Factoring a polynomial. Methods for factoring a polynomial of degree higher than two Expansion of square brackets
This is one of the most basic ways to simplify an expression. To apply this method, let’s remember the distributive law of multiplication relative to addition (don’t 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 resulting 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 a very elementary example, just like the example given earlier, with the decomposition of a number, because everyone knows that numbers are divisible by, but what if you get a more complicated expression:
How do you find out what, for example, a number is divisible by? No, anyone can do it with a calculator, but without it it’s difficult? And for this there are signs of divisibility, these signs are really worth knowing, they will help you quickly understand whether the common factor can be taken out of the bracket.
Signs of divisibility
It’s not so difficult to remember them; most likely, most of them were already familiar to you, and some will be a new useful discovery, more details in the table:
Note: The table is missing the divisibility test by 4. If the last two digits are divisible by 4, then the entire number is divisible by 4.
Well, how do you like the sign? I advise you to remember it!
Well, let's return to the expression, maybe he can take it out of the bracket and that's enough? No, mathematicians tend to simplify, so to the fullest, endure EVERYTHING that is endured!
And so, everything is clear with the game, but what about the numerical part of the expression? Both numbers are odd, so you can't divide by
You can use the divisibility test: the sum of the digits, and, that make up the number is equal, and divisible by, means divisible by.
Knowing this, you can safely divide into a column, and as a result of dividing by we get (the signs of divisibility are useful!). Thus, we can take the number out of brackets, just like y, and as a result we have:
To make sure that everything has been expanded correctly, you can check the expansion by multiplying!
The common factor can also be expressed in power terms. Here, for example, do you see the common multiplier?
All members of this expression have xes - we take them out, they are all divided by - we take them out again, look at what happened: .
2. Abbreviated multiplication formulas
Abbreviated multiplication formulas have already been mentioned in theory; if you have difficulty remembering what they are, then you should refresh your memory.
Well, if you consider yourself very smart and 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 a certain formula in the expression in front of you, apply it and thus obtain the product of something and something, that’s all the decomposition. The following are the formulas:
Now try, factor the following expressions using the above formulas:
Here's 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:
So what are you going to do with it? It seems that something is divided into and into, and something into and into
But you can’t divide everything together into one thing, well there is no common factor here, no matter how you look, what should you leave it like that, without factoring it into factors?
Here you need to show ingenuity, and the name of this ingenuity is grouping!
It is used precisely when not all members have common divisors. For grouping you need find groups of terms that have common factors and rearrange them so that the same factor can be obtained from each group.
Of course, it is not necessary to rearrange them, but this gives clarity; for clarity, you can put 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.
Is all this not very clear? Let me explain with an example:
In a polynomial - we put the term - after the term - we get
we group the first two terms together in a separate bracket and also group the third and fourth terms, taking the minus sign out of the bracket, we get:
Now we look separately at each of the two “piles” into which we divided the expression with brackets.
The trick is to break it down into piles from which the largest factor can be taken out, or, as in this example, try to group the terms so that after removing the factors from the piles out of brackets, we still have the same expressions inside the brackets.
From both brackets we take out the common factors of the terms, from the first bracket, and from the second, we get:
But this is not decomposition!
Pdonkey decomposition should only remain multiplication, but for now our polynomial is simply divided into two parts...
BUT! This polynomial has a common factor. This
beyond the bracket and we get the final product
Bingo! As you can see, there is already a product here and outside the brackets there is no addition or subtraction, the decomposition is complete, because We have nothing more to take out of brackets.
It may seem like a miracle that after taking the factors out of brackets, we were left with identical expressions in brackets, which we again put out of brackets.
And this is not a miracle at all, the fact is that the examples in textbooks and in the Unified State Examination are specially made so that most expressions in tasks for simplification or factorization with the right approach to them, they are easily simplified and collapse sharply like an umbrella when you press a button, so look for that very button in every expression.
I got distracted, what are we doing with simplification? The intricate polynomial took on a simpler form: .
Agree, it’s not as bulky as it was?
4. Selecting a complete square.
Sometimes, to apply abbreviated multiplication formulas (repeat the topic), it is necessary to transform an existing polynomial, presenting one of its terms as the sum or difference of two terms.
In what case you have to do this, you will learn from the example:
A polynomial in this form cannot be expanded using abbreviated multiplication formulas, so it must be transformed. Perhaps at first it will not be obvious to you which term should be divided into which, but over time you will learn to immediately see the formulas for abbreviated multiplication, even if they are not entirely present, and you will quickly determine what is missing from the full formula, but for now - learn , a student, or rather a schoolboy.
For the complete formula for the squared difference, here you need instead. Let's imagine the third term as a difference, we get: To the expression in parentheses, you can apply the formula for the square of the difference (not to be confused with the difference of squares!!!), we have: , to this expression we can apply the difference of squares formula (not to be confused with the squared difference!!!), imagining how, we get: .
An expression factorized does not always look simpler and smaller than it was before the expansion, but in this form it becomes more flexible, in the sense that you don’t have to worry about changing signs and other mathematical nonsense. Well, for you to decide for yourself, the following expressions need to be factorized.
Examples:
Answers:
5. Factoring a quadratic trinomial
For the decomposition of a quadratic trinomial into factors, see further examples of decomposition.
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 the rule:
Example:
Factor the polynomial.
Solution:
Another example:
Factor it out.
Solution:
If the entire term is taken out of the brackets, a unit remains in the brackets instead!
2. Abbreviated multiplication formulas. Examples.
The formulas we most often use are difference of squares, difference of cubes and sum of cubes. Do you remember these formulas? If not, repeat the topic urgently!
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 you can swap terms so that the same factor can be extracted from each pair of adjacent terms. This common factor can be taken out of the bracket and the original polynomial will turn into a product.
Example:
Factor the polynomial.
Solution:
Let's 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. Method for selecting a complete 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 the polynomial.
Solution:Example:
\begin(array)(*(35)(l))
((x)^(2))+6(x)-7=\underbrace(((x)^(2))+2\cdot 3\cdot x+9)_(square\ sum\ ((\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 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. Factoring a quadratic trinomial. Example.
A square trinomial is a polynomial of the form, where - the unknown, - some numbers, and.
The values of the variable that make the quadratic trinomial vanish are called the roots of the trinomial. Therefore, the roots of a trinomial are the roots of a quadratic equation.
Theorem.
Example:
Let's factorize the quadratic trinomial: .
First, let's solve the quadratic equation: Now we can write the factorization of this quadratic trinomial:
Now your opinion...
We have described in detail how and why to factor a polynomial.
We gave a lot of examples of how to do this in practice, pointed out pitfalls, gave solutions...
What do you say?
What do you think of this article? Do you use these techniques? Do you understand their essence?
Write in the comments and... prepare for the exam!
So far he is the most important in your life.
A polynomial is an expression consisting of the sum of monomials. The latter are the product of a constant (number) and the root (or roots) of the expression to the power of k. In this case, we speak of a polynomial of degree k. The expansion of a polynomial involves a transformation of the expression in which the terms are replaced by factors. Let's consider the main ways to carry out this kind of transformation.
Method of expanding a polynomial by isolating a common factor
This method is based on the laws of the distribution law. So, mn + mk = m * (n + k).
- Example: expand 7y 2 + 2uy and 2m 3 – 12m 2 + 4lm.
7y 2 + 2uy = y * (7y + 2u),
2m 3 – 12m 2 + 4lm = 2m(m 2 – 6m + 2l).
However, the factor that is necessarily present in each polynomial may not always be found, so this method is not universal.
Polynomial expansion method based on abbreviated multiplication formulas
Abbreviated multiplication formulas are valid for polynomials of any degree. In general, the transformation expression looks like this:
u k – l k = (u – l)(u k-1 + u k-2 * l + u k-3 *l 2 + … u * l k-2 + l k-1), where k is a representative of natural numbers .
The formulas most often used in practice are for polynomials of the second and third orders:
u 2 – l 2 = (u – l)(u + l),
u 3 – l 3 = (u – l)(u 2 + ul + l 2),
u 3 + l 3 = (u + l)(u 2 – ul + l 2).
- Example: expand 25p 2 – 144b 2 and 64m 3 – 8l 3.
25p 2 – 144b 2 = (5p – 12b)(5p + 12b),
64m 3 – 8l 3 = (4m) 3 – (2l) 3 = (4m – 2l)((4m) 2 + 4m * 2l + (2l) 2) = (4m – 2l)(16m 2 + 8ml + 4l 2 ).
Polynomial expansion method - grouping terms of an expression
This method in some way has something in common with the technique of deriving the common factor, but has some differences. In particular, before isolating a common factor, the monomials should be grouped. The grouping is based on the rules of combinational and commutative laws.
All monomials presented in the expression are divided into groups, in each of which a common value is given such that the second factor will be the same in all groups. In general, this decomposition method can be represented as the expression:
pl + ks + kl + ps = (pl + ps) + (ks + kl) ⇒ pl + ks + kl + ps = p(l + s) + k(l + s),
pl + ks + kl + ps = (p + k)(l + s).
- Example: spread out 14mn + 16ln – 49m – 56l.
14mn + 16ln – 49m – 56l = (14mn – 49m) + (16ln – 56l) = 7m * (2n – 7) + 8l * (2n – 7) = (7m + 8l)(2n – 7).
Polynomial expansion method - forming a perfect square
This method is one of the most effective in the expansion of a polynomial. At the initial stage, it is necessary to determine monomials that can be “collapsed” into the square of the difference or sum. To do this, use one of the relations:
(p – b) 2 = p 2 – 2pb + b 2 ,
- Example: expand the expression u 4 + 4u 2 – 1.
Among its monomials, we select the terms that form a complete square: u 4 + 4u 2 – 1 = u 4 + 2 * 2u 2 + 4 – 4 – 1 =
= (u 4 + 2 * 2u 2 + 4) – 4 – 1 = (u 4 + 2 * 2u 2 + 4) – 5.
Complete the transformation using the abbreviated multiplication rules: (u 2 + 2) 2 – 5 = (u 2 + 2 – √5)(u 2 + 2 + √5).
That. u 4 + 4u 2 – 1 = (u 2 + 2 – √5)(u 2 + 2 + √5).
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 some 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.
This is a summary of the topic "Factorization". Choose next steps:
- Go to the next abstract:
Any algebraic polynomial of degree n can be represented as a product of n-linear factors of the form and a constant number, which is the coefficients of the polynomial at the highest degree x, i.e.
Where 
- are the roots of the polynomial.
The root of a polynomial is a number (real or complex) that turns the polynomial to zero. The roots of a polynomial can be both real roots and complex conjugate roots, then the polynomial can be represented in the following form:
Consider methods for expanding polynomials of degree "n" into the product of factors of the first and second degrees.
Method number 1.Method of undetermined coefficients.
The coefficients of such a transformed expression are determined by the method of indefinite coefficients. The essence of the method is that the type of factors into which the given polynomial is decomposed is known in advance. When using the method of indeterminate coefficients, the following statements are true:
P.1. Two polynomials are identically equal if their coefficients are equal at the same powers of x.
P.2. Any third-degree polynomial decomposes into a product of linear and square factors.
P.3. Any polynomial of the fourth degree decomposes into the product of two polynomials of the second degree.
Example 1.1. It is necessary to factorize the cubic expression:
P.1. In accordance with the accepted statements, the identical equality is true for the cubic expression:
P.2. The right side of the expression can be represented as terms as follows:
P.3. We compose a system of equations from the condition of equality of the coefficients for the corresponding powers of the cubic expression.
This system of equations can be solved by selecting coefficients (if it is a simple academic problem) or methods for solving nonlinear systems of equations can be used. Solving this system of equations, we obtain that the uncertain coefficients are defined as follows:
Thus, the original expression is decomposed into factors in the following form:
This method can be used both in analytical calculations and in computer programming to automate the process of finding the root of an equation.
Method number 2.Vieta formulas
Vieta's formulas are formulas connecting the coefficients of algebraic equations of degree n and its roots. These formulas were implicitly presented in the works of the French mathematician François Vieta (1540 - 1603). Due to the fact that Vieth considered only positive real roots, he therefore did not have the opportunity to write these formulas in a general explicit form.
For any algebraic polynomial of degree n that has n-real roots,
The following relations are valid that connect the roots of a polynomial with its coefficients:
Vieta's formulas are convenient to use to check the correctness of finding the roots of a polynomial, as well as to construct a polynomial from given roots.
Example 2.1. Let us consider how the roots of a polynomial are related to its coefficients using the example of a cubic equation
In accordance with Vieta’s formulas, the relationship between the roots of a polynomial and its coefficients has the following form:
Similar relations can be made for any polynomial of degree n.
Method No. 3. Factoring a quadratic equation with rational roots
From Vieta's last formula it follows that the roots of a polynomial are divisors of its free term and leading coefficient. In this regard, if the problem statement specifies a polynomial of degree n with integer coefficients
then this polynomial has a rational root (irreducible fraction), where p is the divisor of the free term, and q is the divisor of the leading coefficient. In this case, a polynomial of degree n can be represented as (Bezout’s theorem):
A polynomial, the degree of which is 1 less than the degree of the initial polynomial, is determined by dividing a polynomial of degree n binomial, for example, using Horner's scheme or in the simplest way - "column".
Example 3.1. It is necessary to factor the polynomial
P.1. Due to the fact that the coefficient of the highest term is equal to one, the rational roots of this polynomial are divisors of the free term of the expression, i.e. can be whole numbers 
. We substitute each of the presented numbers into the original expression and find that the root of the presented polynomial is equal to .
Let's divide the original polynomial by a binomial:
Let's use Horner's scheme
The coefficients of the original polynomial are set in the top line, while the first cell of the top line remains empty.
In the first cell of the second line, the found root is written (in the example under consideration, the number “2” is written), and the following values in the cells are calculated in a certain way and they are the coefficients of the polynomial, which is obtained by dividing the polynomial by the binomial. The unknown coefficients are determined as follows:
The value from the corresponding cell of the first row is transferred to the second cell of the second row (in the example under consideration, the number “1” is written).
The third cell of the second row contains the value of the product of the first cell and the second cell of the second row plus the value from the third cell of the first row (in the example under consideration 2 ∙ 1 -5 = -3).
The fourth cell of the second row contains the value of the product of the first cell and the third cell of the second row plus the value from the fourth cell of the first row (in the example under consideration, 2 ∙ (-3) +7 = 1).
Thus, the original polynomial is factorized:
Method number 4.Using Shorthand Multiplication Formulas
Abbreviated multiplication formulas are used to simplify calculations, as well as factoring polynomials. Abbreviated multiplication formulas make it possible to simplify the solution of individual problems.
Formulas Used for Factoring
In general, this task requires a creative approach, since there is no universal method for solving it. But let's try to give a few tips.
In the overwhelming majority of cases, the factorization of a polynomial is based on a corollary of Bezout’s theorem, that is, the root is found or selected and the degree of the polynomial is reduced by one by dividing by . The root of the resulting polynomial is sought and the process is repeated until complete expansion.
If the root cannot be found, then specific expansion methods are used: from grouping to introducing additional mutually exclusive terms.
Further presentation is based on the skills of solving equations of higher degrees with integer coefficients.
Bracketing out 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, we can represent the polynomial in the form .
This method is nothing more than putting the common factor out of brackets.
Example.
Factor a third degree polynomial.
Solution.
Obviously, what is the root of the polynomial, that is X can be taken out of brackets:
Let's find the roots of the quadratic trinomial 
Thus, 
Top of page
Factoring a polynomial with rational roots.
First, let's consider a method for expanding a polynomial with integer coefficients of the form , the coefficient of the highest degree is equal to one.
In this case, if a polynomial has integer roots, then they are divisors of the free term.
Example.
Solution.
Let's check if there are intact roots. To do this, write down the divisors of the number -18
: . That is, if a polynomial has integer roots, then they are among the written numbers. Let's check these numbers sequentially using Horner's scheme. Its convenience also lies in the fact that in the end we obtain the expansion coefficients of the polynomial: 
That is, x=2 And x=-3 are the roots of the original polynomial and we can represent it as a product:
It remains to expand the quadratic trinomial.
The discriminant of this trinomial is negative, therefore it has no real roots.
Answer:
Comment:
Instead of Horner's scheme, one could use the selection of the root and subsequent division of the polynomial by a polynomial.
Now consider the expansion of a polynomial with integer coefficients of the form , and the coefficient of the highest degree is not equal to one.
In this case, the polynomial can have fractionally rational roots.
Example.
Factor the expression.
Solution.
By performing a variable change y=2x, let's move on to a polynomial with a coefficient equal to one at the highest degree. To do this, 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:
Let us sequentially calculate the values of the function g(y) at these points until zero is reached. 
That is, y=-5 is the root 
, therefore, is the root of the original function. Let's divide the polynomial by a column (corner) into a binomial. 
Thus, 
It is not advisable to continue checking the remaining divisors, since it is easier to factorize the resulting quadratic trinomial 
Hence, 
Unknown polynomials. The theorem about the distribution of polynomials in additions of unknowns. Canonical layout of a polynomial.