What's happened factorization? This is a way of turning an inconvenient and complex example into a simple and cute one.) A very powerful technique! It is found at every step in both elementary and higher mathematics.
Such transformations in mathematical language are called identical transformations of expressions. For those who are not in the know, take a look at the link. There is very little there, simple and useful.) The meaning of any identity transformation is the recording of the expression in another form while maintaining its essence.
Meaning factorization extremely simple and clear. Right from the name itself. You may forget (or not know) what a multiplier is, but you can figure out that this word comes from the word “multiply”?) Factoring means: represent an expression in the form of multiplying something by something. May mathematics and the Russian language forgive me...) That's all.
For example, you need to expand the number 12. You can safely write:
So we presented the number 12 as a multiplication of 3 by 4. Please note that the numbers on the right (3 and 4) are completely different than on the left (1 and 2). But we understand perfectly well that 12 and 3 4 one and the same. The essence of the number 12 from transformation has not changed.
Is it possible to decompose 12 differently? Easily!
12=3·4=2·6=3·2·2=0.5·24=........
The decomposition options are endless.
Factoring numbers is a useful thing. It helps a lot, for example, when working with roots. But factoring algebraic expressions is not only useful, it is necessary! Just for example:
Simplify:
Those who do not know how to factor an expression rest on the sidelines. Those who know how - simplify and get:
The effect is amazing, right?) By the way, the solution is quite simple. You'll see for yourself below. Or, for example, this task:
Solve the equation:
x 5 - x 4 = 0
It is decided in the mind, by the way. Using factorization. We will solve this example below. Answer: x 1 = 0; x 2 = 1.
Or, the same thing, but for the older ones):
Solve the equation:
In these examples I showed main purpose factorization: simplifying fractional expressions and solving some types of equations. Here's a rule of thumb to remember:
If we have a scary fractional expression in front of us, we can try factoring the numerator and denominator. Very often the fraction is reduced and simplified.
If we have an equation in front of us, where on the right there is zero, and on the left - I don’t understand what, we can try to factorize the left side. Sometimes it helps).
Basic methods of factorization.
Here they are, the most popular methods:
4. Expansion of a quadratic trinomial.
These methods must be remembered. Exactly in that order. Complex examples are checked for all possible decomposition methods. And it’s better to check in order so as not to get confused... So let’s start in order.)
1. Taking the common factor out of brackets.
A simple and reliable way. Nothing bad comes from him! It happens either well or not at all.) That’s why he comes first. Let's figure it out.
Everyone knows (I believe!) the rule:
a(b+c) = ab+ac
Or, more generally:
a(b+c+d+.....) = ab+ac+ad+....
All equalities work both from left to right and vice versa, from right to left. You can write:
ab+ac = a(b+c)
ab+ac+ad+.... = a(b+c+d+.....)
That's the whole point of taking the common factor out of brackets.
On the left side A - common multiplier for all terms. Multiplied by everything that exists). On the right is the most A is already located outside the brackets.
We will consider the practical application of the method using examples. At first the option is simple, even primitive.) But in this option I will mark (in green) very important points for any factorization.
Factorize:
ah+9x
Which general does the multiplier appear in both terms? X, of course! We will put it out of brackets. Let's do this. We immediately write X outside the brackets:
ax+9x=x(
And in parentheses we write the result of division each term on this very X. In order:
That's it. Of course, there is no need to describe it in such detail, this is done in the mind. But it is advisable to understand what’s what). We record in memory:
We write the common factor outside the brackets. In parentheses we write the results of dividing all terms by this common factor. In order.
So we have expanded the expression ah+9x by multipliers. Turned it into multiplying x by (a+9). I note that in the original expression there was also multiplication, even two: a·x and 9·x. But it was not factorized! Because in addition to multiplication, this expression also contained addition, the “+” sign! And in expression x(a+9) There is nothing but multiplication!
How so!? - I hear the indignant voice of the people - And in brackets!?)
Yes, there is addition inside the parentheses. But the trick is that while the brackets are not opened, we consider them like one letter. And we do all the actions with brackets entirely, as with one letter. In this sense, in the expression x(a+9) There is nothing except multiplication. This is the whole point of factorization.
By the way, is it possible to somehow check whether we did everything correctly? Easily! It’s enough to multiply back what you put out (x) by brackets and see if it worked original expression? If it works, everything is great!)
x(a+9)=ax+9x
It worked.)
There are no problems in this primitive example. But if there are several terms, and even with different signs... In short, every third student messes up). Therefore:
If necessary, check the factorization by inverse multiplication.
Factorize:
3ax+9x
We are looking for a common factor. Well, everything is clear with X, it can be taken out. Is there more general factor? Yes! This is a three. You can write the expression like this:
3ax+3 3x
Here it is immediately clear that the common factor will be 3x. Here we take it out:
3ax+3 3x=3x(a+3)
Spread out.
What happens if you take it out only x? Nothing special:
3ax+9x=x(3a+9)
This will also be a factorization. But in this fascinating process, it is customary to lay out everything to the limit while there is an opportunity. Here in brackets there is an opportunity to put out a three. It will turn out:
3ax+9x=x(3a+9)=3x(a+3)
The same thing, only with one extra action.) Remember:
When taking the common factor out of brackets, we try to take out maximum common factor.
Shall we continue the fun?)
Factor the expression:
3akh+9х-8а-24
What will we take away? Three, X? Nope... You can't. I remind you that you can only take out general multiplier that is in all terms of the expression. That's why he general. There is no such multiplier here... What, you don’t have to expand it!? Well, yes, we were so happy... Meet:
2. Grouping.
Actually, grouping can hardly be called an independent method of factorization. This is, rather, a way to get out of a complex example.) You need to group the terms so that everything works out. This can only be shown by example. So, we have the expression:
3akh+9х-8а-24
It can be seen that there are some common letters and numbers. But... General there is no multiplier to be in all terms. Let's not lose heart and break the expression into pieces. Let's group. So that each piece has a common factor, there is something to take away. How do we break it? Yes, we just put parentheses.
Let me remind you that parentheses can be placed anywhere and however you want. Just the essence of the example hasn't changed. For example, you can do this:
3akh+9х-8а-24=(3ах+9х)-(8а+24)
Please pay attention to the second brackets! They are preceded by a minus sign, and 8a And 24 turned positive! If, to check, we open the brackets back, the signs will change, and we get original expression. Those. the essence of the expression from the brackets has not changed.
But if you just inserted parentheses without taking into account the change of sign, for example, like this:
3akh+9х-8а-24=(3ax+9x) -(8a-24 )
it would be a mistake. On the right - already other expression. Open the brackets and everything will become visible. You don’t have to decide further, yes...)
But let's return to factorization. Let's look at the first brackets (3ax+9x) and we think, is there anything we can take out? Well, we solved this example above, we can take it 3x:
(3ax+9x)=3x(a+3)
Let's study the second brackets, we can add an eight there:
(8a+24)=8(a+3)
Our entire expression will be:
(3ax+9x)-(8a+24)=3x(a+3)-8(a+3)
Factored? No. The result of decomposition should be only multiplication but with us the minus sign spoils everything. But... Both terms have a common factor! This (a+3). It was not for nothing that I said that the entire brackets are, as it were, one letter. This means that these brackets can be taken out of brackets. Yes, that's exactly what it sounds like.)
We do as described above. We write the common factor (a+3), in the second brackets we write the results of dividing the terms by (a+3):
3x(a+3)-8(a+3)=(a+3)(3x-8)
All! There is nothing on the right except multiplication! This means that factorization has been completed successfully!) Here it is:
3ax+9x-8a-24=(a+3)(3x-8)
Let us briefly repeat the essence of the group.
If the expression does not general multiplier for everyone terms, we break the expression into brackets so that inside the brackets the common factor was. We take it out and see what happens. If you are lucky and there are absolutely identical expressions left in the brackets, we move these brackets out of brackets.
I will add that grouping is a creative process). It doesn't always work out the first time. It's OK. Sometimes you have to swap terms and consider different grouping options until you find a successful one. The main thing here is not to lose heart!)
Examples.
Now, having enriched yourself with knowledge, you can solve tricky examples.) At the beginning of the lesson there were three of these...
Simplify:
In essence, we have already solved this example. Unbeknownst to ourselves.) I remind you: if we are given a terrible fraction, we try to factor the numerator and denominator. Other simplification options just no.
Well, the denominator here is not expanded, but the numerator... We have already expanded the numerator during the lesson! Like this:
3ax+9x-8a-24=(a+3)(3x-8)
We write the result of the expansion into the numerator of the fraction:
According to the rule of reducing fractions (the main property of a fraction), we can divide (at the same time!) the numerator and denominator by the same number, or expression. Fraction from this doesn't change. So we divide the numerator and denominator by the expression (3x-8). And here and there we will get ones. The final result of the simplification:
I would like to especially emphasize: reducing a fraction is possible if and only if in the numerator and denominator, in addition to multiplying expressions there is nothing. That is why the transformation of the sum (difference) into multiplication so important for simplification. Of course, if the expressions different, then nothing will be reduced. It will happen. But factorization gives a chance. This chance without decomposition is simply not there.
Example with equation:
Solve the equation:
x 5 - x 4 = 0
We take out the common factor x 4 out of brackets. We get:
x 4 (x-1)=0
We realize that the product of factors is equal to zero then and only then, when any of them is zero. If in doubt, find me a couple of non-zero numbers that, when multiplied, will give zero.) So we write, first the first factor:
With such an equality, the second factor does not concern us. Anyone can be, but in the end it will still be zero. What number to the fourth power does zero give? Only zero! And no other... Therefore:
We figured out the first factor and found one root. Let's look at the second factor. Now we don’t care about the first factor anymore.):
Here we found a solution: x 1 = 0; x 2 = 1. Any of these roots fits our equation.
Very important note. Please note that we solved the equation piece by piece! Each factor was equal to zero, regardless of other factors. By the way, if in such an equation there are not two factors, like ours, but three, five, as many as you like, we will solve exactly the same. Piece by piece. For example:
(x-1)(x+5)(x-3)(x+2)=0
Anyone who opens the brackets and multiplies everything will be stuck on this equation forever.) A correct student will immediately see that there is nothing on the left except multiplication, and zero on the right. And he will begin (in his mind!) to equate all brackets in order to zero. And he will get (in 10 seconds!) the correct solution: x 1 = 1; x 2 = -5; x 3 = 3; x 4 = -2.
Cool, right?) Such an elegant solution is possible if the left side of the equation factorized. Got the hint?)
Well, one last example, for the older ones):
Solve the equation:
It’s somewhat similar to the previous one, don’t you think?) Of course. It's time to remember that in seventh grade algebra, sines, logarithms, and anything else can be hidden under the letters! Factoring works throughout mathematics.
We take out the common factor lg 4 x out of brackets. We get:
log 4 x=0
This is one root. Let's look at the second factor.
Here is the final answer: x 1 = 1; x 2 = 10.
I hope you've realized the power of factoring in simplifying fractions and solving equations.)
In this lesson we learned about common factoring and grouping. It remains to deal with the formulas for abbreviated multiplication and the quadratic trinomial.
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What should you do if, in the process of solving a problem from the Unified State Exam or on an entrance exam in mathematics, you received a polynomial that cannot be factorized using the standard methods that you learned at school? In this article, a math tutor will tell you about one effective method, the study of which is outside the scope of the school curriculum, but with the help of which factoring a polynomial is not difficult. Read this article to the end and watch the attached video tutorial. The knowledge you gain will help you in the exam.
Factoring a polynomial using the division method
In the event that you received a polynomial greater than the second degree and were able to guess the value of the variable at which this polynomial becomes equal to zero (for example, this value is equal to ), know! This polynomial can be divided by .
For example, it is easy to see that a fourth-degree polynomial vanishes at . This means that it can be divided without a remainder by , thereby obtaining a polynomial of third degree (less by one). That is, present it in the form:
Where A, B, C And D- some numbers. Let's expand the brackets:
Since the coefficients for the same degrees must be the same, we get:
So, we got:
Let's move on. It is enough to go through several small integers to see that the third-degree polynomial is again divisible by . This results in a polynomial of the second degree (less by one). Then move on to a new entry:
Where E, F And G- some numbers. We open the brackets again and arrive at the following expression:
Again, from the condition of equality of coefficients for the same degrees, we obtain:
Then we get:
That is, the original polynomial can be factorized as follows:
In principle, if desired, using the difference of squares formula, the result can also be represented in the following form:
Here is a simple and effective way to factor polynomials. Remember it, it may be useful to you in an exam or math competition. Check if you have learned how to use this method. Try to solve the following task yourself.
Factor the polynomial:
Write your answers in the comments.
Material prepared by Sergey Valerievich
8 examples of factoring polynomials are given. They include examples of solving quadratic and biquadratic equations, examples of reciprocal polynomials, and examples of finding integer roots of third- and fourth-degree polynomials.
1. Examples with solving a quadratic equation
Example 1.1
x 4 + x 3 - 6 x 2.
Solution
We take out x 2
outside the brackets:
.
2 + x - 6 = 0:
.
Roots of the equation:
, .
.
Answer
Example 1.2
Factor the third degree polynomial:
x 3 + 6 x 2 + 9 x.
Solution
Let's take x out of brackets:
.
Solving the quadratic equation x 2 + 6 x + 9 = 0:
Its discriminant: .
Since the discriminant is zero, the roots of the equation are multiples: ;
.
From this we obtain the factorization of the polynomial:
.
Answer
Example 1.3
Factor the fifth degree polynomial:
x 5 - 2 x 4 + 10 x 3.
Solution
We take out x 3
outside the brackets:
.
Solving the quadratic equation x 2 - 2 x + 10 = 0.
Its discriminant: .
Since the discriminant is less than zero, the roots of the equation are complex: ;
, .
The factorization of the polynomial has the form:
.
If we are interested in factorization with real coefficients, then:
.
Answer
Examples of factoring polynomials using formulas
Examples with biquadratic polynomials
Example 2.1
Factor the biquadratic polynomial:
x 4 + x 2 - 20.
Solution
Let's apply the formulas:
a 2 + 2 ab + b 2 = (a + b) 2;
a 2 - b 2 = (a - b)(a + b).
;
.
Answer
Example 2.2
Factor the polynomial that reduces to a biquadratic one:
x 8 + x 4 + 1.
Solution
Let's apply the formulas:
a 2 + 2 ab + b 2 = (a + b) 2;
a 2 - b 2 = (a - b)(a + b):
;
;
.
Answer
Example 2.3 with recurrent polynomial
Factor the reciprocal polynomial:
.
Solution
A reciprocal polynomial has odd degree. Therefore it has root x = - 1
. Divide the polynomial by x - (-1) = x + 1. As a result we get:
.
Let's make a substitution:
, ;
;
;
.
Answer
Examples of factoring polynomials with integer roots
Example 3.1
Factor the polynomial:
.
Solution
Let's assume that the equation
6
-6, -3, -2, -1, 1, 2, 3, 6
.
(-6) 3 - 6·(-6) 2 + 11·(-6) - 6 = -504;
(-3) 3 - 6·(-3) 2 + 11·(-3) - 6 = -120;
(-2) 3 - 6·(-2) 2 + 11·(-2) - 6 = -60;
(-1) 3 - 6·(-1) 2 + 11·(-1) - 6 = -24;
1 3 - 6 1 2 + 11 1 - 6 = 0;
2 3 - 6 2 2 + 11 2 - 6 = 0;
3 3 - 6 3 2 + 11 3 - 6 = 0;
6 3 - 6 6 2 + 11 6 - 6 = 60.
So, we found three roots:
x 1 = 1
, x 2 = 2
, x 3 = 3
.
Since the original polynomial is of the third degree, it has no more than three roots. Since we found three roots, they are simple. Then
.
Answer
Example 3.2
Factor the polynomial:
.
Solution
Let's assume that the equation
has at least one whole root. Then it is a divisor of the number 2
(member without x). That is, the whole root can be one of the numbers:
-2, -1, 1, 2
.
We substitute these values one by one:
(-2) 4 + 2·(-2) 3 + 3·(-2) 3 + 4·(-2) + 2 = 6
;
(-1) 4 + 2·(-1) 3 + 3·(-1) 3 + 4·(-1) + 2 = 0
;
1 4 + 2 1 3 + 3 1 3 + 4 1 + 2 = 12;
2 4 + 2 2 3 + 3 2 3 + 4 2 + 2 = 54
.
If we assume that this equation has an integer root, then it is a divisor of the number 2
(member without x). That is, the whole root can be one of the numbers:
1, 2, -1, -2
.
Let's substitute x = -1
:
.
So, we have found another root x 2
= -1
. It would be possible, as in the previous case, to divide the polynomial by , but we will group the terms:
.
Since the equation x 2 + 2 = 0 has no real roots, then the factorization of the polynomial has the form.
Let's look at specific examples of how to factor a polynomial.
We will expand the polynomials in accordance with .
Factor polynomials:
Let's check if there is a common factor. yes, it is equal to 7cd. Let's take it out of brackets:
The expression in parentheses consists of two terms. There is no longer a common factor, the expression is not a formula for the sum of cubes, which means the decomposition is complete.
Let's check if there is a common factor. No. The polynomial consists of three terms, so we check to see if there is a formula for a complete square. Two terms are the squares of the expressions: 25x²=(5x)², 9y²=(3y)², the third term is equal to the double product of these expressions: 2∙5x∙3y=30xy. This means that this polynomial is a perfect square. Since the double product has a minus sign, it is:
We check whether it is possible to take the common factor out of brackets. There is a common factor, it is equal to a. Let's take it out of brackets:
There are two terms in brackets. We check if there is a formula for difference of squares or difference of cubes. a² is the square of a, 1=1². This means that the expression in brackets can be written using the difference of squares formula:
There is a common factor, it is equal to 5. Let’s take it out of brackets:
in brackets there are three terms. We check whether the expression is a perfect square. Two terms are squares: 16=4² and a² - the square of a, the third term is equal to the double product of 4 and a: 2∙4∙a=8a. Therefore, it is a perfect square. Since all terms have a “+” sign, the expression in parentheses is the perfect square of the sum:
We take the general multiplier -2x out of brackets:
In parentheses is the sum of two terms. We check whether this expression is a sum of cubes. 64=4³, x³- cube x. This means that the binomial can be expanded using the formula:
There is a common multiplier. But, since the polynomial consists of 4 terms, we will first, and only then, take the common factor out of brackets. Let’s group the first term with the fourth, and the second with the third:
From the first brackets we take out the common factor 4a, from the second - 8b:
There is no common multiplier yet. To get it, we take out the “-“ from the second brackets, and each sign in the brackets changes to the opposite:
Now let’s take the common factor (1-3a) out of brackets:
In the second brackets there is a common factor 4 (this is the same factor that we did not put out of brackets at the beginning of the example):
Since the polynomial consists of four terms, we perform grouping. Let’s group the first term with the second, the third with the fourth:
In the first brackets there is no common factor, but there is a formula for the difference of squares, in the second brackets the common factor is -5:
A common multiplier has appeared (4m-3n). Let's take it out of the equation.
Very often, the numerator and denominator of a fraction are algebraic expressions that must first be factored, and then, having found identical ones among them, divide both the numerator and denominator by them, that is, reduce the fraction. An entire chapter of the 7th grade algebra textbook is devoted to the task of factoring a polynomial. Factorization can be done 3 ways, as well as a combination of these methods.
1. Application of abbreviated multiplication formulas
As is known, to multiply a polynomial by a polynomial, you need to multiply each term of one polynomial by each term of the other polynomial and add the resulting products. There are at least 7 (seven) frequently occurring cases of multiplying polynomials that are included in the concept. For example,
Table 1. Factorization in the 1st way
2. Taking the common factor out of brackets
This method is based on the application of the distributive multiplication law. For example,
We divide each term of the original expression by the factor that we take out, and we get an expression in parentheses (that is, the result of dividing what was by what we take out remains in parentheses). First of all you need determine the multiplier correctly, which must be taken out of the bracket.
The common factor can also be a polynomial in brackets:
When performing the “factorize” task, you need to be especially careful with the signs when putting the total factor out of brackets. To change the sign of each term in a parenthesis (b - a), let’s take the common factor out of brackets -1 , and each term in the bracket will be divided by -1: (b - a) = - (a - b) .
If the expression in brackets is squared (or to any even power), then numbers inside brackets can be swapped completely freely, since the minuses taken out of brackets will still turn into a plus when multiplied: (b - a) 2 = (a - b) 2, (b - a) 4 = (a - b) 4 and so on…
3. Grouping method
Sometimes not all terms in an expression have a common factor, but only some. Then you can try group terms in brackets so that a factor can be taken out of each one. Grouping method- this is a double removal of common factors from brackets.
4. Using several methods at once
Sometimes you need to apply not one, but several methods of factoring a polynomial at once.
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