What is factoring a polynomial? Complex cases of factoring polynomials
When solving equations and inequalities, it is often necessary to factor a polynomial whose degree is three or higher. In this article we will look at the easiest way to do this.
As usual, let's turn to theory for help.
Bezout's theorem states that the remainder when dividing a polynomial by a binomial is .
But what is important for us is not the theorem itself, but corollary from it:
If the number is the root of a polynomial, then the polynomial is divisible by the binomial without a remainder.
We are faced with the task of somehow finding at least one root of the polynomial, then dividing the polynomial by , where is the root of the polynomial. As a result, we obtain a polynomial whose degree is one less than the degree of the original one. And then, if necessary, you can repeat the process.
This task breaks down into two: how to find the root of a polynomial, and how to divide a polynomial by a binomial.
Let's take a closer look at these points.
1. How to find the root of a polynomial.
First, we check whether the numbers 1 and -1 are roots of the polynomial.
The following facts will help us here:
If the sum of all the coefficients of a polynomial is zero, then the number is the root of the polynomial.
For example, in a polynomial the sum of the coefficients is zero: . It's easy to check what the root of a polynomial is.
If the sum of the coefficients of a polynomial at even powers is equal to the sum of the coefficients at odd powers, then the number is the root of the polynomial. The free term is considered a coefficient for an even degree, since , a is an even number.
For example, in a polynomial the sum of coefficients for even powers is: , and the sum of coefficients for odd powers is: . It's easy to check what the root of a polynomial is.
If neither 1 nor -1 are roots of the polynomial, then we move on.
For a reduced polynomial of degree (that is, a polynomial in which the leading coefficient - the coefficient at - is equal to unity), the Vieta formula is valid:
Where are the roots of the polynomial.
There are also Vieta formulas concerning the remaining coefficients of the polynomial, but we are interested in this one.
From this Vieta formula it follows that if the roots of a polynomial are integers, then they are divisors of its free term, which is also an integer.
Based on this, we need to factor the free term of the polynomial into factors, and sequentially, from smallest to largest, check which of the factors is the root of the polynomial.
Consider, for example, the polynomial
Divisors of the free term: ; ; ;
The sum of all coefficients of a polynomial is equal to , therefore, the number 1 is not the root of the polynomial.
Sum of coefficients for even powers:
Sum of coefficients for odd powers:
Therefore, the number -1 is also not a root of the polynomial.
Let's check whether the number 2 is the root of the polynomial: therefore, the number 2 is the root of the polynomial. This means, according to Bezout’s theorem, the polynomial is divisible by a binomial without a remainder.
2. How to divide a polynomial into a binomial.
A polynomial can be divided into a binomial by a column.
Divide the polynomial by a binomial using a column:
There is another way to divide a polynomial by a binomial - Horner's scheme.
Watch this video to understand how to divide a polynomial by a binomial with a column, and using Horner's diagram.
I note that if, when dividing by a column, some degree of the unknown is missing in the original polynomial, we write 0 in its place - the same way as when compiling a table for Horner’s scheme.
So, if we need to divide a polynomial by a binomial and as a result of the division we get a polynomial, then we can find the coefficients of the polynomial using Horner’s scheme:
We can also use Horner scheme in order to check whether a given number is the root of a polynomial: if the number is the root of a polynomial, then the remainder when dividing the polynomial by is equal to zero, that is, in the last column of the second row of Horner’s diagram we get 0.
Using Horner's scheme, we "kill two birds with one stone": we simultaneously check whether the number is the root of a polynomial and divide this polynomial by a binomial.
Example. Solve the equation:
1. Let's write down the divisors of the free term and look for the roots of the polynomial among the divisors of the free term.
Divisors of 24:
2. Let's check whether the number 1 is the root of the polynomial.
The sum of the coefficients of a polynomial, therefore, the number 1 is the root of the polynomial.
3. Divide the original polynomial into a binomial using Horner's scheme.
A) Let’s write down the coefficients of the original polynomial in the first row of the table.
Since the containing term is missing, in the column of the table in which the coefficient should be written we write 0. On the left we write the found root: the number 1.
B) Fill in the first row of the table.
In the last column, as expected, we got zero; we divided the original polynomial by a binomial without a remainder. The coefficients of the polynomial resulting from division are shown in blue in the second row of the table:
It's easy to check that the numbers 1 and -1 are not roots of the polynomial
B) Let's continue the table. Let's check whether the number 2 is the root of the polynomial:
So the degree of the polynomial, which is obtained as a result of division by one, is less than the degree of the original polynomial, therefore, the number of coefficients and the number of columns are one less.
In the last column we got -40 - a number that is not equal to zero, therefore, the polynomial is divisible by a binomial with a remainder, and the number 2 is not the root of the polynomial.
C) Let's check whether the number -2 is the root of the polynomial. Since the previous attempt failed, to avoid confusion with the coefficients, I will erase the line corresponding to this attempt:
Great! We got zero as a remainder, therefore, the polynomial was divided into a binomial without a remainder, therefore, the number -2 is the root of the polynomial. The coefficients of the polynomial that is obtained by dividing a polynomial by a binomial are shown in green in the table.
As a result of division we get a quadratic trinomial 
, whose roots can easily be found using Vieta’s theorem:
So, the roots of the original equation are:
{
}
Answer: ( 
}
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, therefore 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 view The conversion 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 general meaning 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).
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 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. 
The concepts of “polynomial” and “factorization of a polynomial” in algebra are encountered very often, because you need to know them in order to easily carry out calculations with large multi-digit numbers. This article will describe several decomposition methods. All of them are quite easy to use; you just need to choose the right one for each specific case.
The concept of a polynomial
A polynomial is a sum of monomials, that is, expressions containing only the operation of multiplication.
For example, 2 * x * y is a monomial, but 2 * x * y + 25 is a polynomial that consists of 2 monomials: 2 * x * y and 25. Such polynomials are called binomials.
Sometimes, for the convenience of solving examples with multivalued values, an expression needs to be transformed, for example, decomposed into a certain number of factors, that is, numbers or expressions between which the multiplication action is performed. There are a number of ways to factor a polynomial. It is worth considering them, starting with the most primitive one, which is used in primary school.
Grouping (record in general form)
The formula for factoring a polynomial using the grouping method in general looks like this:
ac + bd + bc + ad = (ac + bc) + (ad + bd)
It is necessary to group the monomials so that each group has a common factor. In the first bracket this is the factor c, and in the second - d. This must be done in order to then move it out of the bracket, thereby simplifying the calculations.
Decomposition algorithm using a specific example
The simplest example of factoring a polynomial using the grouping method is given below:
10ac + 14bc - 25a - 35b = (10ac - 25a) + (14bc - 35b)
In the first bracket you need to take the terms with the factor a, which will be common, and in the second - with the factor b. Pay attention to the + and - signs in the finished expression. We put in front of the monomial the sign that was in the initial expression. That is, you need to work not with the expression 25a, but with the expression -25. The minus sign seems to be “glued” to the expression behind it and always taken into account when calculating.
In the next step, you need to take the multiplier, which is common, out of brackets. This is exactly what the grouping is for. To put outside the bracket means to write before the bracket (omitting the multiplication sign) all those factors that are exactly repeated in all the terms that are in the bracket. If there are not 2, but 3 or more terms in a bracket, the common factor must be contained in each of them, otherwise it cannot be taken out of the bracket.
In our case, there are only 2 terms in brackets. The overall multiplier is immediately visible. In the first bracket it is a, in the second it is b. Here you need to pay attention to the digital coefficients. In the first bracket, both coefficients (10 and 25) are multiples of 5. This means that not only a, but also 5a can be taken out of the bracket. Before the bracket, write 5a, and then divide each of the terms in brackets by the common factor that was taken out, and also write the quotient in brackets, not forgetting about the signs + and - Do the same with the second bracket, take out 7b, as well as 14 and 35 multiple of 7.
10ac + 14bc - 25a - 35b = (10ac - 25a) + (14bc - 35b) = 5a(2c - 5) + 7b(2c - 5).
We got 2 terms: 5a(2c - 5) and 7b(2c - 5). Each of them contains a common factor (the entire expression in brackets is the same here, which means it is a common factor): 2c - 5. It also needs to be taken out of the bracket, that is, terms 5a and 7b remain in the second bracket:
5a(2c - 5) + 7b(2c - 5) = (2c - 5)*(5a + 7b).
So the full expression is:
10ac + 14bc - 25a - 35b = (10ac - 25a) + (14bc - 35b) = 5a(2c - 5) + 7b(2c - 5) = (2c - 5)*(5a + 7b).
Thus, the polynomial 10ac + 14bc - 25a - 35b is decomposed into 2 factors: (2c - 5) and (5a + 7b). The multiplication sign between them can be omitted when writing
Sometimes there are expressions of this type: 5a 2 + 50a 3, here you can put out of brackets not only a or 5a, but even 5a 2. You should always try to put the largest common factor out of the bracket. In our case, if we divide each term by a common factor, we get:
5a 2 / 5a 2 = 1; 50a 3 / 5a 2 = 10a(when calculating the quotient of several powers with equally The base is preserved and the exponent is subtracted). Thus, the unit remains in the bracket (in no case do you forget to write one if you take one of the terms out of the bracket) and the quotient of division: 10a. It turns out that:
5a 2 + 50a 3 = 5a 2 (1 + 10a)
Square formulas
For ease of calculation, several formulas were derived. These are called abbreviated multiplication formulas and are used quite often. These formulas help factor polynomials containing powers. This is another one effective way factorization. So here they are:
- a 2 + 2ab + b 2 = (a + b) 2 - a formula called the “square of the sum”, since as a result of decomposition into a square, the sum of numbers enclosed in brackets is taken, that is, the value of this sum is multiplied by itself 2 times, and therefore is a multiplier.
- a 2 + 2ab - b 2 = (a - b) 2 - the formula for the square of the difference, it is similar to the previous one. The result is the difference, enclosed in parentheses, contained in the square power.
- a 2 - b 2 = (a + b)(a - b)- this is a formula for the difference of squares, since initially the polynomial consists of 2 squares of numbers or expressions, between which subtraction is performed. Perhaps, of the three mentioned, it is used most often.
Examples for calculations using square formulas
The calculations for them are quite simple. For example:
- 25x 2 + 20xy + 4y 2 - use the formula “square of the sum”.
- 25x 2 is the square of 5x. 20xy is the double product of 2*(5x*2y), and 4y 2 is the square of 2y.
- Thus, 25x 2 + 20xy + 4y 2 = (5x + 2y) 2 = (5x + 2y)(5x + 2y). This polynomial is decomposed into 2 factors (the factors are the same, so it is written as an expression with a square power).
Actions using the squared difference formula are carried out similarly to these. The remaining formula is difference of squares. Examples of this formula are very easy to define and find among other expressions. For example:
- 25a 2 - 400 = (5a - 20)(5a + 20). Since 25a 2 = (5a) 2, and 400 = 20 2
- 36x 2 - 25y 2 = (6x - 5y) (6x + 5y). Since 36x 2 = (6x) 2, and 25y 2 = (5y 2)
- c 2 - 169b 2 = (c - 13b)(c + 13b). Since 169b 2 = (13b) 2
It is important that each of the terms is a square of some expression. Then this polynomial must be factorized using the difference of squares formula. For this, it is not necessary that the second degree be above the number. There are polynomials that contain large degrees, but still fit these formulas.
a 8 +10a 4 +25 = (a 4) 2 + 2*a 4 *5 + 5 2 = (a 4 +5) 2
In this example, a 8 can be represented as (a 4) 2, that is, the square of a certain expression. 25 is 5 2, and 10a is 4 - this is the double product of the terms 2 * a 4 * 5. That is, this expression, despite the presence of degrees with large exponents, can be decomposed into 2 factors in order to subsequently work with them.
Cube formulas
The same formulas exist for factoring polynomials containing cubes. They are a little more complicated than those with squares:
- a 3 + b 3 = (a + b)(a 2 - ab + b 2)- this formula is called the sum of cubes, since in its initial form the polynomial is the sum of two expressions or numbers enclosed in a cube.
- a 3 - b 3 = (a - b)(a 2 + ab + b 2) - a formula identical to the previous one is designated as the difference of cubes.
- a 3 + 3a 2 b + 3ab 2 + b 3 = (a + b) 3 - cube of a sum, as a result of calculations, the sum of numbers or expressions is enclosed in brackets and multiplied by itself 3 times, that is, located in a cube
- a 3 - 3a 2 b + 3ab 2 - b 3 = (a - b) 3 - the formula, compiled by analogy with the previous one, changing only some signs of mathematical operations (plus and minus), is called the “difference cube”.
The last two formulas are practically not used for the purpose of factoring a polynomial, since they are complex, and it is rare enough to find polynomials that fully correspond to exactly this structure so that they can be factored using these formulas. But you still need to know them, since they will be required when operating in the opposite direction - when opening parentheses.
Examples on cube formulas
Let's look at an example: 64a 3 − 8b 3 = (4a) 3 − (2b) 3 = (4a − 2b)((4a) 2 + 4a*2b + (2b) 2) = (4a−2b)(16a 2 + 8ab + 4b 2 ).
Quite simple numbers are taken here, so you can immediately see that 64a 3 is (4a) 3, and 8b 3 is (2b) 3. Thus, this polynomial is expanded according to the formula difference of cubes into 2 factors. Actions using the formula for the sum of cubes are carried out by analogy.
It is important to understand that not all polynomials can be expanded in at least one way. But there are expressions that contain greater powers than a square or a cube, but they can also be expanded into abbreviated multiplication forms. For example: x 12 + 125y 3 =(x 4) 3 +(5y) 3 =(x 4 +5y)*((x 4) 2 − x 4 *5y+(5y) 2)=(x 4 + 5y)( x 8 − 5x 4 y + 25y 2).
This example contains as much as the 12th degree. But even it can be factorized using the sum of cubes formula. To do this, you need to imagine x 12 as (x 4) 3, that is, as a cube of some expression. Now, instead of a, you need to substitute it in the formula. Well, the expression 125y 3 is a cube of 5y. Next, you need to compose the product using the formula and perform calculations.
At first, or in case of doubt, you can always check by inverse multiplication. You just need to open the parentheses in the resulting expression and perform actions with similar terms. This method applies to all of the reduction methods listed: both to working with a common factor and grouping, and to working with formulas of cubes and quadratic powers.
Let's look at specific examples, 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 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.