In the general case, this task involves a creative approach, since there is no universal method for solving it. However, let's try to give a few hints.
In the vast majority of cases, the decomposition of a polynomial into factors is based on the consequence of the Bezout theorem, that is, the root is found or selected and the degree of the polynomial is reduced by one by dividing by. The resulting polynomial is searched for a root and the process is repeated until complete expansion.
If the root cannot be found, then specific decomposition methods are used: from grouping to introducing additional mutually exclusive terms.
Further presentation is based on the skills of solving equations higher degrees with integer coefficients.
Bracketing the common factor.
Let's start with the simplest case, when the free term is equal to zero, that is, the polynomial has the form .
Obviously, the root of such a polynomial is , that is, the polynomial can be represented as .
This method is nothing but taking the common factor out of brackets.
Example.
Decompose a polynomial of the third degree into factors.
Solution.
It is obvious that is the root of the polynomial, that is, X can be bracketed:
Let's find the roots square trinomial 
Thus, 
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Factorization of a polynomial with rational roots.
First, consider the method of expanding a polynomial with integer coefficients of the form , the coefficient at the highest degree is equal to one.
In this case, if the polynomial has integer roots, then they are divisors of the free term.
Example.
Solution.
Let's check if there are integer roots. To do this, we write out the divisors of the number -18
: . That is, if the polynomial has integer roots, then they are among the numbers written out. Let's check these numbers sequentially according to Horner's scheme. Its convenience also lies in the fact that in the end we will also obtain the expansion coefficients of the polynomial: 
That is, x=2 And x=-3 are the roots of the original polynomial and it can be represented as a product:
It remains to expand the square trinomial.
The discriminant of this trinomial is negative, hence it has no real roots.
Answer:
Comment:
instead of Horner's scheme, one could use the selection of a root and the subsequent division of a polynomial by a polynomial.
Now consider the expansion of a polynomial with integer coefficients of the form , and the coefficient at the highest degree is not equal to one.
In this case, the polynomial can have fractionally rational roots.
Example.
Factorize the expression.
Solution.
By changing the variable y=2x, we pass to a polynomial with a coefficient equal to one at the highest degree. To do this, we first multiply the expression by 4
.
If the resulting function has integer roots, then they are among the divisors of the free term. Let's write them down:
Calculate sequentially the values of the function g(y) at these points until reaching zero. 
That is, y=-5 is the root 
, therefore, is the root of the original function. Let's carry out the division by a column (corner) of a polynomial by a binomial. 
Thus, 
It is not advisable to continue checking the remaining divisors, since it is easier to factorize the resulting square trinomial 
Hence, 
Unknown polynomials. The theorem about the distribution of the polynomial in dobutok unknowing. Canonical layout of the polynomial.
Very often, the numerator and denominator of a fraction are algebraic expressions that must first be decomposed into factors, and then, having found the same among them, divide both the numerator and the denominator into them, that is, reduce the fraction. A whole chapter of a textbook on algebra in the 7th grade is devoted to tasks to factorize a polynomial. Factoring 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) common cases of multiplication of polynomials that are included in the concept. For example,
Table 1. Factorization in the 1st way
2. Taking the common factor out of the bracket
This method is based on the application of the distributive law of multiplication. For example,
We divide each term of the original expression by the factor that we take out, and at the same time we get the expression in brackets (that is, the result of dividing what was by what we take out remains in brackets). First of all, you need correctly determine the multiplier, which must be bracketed.
The polynomial in brackets can also be a common factor:
When performing the “factorize” task, one must be especially careful with the signs when taking the common factor out of brackets. To change the sign of each term in a parenthesis (b - a), we take out the common factor -1 , while each term in the bracket is divided by -1: (b - a) = - (a - b) .
In the event that the expression in brackets is squared (or any even degree), That numbers inside brackets can be swapped completely free, 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 the expression have a common factor, but only some. Then you can try group terms in brackets so that some factor can be taken out of each. Grouping method is double bracketing of common factors.
4. Using several methods at once
Sometimes you need to apply not one, but several ways to factorize a polynomial into factors at once.
This is a synopsis on the topic. "Factorization". Choose next steps:
- Go to the next abstract:
Expanding polynomials to get a product sometimes seems confusing. But it is not so difficult if you understand the process step by step. The article details how to factorize a square trinomial.
Many do not understand how to factorize a square trinomial, and why this is done. At first it may seem that this is a useless exercise. But in mathematics, nothing is done just like that. The transformation is necessary to simplify the expression and the convenience of calculation.
A polynomial having the form - ax² + bx + c, is called a square trinomial. The term "a" must be negative or positive. In practice, this expression is called a quadratic equation. Therefore, sometimes they say differently: how to expand a quadratic equation.
Interesting! A square polynomial is called because of its largest degree - a square. And a trinomial - because of the 3 component terms.
Some other kinds of polynomials:
- linear binomial (6x+8);
- cubic quadrilateral (x³+4x²-2x+9).
Factorization of a square trinomial
First, the expression is equal to zero, then you need to find the values of the roots x1 and x2. There may be no roots, there may be one or two roots. The presence of roots is determined by the discriminant. Its formula must be known by heart: D=b²-4ac.
If the result of D is negative, there are no roots. If positive, there are two roots. If the result is zero, the root is one. The roots are also calculated by the formula.
If the calculation of the discriminant results in zero, you can apply any of the formulas. In practice, the formula is simply abbreviated: -b / 2a.
Formulas for different values of the discriminant are different.
If D is positive:
If D is zero:
Online calculators
The Internet has online calculator. It can be used to factorize. Some resources provide the opportunity to see the solution step by step. Such services help to better understand the topic, but you need to try to understand well.
Useful video: Factoring a square trinomial
Examples
We invite you to view simple examples how to factorize a quadratic equation.
Example 1
Here it is clearly shown that the result will be two x, because D is positive. They need to be substituted into the formula. If the roots are negative, the sign in the formula is reversed.
We know the formula for factoring a square trinomial: a(x-x1)(x-x2). We put the values in brackets: (x+3)(x+2/3). There is no number before the term in the exponent. This means that there is a unit, it is lowered.
Example 2
This example clearly shows how to solve an equation that has one root.
Substitute the resulting value:
Example 3
Given: 5x²+3x+7
First, we calculate the discriminant, as in the previous cases.
D=9-4*5*7=9-140= -131.
The discriminant is negative, which means there are no roots.
After receiving the result, it is worth opening the brackets and checking the result. The original trinomial should appear.
Alternative solution
Some people have never been able to make friends with the discriminant. There is another way to factorize a square trinomial. For convenience, the method is shown in an example.
Given: x²+3x-10
We know that we should end up with 2 parentheses: (_)(_). When the expression looks like this: x² + bx + c, we put x at the beginning of each bracket: (x_) (x_). The remaining two numbers are the product that gives "c", i.e. -10 in this case. To find out what these numbers are, you can only use the selection method. Substituted numbers must match the remaining term.
For example, multiplying the following numbers gives -10:
- -1, 10;
- -10, 1;
- -5, 2;
- -2, 5.
- (x-1)(x+10) = x2+10x-x-10 = x2+9x-10. No.
- (x-10)(x+1) = x2+x-10x-10 = x2-9x-10. No.
- (x-5)(x+2) = x2+2x-5x-10 = x2-3x-10. No.
- (x-2)(x+5) = x2+5x-2x-10 = x2+3x-10. Fits.
So, the transformation of the expression x2+3x-10 looks like this: (x-2)(x+5).
Important! You should be careful not to confuse the signs.
Decomposition of a complex trinomial
If "a" is greater than one, difficulties begin. But everything is not as difficult as it seems.
In order to factorize, one must first see if it is possible to factor something out.
For example, given the expression: 3x²+9x-30. Here the number 3 is taken out of brackets:
3(x²+3x-10). The result is the already known trinomial. The answer looks like this: 3(x-2)(x+5)
How to decompose if the term that is squared is negative? In this case, the number -1 is taken out of the bracket. For example: -x²-10x-8. The expression will then look like this:
The scheme differs little from the previous one. There are only a few new things. Let's say the expression is given: 2x²+7x+3. The answer is also written in 2 brackets, which must be filled in (_) (_). X is written in the 2nd bracket, and what is left in the 1st. It looks like this: (2x_)(x_). Otherwise, the previous scheme is repeated.
The number 3 gives the numbers:
- -1, -3;
- -3, -1;
- 3, 1;
- 1, 3.
We solve equations by substituting the given numbers. The last option fits. So the transformation of the expression 2x²+7x+3 looks like this: (2x+1)(x+3).
Other cases
It is not always possible to transform an expression. In the second method, the solution of the equation is not required. But the possibility of converting terms into a product is checked only through the discriminant.
Worth the practice of deciding quadratic equations so that there are no difficulties when using formulas.
Useful video: factorization of a trinomial
Conclusion
You can use it in any way. But it is better to work both to automatism. Also, those who are going to connect their lives with mathematics need to learn how to solve quadratic equations well and decompose polynomials into factors. All the following mathematical topics are built on this.
In contact with
This online calculator is designed to factorize a function.
For example, factorize: x 2 /3-3x+12 . Let's write it as x^2/3-3*x+12 . You can also use this service, where all calculations are saved in Word format.
For example, decompose into terms. Let's write it as (1-x^2)/(x^3+x) . To see the progress of the solution, click Show steps . If you need to get the result in Word format, use this service.
Note: the number "pi" (π) is written as pi ; square root as sqrt , e.g. sqrt(3) , the tangent of tg is written as tan . See the Alternative section for a response.
- If a simple expression is given, for example, 8*d+12*c*d , then factoring the expression means to factor the expression. To do this, you need to find common factors. We write this expression as: 4*d*(2+3*c) .
- Express the product as two binomials: x 2 + 21yz + 7xz + 3xy . Here we already need to find several common factors: x(x + 7z) + 3y(x + 7z). We take out (x+7z) and get: (x+7z)(x + 3y) .
see also Division of polynomials by a corner (all steps of division by a column are shown)
Useful in learning the rules of factorization are abbreviated multiplication formulas, with which it will be clear how to open brackets with a square:
- (a+b) 2 = (a+b)(a+b) = a 2 +2ab+b 2
- (a-b) 2 = (a-b)(a-b) = a 2 -2ab+b 2
- (a+b)(a-b) = a 2 - b 2
- a 3 +b 3 = (a+b)(a 2 -ab+b 2)
- a 3 -b 3 = (a-b)(a 2 +ab+b 2)
- (a+b) 3 = (a+b)(a+b) 2 = a 3 +3a 2 b + 3ab 2 +b 3
- (a-b) 3 = (a-b)(a-b) 2 = a 3 -3a 2 b + 3ab 2 -b 3
Factoring methods
After learning a few tricks factorization solutions can be classified as follows:- Using abbreviated multiplication formulas.
- Search for a common factor.
This is one of the most elementary ways to simplify an expression. To apply this method, let's remember the distributive law of multiplication with respect to addition (do not be afraid of these words, you definitely know this law, you just might have forgotten its name).
The law says: in order to multiply the sum of two numbers by a third number, you need to multiply each term by this number and add the results, in other words,.
You can also do the reverse operation, and it is this reverse operation that interests us. As can be seen from the sample, the common factor a, can be taken out of the bracket.
A similar operation can be done both with variables, such as and, for example, and with numbers: .
Yes, this is too elementary an example, just like the example given earlier, with the expansion of a number, because everyone knows what numbers are, and are divisible by, but what if you got a more complicated expression:
How to find out what, for example, a number is divided into, no, with a calculator, anyone can, but without it it’s weak? And for this there are signs of divisibility, these signs are really worth knowing, they will help you quickly understand whether it is possible to take the common factor out of the bracket.
Signs of divisibility
It is not so difficult to remember them, most likely, most of them were already familiar to you, and something will be a new useful discovery, more details in the table:
Note: The table lacks a sign of divisibility by 4. If the last two digits are divisible by 4, then the whole number is divisible by 4.
Well, how do you like the sign? I advise you to remember it!
Well, let's get back to the expression, maybe take it out of the bracket and that's enough from it? No, it is customary for mathematicians to simplify, so to the fullest, take out EVERYTHING that is taken out!
And so, everything is clear with the player, but what about the numerical part of the expression? Both numbers are odd, so you can't divide by
You can use the sign of divisibility by, the sum of the digits, and, of which the number consists, is equal, and is divisible by, which means it is divisible by.
Knowing this, you can safely divide into a column, as a result of dividing by we get (signs of divisibility came in handy!). Thus, we can take the number out of the bracket, just like y, and as a result we have:
To make sure that everything is decomposed correctly, you can check the expansion by multiplication!
Also, the common factor can be taken out in power expressions. Here, for example, do you see the common factor?
All members of this expression have x's - we take out, all are divided by - we take out again, we look at what happened: .
2. Abbreviated multiplication formulas
Abbreviated multiplication formulas have already been mentioned in theory, if you can hardly remember what it is, then you should refresh them in your memory.
Well, if you consider yourself very smart and you are too lazy to read such a cloud of information, then just read on, look at the formulas and immediately take on the examples.
The essence of this decomposition is to notice some definite formula in the expression before you, apply it and thus obtain the product of something and something, that's all the decomposition. Following are the formulas:
Now try factoring the following expressions using the above formulas:
And here is what should have happened:
As you have noticed, these formulas are a very effective way of factoring, it is not always suitable, but it can be very useful!
3. Grouping or grouping method
Here's another example for you:
Well, what are you going to do with it? It seems to be divisible by and into something, and something into and into
But you can’t divide everything together into one thing, well there is no common factor, how not to look for what, and leave it without factoring?
Here you need to show ingenuity, and the name of this ingenuity is a grouping!
It is applied when common divisors Not all members have. For grouping you need find groups of terms that have common divisors and rearrange them so that the same multiplier can be obtained from each group.
Of course, it is not necessary to rearrange in places, but this gives visibility, for clarity, you can take individual parts of the expression in brackets, it is not forbidden to put them as much as you like, the main thing is not to confuse the signs.
All this is not very clear? Let me explain with an example:
In a polynomial - put a member - after the member - we get
we group the first two terms together in a separate bracket and group the third and fourth terms in the same way, leaving the minus sign out of the bracket, we get:
And now we look separately at each of the two "heaps" into which we have broken the expression with brackets.
The trick is to break it into such piles from which it will be possible to take out the largest possible factor, or, as in this example, try to group the members so that after taking the factors out of the brackets from the piles, we have the same expressions inside the brackets.
From both brackets we take out the common factors of the members, from the first bracket, and from the second bracket, we get:
But it's not decomposition!
Pdonkey decomposition should remain only multiplication, but for now we have a polynomial simply divided into two parts ...
BUT! This polynomial has a common factor. This
outside the bracket and we get the final product
Bingo! As you can see, there is already a product and outside the brackets there is neither addition nor subtraction, the decomposition is completed, because we have nothing more to take out of the brackets.
It may seem like a miracle that after taking the factors out of the brackets, we still have the same expressions in the brackets, which, again, we took out of the brackets.
And this is not a miracle at all, the fact is that the examples in textbooks and in the exam are specially made in such a way that most of the expressions in tasks for simplification or factorization with the right approach to them, they are easily simplified and abruptly collapse like an umbrella when you press a button, so look for that very button in each expression.
Something I digress, what do we have there with simplification? The intricate polynomial took on a simpler form: .
Agree, not as bulky as it used to be?
4. Selection of a full square.
Sometimes, in order to apply the formulas for abbreviated multiplication (repeat the topic), it is necessary to transform the existing polynomialby presenting one of its terms as the sum or difference of two terms.
In which case you have to do this, you will learn from the example:
A polynomial in this form cannot be decomposed using abbreviated multiplication formulas, so it must be converted. Perhaps at first it will not be obvious to you which term to divide into which, but over time you will learn to immediately see the formulas for abbreviated multiplication, even if they are not present in their entirety, and you will quickly determine what is missing here to the full formula, but for now - learn , a student, more precisely a schoolboy.
For the full formula of the square of the difference, here you need instead. Let's represent the third term as a difference, we get: We can apply the difference square formula to the expression in brackets (not to be confused with the difference of squares!!!), we have: , to this expression, we can apply the formula for the difference of squares (not to be confused with the squared difference!!!), imagining how, we get: .
An expression not always factored into factors looks simpler and smaller than it was before decomposition, but in this form it becomes more mobile, in the sense that you can not worry about changing signs and other mathematical nonsense. Well, here's for you independent decision, the following expressions must be factored.
Examples:
Answers:
5. Factorization of a square trinomial
For the factorization of a square trinomial, see below in the decomposition examples.
Examples of 5 Methods for Factoring a Polynomial
1. Taking the common factor out of brackets. Examples.
Do you remember what the distributive law is? This is such a rule:
Example:
Factorize a polynomial.
Solution:
Another example:
Multiply.
Solution:
If the whole term is taken out of brackets, one remains in brackets instead of it!
2. Formulas for abbreviated multiplication. Examples.
The most commonly used formulas are the difference of squares, the difference of cubes and the sum of cubes. Remember these formulas? If not, urgently repeat the topic!
Example:
Factor the expression.
Solution:
In this expression, it is easy to find out the difference of cubes:
Example:
Solution:
3. Grouping method. Examples
Sometimes it is possible to interchange the terms in such a way that one and the same factor can be extracted from each pair of neighboring terms. This common factor can be taken out of the bracket and the original polynomial will turn into a product.
Example:
Factor out the polynomial.
Solution:
We group the terms as follows:
.
In the first group, we take the common factor out of brackets, and in the second - :
.
Now the common factor can also be taken out of brackets:
.
4. The method of selection of a full square. Examples.
If the polynomial can be represented as the difference of the squares of two expressions, all that remains is to apply the abbreviated multiplication formula (difference of squares).
Example:
Factor out the polynomial.
Solution:Example:
\begin(array)(*(35)(l))
((x)^(2))+6(x)-7=\underbrace(((x)^(2))+2\cdot 3\cdot x+9)_(square\ sums\ ((\left (x+3 \right))^(2)))-9-7=((\left(x+3 \right))^(2))-16= \\
=\left(x+3+4 \right)\left(x+3-4 \right)=\left(x+7 \right)\left(x-1 \right) \\
\end(array)
Factor out the polynomial.
Solution:
\begin(array)(*(35)(l))
((x)^(4))-4((x)^(2))-1=\underbrace(((x)^(4))-2\cdot 2\cdot ((x)^(2) )+4)_(square\ differences((\left(((x)^(2))-2 \right))^(2)))-4-1=((\left(((x)^ (2))-2 \right))^(2))-5= \\
=\left(((x)^(2))-2+\sqrt(5) \right)\left(((x)^(2))-2-\sqrt(5) \right) \\
\end(array)
5. Factorization of a square trinomial. Example.
A square trinomial is a polynomial of the form, where is an unknown, are some numbers, moreover.
Variable values that turn the square trinomial to zero are called roots of the trinomial. Therefore, the roots of a trinomial are the roots of a quadratic equation.
Theorem.
Example:
Let's factorize the square trinomial: .
First, we solve the quadratic equation: Now we can write the factorization of this square trinomial into factors:
Now your opinion...
We have described in detail how and why to factorize a polynomial.
We gave a lot of examples of how to do it in practice, pointed out the pitfalls, gave solutions ...
What do you say?
How do you like this article? Do you use these tricks? Do you understand their essence?
Write in the comments and... get ready for the exam!
So far, it's the most important thing in your life.
