How to factorize a square trinomial. Factoring a polynomial Factoring a quadratic polynomial

He has 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 quaternary
\(2x+1\) - linear binomial

The root of the square 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 the roots for the square 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 square 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 the 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 the roots are equal to \(-5\) and \(\frac(2)(3)\). So \(3x^2+13x-10=3(x+5)(x-\frac(2)(3))\). It is easy to verify the correctness of this statement - if we , then we get 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 equal to zero.

The square trinomial \(ax^2+bx+c\) does not factorize 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 decompose them into factors.

Example . Factor \(2x^2-11x+12\).
Solution :
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 received answer may be written in a different way: \((2x-3)(x-4)\).

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 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, therefore, in this case, factoring according to 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 .

This online calculator is designed to factorize a function.

For example, factorize: x 2 /3-3x+12 . Let's write it as x^2/3-3*x+12 . You can also use this service, where all calculations are saved in Word format.

For example, decompose into terms. Let's write it as (1-x^2)/(x^3+x) . To see the progress of the solution, click Show steps . If you need to get the result in Word format, use this service.

Note: the number "pi" (π) is written as pi ; square root as sqrt , e.g. sqrt(3) , the tangent of tg is written as tan . See the Alternative section for a response.

1. If a simple expression is given, for example, 8*d+12*c*d , then factoring the expression means to factor the expression. To do this, you need to find common factors. We write this expression as: 4*d*(2+3*c) .
2. Express the product as two binomials: x 2 + 21yz + 7xz + 3xy . Here we already need to find several common factors: x(x + 7z) + 3y(x + 7z). We take out (x+7z) and get: (x+7z)(x + 3y) .

see also Division of polynomials by a corner (all steps of division by a column are shown)

Useful in learning the rules of factorization are abbreviated multiplication formulas, with which it will be clear how to open brackets with a square:

1. (a+b) 2 = (a+b)(a+b) = a 2 +2ab+b 2
2. (a-b) 2 = (a-b)(a-b) = a 2 -2ab+b 2
3. (a+b)(a-b) = a 2 - b 2
4. a 3 +b 3 = (a+b)(a 2 -ab+b 2)
5. a 3 -b 3 = (a-b)(a 2 +ab+b 2)
6. (a+b) 3 = (a+b)(a+b) 2 = a 3 +3a 2 b + 3ab 2 +b 3
7. (a-b) 3 = (a-b)(a-b) 2 = a 3 -3a 2 b + 3ab 2 -b 3

Factoring methods

1. Using abbreviated multiplication formulas.
2. Search for a common factor.

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, therefore, in this case, factoring according to 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 .

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 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):

1. 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.
2. Consider all possible options for 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`.
3. Estimate from which pair you can get the average coefficient. Obviously it's `1, 4`.
4. Write $$x^2+5x+4=(x \quad 4)(x \quad 1)$$.
5. 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.

1. $$x^2-x-2=(x \ldots) (x \ldots).$$
2. There is only one integer factorization of 2: `2 · 1`.
3. We skip the point - there is nothing to choose from.
4. $$x^2-x-2=(x \quad 2) (x \quad 1).$$
5. 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`.

1. $$x+ 5x-84=(x \ldots) (x \ldots).$$
2. Decomposition of 84 into integer factors: `4 21, 6 14, 12 7, 2 42`.
3. Since we need the difference (or sum) of the numbers to be 5, the pair `7, 12` will do.
4. $$x+ 5x-84=(x\quad 12) (x \quad 7).$$
5. $$x+ 5x-84=(x + 12) (x - 7).$$

Hope, decomposition of this square trinomial into brackets It's clear.

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.)

1. `x^2+x-2=0`
2. `x^2-x-2=0`
3. `x^2+x-6=0`
4. `x^2-x-6=0`
5. `x^2+x-12=0`
6. `x^2-x-12=0`
7. `x^2+x-20=0`
8. `x^2-x-20=0`
9. `x^2+x-42=0`
10. `x^2-x-42=0`
11. `x^2+x-56=0`
12. `x^2-x-56=0`
13. `x^2+x-72=0`
14. `x^2-x-72=0`
15. `x^2+x-110=0`
16. `x^2-x-110=0`
17. `x^2+x-420=0`
18. `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.

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