Bracketing the common factor, rule, examples. Bracketing the common factor: rule, examples

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 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 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 the right side with 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 the common one. 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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\(5x+xy\) 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 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 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

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.

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 analyzed, 2 examples are offered for independent solutions, 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 amount 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 that 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:

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

We continue to understand the basics of algebra. Today we will work with, namely, we will consider an action such as putting the common factor out of brackets.

Lesson content

The basic 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 that is 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 just 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 for the numbers 12 and 15. Recall that to find the GCD, you need 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.

GCD for 12 and 15 is the number 3. This number is a common factor for the terms 12 and 15. It must 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

Let's find the gcd for terms 15 and 20

GCD for 15 and 20 is the number 5. This number is the common factor for the terms 15 and 20. Let’s take it out of brackets:

We got the expression 5(3 + 4). The resulting expression can be checked. 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 for terms 18, 24 and 36. To find , you need to factor these numbers into prime factors, then find the product of common factors:

GCD for 18, 24 and 36 is the number 6. This number is the common factor for the 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 for terms 195, 156 and 260

GCD for 195, 156 and 260 is the number 13. This number is the 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. For example, let's take the common factor out of brackets in the expression 16 − 12 − 4. The greatest common factor for the numbers 16, 12 and 4 is the number 4. Let's 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 GCD for numbers 72, 96 and 120

GCD for 72, 96 and 120 is the number 24. This number is the 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 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 a common factor for 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 put 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:

−2 × 3 = −6

−1 × 3 = −3

in the second case, the multiplicand can be made positive and the multiplier negative:

2 × (−3) = −6

1 × (−3) = −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 factor for the terms −20, −16, and −2 is the number 2. This number is the common factor for these terms. Let's see what it looks like:

−10 × 2 = −20

−8 × 2 = −16

−1 × 2 = −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

10 × (−2) = −20

8 × (−2) = −16

1 × (−2) = −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 take out not 6, but −6, since the numbers −30, −36 and −42 can be represented as follows:

5 × (−6) = −30

6 × (−6) = −36

7 × (−6) = −42

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 put it in order.

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 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 carry the same meaning:

−15 + (−5) + (−3) = −(15 + 5 + 3)

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 have already said 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 putting 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 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 numeric expression, in a literal expression the common factor can be negative.

For example, let’s take the general 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 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