In order to factorize, it is necessary to simplify the expressions. This is necessary so that it can be further reduced. The expansion of a polynomial makes sense when its degree is not lower than two. A polynomial with the first degree is called linear.
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The article will cover all the concepts of decomposition, theoretical foundations and methods of factoring a polynomial.
Theory
Theorem 1When any polynomial with degree n, having the form P n x = a n x n + a n - 1 x n - 1 + . . . + a 1 x + a 0, are represented as a product with a constant factor with the highest degree a n and n linear factors (x - x i), i = 1, 2, ..., n, then P n (x) = a n (x - x n) (x - x n - 1) · . . . · (x - x 1) , where x i, i = 1, 2, …, n are the roots of the polynomial.
The theorem is intended for roots of complex type x i, i = 1, 2, …, n and for complex coefficients a k, k = 0, 1, 2, …, n. This is the basis of any decomposition.
When coefficients of the form a k, k = 0, 1, 2, …, n are real numbers, then the complex roots will occur in conjugate pairs. For example, roots x 1 and x 2 related to a polynomial of the form P n x = a n x n + a n - 1 x n - 1 + . . . + a 1 x + a 0 are considered complex conjugate, then the other roots are real, from which we obtain that the polynomial takes the form P n (x) = a n (x - x n) (x - x n - 1) · . . . · (x - x 3) x 2 + p x + q, where x 2 + p x + q = (x - x 1) (x - x 2) .
Comment
The roots of a polynomial can be repeated. Let's consider the proof of the algebra theorem, a consequence of Bezout's theorem.
Fundamental theorem of algebra
Theorem 2Any polynomial with degree n has at least one root.
Bezout's theorem
After dividing a polynomial of the form P n x = a n x n + a n - 1 x n - 1 + . . . + a 1 x + a 0 on (x - s), then we get the remainder, which is equal to the polynomial at point s, then we get
P n x = a n x n + a n - 1 x n - 1 + . . . + a 1 x + a 0 = (x - s) · Q n - 1 (x) + P n (s) , where Q n - 1 (x) is a polynomial with degree n - 1.
Corollary to Bezout's theorem
When the root of the polynomial P n (x) is considered to be s, then P n x = a n x n + a n - 1 x n - 1 + . . . + a 1 x + a 0 = (x - s) · Q n - 1 (x) . This corollary is sufficient when used to describe the solution.
Factoring a quadratic trinomial
A square trinomial of the form a x 2 + b x + c can be factorized into linear factors. then we get that a x 2 + b x + c = a (x - x 1) (x - x 2) , where x 1 and x 2 are roots (complex or real).
From this it is clear that the expansion itself reduces to the solution quadratic equation subsequently.
Example 1
Factor the quadratic trinomial.
Solution
It is necessary to find the roots of the equation 4 x 2 - 5 x + 1 = 0. To do this, you need to find the value of the discriminant using the formula, then we get D = (- 5) 2 - 4 · 4 · 1 = 9. From here we have that
x 1 = 5 - 9 2 4 = 1 4 x 2 = 5 + 9 2 4 = 1
From this we get that 4 x 2 - 5 x + 1 = 4 x - 1 4 x - 1.
To perform the check, you need to open the parentheses. Then we get an expression of the form:
4 x - 1 4 x - 1 = 4 x 2 - x - 1 4 x + 1 4 = 4 x 2 - 5 x + 1
After checking, we arrive at the original expression. That is, we can conclude that the decomposition was performed correctly.
Example 2
Factor the quadratic trinomial of the form 3 x 2 - 7 x - 11 .
Solution
We find that it is necessary to calculate the resulting quadratic equation of the form 3 x 2 - 7 x - 11 = 0.
To find the roots, you need to determine the value of the discriminant. We get that
3 x 2 - 7 x - 11 = 0 D = (- 7) 2 - 4 3 (- 11) = 181 x 1 = 7 + D 2 3 = 7 + 181 6 x 2 = 7 - D 2 3 = 7 - 181 6
From this we get that 3 x 2 - 7 x - 11 = 3 x - 7 + 181 6 x - 7 - 181 6.
Example 3
Factor the polynomial 2 x 2 + 1.
Solution
Now we need to solve the quadratic equation 2 x 2 + 1 = 0 and find its roots. We get that
2 x 2 + 1 = 0 x 2 = - 1 2 x 1 = - 1 2 = 1 2 i x 2 = - 1 2 = - 1 2 i
These roots are called complex conjugate, which means the expansion itself can be depicted as 2 x 2 + 1 = 2 x - 1 2 · i x + 1 2 · i.
Example 4
Decompose the quadratic trinomial x 2 + 1 3 x + 1 .
Solution
First you need to solve a quadratic equation of the form x 2 + 1 3 x + 1 = 0 and find its roots.
x 2 + 1 3 x + 1 = 0 D = 1 3 2 - 4 1 1 = - 35 9 x 1 = - 1 3 + D 2 1 = - 1 3 + 35 3 i 2 = - 1 + 35 i 6 = - 1 6 + 35 6 i x 2 = - 1 3 - D 2 1 = - 1 3 - 35 3 i 2 = - 1 - 35 i 6 = - 1 6 - 35 6 i
Having obtained the roots, we write
x 2 + 1 3 x + 1 = x - - 1 6 + 35 6 i x - - 1 6 - 35 6 i = = x + 1 6 - 35 6 i x + 1 6 + 35 6 i
Comment
If the discriminant value is negative, then the polynomials will remain second-order polynomials. It follows from this that we will not expand them into linear factors.
Methods for factoring a polynomial of degree higher than two
When decomposing it is assumed universal method. Most of all cases are based on a corollary of Bezout's theorem. To do this, you need to select the value of the root x 1 and reduce its degree by dividing by a polynomial by 1 by dividing by (x - x 1). The resulting polynomial needs to find the root x 2, and the search process is cyclical until we obtain a complete expansion.
If the root is not found, then other methods of factorization are used: grouping, additional terms. This topic posits the solution of equations with higher powers and integer coefficients.
Taking the common factor out of brackets
Consider the case when the free term is equal to zero, then the form of the polynomial becomes P n (x) = a n x n + a n - 1 x n - 1 + . . . + a 1 x .
It can be seen that the root of such a polynomial will be equal to x 1 = 0, then the polynomial can be represented as the expression P n (x) = a n x n + a n - 1 x n - 1 +. . . + a 1 x = = x (a n x n - 1 + a n - 1 x n - 2 + . . . + a 1)
This method is considered to be taking the common factor out of brackets.
Example 5
Factor the third degree polynomial 4 x 3 + 8 x 2 - x.
Solution
We see that x 1 = 0 is the root of the given polynomial, then we can remove x from the brackets of the entire expression. We get:
4 x 3 + 8 x 2 - x = x (4 x 2 + 8 x - 1)
Let's move on to finding the roots of the square trinomial 4 x 2 + 8 x - 1. Let's find the discriminant and roots:
D = 8 2 - 4 4 (- 1) = 80 x 1 = - 8 + D 2 4 = - 1 + 5 2 x 2 = - 8 - D 2 4 = - 1 - 5 2
Then it follows that
4 x 3 + 8 x 2 - x = x 4 x 2 + 8 x - 1 = = 4 x x - - 1 + 5 2 x - - 1 - 5 2 = = 4 x x + 1 - 5 2 x + 1 + 5 2
To begin with, let us take into consideration a decomposition method containing integer coefficients of the form P n (x) = x n + a n - 1 x n - 1 + . . . + a 1 x + a 0, where the coefficient of the highest degree is 1.
When a polynomial has integer roots, then they are considered divisors of the free term.
Example 6
Expand the expression f (x) = x 4 + 3 x 3 - x 2 - 9 x - 18.
Solution
Let's consider whether there are complete roots. It is necessary to write down the divisors of the number - 18. We get that ±1, ±2, ±3, ±6, ±9, ±18. It follows that this polynomial has integer roots. You can check using Horner's scheme. It is very convenient and allows you to quickly obtain the expansion coefficients of a polynomial:
It follows that x = 2 and x = - 3 are the roots of the original polynomial, which can be represented as a product of the form:
f (x) = x 4 + 3 x 3 - x 2 - 9 x - 18 = (x - 2) (x 3 + 5 x 2 + 9 x + 9) = = (x - 2) (x + 3) (x 2 + 2 x + 3)
We proceed to the expansion of a quadratic trinomial of the form x 2 + 2 x + 3.
Since the discriminant is negative, it means there are no real roots.
Answer: f (x) = x 4 + 3 x 3 - x 2 - 9 x - 18 = (x - 2) (x + 3) (x 2 + 2 x + 3)
Comment
It is allowed to use root selection and division of a polynomial by a polynomial instead of Horner's scheme. Let's move on to considering the expansion of a polynomial containing integer coefficients of the form P n (x) = x n + a n - 1 x n - 1 + . . . + a 1 x + a 0 , the highest of which is equal to one.
This case occurs for rational fractions.
Example 7
Factorize f (x) = 2 x 3 + 19 x 2 + 41 x + 15 .
Solution
It is necessary to replace the variable y = 2 x, you should move on to a polynomial with coefficients equal to 1 at the highest degree. You need to start by multiplying the expression by 4. We get that
4 f (x) = 2 3 x 3 + 19 2 2 x 2 + 82 2 x + 60 = = y 3 + 19 y 2 + 82 y + 60 = g (y)
When the resulting function of the form g (y) = y 3 + 19 y 2 + 82 y + 60 has integer roots, then their location is among the divisors of the free term. The entry will look like:
±1, ±2, ±3, ±4, ±5, ±6, ±10, ±12, ±15, ±20, ±30, ±60
Let's move on to calculating the function g (y) at these points in order to get zero as a result. We get that
g (1) = 1 3 + 19 1 2 + 82 1 + 60 = 162 g (- 1) = (- 1) 3 + 19 (- 1) 2 + 82 (- 1) + 60 = - 4 g (2) = 2 3 + 19 2 2 + 82 2 + 60 = 308 g (- 2) = (- 2) 3 + 19 (- 2) 2 + 82 (- 2) + 60 = - 36 g (3) = 3 3 + 19 3 2 + 82 3 + 60 = 504 g (- 3) = (- 3) 3 + 19 (- 3) 2 + 82 (- 3) + 60 = - 42 g (4) = 4 3 + 19 · 4 2 + 82 · 4 + 60 = 756 g (- 4) = (- 4) 3 + 19 · (- 4) 2 + 82 · (- 4) + 60 = - 28 g (5) = 5 3 + 19 5 2 + 82 5 + 60 = 1070 g (- 5) = (- 5) 3 + 19 (- 5) 2 + 82 (- 5) + 60
We find that y = - 5 is the root of an equation of the form y 3 + 19 y 2 + 82 y + 60, which means x = y 2 = - 5 2 is the root of the original function.
Example 8
It is necessary to divide with a column 2 x 3 + 19 x 2 + 41 x + 15 by x + 5 2.
Solution
Let's write it down and get:
2 x 3 + 19 x 2 + 41 x + 15 = x + 5 2 (2 x 2 + 14 x + 6) = = 2 x + 5 2 (x 2 + 7 x + 3)
Checking the divisors will take a lot of time, so it is more profitable to factorize the resulting quadratic trinomial of the form x 2 + 7 x + 3. By equating to zero we find the discriminant.
x 2 + 7 x + 3 = 0 D = 7 2 - 4 1 3 = 37 x 1 = - 7 + 37 2 x 2 = - 7 - 37 2 ⇒ x 2 + 7 x + 3 = x + 7 2 - 37 2 x + 7 2 + 37 2
It follows that
2 x 3 + 19 x 2 + 41 x + 15 = 2 x + 5 2 x 2 + 7 x + 3 = = 2 x + 5 2 x + 7 2 - 37 2 x + 7 2 + 37 2
Artificial techniques for factoring a polynomial
Rational roots are not inherent in all polynomials. To do this, you need to use special methods to find factors. But not all polynomials can be expanded or represented as a product.
Grouping method
There are cases when you can group the terms of a polynomial to find a common factor and put it out of brackets.
Example 9
Factor the polynomial x 4 + 4 x 3 - x 2 - 8 x - 2.
Solution
Because the coefficients are integers, then the roots can presumably also be integers. To check, take the values 1, - 1, 2 and - 2 in order to calculate the value of the polynomial at these points. We get that
1 4 + 4 1 3 - 1 2 - 8 1 - 2 = - 6 ≠ 0 (- 1) 4 + 4 (- 1) 3 - (- 1) 2 - 8 (- 1) - 2 = 2 ≠ 0 2 4 + 4 2 3 - 2 2 - 8 2 - 2 = 26 ≠ 0 (- 2) 4 + 4 (- 2) 3 - (- 2) 2 - 8 (- 2) - 2 = - 6 ≠ 0
This shows that there are no roots; it is necessary to use another method of expansion and solution.
It is necessary to group:
x 4 + 4 x 3 - x 2 - 8 x - 2 = x 4 + 4 x 3 - 2 x 2 + x 2 - 8 x - 2 = = (x 4 - 2 x 2) + (4 x 3 - 8 x) + x 2 - 2 = = x 2 (x 2 - 2) + 4 x (x 2 - 2) + x 2 - 2 = = (x 2 - 2) (x 2 + 4 x + 1)
After grouping the original polynomial, you need to represent it as the product of two square trinomials. To do this, we need to factorize. we get that
x 2 - 2 = 0 x 2 = 2 x 1 = 2 x 2 = - 2 ⇒ x 2 - 2 = x - 2 x + 2 x 2 + 4 x + 1 = 0 D = 4 2 - 4 1 1 = 12 x 1 = - 4 - D 2 1 = - 2 - 3 x 2 = - 4 - D 2 1 = - 2 - 3 ⇒ x 2 + 4 x + 1 = x + 2 - 3 x + 2 + 3
x 4 + 4 x 3 - x 2 - 8 x - 2 = x 2 - 2 x 2 + 4 x + 1 = = x - 2 x + 2 x + 2 - 3 x + 2 + 3
Comment
The simplicity of grouping does not mean that choosing terms is easy enough. There is no specific solution method, so it is necessary to use special theorems and rules.
Example 10
Factor the polynomial x 4 + 3 x 3 - x 2 - 4 x + 2 .
Solution
The given polynomial has no integer roots. The terms should be grouped. We get that
x 4 + 3 x 3 - x 2 - 4 x + 2 = = (x 4 + x 3) + (2 x 3 + 2 x 2) + (- 2 x 2 - 2 x) - x 2 - 2 x + 2 = = x 2 (x 2 + x) + 2 x (x 2 + x) - 2 (x 2 + x) - (x 2 + 2 x - 2) = = (x 2 + x) (x 2 + 2 x - 2) - (x 2 + 2 x - 2) = (x 2 + x - 1) (x 2 + 2 x - 2)
After factorization we get that
x 4 + 3 x 3 - x 2 - 4 x + 2 = x 2 + x - 1 x 2 + 2 x - 2 = = x + 1 + 3 x + 1 - 3 x + 1 2 + 5 2 x + 1 2 - 5 2
Using abbreviated multiplication formulas and Newton's binomial to factor a polynomial
Appearance often does not always make it clear which method should be used during decomposition. After the transformations have been made, you can build a line consisting of Pascal’s triangle, otherwise they are called Newton’s binomial.
Example 11
Factor the polynomial x 4 + 4 x 3 + 6 x 2 + 4 x - 2.
Solution
It is necessary to convert the expression to the form
x 4 + 4 x 3 + 6 x 2 + 4 x - 2 = x 4 + 4 x 3 + 6 x 2 + 4 x + 1 - 3
The sequence of coefficients of the sum in parentheses is indicated by the expression x + 1 4 .
This means we have x 4 + 4 x 3 + 6 x 2 + 4 x - 2 = x 4 + 4 x 3 + 6 x 2 + 4 x + 1 - 3 = x + 1 4 - 3.
After applying the difference of squares, we get
x 4 + 4 x 3 + 6 x 2 + 4 x - 2 = x 4 + 4 x 3 + 6 x 2 + 4 x + 1 - 3 = x + 1 4 - 3 = = x + 1 4 - 3 = x + 1 2 - 3 x + 1 2 + 3
Consider the expression that is in the second bracket. It is clear that there are no knights there, so we should apply the difference of squares formula again. We get an expression of the form
x 4 + 4 x 3 + 6 x 2 + 4 x - 2 = x 4 + 4 x 3 + 6 x 2 + 4 x + 1 - 3 = x + 1 4 - 3 = = x + 1 4 - 3 = x + 1 2 - 3 x + 1 2 + 3 = = x + 1 - 3 4 x + 1 + 3 4 x 2 + 2 x + 1 + 3
Example 12
Factorize x 3 + 6 x 2 + 12 x + 6 .
Solution
Let's start transforming the expression. We get that
x 3 + 6 x 2 + 12 x + 6 = x 3 + 3 2 x 2 + 3 2 2 x + 2 3 - 2 = (x + 2) 3 - 2
It is necessary to apply the formula for abbreviated multiplication of the difference of cubes. We get:
x 3 + 6 x 2 + 12 x + 6 = = (x + 2) 3 - 2 = = x + 2 - 2 3 x + 2 2 + 2 3 x + 2 + 4 3 = = x + 2 - 2 3 x 2 + x 2 + 2 3 + 4 + 2 2 3 + 4 3
A method for replacing a variable when factoring a polynomial
When replacing a variable, the degree is reduced and the polynomial is factored.
Example 13
Factor the polynomial of the form x 6 + 5 x 3 + 6 .
Solution
According to the condition, it is clear that it is necessary to make the replacement y = x 3. We get:
x 6 + 5 x 3 + 6 = y = x 3 = y 2 + 5 y + 6
The roots of the resulting quadratic equation are y = - 2 and y = - 3, then
x 6 + 5 x 3 + 6 = y = x 3 = y 2 + 5 y + 6 = = y + 2 y + 3 = x 3 + 2 x 3 + 3
It is necessary to apply the formula for abbreviated multiplication of the sum of cubes. We get expressions of the form:
x 6 + 5 x 3 + 6 = y = x 3 = y 2 + 5 y + 6 = = y + 2 y + 3 = x 3 + 2 x 3 + 3 = = x + 2 3 x 2 - 2 3 x + 4 3 x + 3 3 x 2 - 3 3 x + 9 3
That is, we obtained the desired decomposition.
The cases discussed above will help in considering and factoring a polynomial in different ways.
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Any algebraic polynomial of degree n can be represented as a product n-linear factors type 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 the number (real or complex) that makes the polynomial vanish. The roots of a polynomial can be either real roots or complex conjugate roots, then the polynomial can be represented in the following form:
Let's consider methods for decomposing 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 a given polynomial is decomposed is known in advance. When using the method of uncertain coefficients, the following statements are true:
P.1. Two polynomials are identically equal if their coefficients are equal for the same powers of x.
P.2. Any polynomial of the third degree is decomposed into the product of linear and quadratic factors.
P.3. Any fourth-degree polynomial can be decomposed into the product of two second-degree polynomials.
Example 1.1. It is necessary to factorize the cubic expression:
P.1. In accordance with the accepted statements, the identical equality holds for the cubic expression:
P.2. Right side expressions can be represented as terms as follows:
P.3. We compose a system of equations from the condition of equality of coefficients at 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 solution methods can be used nonlinear systems equations. Solving this system of equations, we find that the uncertain coefficients are determined as follows:
Thus, the original expression is factorized 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 Francois 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 for checking the correctness of finding the roots of a polynomial, as well as for composing 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 whose degree 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 most in a simple 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 equal to one, then the rational roots of this polynomial are divisors of the free term of the expression, i.e. can be integers 
. 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 abbreviated multiplication formulas
Abbreviated multiplication formulas are used to simplify calculations, as well as factoring polynomials. Abbreviated multiplication formulas allow you to simplify the solution of individual problems.
Formulas used to factorize
Factoring polynomials is an identity transformation, as a result of which a polynomial is transformed into the product of several factors - polynomials or monomials.
There are several ways to factor polynomials.
Method 1. Taking the common factor out of brackets.
This transformation is based on the distributive law of multiplication: ac + bc = c(a + b). The essence of the transformation is to isolate the common factor in the two components under consideration and “take” it out of brackets.
Let us factor the polynomial 28x 3 – 35x 4.
Solution.
1. Find a common divisor for the elements 28x3 and 35x4. For 28 and 35 it will be 7; for x 3 and x 4 – x 3. In other words, our common factor is 7x 3.
2. We represent each of the elements as a product of factors, one of which
7x 3: 28x 3 – 35x 4 = 7x 3 ∙ 4 – 7x 3 ∙ 5x.
3. We take the common factor out of brackets
7x 3: 28x 3 – 35x 4 = 7x 3 ∙ 4 – 7x 3 ∙ 5x = 7x 3 (4 – 5x).
Method 2. Using abbreviated multiplication formulas. The “mastery” of using this method is to notice one of the abbreviated multiplication formulas in the expression.
Let us factor the polynomial x 6 – 1.
Solution.
1. We can apply the difference of squares formula to this expression. To do this, imagine x 6 as (x 3) 2, and 1 as 1 2, i.e. 1. The expression will take the form:
(x 3) 2 – 1 = (x 3 + 1) ∙ (x 3 – 1).
2. We can apply the formula for the sum and difference of cubes to the resulting expression:
(x 3 + 1) ∙ (x 3 – 1) = (x + 1) ∙ (x 2 – x + 1) ∙ (x – 1) ∙ (x 2 + x + 1).
So,
x 6 – 1 = (x 3) 2 – 1 = (x 3 + 1) ∙ (x 3 – 1) = (x + 1) ∙ (x 2 – x + 1) ∙ (x – 1) ∙ (x 2 + x + 1).
Method 3. Grouping. The grouping method involves combining the components of a polynomial in such a way that it is easy to perform operations on them (addition, subtraction, subtraction of a common factor).
Let's factor the polynomial x 3 – 3x 2 + 5x – 15.
Solution.
1. Let's group the components in this way: 1st with 2nd, and 3rd with 4th
(x 3 – 3x 2) + (5x – 15).
2. In the resulting expression, we take the common factors out of brackets: x 2 in the first case and 5 in the second.
(x 3 – 3x 2) + (5x – 15) = x 2 (x – 3) + 5(x – 3).
3. We take the common factor x – 3 out of brackets and get:
x 2 (x – 3) + 5(x – 3) = (x – 3)(x 2 + 5).
So,
x 3 – 3x 2 + 5x – 15 = (x 3 – 3x 2) + (5x – 15) = x 2 (x – 3) + 5(x – 3) = (x – 3) ∙ (x 2 + 5 ).
Let's secure the material.
Factor the polynomial a 2 – 7ab + 12b 2 .
Solution.
1. Let us represent the monomial 7ab as the sum 3ab + 4ab. The expression will take the form:
a 2 – (3ab + 4ab) + 12b 2. 
Let's open the brackets and get:
a 2 – 3ab – 4ab + 12b 2.
2. Let's group the components of the polynomial in this way: 1st with 2nd and 3rd with 4th. We get:
(a 2 – 3ab) – (4ab – 12b 2).
3. Let’s take the common factors out of brackets:
(a 2 – 3ab) – (4ab – 12b 2) = a(a – 3b) – 4b(a – 3b).
4. Let’s take the common factor (a – 3b) out of brackets:
a(a – 3b) – 4b(a – 3b) = (a – 3 b) ∙ (a – 4b).
So,
a 2 – 7ab + 12b 2 =
= a 2 – (3ab + 4ab) + 12b 2 =
= a 2 – 3ab – 4ab + 12b 2 =
= (a 2 – 3ab) – (4ab – 12b 2) =
= a(a – 3b) – 4b(a – 3b) =
= (a – 3 b) ∙ (a – 4b).
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Factoring a polynomial. Part 1
Factorization is a universal technique that helps solve complex equations and inequalities. The first thought that should come to mind when solving equations and inequalities in which there is a zero on the right side is to try to factor the left side.
Let's list the main ways to factor a polynomial:
- putting the common factor out of brackets
- using abbreviated multiplication formulas
- using the formula for factoring a quadratic trinomial
- grouping method
- dividing a polynomial by a binomial
- method of uncertain coefficients
In this article we will dwell in detail on the first three methods, and we will consider the rest in subsequent articles.
1. Taking the common factor out of brackets.
To take the common factor out of brackets, you must first find it. Common multiplier factor equal to the greatest common divisor of all coefficients.
Letter part the common factor is equal to the product of the expressions included in each term with the smallest exponent.
The scheme for adding a common multiplier looks like this:
Attention!
The number of terms in brackets is equal to the number of terms in the original expression. If one of the terms coincides with the common factor, then when dividing it by the common factor, we get one.
Example 1.
Factor the polynomial:
Let's take the common factor out of brackets. To do this, we will first find it.
1. Find the greatest common divisor of all coefficients of the polynomial, i.e. numbers 20, 35 and 15. It is equal to 5.
2. We establish that the variable is contained in all terms, and the smallest of its exponents is equal to 2. The variable is contained in all terms, and the smallest of its exponents is 3.
The variable is contained only in the second term, so it is not part of the common factor.
So the total factor is
3. We take the multiplier out of brackets using the diagram given above:
Example 2. Solve the equation:
Solution. Let's factorize the left side of the equation. Let's take the factor out of brackets:
So we get the equation 
Let's equate each factor to zero:
We get - the root of the first equation.
Roots:
Answer: -1, 2, 4
2. Factorization using abbreviated multiplication formulas.
If the number of terms in the polynomial we are going to factor is less than or equal to three, then we try to apply the abbreviated multiplication formulas.
1. If the polynomial isdifference of two terms, then we try to apply square difference formula:
or difference of cubes formula:
Here are the letters and denote a number or algebraic expression.
2. If a polynomial is the sum of two terms, then perhaps it can be factored using sum of cubes formulas:
3. If a polynomial consists of three terms, then we try to apply square sum formula:
or squared difference formula:
Or we try to factorize by formula for factoring a quadratic trinomial:
Here and are the roots of the quadratic equation 
Example 3.Factor the expression:
Solution. We have before us the sum of two terms. Let's try to apply the formula for the sum of cubes. To do this, you first need to represent each term as a cube of some expression, and then apply the formula for the sum of the cubes:
Example 4. Factor the expression: 
Decision. Here we have the difference of the squares of two expressions. First expression: , second expression:
Let's apply the formula for the difference of squares:
Let's open the brackets and add similar terms, we get:
On this lesson we will recall all the previously studied methods of factoring a polynomial and consider examples of their application, in addition, we will study a new method - the method of isolating a complete square and learn how to use it in solving various problems.
Subject:Factoring polynomials
Lesson:Factoring polynomials. Method for selecting a complete square. Combination of methods
Let us recall the basic methods of factoring a polynomial that were studied earlier:
The method of putting a common factor out of brackets, that is, a factor that is present in all terms of the polynomial. Let's look at an example:
Recall that a monomial is the product of powers and numbers. In our example, both terms have some common, identical elements.
So, let's take the common factor out of brackets:
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Let us remind you that by multiplying the factor taken out by a parenthesis, you can check the correctness of the factor taken out.
Grouping method. It is not always possible to extract a common factor in a polynomial. In this case, you need to divide its members into groups in such a way that in each group you can take out a common factor and try to break it down so that after taking out the factors in the groups, a common factor appears in the entire expression, and you can continue the decomposition. Let's look at an example:
Let's group the first term with the fourth, the second with the fifth, and the third with the sixth:
Let's take out the common factors in the groups:
The expression now has a common factor. Let's take it out:
Application of abbreviated multiplication formulas. Let's look at an example:
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Let's write the expression in detail:
Obviously, we have before us the formula for the squared difference, since it is the sum of the squares of two expressions and their double product is subtracted from it. Let's use the formula:
Today we will learn another method - the method of selecting a complete square. It is based on the formulas of the square of the sum and the square of the difference. Let's remind them:
Formula for the square of the sum (difference);
The peculiarity of these formulas is that they contain the squares of two expressions and their double product. Let's look at an example:
Let's write down the expression:
So, the first expression is , and the second is .
In order to create a formula for the square of a sum or difference, twice the product of expressions is not enough. It needs to be added and subtracted:
Let's complete the square of the sum:
Let's transform the resulting expression:
Let's apply the formula for the difference of squares, recall that the difference of the squares of two expressions is the product of and the sum of their difference:
So, this method consists, first of all, in the fact that you need to identify the expressions a and b that are squared, that is, determine which expressions squares are squared in this example. After this, you need to check for the presence of a doubled product and if it is not there, then add and subtract it, this will not change the meaning of the example, but the polynomial can be factorized using the formulas for the square of the sum or difference and difference of squares, if possible.
Let's move on to solving examples.
Example 1 - factorize:
Let's find expressions that are squared:
Let us write down what their double product should be:
Let's add and subtract double the product:
Let's complete the square of the sum and give similar ones:
Let's write it using the difference of squares formula:
Example 2 - solve the equation:
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On the left side of the equation is a trinomial. You need to factor it into factors. We use the squared difference formula:
We have the square of the first expression and the double product, the square of the second expression is missing, let’s add and subtract it:
Let's fold a complete square and give similar terms:
Let's apply the difference of squares formula:
So we have the equation 
We know that a product is equal to zero only if at least one of the factors is equal to zero. Let's create the following equations based on this:
Let's solve the first equation:
Let's solve the second equation:
Answer: or
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We proceed similarly to the previous example - select the square of the difference.