – 6)*5

2) 3*(x – y) 5) -4*(y + 5)

3) a*(4 + x) 6) -2*(c – a)

The tasks are written on a closed board, the guys solve and write the result on the board. Problems on multiplying a monomial by a polynomial.

To begin with, I offer you an example of multiplying a monomial by a polynomial:

2 x (x 2 +4 x y – 3) = 2x 3 + 8x 2 y – 6x Don’t wash!

Write the rule for multiplying a monomial by a polynomial in the form of a diagram.

A note appears on the board:

I can write this property as:

In this form we have already used the recording for simple way expression calculations.

a) 23 * 15 + 15 * 77 = (23 + 77) * 15 = 100 * 15 = 1500

The rest are oral, check the answers:

e) 55*682 – 45*682 = 6820

g) 7300*3 + 730*70 = 73000

h) 500*38 – 50*80 = 15000

What law helped you find a simple way to calculate? (Distribution)

Indeed, the distributive law helps to simplify expressions.

4 . Setting the goal and topic of the lesson. Verbal counting. Guess the topic of the lesson.

Work in pairs.

Cards for couples.

It turns out that factoring an expression is the inverse operation of term-by-term multiplication of a monomial by a polynomial.

Let's look at the same example that the student solved, but in reverse order. Factoring means taking the common factor out of brackets.

2 x 3 + 8 x 2 y – 6 x = 2 x (x 2 + 4 xy – 3).

Today in the lesson we will look at the concepts of factoring a polynomial and taking the common factor out of brackets, and we will learn to apply these concepts when doing exercises.

Algorithm for taking the common factor out of brackets

    The greatest common divisor of the coefficients.

    Same letter variables.

    Add the smallest degree to the removed variables.

    Then the remaining monomials of the polynomial are written in parentheses.

The greatest common divisor was found in the lower grades, the common variable to the least degree can be immediately seen. And in order to quickly find the polynomial remaining in brackets, you need to practice using number 657.

5. Primary learning with speaking out loud.

No. 657 (1 column)

Module 2 (30 min).

1. The result of the first 30 minutes.

A) What transformation is called factorization of a polynomial?

B) What property is based on taking the common factor out of brackets?

Q) How is the common factor taken out of brackets?

2. Primary consolidation.

Expressions are written on the board. Find errors in these equalities, if any, and correct them.

1) 2 x 3 – 3 x 2 – x = x (2 x 2 – 3 x).

2) 2 x + 6 = 2 (x + 3).

3) 8 x + 12 y = 4 (2 x - 3y).

4) a 6 – a 2 = a 2 (a 2 – 1).

5) 4 -2a = – 2 (2 – a).

3. Initial check of understanding.

Working with self-test. 2 people per back side

Take the common factor out of brackets:

Verbally check by multiplication.

4. Preparing students for general activities.

Let's take the polynomial factor out of brackets (teacher's explanation).

Factor the polynomial.

In this expression we see that there is one and the same factor, which can be taken out of brackets. So, we get:

The expressions and are opposite, so in some cases you can use this equality . We change the sign twice! Factor the polynomial

There are opposite expressions here and, using the previous identity, we get the following entry: .

And now we see that the common factor can be taken out of brackets.

In this article we will focus on taking the common factor out of brackets. First, let's figure out what this expression transformation consists of. Next, we will present the rule for placing the common factor out of brackets and consider in detail examples of its application.

Page navigation.

For example, the terms in the expression 6 x + 4 y have a common factor 2, which is not written down explicitly. It can be seen only after representing the number 6 as a product of 2·3, and 4 as a product of 2·2. So, 6 x+4 y=2 3 x+2 2 y=2 (3 x+2 y). Another example: in the expression x 3 +x 2 +3 x the terms have a common factor x, which becomes clearly visible after replacing x 3 with x x 2 (in this case we used) and x 2 with x x. After taking it out of brackets, we get x·(x 2 +x+3) .

Let’s separately say about putting the minus out of brackets. In fact, putting the minus out of the brackets means putting the minus one out of the brackets. For example, let’s take out the minus in the expression −5−12·x+4·x·y. The original expression can be rewritten as (−1) 5+(−1) 12 x−(−1) 4 x y, from where the common factor −1 is clearly visible, which we take out of the brackets. As a result, we arrive at the expression (−1)·(5+12·x−4·x·y) in which the coefficient −1 is replaced simply by a minus before the brackets, as a result we have −(5+12·x−4·x· y) . From here it is clearly seen that when the minus is taken out of brackets, the original sum remains in brackets, in which the signs of all its terms have been changed to the opposite.

In conclusion of this article, we note that bracketing the common factor is used very widely. For example, it can be used to more efficiently calculate the values ​​of numeric expressions. Also, putting a common factor out of brackets allows you to represent expressions in the form of a product; in particular, one of the methods for factoring a polynomial is based on bracketing out.

Bibliography.

  • Mathematics. 6th grade: educational. for general education institutions / [N. Ya. Vilenkin and others]. - 22nd ed., rev. - M.: Mnemosyne, 2008. - 288 p.: ill. ISBN 978-5-346-00897-2.


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