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 common multiplier is the case when the free member 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 at senior degree is equal to 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. polynomial does not have 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 as an 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 in the following way: x=b/a. If you find it difficult to operate with abstract quantities, try representing this problem in the form specific items. 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.
Video on the topic
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 maximum degree 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.
Video on the topic
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 behind brackets.
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:
- Adding and subtracting fractions with different denominators. There is no other way to perform this operation;
- Comparing fractions. Sometimes reduction to a common denominator greatly simplifies this task;
- 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 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 this method- you have to count a lot, because the denominators are multiplied “throughout”, 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:
- 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.
- The number resulting from this division will be an additional factor for the fraction with a smaller denominator.
- 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 of the denominators. 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 equal direct product denominators of the original fractions, as 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.
Nai smaller 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:
- Having discovered identical factors, we immediately arrived at the least common multiple, which, generally speaking, is a non-trivial problem;
- 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 is 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.
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 discussed, 2 examples are offered for independent decision, to which there are answers. The work will help you study this topic those students who, for some reason, did not learn it when going through 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
average comprehensive school №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 is based on distributive properties multiplication can be represented as a product of two factors. One of them is the common multiplier 3b , and the other is the sum 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 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: Of two degrees with the same basis we take out the degree with a lower 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 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).
To reduce fractions to the least common denominator, you need to: 1) find the least common multiple of the denominators of the given fractions, it will be the least common denominator. 2) find an additional factor for each fraction, why divide new denominator to the denominator of each fraction. 3) multiply the numerator and denominator of each fraction by its additional factor.
Examples. Reduce the following fractions to their lowest common denominator.
We find the least common multiple of the denominators: LCM(5; 4) = 20, since 20 is the smallest number that is divisible by both 5 and 4. Find for the 1st fraction an additional factor 4 (20 : 5=4). For the 2nd fraction the additional factor is 5 (20 : 4=5). We multiply the numerator and denominator of the 1st fraction by 4, and the numerator and denominator of the 2nd fraction by 5. We have reduced these fractions to the lowest common denominator ( 20 ).
The lowest common denominator of these fractions is the number 8, since 8 is divisible by 4 and itself. There will be no additional factor to the 1st fraction (or we can say that it equal to one), to the 2nd fraction the additional factor is 2 (8 : 4=2). We multiply the numerator and denominator of the 2nd fraction by 2. We have reduced these fractions to the lowest common denominator ( 8 ).
These fractions are not irreducible.
Let's reduce the 1st fraction by 4, and reduce the 2nd fraction by 2. ( see examples for abbreviation ordinary fractions: Sitemap → 5.4.2. Examples of reducing common fractions). Find the LOC(16 ; 20)=2 4 · 5=16· 5=80. The additional multiplier for the 1st fraction is 5 (80 : 16=5). The additional factor for the 2nd fraction is 4 (80 : 20=4). We multiply the numerator and denominator of the 1st fraction by 5, and the numerator and denominator of the 2nd fraction by 4. We have reduced these fractions to the lowest common denominator ( 80 ).
We find the lowest common denominator NCD(5 ; 6 and 15)=NOK(5 ; 6 and 15)=30. The additional factor to the 1st fraction is 6 (30 : 5=6), the additional factor to the 2nd fraction is 5 (30 : 6=5), the additional factor to the 3rd fraction is 2 (30 : 15=2). We multiply the numerator and denominator of the 1st fraction by 6, the numerator and denominator of the 2nd fraction by 5, the numerator and denominator of the 3rd fraction by 2. We have reduced these fractions to the lowest common denominator ( 30 ).
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\(5x+xy\) can be represented as \(x(5+y)\). This is in fact 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 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