The denominator of the arithmetic fraction a / b is the number b, which shows the size of the fractions of a unit from which the fraction is composed. The denominator of an algebraic fraction A / B is the algebraic expression B. To perform arithmetic operations with fractions, they must be reduced to the lowest common denominator.
You will need
- To work with algebraic fractions and find the lowest common denominator, you need to know how to factor polynomials.
Instructions
Let's consider reducing two arithmetic fractions n/m and s/t to the least common denominator, where n, m, s, t are integers. It is clear that these two fractions can be reduced to any denominator divisible by m and t. But they try to lead to the lowest common denominator. It is equal to the least common multiple of the denominators m and t of the given fractions. The least multiple (LMK) of a number is the smallest divisible by all given numbers at the same time. Those. in our case, we need to find the least common multiple of the numbers m and t. Denoted as LCM (m, t). Next, the fractions are multiplied by the corresponding ones: (n/m) * (LCM (m, t) / m), (s/t) * (LCM (m, t) / t).
Let's find the lowest common denominator of three fractions: 4/5, 7/8, 11/14. First, let's expand the denominators 5, 8, 14: 5 = 1 * 5, 8 = 2 * 2 * 2 = 2^3, 14 = 2 * 7. Next, calculate the LCM (5, 8, 14) by multiplying all the numbers included into at least one of the expansions. LCM (5, 8, 14) = 5 * 2^3 * 7 = 280. Note that if a factor occurs in the expansion of several numbers (factor 2 in the expansion of denominators 8 and 14), then we take the factor to a greater degree (2^3 in our case).
So, the general one is received. It is equal to 280 = 5 * 56 = 8 * 35 = 14 * 20. Here we get the numbers by which we need to multiply the fractions with the corresponding denominators in order to bring them to the lowest common denominator. We get 4/5 = 56 * (4/5) = 224/280, 7/8 = 35 * (7/8) = 245/280, 11/14 = 20 * (11/14) = 220/280.
Reduction to the lowest common denominator algebraic fractions performed by analogy with arithmetic. For clarity, let's look at the problem using an example. Let two fractions (2 * x) / (9 * y^2 + 6 * y + 1) and (x^2 + 1) / (3 * y^2 + 4 * y + 1) be given. Let's factorize both denominators. Note that the denominator of the first fraction is perfect square: 9 * y^2 + 6 * y + 1 = (3 * y + 1)^2. For
Within the framework of the study of identity transformations, the topic of rendering common multiplier out of brackets. 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 should be obtained 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 the 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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In this lesson we will get acquainted with the rules for putting a common factor out of brackets and learn how to find it in various examples and expressions. Let's talk about how simple operation, placing the common factor out of brackets allows you to simplify the calculations. We will consolidate the acquired knowledge and skills by looking at examples of various complexities.
What is a common factor, why look for it and for what purpose is it taken out of brackets? Let's answer these questions by looking at a simple example.
Let's solve the equation. Left side equation is a polynomial consisting of similar terms. The letter part is common to these terms, which means it will be the common factor. Let's put it out of brackets:
In this case, taking the common factor out of brackets helped us convert the polynomial to a monomial. Thus, we were able to simplify the polynomial and its transformation helped us solve the equation.
In the example considered, the common factor was obvious, but would it be so easy to find it in an arbitrary polynomial?
Let's find the meaning of the expression: .
IN in this example placing the common factor out of brackets greatly simplified the calculation.
Let's solve one more example. Let's prove divisibility into expressions.
The resulting expression is divisible by , as required to be proved. Once again, taking the common factor allowed us to solve the problem.
Let's solve one more example. Let us prove that the expression is divisible by for any natural number: 
.
The expression is the product of two adjacent natural numbers. One of the two numbers will definitely be even, which means the expression will be divisible by .
We've sorted it out different examples, but they used the same solution method: they took the common factor out of brackets. We see that this simple operation greatly simplifies the calculations. It was easy to find a common factor for these special cases, but what to do in general case, for an arbitrary polynomial?
Recall that a polynomial is a sum of monomials.
Consider the polynomial 
. This polynomial is the sum of two monomials. A monomial is the product of a number, a coefficient, and a letter part. Thus, in our polynomial, each monomial is represented by the product of a number and powers, the product of factors. The factors can be the same for all monomials. It is these factors that need to be determined and taken out of the bracket. First, we find the common factor for the coefficients, which are integer ones.
It was easy to find the common factor, but let's define the gcd of the coefficients: 
.
Let's look at another example: .
Let's find , which will allow us to determine the common factor for this expression: .
We have derived a rule for integer coefficients. You need to find their gcd and take it out of the bracket. Let's consolidate this rule by solving one more example.
We have looked at the rule for assigning a common factor for integer coefficients, let's move on to the letter part. First, we look for those letters that are included in all monomials, and then we determine the highest degree of the letter that is included in all monomials: .
In this example there was only one common letter variable, but there can be several, as in the following example:
Let's complicate the example by increasing the number of monomials:
After taking out the common factor, we converted the algebraic sum into a product.
We looked at the subtraction rules for integer coefficients and letter variables separately, but most often you need to apply them together to solve the example. Let's look at an example:
Sometimes it can be difficult to determine which expression is left in parentheses, let's look at an easy trick that will allow you to quickly solve this problem.
The common factor can also be the desired value:
The common factor can be not only a number or a monomial, but also any expression, such as in the following equation.
\(5x+xy\) can be represented as \(x(5+y)\). These are indeed identical expressions, we can verify this if we open the brackets: \(x(5+y)=x \cdot 5+x \cdot y=5x+xy\). As you can see, as a result we get the original expression. This means that \(5x+xy\) is indeed equal to \(x(5+y)\). By the way, this reliable way to check the correctness of the common factors - open the resulting bracket and compare the result with the original expression.
The main rule for bracketing:
For example, in the expression \(3ab+5bc-abc\) only \(b\) can be taken out of the bracket, because it is the only one that is present in all three terms. The process of taking common factors out of brackets is shown in the diagram below:
Bracketing rules
In mathematics, it is customary to take out all common factors at once.
Example:\(3xy-3xz=3x(y-z)\)
Please note that here we could expand like this: \(3(xy-xz)\) or like this: \(x(3y-3z)\). However, these would be incomplete decompositions. Both the C and the X must be taken out.
Sometimes the common members are not immediately visible.
Example:\(10x-15y=2·5·x-3·5·y=5(2x-3y)\)
In this case, the common term (five) was hidden. However, having expanded \(10\) as \(2\) multiplied by \(5\), and \(15\) as \(3\) multiplied by \(5\) - we “pulled the five into the light of God”, after which they were easily able to take it out of the bracket.
If a monomial is removed completely, one remains from it.
Example: \(5xy+axy-x=x(5y+ay-1)\)
We put \(x\) out of brackets, and the third monomial consists only of x. Why does one remain from 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
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 frighten schoolchildren 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 becomes in the form of a fraction priority. 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 so that they all have the same denominators. How to do it?
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 NOZ.
If not, then the largest denominator is multiplied by 3, 4, 5, and so on, until the least common multiple of the lower parts of all fractions is found. 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 equal to 85/40 or 2 whole 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 quickly solve school problems.
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