Bracketing the common factor, rule, examples. Factoring polynomials. Taking the common factor out of brackets
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.
Yandex.RTB R-A-339285-1
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 you want to get 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 parentheses, 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 the right side with the left, we will see a scheme for taking the common factor out of brackets.
The rule for putting 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.
If you notice an error in the text, please highlight it and press Ctrl+Enter
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.
In this lesson, we will get acquainted with the rules for bracketing the common factor and learn how to find it in various examples and expressions. Let's talk about how a simple operation, taking the common factor out of brackets, allows you to simplify calculations. We will consolidate the acquired knowledge and skills by looking at examples of various complexities.
What is a common factor, why look for it and for what purpose is it taken out of brackets? Let's answer these questions by looking at a simple example.
Let's solve the equation. The left side of the equation is a polynomial consisting of similar terms. The letter part is common to these terms, which means it will be the common factor. Let's put it out of brackets:
In this case, taking the common factor out of brackets helped us convert the polynomial to a monomial. Thus, we were able to simplify the polynomial and its transformation helped us solve the equation.
In the example considered, the common factor was obvious, but would it be so easy to find it in an arbitrary polynomial?
Let's find the meaning of the expression: .
In this example, putting the common factor out of brackets greatly simplified the calculation.
Let's solve one more example. Let's prove divisibility into expressions.
The resulting expression is divisible by , as required to be proved. Once again, taking the common factor allowed us to solve the problem.
Let's solve one more example. Let us prove that the expression is divisible by for any natural number: 
.
The expression is the product of two adjacent natural numbers. One of the two numbers will definitely be even, which means the expression will be divisible by .
We looked at different examples, but we used the same solution method: we took the common factor out of brackets. We see that this simple operation greatly simplifies the calculations. It was easy to find a common factor for these special cases, but what to do in the general case, for an arbitrary polynomial?
Recall that a polynomial is a sum of monomials.
Consider the polynomial 
. This polynomial is the sum of two monomials. A monomial is the product of a number, a coefficient, and a letter part. Thus, in our polynomial, each monomial is represented by the product of a number and powers, the product of factors. The factors can be the same for all monomials. It is these factors that need to be determined and taken out of the bracket. First, we find the common factor for the coefficients, which are integer ones.
It was easy to find the common factor, but let's define the gcd of the coefficients: 
.
Let's look at another example: .
Let's find , which will allow us to determine the common factor for this expression: .
We have derived a rule for integer coefficients. You need to find their gcd and take it out of the bracket. Let's consolidate this rule by solving one more example.
We have looked at the rule for assigning a common factor for integer coefficients, let's move on to the letter part. First, we look for those letters that are included in all monomials, and then we determine the highest degree of the letter that is included in all monomials: .
In this example there was only one common letter variable, but there can be several, as in the following example:
Let's complicate the example by increasing the number of monomials:
After taking out the common factor, we converted the algebraic sum into a product.
We looked at the subtraction rules for integer coefficients and letter variables separately, but most often you need to apply them together to solve the example. Let's look at an example:
Sometimes it can be difficult to determine which expression is left in parentheses, let's look at an easy trick that will allow you to quickly solve this problem.
The common factor can also be the desired value:
The common factor can be not only a number or a monomial, but also any expression, such as in the following equation.
Math lesson in 7th grade
| 1. | Full name (full name) | Trofimenko Nadezhda Pavlovna |
| 2. | Place of work | Municipal educational institution "Miloslavskaya school" |
| 3. | Job title | Mathematic teacher |
| 4. | Item | |
| 5. | Class | |
| 6. | Topic and lesson number in the topic | Taking the common factor out of brackets (1 lesson per topic) |
| 7. | Basic tutorial | Yu.M. Kolyagin, M.V. Tkacheva, N.E. Fedorova, M.I. Shabunin. "Algebra 7th grade" textbook for general education organizations. M. Prosveshchenie. 2016. |
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:
Independently determine the goals of 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 Point, handouts
Lesson structure and flow
| Lesson steps | Teacher activities | Student activities | Educational | ||
| Organizational | Hello guys! I'm very glad to see you! Our lesson motto: I hear and forget. Let's give our lesson an unusual coloring (the emblem of a green tree and a red heart), the emblem on the board. At the end of the lesson we will reveal the secret of this emblem | They check the workplace, greet the teacher, and get into the working rhythm of the lesson. | |||
| Updating knowledge and motivation | Today in class you will learn new material. But first, let's work verbally. 1.Multiply monomials: 2a 2 *3av; 2av*(-a 4) ; 6x 2 *(-2x); -3s*5x; -3x*(-xy 2);-4a 2 b*(-0.2av 2) If the answer is correct, open the first letter 2) Which monomials should be put in place of * to get the correct equality: x 3 * = x 6; - a 6 = a 4 *; *y 7 = y 8; -2a 3 * = 8a 5 ; 5xy 4 * = 25x 2 y 6. If the answer is correct, open the second letter 3) Introduce a monomial 12x 3 at 4 as a product of two factors, one of which is equal 2x 3 ; 3u 3 ; -4x ; 6xy ; -2x 3 at ; 6x 2 at 2 . If the answer is correct, the third letter is revealed 4) Present the monomial in different ways 6x 2 at as a product of two factors. Open the 4th letter 5) The student multiplied a monomial by a polynomial, after which the monomial was erased. Restore it …*(x – y) = 3ax – 3ay …*(-x + y 2 – 1) = xy 2 – y 4 + y …*(a +b – 1) = 2ah +2in – 2x …*(a – b) = a 2 c – a 3 …*(2у 2 – 3) = 10у 4 – 15у 2. Open the 5th letter 6.Calculate 768*95 – 668*95 = 76,8*9,5 + 23,2*9,5 = Open the 6th letter. The letters formed the name of a German mathematician. | Perform the task orally Comment on the solution using the rules Open the letters on the board Student (received the task in advance) Historical reference : Michel Stiefel (1487-1567), German mathematician and itinerant preacher; author of the book “Complete Arithmetic”, he introduced the term “exponent”, and also considered the properties of polynomials and made a significant contribution to the development of algebra. (photo) | |||
| 3. Goal setting and motivation | Providing motivation for children to learn and their acceptance of lesson goals. | On the board: Find expression value A 2 – 3av at a = 106.45; in = 2.15 . How to do it? a) You can substitute numerical values A And V and find the meaning of the expression, but it's difficult. c) Is it possible to do otherwise? How? On the board we write down the topic of the lesson: “Putting the common factor out of brackets.” Guys, write carefully! Remember that to produce a ton of paper, you need to cut down about 17 mature trees. Let's try to set lesson goals according to the following scheme: What concepts will you become familiar with? What skills and abilities will we master? | Offer their own solutions | ||
| 4. Assimilation of new knowledge and methods of assimilation (initial acquaintance with the material) | Ensuring children’s perception, comprehension and primary memorization of the topic studied | Open the textbook pp. 120-121, read and answer the questions on pp. 121. Highlight the points of the algorithm Algorithm for taking the common factor out of brackets Find the common factor of the coefficients of polynomials Bring it out of the bracket 3.Teacher: I will give an example of taking a multiplier out of brackets in Russian. In the expression “Take a book, take a pen, take a notebook,” the function of a common factor is performed by the verb “take,” and the book, notebook and pen are complements. 4 I wrote the rule for multiplying a monomial by a polynomial in the form of a diagram. Try to draw a schematic rule for subtracting a common factor | Read the material Answer questions Find a sheet with an algorithm
Oh, now you try: Draw a reverse diagram on the board | ||
| 5. Relaxation | Includes the cartoon "Summer Assignment" From winter weather we find ourselves in warm summer. But the fragment is instructive, try to catch the main idea | They watch a fragment of a cartoon and draw conclusions about the beauty of their native land | Cartoon fragment "Summer Assignment" | ||
| 6.Primary consolidation | Establishing the correctness and awareness of studying the topic. Identifying gaps in the initial understanding of the studied material, correcting the identified gaps, ensuring that the knowledge and methods of action that they need to independently work on new material are consolidated in children’s memory. | Frontal to the board: № 318, 319, 320,321,324,325,328 Take turns, as desired | Solve at the board with comments | ||
| 6. Organization of primary control | Identification of the quality and level of assimilation of knowledge and methods of action, as well as identification of shortcomings in knowledge and methods of action, establishing the causes of identified shortcomings | Solve independently based on the text on pieces of paper and check the answers on the board: INDEPENDENT WORK (differentiated) 1 option Complete the factorization of the polynomial: 5akh – 30ау = 5а(…………..) x 4 – 5x 3 – x 2 = x 2 (…………..) Factor the polynomial - 5ав + 15а 2 в, taking the factor out of brackets: a) 5а; b) -5a. Factor it out: 5x + 5y = 7av + 14ac = 20a – 4b= 5mn – 5= ah – ay= 3x 2 – 6x= 2a – 10ау= 15a 2 + 5a 3 = 2 option Finish the entry: 18av +16v= 2v(…………) 4a 2 s – 8ac= 4ac(………..) Factor the polynomial -15a 2 in + 5ab 4 in two ways: a) taking the factor 5ab out of brackets; b) taking the factor -5av out of brackets. 5х+6ху= 2ав – 3а 3 в= 12av – 9v= x 3 -4x 2 +6x= 6a 4 – 4a 2 = 4a 4 -8a 3 +12a 2 = 24x 2 y -12xy= 9v 2 -6v 4 +3v= 4. Find the value of the expression by factoring it: xy 2 +y 3 with x=97, y=3. Option 3 Take the common factor out of brackets and check by multiplying the monomial by the polynomial: a) 12xy+ 18x= b) 36ab 2 – 12a 2 c= 2. Finish the recording: 18a 3 in 2 +36av = 18av(…………) 18a 3 in 2 +36av = -18av(…………) 3. Take the common factor out of brackets: 12a 2 +16a= -11x 2 y 2 +22xy= 2a 4 -6a 2 = -12a 3 in 3 +6av= 30a 4 b- 6av 4 = x 8 -8x 4 + x 2 = 4. Replace M with a polynomial or monomial so that the resulting equality is the identity: 12a 2 b-8av 2 +6av=M*(6a-4b+3) 15x 2 y-10x3y2+25x 4 y 3 =5x 2 y*M 5. Find the meaning of the expression: a) 2.76a-ab at a=1.25 and b=0.76; b) 2xy + 2y 2 at x=0.27 and b=0.73. | They do their work, after completion they receive the keys and check, put + or minus, evaluate their work according to the criteria on the board: (answers on the board) 10-12 points - “5” 8-9 points - “4” 6-7 points - “3” Less than 6 - you need to work more. | Differentiated task sheets | |
| 7. Summing up the lesson. | Provide a qualitative assessment of the work of the class and individual students | Mark actively working students and summarize the results of independent work: Raise your hands who has 5,4,3. | Analyze their work | ||
| 8. Information about homework | Ensuring that children understand the purpose, content and methods of completing homework. | Paragraph No. 19 We do it according to sample assignments in class work | Record tasks in a diary | ||
| 9. Reflection | Teacher: It was a lesson - a search. You and I looked for common ground with each other, learned to communicate, and also revealed one of the methods of explaining and consolidating the topic. Let's return to the lesson goals and analyze how we achieved them Oh, what else did we talk about, besides taking the common factor out of brackets? Let's return to the lesson logo. | Read out the goals and analyze their implementation On the connection between mathematics and the Russian language, About the beauty of our native land, about ecology |
In this lesson, we will get acquainted with the rules for bracketing the common factor and learn how to find it in various examples and expressions. Let's talk about how a simple operation, taking the common factor out of brackets, allows you to simplify calculations. We will consolidate the acquired knowledge and skills by looking at examples of various complexities.
What is a common factor, why look for it and for what purpose is it taken out of brackets? Let's answer these questions by looking at a simple example.
Let's solve the equation. The left side of the equation is a polynomial consisting of similar terms. The letter part is common to these terms, which means it will be the common factor. Let's put it out of brackets:
In this case, taking the common factor out of brackets helped us convert the polynomial to a monomial. Thus, we were able to simplify the polynomial and its transformation helped us solve the equation.
In the example considered, the common factor was obvious, but would it be so easy to find it in an arbitrary polynomial?
Let's find the meaning of the expression: .
In this example, putting the common factor out of brackets greatly simplified the calculation.
Let's solve one more example. Let's prove divisibility into expressions.
The resulting expression is divisible by , as required to be proved. Once again, taking the common factor allowed us to solve the problem.
Let's solve one more example. Let us prove that the expression is divisible by for any natural number: .
The expression is the product of two adjacent natural numbers. One of the two numbers will definitely be even, which means the expression will be divisible by .
We looked at different examples, but we used the same solution method: we took the common factor out of brackets. We see that this simple operation greatly simplifies the calculations. It was easy to find a common factor for these special cases, but what to do in the general case, for an arbitrary polynomial?
Recall that a polynomial is a sum of monomials.
Consider the polynomial . This polynomial is the sum of two monomials. A monomial is the product of a number, a coefficient, and a letter part. Thus, in our polynomial, each monomial is represented by the product of a number and powers, the product of factors. The factors can be the same for all monomials. It is these factors that need to be determined and taken out of the bracket. First, we find the common factor for the coefficients, which are integer ones.
It was easy to find the common factor, but let's define the gcd of the coefficients: .
Let's look at another example: .
Let's find , which will allow us to determine the common factor for this expression: .
We have derived a rule for integer coefficients. You need to find their gcd and take it out of the bracket. Let's consolidate this rule by solving one more example.
We have looked at the rule for assigning a common factor for integer coefficients, let's move on to the letter part. First, we look for those letters that are included in all monomials, and then we determine the highest degree of the letter that is included in all monomials: .
In this example there was only one common letter variable, but there can be several, as in the following example:
Let's complicate the example by increasing the number of monomials:
After taking out the common factor, we converted the algebraic sum into a product.
We looked at the subtraction rules for integer coefficients and letter variables separately, but most often you need to apply them together to solve the example. Let's look at an example:
Sometimes it can be difficult to determine which expression is left in parentheses, let's look at an easy trick that will allow you to quickly solve this problem.
The common factor can also be the desired value:
The common factor can be not only a number or a monomial, but also any expression, such as in the following equation.
