Expanding polynomials to get a product sometimes seems confusing. But it is not so difficult if you understand the process step by step. The article details how to factorize a square trinomial.

Many do not understand how to factorize a square trinomial, and why this is done. At first it may seem that this is a useless exercise. But in mathematics, nothing is done just like that. The transformation is necessary to simplify the expression and the convenience of calculation.

A polynomial having the form - ax² + bx + c, is called a square trinomial. The term "a" must be negative or positive. In practice, this expression is called a quadratic equation. Therefore, sometimes they say differently: how to expand a quadratic equation.

Interesting! A square polynomial is called because of its largest degree - a square. And a trinomial - because of the 3 component terms.

Some other kinds of polynomials:

• linear binomial (6x+8);
• cubic quadrilateral (x³+4x²-2x+9).

Factorization of a square trinomial

First, the expression is equal to zero, then you need to find the values ​​of the roots x1 and x2. There may be no roots, there may be one or two roots. The presence of roots is determined by the discriminant. Its formula must be known by heart: D=b²-4ac.

If the result of D is negative, there are no roots. If positive, there are two roots. If the result is zero, the root is one. The roots are also calculated by the formula.

If the calculation of the discriminant results in zero, you can apply any of the formulas. In practice, the formula is simply abbreviated: -b / 2a.

Formulas for different values discriminant are different.

If D is positive:

If D is zero:

Online calculators

The Internet has online calculator. It can be used to factorize. Some resources provide the opportunity to see the solution step by step. Such services help to better understand the topic, but you need to try to understand well.

Useful video: Factoring a square trinomial

Examples

We suggest looking at simple examples of how to factorize a quadratic equation.

Example 1

Here it is clearly shown that the result will be two x, because D is positive. They need to be substituted into the formula. If the roots are negative, the sign in the formula is reversed.

We know the decomposition formula square trinomial multipliers: a(x-x1)(x-x2). We put the values ​​in brackets: (x+3)(x+2/3). There is no number before the term in the exponent. This means that there is a unit, it is lowered.

Example 2

This example clearly shows how to solve an equation that has one root.

Substitute the resulting value:

Example 3

Given: 5x²+3x+7

First, we calculate the discriminant, as in the previous cases.

D=9-4*5*7=9-140= -131.

The discriminant is negative, which means there are no roots.

After receiving the result, it is worth opening the brackets and checking the result. The original trinomial should appear.

Alternative solution

Some people have never been able to make friends with the discriminant. There is another way to factorize a square trinomial. For convenience, the method is shown in an example.

Given: x²+3x-10

We know that we should end up with 2 parentheses: (_)(_). When the expression looks like this: x² + bx + c, we put x at the beginning of each bracket: (x_) (x_). The remaining two numbers are the product that gives "c", i.e. -10 in this case. To find out what these numbers are, you can only use the selection method. Substituted numbers must match the remaining term.

For example, multiplying the following numbers gives -10:

• -1, 10;
• -10, 1;
• -5, 2;
• -2, 5.

1. (x-1)(x+10) = x2+10x-x-10 = x2+9x-10. No.
2. (x-10)(x+1) = x2+x-10x-10 = x2-9x-10. No.
3. (x-5)(x+2) = x2+2x-5x-10 = x2-3x-10. No.
4. (x-2)(x+5) = x2+5x-2x-10 = x2+3x-10. Fits.

So, the transformation of the expression x2+3x-10 looks like this: (x-2)(x+5).

Important! You should be careful not to confuse the signs.

Decomposition of a complex trinomial

If "a" is greater than one, difficulties begin. But everything is not as difficult as it seems.

In order to factorize, one must first see if it is possible to factor something out.

For example, given the expression: 3x²+9x-30. Here the number 3 is taken out of brackets:

3(x²+3x-10). The result is the already known trinomial. The answer looks like this: 3(x-2)(x+5)

How to decompose if the term that is squared is negative? AT this case the number -1 is taken out of the bracket. For example: -x²-10x-8. The expression will then look like this:

The scheme differs little from the previous one. There are only a few new things. Let's say the expression is given: 2x²+7x+3. The answer is also written in 2 brackets, which must be filled in (_) (_). X is written in the 2nd bracket, and what is left in the 1st. It looks like this: (2x_)(x_). Otherwise, the previous scheme is repeated.

The number 3 gives the numbers:

• -1, -3;
• -3, -1;
• 3, 1;
• 1, 3.

We solve equations by substituting the given numbers. The last option fits. So the transformation of the expression 2x²+7x+3 looks like this: (2x+1)(x+3).

Other cases

It is not always possible to transform an expression. In the second method, the solution of the equation is not required. But the possibility of converting terms into a product is checked only through the discriminant.

It is worth practicing solving quadratic equations so that there are no difficulties when using formulas.

Useful video: factorization of a trinomial

Conclusion

You can use it in any way. But it is better to work both to automatism. Also, those who are going to connect their lives with mathematics need to learn how to solve quadratic equations well and decompose polynomials into factors. All the following mathematical topics are built on this.

In contact with

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 control work and exams, when testing knowledge before the exam, 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 training your younger brothers or sisters, while the level of education in the field of tasks being solved increases.

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:$$2x^2+2x-4 = 2 \left(x -1 \right) \left(x +2 \right) $$

Decide

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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.

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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 .

Plan - lesson summary (MBOU "Chernomorsk secondary school No. 2"

Name of the teacher

Ponomarenko Vladislav Vadimovich

Thing

Algebra

Date of the lesson

19.09.2018

№ lesson

Class

9B

Lesson topic

(according to KTP)

"Decomposition of a square trinomial into factors"

goal setting

- educational: teach students how to factorize a square trinomial, teach how to apply the factorization algorithm of a square trinomial when solving examples, consider tasks in the GIA database that use the algorithm for factoring a square trinomial into factors

- developing: to develop in schoolchildren the ability to formulate problems, to propose ways to solve them, to promote the development of schoolchildren's skills to highlight the main thing in a cognitive object.

- educational: to help students realize the value of joint activities, to promote the development of children's skills to exercise self-control, self-assessment and self-correction of educational activities.

Lesson type

learning and primary consolidation of new knowledge.

Equipment:

multimedia projector, screen, computer, didactic material, textbooks, notebooks, presentationto the lesson

During the classes

1. Organizing time: the teacher greets the students, checks the readiness for the lesson.

Motivates students:

Today in the lesson in joint activity we will confirm the words of Poya (Slide 1). (“The problem you solve may be very modest, but if it challenges your curiosity, and if you solve it on your own, then you can experience leading to open the tension of the mind and enjoy the joy of victory.” Poya door.)

Message about Poya (Slide 2)

I want to challenge your curiosity. Consider the task from the GIA. Plot the Function .

Can we enjoy the joy of victory and complete this task? (problem situation).

How to solve this problem?

- Outline an action plan to address this issue.

Corrects the lesson plan, comments on the principle of independent work.

Independent work(hand out leaflets with the text of independent work to the class) (Appendix 1)

Independent work

Multiply:

x 2 – 3x;

x 2 – 9;

x 2 – 8x + 16;

2a 2 – 2b 2 –a + b;

2x 2 – 7x – 4.

Reduce fraction:

SlideWith answers for self-examination.

Question to the class:

What methods of factoring a polynomial did you use?

Have you been able to factorize all polynomials?

Can all fractions be reduced?

Problem2:Slide

How to factorize a polynomial

2 x 2 – 7 x – 4?

How to reduce a fraction?

Frontal survey:

What are polynomials

2 x 2 – 7 x– 4 andx 2 – 5 x +6?

Define a square trinomial.

What do we know about the square trinomial?

How to find its roots?

What determines the number of roots?

Compare this knowledge with what we need to know and formulate the topic of the lesson. (After that, the topic of the lesson is displayed on the screen)Slide

Set the goal of the lessonSlide

Let's see the end resultSlide

Question to the class:How to solve this problem?

The class works in groups.

Task for groups:

find the desired page in the table of contents, read item 4 with a pencil in hand, highlight the main idea, draw up an algorithm by which any square trinomial can be factorized.

Checking the performance of the task by the class (frontal work):

What is the main idea of ​​paragraph 4?Slide(on the screen, the formula for factoring a square trinomial into factors).

algorithm on the screen.Slide

1. Equate the square trinomial to zero.

2. Find the discriminant.

3. Find the roots of a square trinomial.

4. Substitute the found roots into the formula.

5. If necessary, enter the leading coefficient in brackets.

Another onelittle problem : if D=0, is it possible to factorize a square trinomial, and if so, how?

(Research in groups).

Slide(on the screen:

If D = 0, then
.

If the square trinomial has no roots,

then it cannot be factored.)

Let's return to the task in independent work. Can we now factorize square trinomials2 x 2 – 7 x– 4 andx 2 – 5 x +6?

The class works independently, multiplies, I work individually with weak students.

Slide(with solution)Mutual check

Can we reduce the fraction?

Reduce the fraction, I call a strong student to the board.

Let's get back to the taskfrom GIA. Can we now graph the function?

What is the graph of this function?

Draw a graph of the function in your notebook.

Test (withindependent work)Annex 2

Self-examination and self-assessmentThe students were given leaflets (Appendix 3) in which they need to write down their answers. They provide evaluation criteria.

Evaluation criteria:

3 tasks - assessment "4"

4 tasks - grade "5"

Reflection:(slide)

1. Today at the lesson I learned ...

2. Today in the lesson I repeated ...

3. I fixed…

4. I liked ...

5. I gave myself a grade for the activity in the lesson ...

6. What types of work caused difficulties and require repetition ...

7. Have we achieved the intended result?

Slide: Thanks for the lesson!

Appendix 1

Independent work

Multiply:

x 2 – 3x;

x 2 – 9;

x 2 – 8x + 16;

x 2 + x - 2;

2a 2 – 2b 2 –a + b;

2 x 2 – 7 x – 4.

Reduce fraction:

Annex 2

Test

1 option

factorize?

x 2 – 8x+ 7;

x 2 – 8x+ 16 ;

x 2 – 8x+ 9;

x 2 – 8x+ 1 7.

2 x 2 – 9 x – 5 = 2( x – 5)(…)?

Answer:_________ .

Reduce the fraction:

x – 3;

x + 3;

x – 4;

another answer.

Test

Option 2

What square trinomial cannot be pfactorize?

5 x 2 + x+ 1;

⅓ x 2 –8x+ 2;

0,1 x 2 + 3 x - 5;

x 2 + 4 x+ 5.

Which polynomial should be substituted for the ellipsis in order to have equality:2 x 2 + 5 x – 3 = 2( x + 3)(…)?

Answer:_________ .

Reduce the fraction:

3 x 2 – 6 x – 15;

0,25(3 x - 1);

0,25( x - 1);

another answer.

Annex 3

Write down the answers.

Evaluation criteria:

Correctly done: task 2 - grade "3"

3 tasks - assessment "4"

4 tasks - grade "5"

Task number 1

Task number 2

Task number 3

1 option

Option 2

A square trinomial is a polynomial of the form ax^2+bx+c, where x is a variable, a, b and c are some numbers, and a is not equal to zero.
Actually, the first thing we need to know in order to factorize the ill-fated trinomial is the theorem. It looks like this: “If x1 and x2 are the roots of the square trinomial ax^2+bx+c, then ax^2+bx+c=a(x-x1)(x-x2)”. Of course, there is also a proof of this theorem, but it requires some theoretical knowledge (if we take out the factor a in the polynomial ax^2+bx+c we get ax^2+bx+c=a(x^2+(b/a) x + c/a) By Viette's theorem x1+x2=-(b/a), x1*x2=c/a, hence b/a=-(x1+x2), c/a=x1*x2. , x^2+ (b/a)x+c/a= x^2- (x1+x2)x+ x1x2=x^2-x1x-x2x+x1x2=x(x-x1)-x2(x-x1 )= (x-x1)(x-x2), so ax^2+bx+c=a(x-x1)(x-x2) Sometimes teachers make you learn the proof, but if it is not required, I advise you to just remember final formula.

2 step

Let's take as an example the trinomial 3x^2-24x+21. The first thing we need to do is equate the trinomial to zero: 3x^2-24x+21=0. The roots of the resulting quadratic equation will be the roots of the trinomial, respectively.

3 step

Solve the equation 3x^2-24x+21=0. a=3, b=-24, c=21. So, let's decide. Who does not know how to solve quadratic equations, look at my instructions with 2 ways to solve them using the example of the same equation. We got the roots x1=7, x2=1.

4 step

Now that we have the trinomial roots, we can safely substitute them into the formula =) ax^2+bx+c=a(x-x1)(x-x2)
we get: 3x^2-24x+21=3(x-7)(x-1)
You can get rid of the term a by putting it in brackets: 3x^2-24x+21=(x-7)(x*3-1*3)
as a result we get: 3x^2-24x+21=(x-7)(3x-3). Note: each of the obtained factors ((x-7), (3x-3) are polynomials of the first degree. That's the whole decomposition =) If you doubt the answer you got, you can always check it by multiplying the brackets.

5 step

Verification of the solution. 3x^2-24x+21=3(x-7)(x-3)
(x-7)(3x-3)=3x^2-3x-21x+21=3x^2-24x+21. Now we know for sure that our solution is correct! I hope my instructions help someone =) Good luck with your studies!

• In our case, in the equation D > 0 and we got 2 roots each. If it were D<0, то уравнение, как и многочлен, соответственно, корней бы не имело.
• If a square trinomial has no roots, then it cannot be factored into factors that are polynomials of the first degree.