Taking the common factor out of brackets online calculator. Bracketing the common factor, rule, examples
>>Math: Taking the common factor out of brackets
Before starting to study this section, return to § 15. There we 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, it was possible to compose such a product, then they usually say that the polynomial is factorized by means of the general removal of the common factor from 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.
We looked at similar examples in § 18; Please look at 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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Lesson contentBasic principle
The distributive law of multiplication allows you to multiply a number by an amount (or an amount by a number). For example, to find the value of the expression 3 × (4 + 5), you can multiply the number 3 by each term in parentheses and add the results:
3 × (4 + 5) = 3 × 4 + 3 × 5 = 12 + 15
The number 3 and the expression in brackets can be swapped (this follows from the commutative law of multiplication). Then each term in parentheses will be multiplied by the number 3
(4 + 5) × 3 = 4 × 3 + 5 × 3 = 12 + 15
For now, we will not calculate the construction 3 × 4 + 3 × 5 and add the results obtained 12 and 15. Let us leave the expression in the form 3 (4 + 5) = 3 × 4 + 3 × 5. Below we will need it in exactly this form in order to understand the essence of taking the common factor out of brackets.
The distributive law of multiplication is sometimes called placing a factor inside parentheses. In the expression 3 × (4 + 5), the factor 3 was left out of brackets. By multiplying it by each term in the brackets, we essentially brought it inside the brackets. For clarity, you can write it this way, although it is not customary to write it this way:
3 (4 + 5) = (3 × 4 + 3 × 5)
Since in the expression 3 × (4 + 5) the number 3 is multiplied by each term in brackets, this number is a common factor for terms 4 and 5
As mentioned earlier, by multiplying this common factor by each term in the parentheses, we put it inside the parentheses. But the reverse process is also possible - the common factor can be taken back out of brackets. In this case, in the expression 3×4 + 3×5 the general multiplier is clearly visible - this is a multiplier of 3. It needs to be taken out of the equation. To do this, first write down the factor 3 itself
and next to it in parentheses the expression is written 3×4 + 3×5 but without the common factor 3, since it is taken out of brackets
3 (4 + 5)
As a result of taking the common factor out of brackets, we obtain the expression 3 (4 + 5) . This expression is identical to the previous expression 3×4 + 3×5
3(4 + 5) = 3 × 4 + 3 × 5
If we calculate both sides of the resulting equality, we obtain the identity:
3(4 + 5) = 3 × 4 + 3 × 5
27 = 27
How does the common factor go out of brackets?
Placing the common factor outside the brackets is essentially the reverse operation of putting the common factor inside the brackets.
If, when introducing a common factor inside brackets, we multiply this factor by each term in parentheses, then when moving this factor back outside the brackets, we must divide each term in parentheses by this factor.
In expression 3×4 + 3×5, which was discussed above, this is what happened. Each term was divided by a common factor of 3. The products 3 × 4 and 3 × 5 are terms, because if we calculate them, we get the sum 12 + 15
Now we can see in detail how the general factor is taken out of brackets:
It can be seen that the common factor 3 is first taken out of brackets, then in brackets each term is divided by this common factor.
Dividing each term by a common factor can be done not only by dividing the numerator by the denominator, as shown above, but also by reducing these fractions. In both cases you will get the same result:
We looked at the simplest example of taking a common factor out of brackets to understand the basic principle.
But not everything is as simple as it seems at first glance. After the number is multiplied by each term in parentheses, the results are added together, and the common factor is lost from view.
Let's return to our example 3 (4 + 5). Let's apply the distributive law of multiplication, that is, multiply the number 3 by each term in brackets and add the results:
3 × (4 + 5) = 3 × 4 + 3 × 5 = 12 + 15
After the construction 3 × 4 + 3 × 5 is calculated, we get the new expression 12 + 15. We see that the common factor of 3 has disappeared from view. Now, in the resulting expression 12 + 15, let’s try to take the common factor back out of brackets, but in order to take this common factor out, we first need to find it.
Usually, when solving problems, we encounter precisely such expressions in which the common factor must first be found before it can be taken out.
In order to take the common factor out of brackets in the expression 12 + 15, you need to find the greatest common factor (GCD) of terms 12 and 15. The found GCD will be the common factor.
So, let’s find the GCD of terms 12 and 15. Let us recall that to find the GCD it is necessary to decompose the original numbers into prime factors, then write out the first decomposition and remove from it the factors that are not included in the decomposition of the second number. The remaining factors need to be multiplied to obtain the desired gcd. If you have difficulty at this point, be sure to repeat.
The gcd of terms 12 and 15 is the number 3. This number is the common factor of terms 12 and 15. It should be taken out of brackets. To do this, we first write down the factor 3 itself and next to it in parentheses we write a new expression in which each term of the expression 12 + 15 is divided by a common factor 3
Well, further calculation is not difficult. The expression in parentheses is easy to calculate - twelve divided by three is four, A fifteen divided by three is five:
Thus, when taking the common factor out of brackets in the expression 12 + 15, the expression 3(4 + 5) is obtained. The detailed solution is as follows:
The short solution skips the notation showing how each term is divided by a common factor:
Example 2. 15 + 20
The greatest common divisor of the terms 15 and 20 is the number 5. This number is the common factor of the terms 15 and 20. Let’s take it out of brackets:
We got the expression 5(3 + 4).
The resulting expression 5(3 + 4) can be verified. To do this, just multiply the five by each term in brackets. If we did everything correctly, we should get the expression 15 + 20
Example 3. Take the common factor out of brackets in the expression 18 + 24 + 36
Let's find the gcd of terms 18, 24 and 36. To find , you need to factor these numbers into prime factors, then find the product of common factors:
The gcd of terms 18, 24 and 36 is the number 6. This number is the common factor of terms 18, 24 and 36. Let’s take it out of brackets:
Let's check the resulting expression. To do this, multiply the number 6 by each term in brackets. If we did everything correctly, we should get the expression 18 + 24 + 36
Example 4. Take the common factor out of brackets in the expression 13 + 5
The terms 13 and 5 are prime numbers. They decompose only into one and themselves:
This means that terms 13 and 5 have no common factors other than one. Accordingly, there is no point in putting this unit out of brackets, since it will not give anything. Let's show this:
Example 5. Take the common factor out of brackets in the expression 195 + 156 + 260
Let's find the gcd of terms 195, 156 and 260
The gcd of the terms 195, 156 and 260 is the number 13. This number is a common factor for the terms 195, 156 and 260. Let’s take it out of brackets:
Let's check the resulting expression. To do this, multiply 13 by each term in brackets. If we did everything correctly, we should get the expression 195 + 156 + 260
An expression in which you need to take the common factor out of brackets can be not only a sum of numbers, but also a difference. 16 − 12 − 4. The greatest common divisor of the numbers 16, 12 and 4 is the number 4. We will take this number out of brackets:
Let's check the resulting expression. To do this, multiply four by each number in brackets. If we did everything correctly, we should get the expression 16 − 12 − 4
Example 6. Take the common factor out of brackets in the expression 72 + 96 − 120
Let's find the gcd of numbers 72, 96 and 120
GCD for 72, 96 and 120 is the number 24. This number is the common factor of the terms 195, 156 and 260. Let’s take it out of brackets:
Let's check the resulting expression. To do this, multiply 24 by each number in brackets. If we did everything correctly, we should get the expression 72+96−120
The overall factor taken out of brackets can also be negative. For example, let's take the common factor out of brackets in the expression −6 − 3. There are two ways to take the common factor out of brackets in this expression. Let's look at each of them.
Method 1.
Let's replace subtraction with addition:
−6 + (−3)
Now we find the common factor. The common factor of this expression will be the greatest common divisor of the terms −6 and −3.
The modulus of the first term is 6. And the modulus of the second term is 3. GCD(6 and 3) is equal to 3. This number is the common factor of the terms 6 and 3. Let’s take it out of brackets:
The expression obtained in this way was not very accurate. Lots of parentheses and negative numbers do not make the expression simple. Therefore, you can use the second method, the essence of which is to put out of brackets not 3, but −3.
Method 2.
Just like last time, we replace subtraction with addition.
−6 + (−3)
This time we will take out of brackets not 3, but −3
The expression obtained this time looks much simpler. Let's write the solution shorter to make it even simpler:
Allowing a negative factor to be taken out of brackets is due to the fact that the expansion of the numbers −6 and (−3) can be written in two ways: first make the multiplicand negative and the multiplier positive:
−6 = −2 × 3
−3 = −1 × 3
in the second case, the multiplicand can be made positive and the multiplier negative:
−6 = 2 × (−3)
−3 = 1 × (−3)
This means we are free to put out of brackets the factor we want.
Example 8. Take the common factor out of brackets in the expression −20 − 16 − 2
Let's replace subtraction with addition
−20 − 16 − 2 = −20 + (−16) + (−2)
The greatest common divisor of the terms −20, −16, and −2 is the number 2. This number is the common factor of these terms. Let's see what it looks like:
−20 = −10 × 2
−16 = −8 × 2
−2 = −1 × 2
But the given expansions can be replaced by identically equal expansions. The difference will be that the common factor will not be 2, but −2
−20 = 10 × (−2)
−16 = 8 × (−2)
−2 = 1 × (−2)
Therefore, for convenience, we can put out of brackets not 2, but −2
Let's write down the above solution in short:
And if we took 2 out of brackets, we would get a not entirely accurate expression:
Example 9. Take the common factor out of brackets in the expression −30 − 36 − 42
Let's replace subtraction with addition:
−30 + (−36) + (−42)
The greatest common divisor of the terms −30, −36 and −42 is the number 6. This number is the common factor for these terms. But we will put out of brackets not 6, but −6, since the numbers −30, −36 and −42 can be represented as follows:
−30 = 5 × (−6)
−36 = 6 × (−6)
−42 = 7 × (−6)
Taking the minus out of brackets
When solving problems, it can sometimes be useful to put the minus sign out of brackets. This allows you to simplify the expression and make it simpler.
Consider the following example. Take the minus out of brackets in the expression −15 + (−5) + (−3)
For clarity, let’s enclose this expression in brackets, because we are talking about taking the minus out of these brackets
(−15 + (−5) + (−3))
So, to take the minus out of the brackets, you need to write the minus before the brackets and write all the terms in the brackets, but with opposite signs. We leave the operation signs (that is, pluses) unchanged:
−(15 + 5 + 3)
We took the minus out of the brackets in the expression −15 + (−5) + (−3) and got −(15 + 5 + 3) . Both expressions equal the same value −23
−15 + (−5) + (−3) = −23
−(15 + 5 + 3) = −(23) = −23
Therefore, we can put an equal sign between the expressions −15 + (−5) + (−3) and −(15 + 5 + 3), because they are equal to the same value:
−15 + (−5) + (−3) = −(15 + 5 + 3)
−23 = −23
In fact, when the minus is taken out of brackets, the distributive law of multiplication again works:
a(b + c) = ab + ac
If we swap the left and right sides of this identity, it turns out that the factor a bracketed
ab + ac = a(b+c)
The same thing happens when we take out the common factor in other expressions and when we take the minus out of brackets.
Obviously, when taking a minus out of brackets, it is not a minus that is taken out, but a minus one. We said earlier that it is customary not to write down coefficient 1.
Therefore, a minus is formed in front of the brackets, and the signs of the terms that were in the brackets change their sign to the opposite, since each term is divided by minus one.
Let's return to the previous example and see in detail how the minus was actually taken out of brackets
Example 2. Place the minus out of brackets in the expression −3 + 5 + 11
We put a minus and next to it in parentheses we write the expression −3 + 5 + 11 with the opposite sign for each term:
−3 + 5 + 11 = −(3 − 5 − 11)
As in the previous example, here it is not minus that is taken out of brackets, but minus one. The detailed solution is as follows:
At first we got the expression −1(3 + (−5) + (−11)), but we opened the inner brackets in it and got the expression −(3 − 5 − 11) . Expanding parentheses is the topic of the next lesson, so if this example is difficult for you, you can skip it for now.
Taking the common factor out of brackets in literal expression
Taking the common factor out of brackets in literal terms is much more interesting.
First, let's look at a simple example. Let there be an expression 3 a + 2 a. Let's take the common factor out of brackets.
In this case, the total multiplier is visible to the naked eye - this is the multiplier a. Let’s take it out of the brackets. To do this, we write down the multiplier itself a and next to it in parentheses we write the expression 3a + 2a, but without a multiplier a since it is taken out of brackets:
As in the case of a numerical expression, here each term is divided by the common factor taken out. It looks like this:
Variables in both fractions a were reduced by a. Instead, the numerator and denominator have units. The units were obtained due to the fact that instead of a variable a can be any number. This variable was located in both the numerator and denominator. And if the numerator and denominator have the same numbers, then the greatest common divisor for them will be this number itself.
For example, if instead of a variable a substitute the number 4
, then the construction will take the following form: 
. Then the fours in both fractions can be reduced by 4:
It turns out the same as before, when instead of fours there was a variable a .
Therefore, you should not be alarmed by the reduction of variables. A variable is a full-fledged multiplier, even if expressed by a letter. Such a multiplier can be taken out of brackets, reduced, and other actions that are permissible for ordinary numbers.
A literal expression contains not only numbers, but also letters (variables). Therefore, the common factor that is taken out of brackets is often a letter factor, consisting of a number and a letter (coefficient and variable). For example, the following expressions are literal factors:
3a, 6b, 7ab, a, b, c
Before taking such a factor out of brackets, you need to decide which number will be in the numerical part of the common factor and which variable will be in the letter part of the common factor. In other words, you need to find out what coefficient the common factor will have and what variable will be included in it.
Consider the expression 10 a+ 15a. Let's try to take the common factor out of brackets. First, let’s decide what the common factor will consist of, that is, we’ll find out its coefficient and what variable will be included in it.
The coefficient of the common multiplier must be the greatest common divisor of the coefficients of the literal expression 10 a+ 15a. 10 and 15, and their greatest common divisor is the number 5. This means that the number 5 will be the coefficient of the common factor taken out of brackets.
Now let’s decide which variable will be included in the common factor. To do this you need to look at expression 10 a+ 15a and find the letter factor that is included in all terms. In this case, it is a factor a. This factor is included in each term of the expression 10 a+ 15a. So the variable a will be included in the literal part of the common factor taken out of brackets:
Now all that remains is to calculate the common factor 5a out of brackets. To do this, we divide each term of the expression 10a + 15a on 5a. For clarity, we will separate coefficients and numbers with a multiplication sign (×)
Let's check the resulting expression. To do this, let's multiply 5a for each term in parentheses. If we did everything correctly, we will get the expression 10a + 15a
The letter factor cannot always be taken out of brackets. Sometimes the common factor consists only of a number, since there is nothing suitable for the letter part in the expression.
For example, let’s take the common factor out of brackets in the expression 2a−2b. Here the common factor will be only the number 2 , and among the letter factors there are no common factors in the expression. Therefore, in this case only the multiplier will be taken out 2
Example 2. Extract the common factor from the expression 3x + 9y + 12
The coefficients of this expression are numbers 3, 9 And 12, their gcd is equal 3 3 . And among the letter factors (variables) there is no common factor. Therefore the final common factor is 3
Example 3. Place the common factor out of brackets in the expression 8x + 6y + 4z + 10 + 2
The coefficients of this expression are numbers 8, 6, 4, 10 And 2, their gcd is equal 2 . This means that the coefficient of the common factor taken out of brackets will be the number 2 . And among the letter factors there is no common factor. Therefore the final common factor is 2
Example 4. Take out the common factor 6ab + 18ab + 3abc
The coefficients of this expression are numbers 6, 18 and 3, their gcd is equal 3 . This means that the coefficient of the common factor taken out of brackets will be the number 3 . The literal part of the common factor will include variables a And b, since in the expression 6ab + 18ab + 3abc these two variables are included in each term. Therefore the final common factor is 3ab
With a detailed solution, the expression becomes cumbersome and even incomprehensible. In this example this is more than noticeable. This is due to the fact that we cancel the factors in the numerator and denominator. It is best to do this in your head and immediately write down the division results. Then the expression becomes short and neat:
As in the case of a numerical expression, in a literal expression the common factor can be negative.
For example, let’s take the common factor out of brackets in the expression −3a − 2a.
For convenience, we replace subtraction with addition
−3a − 2a = −3a + (−2a )
The common factor in this expression is the factor a. But not only can we take into account a, but also −a. Let’s take it out of brackets:
It turned out to be a neat expression −a (3+2). It should not be forgotten that the multiplier −a actually looked like −1a and after reduction in both fractions of variables a, minus one remains in the denominators. Therefore, in the end we get positive answers in brackets
Example 6. Place the common factor out of brackets in the expression −6x − 6y
Let's replace subtraction with addition
−6x−6y = −6x+(−6y)
Let's put it out of brackets −6
Let's write down the solution briefly:
−6x − 6y = −6(x + y)
Example 7. Place the common factor out of brackets in the expression −2a − 4b − 6c
Let's replace subtraction with addition
−2a-4b-6c = −2a + (−4b) + (−6c)
Let's put it out of brackets −2
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Algebra lesson in 7th grade.
Topic: “Putting the common factor out of brackets.”
Textbook Makarychev Yu.N., Mindyuk N.G. 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. Organizational moment.
greetings;
preparing students for work.
2. Checking 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 lowest 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 involving 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 notation for a simple way to evaluate expressions.
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. Oral 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 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 on the 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.
Representation of a polynomial as a product of several polynomials (or monomials)
For example,
Taking the common factor out of brackets
It is necessary to analyze each term of the polynomial and find the common part (if any). For example, in an expression, each term has y. Variable y can be taken out of brackets.
The variables included in each term of the polynomial are taken out of brackets in powers with the smallest exponent that occurs. In the example there is y 2, y 5 And y 4. Let's put it out of brackets y 2.
What remains of each term after taking the common factor out of brackets? What should I write in parentheses? It is necessary to divide each term by a common factor, which is taken out of brackets. For example, when making y 2 outside the brackets in our example
If the numerical coefficients of each term of the polynomial have a greatest common divisor, then it can also be taken out of brackets. In our example GCD(18; 30; 6)=6
If the factor “-1” is taken out of brackets (they also say “minus is taken out”), then in brackets the sign of each term changes to the opposite
Polynomials can also be a common factor. For example, for the expression the common factor is the polynomial
Taking it out of brackets, we get 
You can always check whether the removal of the common factor from brackets is correct. To do this, you need to multiply the common factor by the polynomial in brackets and check that the resulting expression completely coincides with the original one.
Grouping method
If the terms of a polynomial do not have a common factor, then you should try to expand it using the grouping method.
To do this, you need to combine into groups those members that have common factors, and put the common factor of each group out of brackets. After this, the resulting groups may have a common factor, a polynomial, which is taken out of brackets.
The terms of a polynomial can be grouped in different ways. Not with every grouping it will be possible to factor a polynomial.
Expansion of a polynomial is sometimes impossible using known methods. Then it is possible to expand the polynomial by finding one root and
\(5x+xy\) 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 is a 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, 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 it? 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 correct way to extract it, because if we do not leave one, then when opening 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