It is a square, and it consists of three terms (). So it turns out - a square trinomial.
Examples Not square trinomials:
\(x^3-3x^2-5x+6\) - cubic quadrinomial
\(2x+1\) - linear binomial
Square root of the trinomial:
Example:
The trinomial \(x^2-2x+1\) has a root \(1\), because \(1^2-2 1+1=0\)
The trinomial \(x^2+2x-3\) has roots \(1\) and \(-3\), because \(1^2+2-3=0\) and \((-3)^ 2-6-3=9-9=0\)
For example: if you need to find roots for the quadratic trinomial \(x^2-2x+1\), we equate it to zero and solve the equation \(x^2-2x+1=0\).
\(D=4-4\cdot1=0\)
\(x=\frac(2-0)(2)=\frac(2)(2)=1\)
Ready. The root is \(1\).
Decomposition of a quadratic trinomial into:
The square trinomial \(ax^2+bx+c\) can be expanded as \(a(x-x_1)(x-x_2)\) if the equations \(ax^2+bx+c=0\) are greater than zero \ (x_1\) and \(x_2\) are roots of the same equation).
For example, consider the trinomial \(3x^2+13x-10\).
The quadratic equation \(3x^2+13x-10=0\) has a discriminant equal to 289 (greater than zero) and roots equal to \(-5\) and \(\frac(2)(3)\). Therefore \(3x^2+13x-10=3(x+5)(x-\frac(2)(3))\). It is easy to verify the correctness of this statement - if we , then we will obtain the original trinomial.
The square trinomial \(ax^2+bx+c\) can be represented as \(a(x-x_1)^2\) if the discriminant of the equation \(ax^2+bx+c=0\) is zero.
For example, consider the trinomial \(x^2+6x+9\).The quadratic equation \(x^2+6x+9=0\) has a discriminant equal to \(0\) and a unique root equal to \(-3\). This means \(x^2+6x+9=(x+3)^2\) (here the coefficient is \(a=1\), so it is not written before the bracket - there is no need). Please note that the same conversion can be done by .
The square trinomial \(ax^2+bx+c\) cannot be factorized if the discriminant of the equation \(ax^2+bx+c=0\) is less than zero.
For example, the trinomials \(x^2+x+4\) and \(-5x^2+2x-1\) have a discriminant less than zero. Therefore, it is impossible to factor them.
Example
. Factor \(2x^2-11x+12\).
Solution
:
Let's find the roots of the quadratic equation \(2x^2-11x+12=0\)
\(D=11^2-4 \cdot 2 \cdot 12=121-96=25>0\)
\(x_1=\frac(11-5)(4)=1.5;\) \(x_2=\frac(11+5)(4)=4.\)
So, \(2x^2-11x+12=2(x-1.5)(x-4)\)
Answer
: \(2(x-1.5)(x-4)\)
The resulting answer may be written differently: \((2x-3)(x-4)\).
Example
. (Assignment from the OGE) The square trinomial is factored \(5x^2+33x+40=5(x++ 5)(x-a)\). Find \(a\).
Solution:
\(5x^2+33x+40=0\)
\(D=33^2-4 \cdot 5 \cdot 40=1089-800=289=17^2\)
\(x_1=\frac(-33-17)(10)=-5\)
\(x_2=\frac(-33+17)(10)=-1.6\)
\(5x^2+33x+40=5(x+5)(x+1.6)\)
Answer
: \(-1,6\)
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.
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).
This shows that the expansion itself reduces to solving the 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, a universal method is assumed. 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 involves solving 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
Decompose 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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It is necessary to factor polynomials when simplifying expressions (so that reduction can be carried out), when solving equations, or when decomposing a fractionally rational function into simple fractions.
It makes sense to talk about factoring a polynomial if its degree is not lower than two.
A polynomial of the first degree is called linear.
Let's first consider the theoretical foundations, then move directly to methods of factoring a polynomial.
Page navigation.
Necessary theory.
Theorem.
Any polynomial of degree n type is represented by the product of a constant factor at the highest power and n linear multipliers, i=1, 2, …, n, that is, and, i=1, 2, …, n are the roots of the polynomial.
This theorem is formulated for complex roots, i=1, 2, …, n and complex coefficients, k=0, 1, 2, …, n. It is the basis for factoring any polynomial.
If the coefficients k=0, 1, 2, …, n are real numbers, then the complex roots of the polynomial MUST occur in complex conjugate pairs.
For example, if the roots of the polynomial are complex conjugate, and the remaining roots are real, then the polynomial will be represented in the form , where
Comment.
Among the roots of a polynomial there may be repeating ones.
The proof of the theorem is carried out using fundamental theorem of algebra And corollaries of Bezout's theorem.
Fundamental theorem of algebra.
Any polynomial of degree n has at least one root (complex or real).
Bezout's theorem.
When dividing a polynomial by (x-s) the remainder is equal to the value of the polynomial at the point s, that is, where there is a polynomial of degree n-1.
Corollary to Bezout's theorem.
If s is the root of the polynomial, then .
We will use this corollary quite often when describing solutions to examples.
Factoring a quadratic trinomial.
The square trinomial is decomposed into two linear factors: , where and are roots (complex or real).
Thus, factoring a quadratic trinomial is reduced to solving a quadratic equation.
Example.
Factor a quadratic trinomial.
Solution.
Let's find the roots of the quadratic equation 
.
The discriminant of the equation is equal, therefore, 
Thus, 
.
To check, you can expand the brackets: . When checking, we arrived at the original trinomial, so the decomposition was correct.
Example.
Solution.
The corresponding quadratic equation is 
.
Let's find its roots. 
That's why, 
.
Example.
Factor the polynomial.
Solution.
Let's find the roots of the quadratic equation. 
We obtained a pair of complex conjugate roots.
The expansion of the polynomial will have the form 
.
Example.
Factor the quadratic trinomial.
Solution.
Let's solve a quadratic equation 
.
That's why, 
Comment:
In what follows, with a negative discriminant, we will leave second-order polynomials in their original form, that is, we will not decompose them into linear factors with complex free terms.
Methods for factoring a polynomial of degree higher than two.
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.
The further presentation is based on skills 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, 
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, 
Artificial techniques for factoring a polynomial.
Polynomials do not always have rational roots. In this case, when factoring, you have to look for special methods. But, no matter how much we would like, some polynomials (or rather the vast majority) cannot be represented as a product.
Grouping method.
Sometimes it turns out to group the terms of a polynomial, which allows you to find a common factor and take it out of brackets.
Example.
Expand polynomial 
by multipliers.
Solution.
Since the coefficients are integers, there can be integer roots among the divisors of the free term. Let's check the values 1
, -1
, 2
And -2
, calculating the value of the polynomial at these points. 
That is, there are no whole roots. Let's look for another way of decomposition.
Let's group: 
After grouping, the original polynomial was represented as the product of two square trinomials. Let's factor them. 
First of all, let's point out some common names. Let us consider polynomials that contain only one letter, for example, the letter x. Then the simplest is a polynomial in which there are two terms, and one of them contains the letter x to the first degree, and the other does not have the letter x at all, for example, 3x – 5 or 15 – 7x or 8z + 7 (here instead of letter x is taken letter z), etc. Such polynomials are called linear binomials .
3x² – 5x + 7 or x² + 2x – 1
or 5y² + 7y + 8 or z² – 5z – 2, etc.
Such polynomials are called square trinomials.
Then, we can form a cubic quadrinomial, for example:
x³ + 2x² – x + 1 or 3x³ – 5x² – 2x – 3 etc.,
polynomial of the fourth degree, for example:
x 4 – 2x³ – 3x² + 4x – 5, etc.
It is possible to denote the coefficients at x, at x², at x³, etc. also by letters, for example, by the letters a, b, c, etc. Then we get:
1) the general form of the binomial ax + b, linear with respect to x,
2) general form of the quadratic trinomial (relative to x): ax² + bx + c,
3) general form of the cubic trinomial (relative to x): ax³ + bx² + cx + d, etc.
By replacing the letters a, b, c, d... in these formulas with different numbers, we get all kinds of linear binomials, square trinomials, etc. For example, in the formula ax² + bx + c, which expresses the general form of a quadratic trinomial, we replace the letter a with the number + 3, the letter b with the number –2 and the letter with the number –1, we get the square trinomial 3x² – 2x – 1. In a particular case, it is also possible to obtain a binomial by replacing one of the letters with zero, for example, if a = +1, b = 0 and c = –3, then we get the quadratic binomial x² – 3.
You can learn to factor some quadratic trinomials fairly quickly into linear factors. We will, however, limit ourselves to considering only those quadratic trinomials that satisfy the following conditions:
1) the coefficient for the leading term (for x²) is +1,
2) you can find two integers (with signs, or two relative integers) such that their sum is equal to the coefficient of x to the first power and their product is equal to the term free of x (where there is no letter x at all).
Examples. 1. x² + 5x + 6; It’s easy to mentally find two numbers (with signs) so that their sum is equal to +5 (the coefficient of x) and so that their product = +6 (the term free of x) - these numbers are: +2 and +3 [in fact In fact, +2 + 3 = +5 and (+2) ∙ (+3) = +6]. Using these two numbers, we replace the +5x term with two terms, namely: +2x + 3x (of course, +2x + 3x = +5x); then our technical term will be artificially converted into a four-term x² + 2x + 3x + 6. Let us now apply the grouping technique to it, assigning the first two terms to one group and the last two to another:
x² + 5x + 6 = x² + 2x + 3x + 6 = x (x + 2) + 3 (x + 2) = (x + 2) (x + 3).
In the first group we took x out of the bracket and in the second +3, we got two terms that had a common factor (x + 2), which we also took out of the bracket, and our trinomial x² + 5x + 6 decomposed into 2 linear factors: x + 2 and x + 3.
2. x² – x – 12. Here you need to find two numbers (relative) so that their sum is equal to –1 and so that their product is equal to –12. These numbers are: –4 and +3.
Check: –4 + 3 = –1; (–4) (+3) = –12. Using these numbers, we replace the term –x with two terms: –x = –4x + 3x, – we get:
x² – x – 12 = x² – 4x + 3x – 12 = x (x – 4) + 3 (x – 4) = (x – 4) (x + 3).
3. x² – 7x + 6; here the required numbers are: –6 and –1. [Check: –6 + (–1) = –7; (–6) (–1) = +6].
x² – 7x + 6 = x² – 6x – x + 6 = x (x – 6) – (x – 6) = (x – 6) (x – 1).
Here the members of the second group –x + 6 had to be enclosed in parentheses, with a minus sign in front of them.
4. x² + 8x – 48. Here you need to find two numbers so that their sum is +8 and their product is –48. Since the product must have a minus sign, the required numbers must have different signs, since the sum of our numbers has a + sign, then the absolute value of the positive number must be greater. Expanding the arithmetic number 48 into two factors (and this can be done in different ways), we get: 48 = 1 ∙ 48 = 2 ∙ 24 = 3 ∙ 16 = 4 ∙ 12 = 6 ∙ 8. From these expansions it is easy to choose the one that suits our requirements , namely: 48 = 4 ∙ 12. Then our numbers are: +12 and –4. The rest is simple:
x² + 8x – 48 = x² + 12x – 4x – 48 = x (x + 12) – 4 (x + 12) = (x + 12) (x – 4).
5. x² + 7x – 12. Here you need to find 2 numbers so that their sum is +7 and their product = –12; 12 = 1 ∙ 12 = 2 ∙ 6 = 3 ∙ 4. Apparently, 3 and 4 would be suitable numbers, but they must be taken with different signs so that their product is equal to –12, and then their sum in no case may be +7 [–3 + (+4) = +1, +3 + (–4) = –1]. Other factorizations also do not give the required numbers; Therefore, we come to the conclusion that we are not yet able to decompose these quadratic trinomials into linear factors, since our technique is not applicable to it (it does not satisfy the second of the conditions that were established at the beginning).
SQUARE TRIPLE III
§ 54. Decomposition of a quadratic trinomial into linear factors
In this section we will consider the following question: in what case is the quadratic trinomial ax 2 + bx + c can be represented as a product
(a 1 x+b 1) (a 2 x+b 2)
two linear relative X multipliers with real coefficients a 1 , b 1 , a 2 , b 2 (a 1 =/=0, a 2 =/=0) ?
1. Suppose that the given quadratic trinomial ax 2 + bx + c let's represent it in the form
ax 2 + bx + c = (a 1 x+b 1) (a 2 x+b 2). (1)
The right-hand side of formula (1) vanishes when X = - b 1 / a 1 and X = - b 2 / a 2 (a 1 and a 2 are not equal to zero by condition). But in this case the numbers are b 1 / a 1 and - b 2 / a 2 are the roots of the equation
ax 2 + bx + c = 0.
Therefore, the discriminant of the quadratic trinomial ax 2 + bx + c must be non-negative.
2. Conversely, suppose that the discriminant D = b 2 - 4ac quadratic trinomial ax 2 + bx + c non-negative. Then this trinomial has real roots x 1 and x 2. Using Vieta's theorem, we obtain:
ax 2 + bx + c =A (x 2 + b / a X + c / a ) = A [x 2 - (x 1 + x 2) X + x 1 x 2 ] =
= A [(x 2 - x 1 x ) - (x 2 x - x 1 x 2)] = A [X (X - x 1) - x 2 (X - x 1) =
=a (X - x 1)(X - x 2).
ax 2 + bx + c = a (X - x 1)(X - x 2), (2)
Where x 1 and x 2 - roots of the trinomial ax 2 + bx + c . Coefficient A can be attributed to either of two linear factors, for example,
a (X - x 1)(X - x 2) = (ah - ax 1)(X - x 2).
But this means that in the case under consideration the square trinomial ax 2 + bx + c represent it as a product of two linear factors with real coefficients.
Combining the results obtained in paragraphs 1 and 2, we arrive at the following theorem.
Theorem. Square trinomial ax 2 + bx + c then and only then can be represented as a product of two linear factors with real coefficients,
ax 2 + bx + c = (ah - ax 1)(X - x 2),
when the discriminant of this quadratic trinomial is non-negative (that is, when this trinomial has real roots).
Example 1. Linear factorize 6 x 2 - X -1.
The roots of this quadratic trinomial are equal x 1 = 1/2 and x 2 = - 1 / 3 .
Therefore, according to formula (2)
6x 2 - X -1 = 6 (X - 1 / 2)(X + 1 / 3) = (2X - 1) (3x + 1).
Example 2. Linear factorization x 2 + X + 1. The discriminant of this quadratic trinomial is negative:
D = 1 2 - 4 1 1 = - 3< 0.
Therefore, this quadratic trinomial cannot be expanded into linear factors with real coefficients.
Exercises
Factor the following expressions into linear factors (No. 403 - 406):
403. 6x 2 - 7X + 2. 405. x 2 - X + 1.
404. 2x 2 - 7Oh + 6A 2 . 406. x 2 - 3Oh + 2A 2 - ab - b 2 .
Reduce fractions (No. 407, 408):
Solve equations:
