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Taking the common factor out of brackets. Reducing fractions to the lowest common denominator, rule, examples, solutions

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

Lesson content

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 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 in this case in 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 have reviewed simplest example taking the 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 for the numbers 12 and 15. Recall that to find the GCD you need to expand original numbers on prime factors, then write out the first expansion and remove from it the factors that are not included in the expansion 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 the 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 a 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 give simplicity to the expression. 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

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 put out of brackets 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.

Let's consider next 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 you swap left and right side 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 sign and write the expression −3 + 5 + 11 s next to it in parentheses 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

Place the common factor out of brackets in literal expression 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 are located 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

At 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 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 beyond the brackets can be taken not only 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

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

Replace subtraction with addition

−2a-4b-6c = −2a + (−4b) + (−6c)

Let's put it out of brackets −2

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I originally wanted to include common denominator techniques in the Adding and Subtracting Fractions section. But there was so much information, and its importance was so great (after all, not only numerical fractions), that it is better to study this issue separately.

So let's say we have two fractions with different denominators. And we want to make sure that the denominators become the same. The basic property of a fraction comes to the rescue, which, let me remind you, sounds like this:

A fraction will not change if its numerator and denominator are multiplied by the same number other than zero.

Thus, if you choose the factors correctly, the denominators of the fractions will become equal - this process is called reduction to a common denominator. And the required numbers, “evening out” the denominators, are called additional factors.

Why do we need to reduce fractions to a common denominator? Here are just a few reasons:

  1. Adding and subtracting fractions with different denominators. There is no other way to perform this operation;
  2. Comparing fractions. Sometimes reduction to a common denominator greatly simplifies this task;
  3. Solving problems involving fractions and percentages. Percentages are, in fact, ordinary expressions that contain fractions.

There are many ways to find numbers that, when multiplied by them, will make the denominators of fractions equal. We will consider only three of them - in order of increasing complexity and, in a sense, effectiveness.

Criss-cross multiplication

The simplest and reliable way, which is guaranteed to equalize the denominators. We will act “in a headlong manner”: we multiply the first fraction by the denominator of the second fraction, and the second by the denominator of the first. As a result, the denominators of both fractions will become equal to the product of the original denominators. Take a look:

As additional factors, consider the denominators of neighboring fractions. We get:

Yes, it's that simple. If you are just starting to study fractions, it is better to work using this method - this way you will insure yourself against many mistakes and are guaranteed to get the result.

The only drawback of this method is that you have to do a lot of counting, because the denominators are multiplied “over and over”, and the result can be very big numbers. This is the price to pay for reliability.

Common Divisor Method

This technique helps to significantly reduce calculations, but, unfortunately, it is used quite rarely. The method is as follows:

  1. Before you go straight ahead (i.e., using the criss-cross method), take a look at the denominators. Perhaps one of them (the one that is larger) is divided into the other.
  2. The number resulting from this division will be an additional factor for the fraction with a smaller denominator.
  3. In this case, a fraction with a large denominator does not need to be multiplied by anything at all - this is where the savings lie. At the same time, the probability of error is sharply reduced.

Task. Find the meanings of the expressions:

Note that 84: 21 = 4; 72: 12 = 6. Since in both cases one denominator is divided without a remainder by the other, we use the method of common factors. We have:

Note that the second fraction was not multiplied by anything at all. In fact, we cut the amount of computation in half!

By the way, I didn’t take the fractions in this example by chance. If you're interested, try counting them using the criss-cross method. After reduction, the answers will be the same, but there will be much more work.

This is the strength of the method common divisors, but, I repeat, it can only be used in the case when one of the denominators is divided by the other without a remainder. Which happens quite rarely.

Least common multiple method

When we reduce fractions to a common denominator, we are essentially trying to find a number that is divisible by each denominator. Then we bring the denominators of both fractions to this number.

There are a lot of such numbers, and the smallest of them will not necessarily be equal to the direct product of the denominators of the original fractions, as is assumed in the “criss-cross” method.

For example, for denominators 8 and 12, the number 24 is quite suitable, since 24: 8 = 3; 24: 12 = 2. This number is much less product 8 12 = 96.

Smallest number, which is divisible by each of the denominators, is called their least common multiple (LCM).

Notation: The least common multiple of a and b is denoted by LCM(a ; b) . For example, LCM(16, 24) = 48 ; LCM(8; 12) = 24 .

If you manage to find such a number, the total amount of calculations will be minimal. Look at the examples:

Task. Find the meanings of the expressions:

Note that 234 = 117 2; 351 = 117 3. Factors 2 and 3 are coprime (have no common factors other than 1), and factor 117 is common. Therefore LCM(234, 351) = 117 2 3 = 702.

Likewise, 15 = 5 3; 20 = 5 · 4. Factors 3 and 4 are coprime, and factor 5 is common. Therefore LCM(15, 20) = 5 3 4 = 60.

Now let's bring the fractions to common denominators:

Notice how useful it was to factorize the original denominators:

  1. Having discovered identical factors, we immediately arrived at the least common multiple, which, generally speaking, is a non-trivial problem;
  2. From the resulting expansion you can find out which factors are “missing” in each fraction. For example, 234 · 3 = 702, therefore, for the first fraction the additional factor is 3.

To appreciate how much of a difference the least common multiple method makes, try calculating these same examples using the criss-cross method. Of course, without a calculator. I think after this comments will be unnecessary.

Don't think that there are such complex fractions will not be the case in real examples. They meet all the time, and the above tasks are not the limit!

The only problem is how to find this very NOC. Sometimes everything can be found in a few seconds, literally “by eye,” but in general this is a complex computational task that requires separate consideration. We won't touch on that here.

\(5x+xy\) 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 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 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

IN real life We need to operate with ordinary fractions. However, to add or subtract fractions with different denominators, such as 2/3 and 5/7, we need to find a common denominator. By bringing fractions to a common denominator, we can easily perform addition or subtraction operations.

Definition

Fractions are one of the most difficult topics in elementary arithmetic, and rational numbers are intimidating for students who encounter them for the first time. We are used to working with numbers written in decimal format. It is much easier to immediately add 0.71 and 0.44 than to add 5/7 and 4/9. After all, to sum fractions, they must be reduced to a common denominator. However, fractions represent the meaning of quantities much more accurately than their decimal equivalents, and in mathematics, the representation of series or ir rational numbers in the form of a fraction becomes a priority task. This task is called “bringing an expression to a closed form.”

If both the numerator and denominator of a fraction are multiplied or divided by the same factor, the value of the fraction does not change. This is one of the most important properties fractional numbers. For example, the fraction 3/4 in decimal form is written as 0.75. If we multiply the numerator and denominator by 3, we get the fraction 9/12, which is exactly the same as 0.75. Thanks to this property, we can multiply different fractions in such a way that they all have same denominators. How to do this?

Finding a common denominator

The least common denominator (LCD) is the smallest common multiple of all denominators in an expression. We can find such a number in three ways.

Using the maximum denominator

This is one of the simplest but most time-consuming methods for searching for NCDs. First, we write out the largest number from the denominators of all fractions and check its divisibility by smaller numbers. If it is divisible, then the largest denominator is the NCD.

If in the previous operation the numbers are divisible with a remainder, then the largest of them must be multiplied by 2 and the divisibility test repeated. If it is divided without a remainder, then the new coefficient becomes NOS.

If not, then the largest denominator is multiplied by 3, 4, 5 and so on until the least common multiple of lower parts all fractions. In practice it looks like this.

Let us have the fractions 1/5, 1/8 and 1/20. We check 20 for divisibility of 5 and 8. 20 is not divisible by 8. Multiply 20 by 2. Check 40 for divisibility of 5 and 8. The numbers are divisible without a remainder, therefore, N3 (1/5, 1/8 and 1/20) = 40 , and the fractions become 8/40, 5/40 and 2/40.

Sequential search of multiples

The second method is a simple search of multiples and choosing the smallest one. To find multiples, we multiply a number by 2, 3, 4, and so on, so the number of multiples goes to infinity. This sequence can be limited by a limit, which is the product of given numbers. For example, for the numbers 12 and 20 the LCM is found as follows:

  • write down numbers that are multiples of 12 - 24, 48, 60, 72, 84, 96, 108, 120;
  • write down numbers that are multiples of 20 - 40, 60, 80, 100, 120;
  • determine common multiples - 60, 120;
  • choose the smallest of them - 60.

Thus, for 1/12 and 1/20, the common denominator is 60, and the fractions are converted to 5/60 and 3/60.

Prime factorization

This method of finding the LOC is the most relevant. This method implies the decomposition of all numbers from the lower parts of fractions into indivisible factors. After this, a number is compiled that contains the factors of all denominators. In practice it works like this. Let's find the LCM for the same pair 12 and 20:

  • factorize 12 - 2 × 2 × 3;
  • lay out 20 - 2 × 2 × 5;
  • we combine the factors so that they contain the numbers both 12 and 20 - 2 × 2 × 3 × 5;
  • multiply the indivisibles and get the result - 60.

In the third point, we combine multipliers without repetitions, that is, two twos are enough to form 12 in combination with a three and 20 with a five.

Our calculator allows you to determine the NOZ for an arbitrary number of fractions written in both ordinary and decimal form. To search for NOS, you just need to enter values ​​separated by tabs or commas, after which the program will calculate the common denominator and display the converted fractions.

Real life example

Adding Fractions

Suppose in an arithmetic problem we need to add five fractions:

0,75 + 1/5 + 0,875 + 1/4 + 1/20

The solution would have to be done manually in the following way. First, we need to represent the numbers in one form of notation:

  • 0,75 = 75/100 = 3/4;
  • 0,875 = 875/1000 = 35/40 = 7/8.

Now we have a series ordinary fractions, which must be reduced to the same denominator:

3/4 + 1/5 + 7/8 + 1/4 + 1/20

Since we have 5 terms, the easiest way is to use the method of searching for NOZ by the largest number. We check 20 for divisibility by other numbers. 20 is not divisible by 8 without a remainder. We multiply 20 by 2, check 40 for divisibility - all numbers divide 40 by a whole. This is our common denominator. Now, to sum the rational numbers, we need to determine additional factors for each fraction, which are defined as the ratio of the LCM to the denominator. Additional multipliers will look like this:

  • 40/4 = 10;
  • 40/5 = 8;
  • 40/8 = 5;
  • 40/4 = 10;
  • 40/20 = 2.

Now we multiply the numerator and denominator of the fractions by the corresponding additional factors:

30/40 + 8/40 + 35/40 + 10/40 + 2/40

For such an expression, we can easily determine the sum to be 85/40 or 2 whole numbers and 1/8. This is a cumbersome calculation, so you can simply enter the problem data into the calculator form and get the answer right away.

Conclusion

Arithmetic operations with fractions are not a very convenient thing, because to find the answer you have to carry out many intermediate calculations. Use our online calculator to convert fractions to a common denominator and quick solution school tasks.

This method makes sense if the degree of the polynomial is not lower than two. In this case, the common factor can be not only a binomial of the first degree, but also of higher degrees.

To find a common factor terms of the polynomial, it is necessary to perform a number of transformations. The simplest binomial or monomial that can be taken out of brackets will be one of the roots of the polynomial. Obviously, in the case when the polynomial does not have a free term, there will be an unknown in the first degree - the polynomial, equal to 0.

More difficult to find a common factor is the case when the free term is not equal to zero. Then methods of simple selection or grouping are applicable. For example, let all the roots of the polynomial be rational, and all the coefficients of the polynomial are integers: y^4 + 3 y³ – y² – 9 y – 18.

Write down all the integer divisors of the free term. If a polynomial has rational roots, then they are among them. As a result of the selection, roots 2 and -3 are obtained. This means that the common factors of this polynomial will be the binomials (y - 2) and (y + 3).

The common factoring method is one of the components of factorization. The method described above is applicable if the coefficient of the highest degree is 1. If this is not the case, then a series of transformations must first be performed. For example: 2y³ + 19 y² + 41 y + 15.

Make a substitution of the form t = 2³·y³. To do this, multiply all the coefficients of the polynomial by 4: 2³·y³ + 19·2²·y² + 82·2·y + 60. After replacement: t³ + 19·t² + 82·t + 60. Now, to find the common factor, we apply the above method .

Besides, effective method Finding a common factor is the elements of a polynomial. It is especially useful when the first method does not, i.e. The polynomial has no rational roots. However, groupings are not always obvious. For example: The polynomial y^4 + 4 y³ – y² – 8 y – 2 has no integer roots.

Use grouping: y^4 + 4 y³ – y² – 8 y – 2 = y^4 + 4 y³ – 2 y² + y² – 8 y – 2 = (y^4 – 2 y²) + ( 4 y³ – 8 y) + y² – 2 = (y² - 2)*(y² + 4 y + 1). The common factor of the elements of this polynomial is (y² - 2).

Multiplication and division, just like addition and subtraction, are basic arithmetic operations. Without learning to solve examples of multiplication and division, a person will encounter many difficulties not only when studying more complex branches of mathematics, but even in the most ordinary everyday affairs. Multiplication and division are closely related, and the unknown components of examples and problems involving one of these operations are calculated using the other operation. At the same time, it is necessary to clearly understand that when solving examples, it makes absolutely no difference which objects you divide or multiply.

You will need

  • - multiplication table;
  • - calculator or sheet of paper and pencil.

Instructions

Write down the example you need. Label the unknown factor like x. An example might look like this: a*x=b. Instead of the factor a and the product b in the example, there can be any or numbers. Remember the basic principle of multiplication: changing the places of the factors does not change the product. So unknown factor x can be placed absolutely anywhere.

To find the unknown factor in an example where there are only two factors, you just need to divide the product by the known factor. That is, this is done as follows: x=b/a. If you find it difficult to operate with abstract quantities, try to imagine this problem in the form of concrete objects. You, you have only apples and how many of them you will eat, but you don’t know how many apples everyone will get. For example, you have 5 family members, and there are 15 apples. Designate the number of apples intended for each as x. Then the equation will look like this: 5(apples)*x=15(apples). Unknown factor is found in the same way as in the equation with letters, that is, divide 15 apples among five family members, in the end it turns out that each of them ate 3 apples.

In the same way the unknown is found factor with the number of factors. For example, the example looks like a*b*c*x*=d. In theory, find with factor it is possible in the same way as in the later example: x=d/a*b*c. But the equation can be reduced to more simple view, denoting the product of known factors with another letter - for example, m. Find what m equals by multiplying numbers a,b and c: m=a*b*c. Then the whole example can be represented as m*x=d, and the unknown quantity will be equal to x=d/m.

If known factor and the product are fractions, the example is solved in exactly the same way as with . But in this case you need to remember the actions. When multiplying fractions, their numerators and denominators are multiplied. When dividing fractions, the numerator of the dividend is multiplied by the denominator of the divisor, and the denominator of the dividend is multiplied by the numerator of the divisor. That is, in this case the example will look like this: a/b*x=c/d. In order to find an unknown quantity, you need to divide the product by the known factor. That is, x=a/b:c/d =a*d/b*c.

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Please note

When solving examples with fractions, the fraction of a known factor can simply be reversed and the action performed as a multiplication of fractions.

A polynomial is the sum of monomials. A monomial is the product of several factors, which are a number or a letter. Degree unknown is the number of times it is multiplied by itself.

Instructions

Please provide it if it has not already been done. Similar monomials are monomials of the same type, that is, monomials with the same unknowns to the same degree.

Take, for example, the polynomial 2*y²*x³+4*y*x+5*x²+3-y²*x³+6*y²*y²-6*y²*y². This polynomial has two unknowns - x and y.

Connect similar monomials. Monomials with the second power of y and the third power of x will come to the form y²*x³, and monomials with the fourth power of y will cancel. It turns out y²*x³+4*y*x+5*x²+3-y²*x³.

Take y as the main unknown letter. Find the maximum degree for unknown y. This is a monomial y²*x³ and, accordingly, degree 2.

Draw a conclusion. Degree polynomial 2*y²*x³+4*y*x+5*x²+3-y²*x³+6*y²*y²-6*y²*y² in x is equal to three, and in y is equal to two.

Find the degree polynomial√x+5*y by y. It is equal to the maximum degree of y, that is, one.

Find the degree polynomial√x+5*y in x. The unknown x is located, which means its degree will be a fraction. Since the root is a square root, the power of x is 1/2.

Draw a conclusion. For polynomial√x+5*y the x power is 1/2 and the y power is 1.

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Simplification algebraic expressions required in many areas of mathematics, including solving equations higher degrees, differentiation and integration. Several methods are used, including factorization. To apply this method, you need to find and make a general factor for brackets.