In this lesson, we will learn how to decompose square trinomials into linear factors. For this, it is necessary to recall Vieta's theorem and its inverse. This skill will help us quickly and conveniently decompose square trinomials into linear factors, and also simplify the reduction of fractions consisting of expressions.
So back to the quadratic equation , where .
What we have on the left side is called the square trinomial.
The theorem is true: If are the roots of a square trinomial, then the identity is true
Where is the leading coefficient, are the roots of the equation.
So, we have a quadratic equation - a square trinomial, where the roots of the quadratic equation are also called the roots of the quadratic trinomial. Therefore, if we have the roots of a square trinomial, then this trinomial is decomposed into linear factors.
Proof:
The proof of this fact is carried out using the Vieta theorem, which we considered in previous lessons.
Let's remember what Vieta's theorem tells us:
If are the roots of a square trinomial for which , then .
This theorem implies the following assertion that .
We see that, according to the Vieta theorem, i.e., substituting these values into the formula above, we get the following expression
Q.E.D.
Recall that we proved the theorem that if are the roots of a square trinomial, then the decomposition is valid.
Now let's recall an example of a quadratic equation, to which we selected the roots using Vieta's theorem. From this fact we can obtain the following equality thanks to the proved theorem:
Now let's check the correctness of this fact by simply expanding the brackets:
We see that we factored correctly, and any trinomial, if it has roots, can be factored according to this theorem into linear factors according to the formula
However, let's check whether for any equation such a factorization is possible:
Let's take the equation for example. First, let's check the sign of the discriminant
And we remember that in order to fulfill the theorem we have learned, D must be greater than 0, so in this case factorization by the studied theorem is impossible.
Therefore, we formulate a new theorem: if a square trinomial has no roots, then it cannot be decomposed into linear factors.
So, we have considered the Vieta theorem, the possibility of decomposing a square trinomial into linear factors, and now we will solve several problems.
Task #1
In this group, we will actually solve the problem inverse to the one posed. We had an equation, and we found its roots, decomposing into factors. Here we will do the opposite. Let's say we have the roots of a quadratic equation
The inverse problem is this: write a quadratic equation so that were its roots.
There are 2 ways to solve this problem.
Since are the roots of the equation, then 
is a quadratic equation whose roots are given numbers. Now let's open the brackets and check:
This was the first way we created a quadratic equation with given roots that does not have any other roots, since any quadratic equation has at most two roots.
This method involves the use of the inverse Vieta theorem.
If are the roots of the equation, then they satisfy the condition that .
For the reduced quadratic equation 
, , i.e. in this case , and .
Thus, we have created a quadratic equation that has the given roots.
Task #2
You need to reduce the fraction.
We have a trinomial in the numerator and a trinomial in the denominator, and the trinomials may or may not be factorized. If both the numerator and the denominator are factorized, then among them there may be equal factors that can be reduced.
First of all, it is necessary to factorize the numerator.
First, you need to check whether this equation can be factored, find the discriminant . Since , then the sign depends on the product ( must be less than 0), in this example , i.e., the given equation has roots.
To solve, we use the Vieta theorem:
In this case, since we are dealing with roots, it will be quite difficult to simply pick up the roots. But we see that the coefficients are balanced, i.e. if we assume that , and substitute this value into the equation, then the following system is obtained: i.e. 5-5=0. Thus, we have chosen one of the roots of this quadratic equation.
We will look for the second root by substituting what is already known into the system of equations, for example, , i.e. .
Thus, we have found both roots of the quadratic equation and can substitute their values into the original equation to factor it:
Recall the original problem, we needed to reduce the fraction.
Let's try to solve the problem by substituting instead of the numerator .
It is necessary not to forget that in this case the denominator cannot be equal to 0, i.e.,.
If these conditions are met, then we have reduced the original fraction to the form .
Task #3 (task with a parameter)
At what values of the parameter is the sum of the roots of the quadratic equation
If the roots of this equation exist, then 
, the question is when .
The factorization of square trinomials is one of the school assignments that everyone faces sooner or later. How to do it? What is the formula for factoring a square trinomial? Let's go through it step by step with examples.
General formula
The factorization of square trinomials is carried out by solving a quadratic equation. This is a simple task that can be solved by several methods - by finding the discriminant, using the Vieta theorem, there is also a graphical way to solve it. The first two methods are studied in high school.
The general formula looks like this:lx 2 +kx+n=l(x-x 1)(x-x 2) (1)
Task execution algorithm
In order to factorize square trinomials, you need to know Wit's theorem, have a program for solving at hand, be able to find a solution graphically or look for the roots of an equation of the second degree through the discriminant formula. If a square trinomial is given and it must be factored, the algorithm of actions is as follows:
1) Equate the original expression to zero to get the equation.
2) Give similar terms (if necessary).
3) Find the roots by any known method. The graphical method is best used if it is known in advance that the roots are integers and small numbers. It must be remembered that the number of roots is equal to the maximum degree of the equation, that is, the quadratic equation has two roots.
4) Substitute value X into expression (1).
5) Write down the factorization of square trinomials.
Examples
Practice allows you to finally understand how this task is performed. Examples illustrate the factorization of a square trinomial:
you need to expand the expression:
Let's use our algorithm:
1) x 2 -17x+32=0
2) similar terms are reduced
3) according to the Vieta formula, it is difficult to find the roots for this example, therefore it is better to use the expression for the discriminant:
D=289-128=161=(12.69) 2
4) Substitute the roots we found in the main formula for expansion:
(x-2.155) * (x-14.845)
5) Then the answer will be:
x 2 -17x + 32 \u003d (x-2.155) (x-14.845)
Let's check whether the solutions found by the discriminant correspond to the Vieta formulas:
14,845 . 2,155=32
For these roots, Vieta's theorem is applied, they were found correctly, which means that the factorization we obtained is also correct.
Similarly, we expand 12x 2 + 7x-6.
x 1 \u003d -7 + (337) 1/2
x 2 \u003d -7- (337) 1/2
In the previous case, the solutions were non-integer, but real numbers, which are easy to find with a calculator in front of you. Now consider a more complex example in which the roots are complex: factorize x 2 + 4x + 9. According to the Vieta formula, the roots cannot be found, and the discriminant is negative. The roots will be on the complex plane.
D=-20
Based on this, we get the roots we are interested in -4 + 2i * 5 1/2 and -4-2i * 5 1/2 because (-20) 1/2 = 2i*5 1/2 .
We obtain the desired expansion by substituting the roots into the general formula.
Another example: you need to factorize the expression 23x 2 -14x + 7.
We have the equation 23x 2 -14x+7 =0
D=-448
So the roots are 14+21,166i and 14-21,166i. The answer will be:
23x 2 -14x+7 =23(x- 14-21,166i )*(X- 14+21.166i ).
Let us give an example that can be solved without the help of the discriminant.
Let it be necessary to decompose the quadratic equation x 2 -32x + 255. Obviously, it can also be solved by the discriminant, but it is faster in this case to find the roots.
x 1 =15
x2=17
Means x 2 -32x + 255 =(x-15)(x-17).
Online calculator.
Selection of the square of the binomial and factorization of the square trinomial.
This math program extracts the square of the binomial from the square trinomial, i.e. makes a transformation of the form: \(ax^2+bx+c \rightarrow a(x+p)^2+q \) and factorizes the square trinomial: \(ax^2+bx+c \rightarrow a(x+n)(x+m) \)
Those. the problems are reduced to finding the numbers \(p, q \) and \(n, m \)
The program not only gives the answer to the problem, but also displays the solution process.
This program can be useful for high school students in preparation for tests and exams, when testing knowledge before the Unified State Exam, for parents to control the solution of many problems in mathematics and algebra. Or maybe it's too expensive for you to hire a tutor or buy new textbooks? Or do you just want to get your math or algebra homework done as quickly as possible? In this case, you can also use our programs with a detailed solution.
In this way, you can conduct your own training and/or the training of your younger brothers or sisters, while the level of education in the field of tasks to be solved is increased.
If you are not familiar with the rules for entering a square trinomial, we recommend that you familiarize yourself with them.
Rules for entering a square polynomial
Any Latin letter can act as a variable.
For example: \(x, y, z, a, b, c, o, p, q \) etc.
Numbers can be entered as integers or fractions.
Moreover, fractional numbers can be entered not only in the form of a decimal, but also in the form of an ordinary fraction.
Rules for entering decimal fractions.
In decimal fractions, the fractional part from the integer can be separated by either a dot or a comma.
For example, you can enter decimals like this: 2.5x - 3.5x^2
Rules for entering ordinary fractions.
Only a whole number can act as the numerator, denominator and integer part of a fraction.
The denominator cannot be negative.
When entering a numerical fraction, the numerator is separated from the denominator by a division sign: /
The integer part is separated from the fraction by an ampersand: &
Input: 3&1/3 - 5&6/5x +1/7x^2
Result: \(3\frac(1)(3) - 5\frac(6)(5) x + \frac(1)(7)x^2 \)
When entering an expression you can use brackets. In this case, when solving, the introduced expression is first simplified.
For example: 1/2(x-1)(x+1)-(5x-10&1/2)
Detailed Solution Example
Selection of the square of the binomial.$$ ax^2+bx+c \rightarrow a(x+p)^2+q $$ $$2x^2+2x-4 = $$ $$2x^2 +2 \cdot 2 \cdot\left( \frac(1)(2) \right)\cdot x+2 \cdot \left(\frac(1)(2) \right)^2-\frac(9)(2) = $$ $$2\left (x^2 + 2 \cdot\left(\frac(1)(2) \right)\cdot x + \left(\frac(1)(2) \right)^2 \right)-\frac(9 )(2) = $$ $$2\left(x+\frac(1)(2) \right)^2-\frac(9)(2) $$ Answer:$$2x^2+2x-4 = 2\left(x+\frac(1)(2) \right)^2-\frac(9)(2) $$ Factorization.$$ ax^2+bx+c \rightarrow a(x+n)(x+m) $$ $$2x^2+2x-4 = $$
$$ 2\left(x^2+x-2 \right) = $$
$$ 2 \left(x^2+2x-1x-1 \cdot 2 \right) = $$ $$ 2 \left(x \left(x +2 \right) -1 \left(x +2 \right ) \right) = $$ $$ 2 \left(x -1 \right) \left(x +2 \right) $$ Answer:$$2x^2+2x-4 = 2 \left(x -1 \right) \left(x +2 \right) $$
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A bit of theory.
Extraction of a square binomial from a square trinomial
If the square trinomial ax 2 + bx + c is represented as a (x + p) 2 + q, where p and q are real numbers, then they say that from square trinomial, the square of the binomial is highlighted.
Let us extract the square of the binomial from the trinomial 2x 2 +12x+14.
\(2x^2+12x+14 = 2(x^2+6x+7) \)
To do this, we represent 6x as a product of 2 * 3 * x, and then add and subtract 3 2 . We get:
$$ 2(x^2+2 \cdot 3 \cdot x + 3^2-3^2+7) = 2((x+3)^2-3^2+7) = $$ $$ = 2 ((x+3)^2-2) = 2(x+3)^2-4 $$
That. we selected the square of the binomial from the square trinomial, and showed that:
$$ 2x^2+12x+14 = 2(x+3)^2-4 $$
Factorization of a square trinomial
If the square trinomial ax 2 +bx+c is represented as a(x+n)(x+m), where n and m are real numbers, then the operation is said to be performed factorizations of a square trinomial.
Let's use an example to show how this transformation is done.
Let's factorize the square trinomial 2x 2 +4x-6.
Let us take the coefficient a out of brackets, i.e. 2:
\(2x^2+4x-6 = 2(x^2+2x-3) \)
Let's transform the expression in brackets.
To do this, we represent 2x as the difference 3x-1x, and -3 as -1*3. We get:
$$ = 2(x^2+3 \cdot x -1 \cdot x -1 \cdot 3) = 2(x(x+3)-1 \cdot (x+3)) = $$
$$ = 2(x-1)(x+3) $$
That. we factorize the square trinomial, and showed that:
$$ 2x^2+4x-6 = 2(x-1)(x+3) $$
Note that the factorization of a square trinomial is possible only when the quadratic equation corresponding to this trinomial has roots.
Those. in our case, factoring the trinomial 2x 2 +4x-6 is possible if the quadratic equation 2x 2 +4x-6 =0 has roots. In the process of factoring, we found that the equation 2x 2 +4x-6 =0 has two roots 1 and -3, because with these values, the equation 2(x-1)(x+3)=0 turns into a true equality.
Factorization of a square trinomial can be useful when solving inequalities from problem C3 or problem with parameter C5. Also, many B13 word problems will be solved much faster if you know Vieta's theorem.
This theorem, of course, can be considered from the standpoint of the 8th grade, in which it is first passed. But our task is to prepare well for the exam and learn how to solve exam tasks as efficiently as possible. Therefore, in this lesson, the approach is slightly different from the school one.
The formula for the roots of the equation according to Vieta's theorem know (or at least have seen) many:
$$x_1+x_2 = -\frac(b)(a), \quad x_1 x_2 = \frac(c)(a),$$
where `a, b` and `c` are the coefficients of the square trinomial `ax^2+bx+c`.
To learn how to use the theorem easily, let's understand where it comes from (it will be really easier to remember this way).
Let us have the equation `ax^2+ bx+ c = 0`. For further convenience, we divide it by `a` and get `x^2+\frac(b)(a) x + \frac(c)(a) = 0`. Such an equation is called a reduced quadratic equation.
Important lesson points: any square polynomial that has roots can be decomposed into brackets. Suppose ours can be represented as `x^2+\frac(b)(a) x + \frac(c)(a) = (x + k)(x+l)`, where `k` and ` l` - some constants.
Let's see how the brackets open:
$$(x + k)(x+l) = x^2 + kx+ lx+kl = x^2 +(k+l)x+kl.$$
Thus, `k+l = \frac(b)(a), kl = \frac(c)(a)`.
This is slightly different from the classical interpretation Vieta's theorems- in it we are looking for the roots of the equation. I propose to look for terms for bracket expansions- so you don't need to remember about the minus from the formula (meaning `x_1+x_2 = -\frac(b)(a)`). It is enough to choose two such numbers, the sum of which is equal to the average coefficient, and the product is equal to the free term.
If we need a solution to the equation, then it is obvious: the roots `x=-k` or `x=-l` (since in these cases one of the brackets will be set to zero, which means that the whole expression will be equal to zero).
For example, I will show the algorithm, how to decompose a square polynomial into brackets.
Example one. Algorithm for Factoring a Square Trinomial
The path we have is the square trinomial `x^2+5x+4`.
It is reduced (the coefficient of `x^2` is equal to one). He has roots. (To be sure, you can estimate the discriminant and make sure that it is greater than zero.)
Further steps (they need to be learned by completing all the training tasks):
- Make the following notation: $$x^2+5x+4=(x \ldots)(x \ldots).$$ Leave free space instead of dots, we will add appropriate numbers and signs there.
- View all possible options, how you can decompose the number `4` into the product of two numbers. We get pairs of "candidates" for the roots of the equation: `2, 2` and `1, 4`.
- Estimate from which pair you can get the average coefficient. Obviously it's `1, 4`.
- Write $$x^2+5x+4=(x \quad 4)(x \quad 1)$$.
- The next step is to place signs in front of the inserted numbers.
How to understand and remember forever what signs should be in front of the numbers in brackets? Try to expand them (brackets). The coefficient before `x` to the first power will be `(± 4 ± 1)` (we don't know the signs yet - we need to choose), and it should equal `5`. Obviously, there will be two pluses here $$x^2+5x+4=(x + 4)(x + 1)$$.
Perform this operation several times (hello, training tasks!) and there will never be more problems with this.
If you need to solve the equation `x^2+5x+4`, then now its solution is not difficult. Its roots are `-4, -1`.
Second example. Factorization of a square trinomial with coefficients of different signs
Let us need to solve the equation `x^2-x-2=0`. Offhand, the discriminant is positive.
We follow the algorithm.
- $$x^2-x-2=(x \ldots) (x \ldots).$$
- There is only one integer factorization of 2: `2 · 1`.
- We skip the point - there is nothing to choose from.
- $$x^2-x-2=(x \quad 2) (x \quad 1).$$
- The product of our numbers is negative (`-2` is a free term), which means that one of them will be negative and the other positive.
Since their sum is equal to `-1` (coefficient of `x`), then `2` will be negative (intuitive explanation - two is the larger of the two numbers, it will "pull" more in the negative direction). We get $$x^2-x-2=(x - 2) (x + 1).$$
Third example. Factorization of a square trinomial
Equation `x^2+5x -84 = 0`.
- $$x+ 5x-84=(x \ldots) (x \ldots).$$
- Decomposition of 84 into integer factors: `4 21, 6 14, 12 7, 2 42`.
- Since we need the difference (or sum) of the numbers to be 5, the pair `7, 12` will do.
- $$x+ 5x-84=(x\quad 12) (x \quad 7).$$
- $$x+ 5x-84=(x + 12) (x - 7).$$
Hope, decomposition of this square trinomial into brackets understandably.
If you need a solution to the equation, then here it is: `12, -7`.
Tasks for training
Here are a few examples that are easy to are solved using Vieta's theorem.(Examples taken from Mathematics, 2002.)
- `x^2+x-2=0`
- `x^2-x-2=0`
- `x^2+x-6=0`
- `x^2-x-6=0`
- `x^2+x-12=0`
- `x^2-x-12=0`
- `x^2+x-20=0`
- `x^2-x-20=0`
- `x^2+x-42=0`
- `x^2-x-42=0`
- `x^2+x-56=0`
- `x^2-x-56=0`
- `x^2+x-72=0`
- `x^2-x-72=0`
- `x^2+x-110=0`
- `x^2-x-110=0`
- `x^2+x-420=0`
- `x^2-x-420=0`
A couple of years after the article was written, a collection of 150 tasks appeared for expanding a quadratic polynomial using the Vieta theorem.
Like and ask questions in the comments!
8 examples of factorization of polynomials are given. They include examples with solving quadratic and biquadratic equations, examples with recurrent polynomials, and examples with finding integer roots of third and fourth degree polynomials.
1. Examples with the solution of a quadratic equation
Example 1.1
x 4 + x 3 - 6 x 2.
Decision
Take out x 2
for brackets:
.
2 + x - 6 = 0:
.
Equation roots:
, .
.
Answer
Example 1.2
Factoring a third-degree polynomial:
x 3 + 6 x 2 + 9 x.
Decision
We take x out of brackets:
.
We solve the quadratic equation x 2 + 6 x + 9 = 0:
Its discriminant is .
Since the discriminant is equal to zero, the roots of the equation are multiples: ;
.
From here we obtain the decomposition of the polynomial into factors:
.
Answer
Example 1.3
Factoring a fifth-degree polynomial:
x 5 - 2 x 4 + 10 x 3.
Decision
Take out x 3
for brackets:
.
We solve the quadratic equation x 2 - 2 x + 10 = 0.
Its discriminant is .
Since the discriminant is less than zero, the roots of the equation are complex: ;
, .
The factorization of a polynomial has the form:
.
If we are interested in factoring with real coefficients, then:
.
Answer
Examples of factoring polynomials using formulas
Examples with biquadratic polynomials
Example 2.1
Factorize the biquadratic polynomial:
x 4 + x 2 - 20.
Decision
Apply the formulas:
a 2 + 2 ab + b 2 = (a + b) 2;
a 2 - b 2 = (a - b)(a + b).
;
.
Answer
Example 2.2
Factoring a polynomial that reduces to a biquadratic:
x 8 + x 4 + 1.
Decision
Apply the formulas:
a 2 + 2 ab + b 2 = (a + b) 2;
a 2 - b 2 = (a - b)(a + b):
;
;
.
Answer
Example 2.3 with recursive polynomial
Factoring the recursive polynomial:
.
Decision
The recursive polynomial has an odd degree. Therefore it has a root x = - 1
. We divide the polynomial by x - (-1) = x + 1. As a result, we get:
.
We make a substitution:
, ;
;
;
.
Answer
Examples of Factoring Polynomials with Integer Roots
Example 3.1
Factoring a polynomial:
.
Decision
Suppose the equation
6
-6, -3, -2, -1, 1, 2, 3, 6
.
(-6) 3 - 6 (-6) 2 + 11 (-6) - 6 = -504;
(-3) 3 - 6 (-3) 2 + 11 (-3) - 6 = -120;
(-2) 3 - 6 (-2) 2 + 11 (-2) - 6 = -60;
(-1) 3 - 6 (-1) 2 + 11 (-1) - 6 = -24;
1 3 - 6 1 2 + 11 1 - 6 = 0;
2 3 - 6 2 2 + 11 2 - 6 = 0;
3 3 - 6 3 2 + 11 3 - 6 = 0;
6 3 - 6 6 2 + 11 6 - 6 = 60.
So, we have found three roots:
x 1 = 1
, x 2 = 2
, x 3 = 3
.
Since the original polynomial is of the third degree, it has no more than three roots. Since we have found three roots, they are simple. Then
.
Answer
Example 3.2
Factoring a polynomial:
.
Decision
Suppose the equation
has at least one integer root. Then it is the divisor of the number 2
(a member without x ). That is, the whole root can be one of the numbers:
-2, -1, 1, 2
.
Substitute these values one by one:
(-2) 4 + 2 (-2) 3 + 3 (-2) 3 + 4 (-2) + 2 = 6
;
(-1) 4 + 2 (-1) 3 + 3 (-1) 3 + 4 (-1) + 2 = 0
;
1 4 + 2 1 3 + 3 1 3 + 4 1 + 2 = 12;
2 4 + 2 2 3 + 3 2 3 + 4 2 + 2 = 54
.
If we assume that this equation has an integer root, then it is a divisor of the number 2
(a member without x ). That is, the whole root can be one of the numbers:
1, 2, -1, -2
.
Substitute x = -1
:
.
So we have found another root x 2
= -1
. It would be possible, as in the previous case, to divide the polynomial by , but we will group the terms:
.
Since the equation x 2 + 2 = 0 has no real roots, then the factorization of the polynomial has the form.
