Trinomial factorization examples. Square trinomial

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 triple square terms into linear factors, and will also simplify the reduction of fractions consisting of expressions.

So back to the quadratic equation, where.

What we have on the left is called a square trinomial.

The theorem is true: If are the roots of a square trinomial, then the identity

Where is the senior coefficient, are the roots of the equation.

So, we have a quadratic equation - a quadratic trinomial, where the roots of the quadratic equation are also called the roots of a quadratic trinomial. Therefore, if we have the roots of a square trinomial, then this trinomial is decomposed into linear factors.

Evidence:

The proof of this fact is carried out using Vieta's theorem, which we considered in the 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 statement that.

We see that, according to Vieta's theorem, i.e., substituting these values \u200b\u200binto the formula above, we get the following expression

q.E.D.

Recall that we have 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 found 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 parentheses:

We see that we have factorized correctly, and any trinomial, if it has roots, can be decomposed by this theorem into linear factors by the formula

However, let's check if such factorization is possible for any equation:

Take an equation, for example. First, let's check the discriminant sign

But we remember that for the theorem we have learned to hold, D must be greater than 0, so in this case the 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 considered Vieta's theorem, the possibility of decomposing a square trinomial into linear factors, and now we will solve several problems.

Problem number 1

In this group, we will in fact solve the opposite problem to the set one. We had an equation, and we found its roots by factoring it. Here we will act in reverse. Let's say we have the roots of a quadratic equation

The inverse problem is this: make a quadratic equation to be 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 expand the brackets and check:

This was the first way that we created a quadratic equation with given roots, in which there are no other roots, since any quadratic equation has at most two roots.

This method assumes the use of Vieta's inverse theorem.

If are the roots of the equation, then they satisfy the condition that.

For the reduced quadratic equation ,, i.e., in this case, a.

Thus, we have created a quadratic equation that has the given roots.

Problem number 2

It is necessary to reduce the fraction.

We have a three-term in the numerator and a three-term in the denominator, and the three-terms can be expanded or not. If both the numerator and the denominator are decomposed into factors, then among them there may be equal factors that can be canceled.

First of all, you need to factor the numerator.

First, it is necessary to check whether the given equation can be factorized, find the discriminant. Since, then the sign depends on the product (must be less than 0), in this example, that is, the given equation has roots.

To solve the problem, we use Vieta's theorem:

In this case, since we are dealing with roots, it will be rather difficult to simply pick up the roots. But we see that the coefficients are balanced, that is, if we assume that, and substitute this value in the equation, then the following system is obtained:, that is, 5-5 \u003d 0. Thus, we have selected one of the roots of this quadratic equation.

We will look for the second root by substituting the 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 \u200b\u200binto the original equation to factor it into factors:

Let's remember the original task, we needed to reduce the fraction.

Let's try to solve the problem by substituting for 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 number 3 (task with a parameter)

For what values \u200b\u200bof the parameter the sum of the roots of the quadratic equation

If the roots of this equation exist, then , question: when.

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 triple square terms into linear factors, and will also simplify the reduction of fractions consisting of expressions.

So back to the quadratic equation, where.

What we have on the left is called a square trinomial.

The theorem is true: If are the roots of a square trinomial, then the identity

Where is the senior coefficient, are the roots of the equation.

So, we have a quadratic equation - a quadratic trinomial, where the roots of the quadratic equation are also called the roots of a quadratic trinomial. Therefore, if we have the roots of a square trinomial, then this trinomial is decomposed into linear factors.

Evidence:

The proof of this fact is carried out using Vieta's theorem, which we considered in the 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 statement that.

We see that, according to Vieta's theorem, i.e., substituting these values \u200b\u200binto the formula above, we get the following expression

q.E.D.

Recall that we have 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 found 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 parentheses:

We see that we have factorized correctly, and any trinomial, if it has roots, can be decomposed by this theorem into linear factors by the formula

However, let's check if such factorization is possible for any equation:

Take an equation, for example. First, let's check the discriminant sign

But we remember that for the theorem we have learned to hold, D must be greater than 0, so in this case the 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 considered Vieta's theorem, the possibility of decomposing a square trinomial into linear factors, and now we will solve several problems.

Problem number 1

In this group, we will in fact solve the opposite problem to the set one. We had an equation, and we found its roots by factoring it. Here we will act in reverse. Let's say we have the roots of a quadratic equation

The inverse problem is this: make a quadratic equation to be 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 expand the brackets and check:

This was the first way that we created a quadratic equation with given roots, in which there are no other roots, since any quadratic equation has at most two roots.

This method assumes the use of Vieta's inverse theorem.

If are the roots of the equation, then they satisfy the condition that.

For the reduced quadratic equation ,, i.e., in this case, a.

Thus, we have created a quadratic equation that has the given roots.

Problem number 2

It is necessary to reduce the fraction.

We have a three-term in the numerator and a three-term in the denominator, and the three-terms can be expanded or not. If both the numerator and the denominator are decomposed into factors, then among them there may be equal factors that can be canceled.

First of all, you need to factor the numerator.

First, it is necessary to check whether the given equation can be factorized, find the discriminant. Since, then the sign depends on the product (must be less than 0), in this example, that is, the given equation has roots.

To solve the problem, we use Vieta's theorem:

In this case, since we are dealing with roots, it will be rather difficult to simply pick up the roots. But we see that the coefficients are balanced, that is, if we assume that, and substitute this value in the equation, then the following system is obtained:, that is, 5-5 \u003d 0. Thus, we have selected one of the roots of this quadratic equation.

We will look for the second root by substituting the 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 \u200b\u200binto the original equation to factor it into factors:

Let's remember the original task, we needed to reduce the fraction.

Let's try to solve the problem by substituting for 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 number 3 (task with a parameter)

For what values \u200b\u200bof the parameter the sum of the roots of the quadratic equation

If the roots of this equation exist, then , question: when.

There are 8 examples of factorization of polynomials. They include examples with solving quadratic and biquadratic equations, examples with reflexive polynomials, and examples with finding integer roots of third and fourth degree polynomials.

Content

See also: Methods for factoring polynomials
Quadratic Roots
Solving cubic equations

1. Examples with solving a quadratic equation

Example 1.1

x 4 + x 3 - 6 x 2.

Take out x 2 outside the brackets:
.
2 + x - 6 \u003d 0:
.
Equation roots:
, .

.

Example 1.2

Factor a third degree polynomial:
x 3 + 6 x 2 + 9 x.

Move x out of parentheses:
.
Solving the quadratic equation x 2 + 6 x + 9 \u003d 0:
Its discriminant:.
Since the discriminant is zero, the roots of the equation are multiple:;
.

From this we get the factorization of the polynomial:
.

Example 1.3

Factor a 5th degree polynomial:
x 5 - 2 x 4 + 10 x 3.

Take out x 3 outside the brackets:
.
Solving the quadratic equation x 2 - 2 x + 10 \u003d 0.
Its discriminant:.
Since the discriminant is less than zero, the roots of the equation are complex:;
, .

The factorization of a polynomial is:
.

If we are interested in factorization with real coefficients, then:
.

Examples of factoring polynomials using formulas

Examples with biquadratic polynomials

Example 2.1

Factor out a biquadratic polynomial:
x 4 + x 2 - 20.

Let's apply the formulas:
a 2 + 2 ab + b 2 \u003d (a + b) 2;
a 2 - b 2 \u003d (a - b) (a + b).

;
.

Example 2.2

Factor a polynomial that reduces to a biquadratic one:
x 8 + x 4 + 1.

Let's apply the formulas:
a 2 + 2 ab + b 2 \u003d (a + b) 2;
a 2 - b 2 \u003d (a - b) (a + b):

;

;
.

Example 2.3 with a returnable polynomial

Factor the return polynomial:
.

The reflexive polynomial has an odd degree. Therefore, it has a root x \u003d - 1 ... We divide the polynomial by x - (-1) \u003d x + 1... As a result, we get:
.
We make the substitution:
, ;
;


;
.

Examples of factoring polynomials with integer roots

Example 3.1

Factor out a polynomial:
.

Suppose the equation

6
-6, -3, -2, -1, 1, 2, 3, 6 .
(-6) 3 - 6 (-6) 2 + 11 (-6) - 6 \u003d -504;
(-3) 3 - 6 (-3) 2 + 11 (-3) - 6 \u003d -120;
(-2) 3 - 6 (-2) 2 + 11 (-2) - 6 \u003d -60;
(-1) 3 - 6 (-1) 2 + 11 (-1) - 6 \u003d -24;
1 3 - 6 1 2 + 11 1 - 6 \u003d 0;
2 3 - 6 2 2 + 11 2 - 6 \u003d 0;
3 3 - 6 3 2 + 11 3 - 6 \u003d 0;
6 3 - 6 6 2 + 11 6 - 6 \u003d 60.

So, we found three roots:
x 1 = 1 , x 2 = 2 , x 3 = 3 .
Since the original polynomial is of the third degree, it has at most three roots. Since we found three roots, they are simple. Then
.

Example 3.2

Factor out a polynomial:
.

Suppose the equation

has at least one whole root. Then it is a divisor of the number 2 (term without x). That is, the whole root can be one of the numbers:
-2, -1, 1, 2 .
We substitute these values \u200b\u200bin turn:
(-2) 4 + 2 (-2) 3 + 3 (-2) 3 + 4 (-2) + 2 \u003d 6 ;
(-1) 4 + 2 (-1) 3 + 3 (-1) 3 + 4 (-1) + 2 \u003d 0 ;
1 4 + 2 1 3 + 3 1 3 + 4 1 + 2 \u003d 12;
2 4 + 2 2 3 + 3 2 3 + 4 2 + 2 \u003d 54.

So, we found one root:
x 1 = -1 .
Divide the polynomial by x - x 1 \u003d x - (-1) \u003d x + 1:


Then,
.

Now we need to solve the third degree equation:
.
If we assume that this equation has an integer root, then it is a divisor of the number 2 (term without x). That is, the whole root can be one of the numbers:
1, 2, -1, -2 .
Substitute x \u003d -1 :
.

So, we 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 members:
.

Square trinomial is called a polynomial of the form ax 2 +bx +cwhere x - variable, a,b,c Are some numbers, and a ≠ 0.

Coefficient and called senior odds, c – free member square trinomial.

Examples of square trinomials:

2 x 2 + 5x + 4 (here a = 2, b = 5, c = 4)

x 2 - 7x + 5 (here a = 1, b = -7, c = 5)

9x 2 + 9x - 9 (here a = 9, b = 9, c = -9)

Coefficient b or coefficient c or both coefficients can simultaneously be equal to zero. For instance:

5 x 2 + 3x(herea \u003d 5,b \u003d 3,c \u003d 0, so there is no c value in the equation).

6x 2 - 8 (here a \u003d 6, b \u003d 0, c \u003d -8)

2x 2 (here a \u003d 2, b \u003d 0, c \u003d 0)

The value of a variable at which the polynomial vanishes is called by the root of the polynomial.

To find the roots of a square trinomialax 2 + bx + c, you need to equate it to zero -
that is, solve the quadratic equationax 2 + bx + c \u003d0 (see section "Quadratic Equation").

Factoring a square trinomial

Example:

Factor the trinomial 2 x 2 + 7x - 4.

We see: the coefficient and = 2.

Now we will find the roots of the trinomial. To do this, equate it to zero and solve the equation

2x 2 + 7x - 4 \u003d 0.

How to solve such an equation - see the section “Formulas for the roots of a quadratic equation. Discriminant ". Here we will immediately name the result of the calculations. Our trinomial has two roots:

x 1 \u003d 1/2, x 2 \u003d –4.

Substitute the values \u200b\u200bof the roots into our formula, taking out the value of the coefficient outside the brackets and, and we get:

2x 2 + 7x - 4 \u003d 2 (x - 1/2) (x + 4).

The result obtained can be written differently by multiplying the coefficient 2 by the binomial x – 1/2:

2x 2 + 7x - 4 \u003d (2x - 1) (x + 4).

The problem is solved: the trinomial is factorized.

Such a decomposition can be obtained for any square trinomial with roots.

ATTENTION!

If the discriminant of a square trinomial is zero, then this trinomial has one root, but when the trinomial is expanded, this root is taken as the value of two roots - that is, as the same value x 1 andx 2 .

For example, a trinomial has one root equal to 3. Then x 1 \u003d 3, x 2 \u003d 3.

Class: 9

Lesson type:a lesson in consolidating and systematizing knowledge.

Lesson type: Verification, assessment and correction of knowledge and methods of action.

Objectives:

• Educational:

- develop students' ability to factor out the square trinomial;
- consolidation of knowledge in the process of solving various tasks on the specified topic;
- the formation of mathematical thinking;
- to increase interest in the subject in the process of repeating the passed material.

Educational:

- education of organization, concentration;
- fostering a positive attitude towards learning;
- education of curiosity.

Developing:

- develop the ability to exercise self-control;
- develop the ability to rationally plan work;
- development of independence, attention.

Equipment: didactic material for oral work, independent work, test tasks to check knowledge, homework cards, textbook on algebra by Yu.N. Makarychev.

Lesson plan.

Lesson steps	Time, min	Techniques and methods
I. The stage of updating knowledge. Motivation of the learning problem	2	Teacher conversation
II. The main content of the lesson. Formation and consolidation of students' ideas about the formula for factoring a square trinomial.	10	Explanation of the teacher. Heuristic conversation
III. Formation of skills and abilities. Consolidation of the studied material	25	Solving problems. Answers to student questions
IV. Testing the assimilation of knowledge. Reflection	5	Teacher's message. Student message
V. Homework	3	Assignment on cards

During the classes

I. The stage of updating knowledge. Learning problem motivation.

Organizing time.

Today in the lesson we will summarize and systematize knowledge on the topic: “Factoring a square trinomial”. As you complete the various exercises, you should note for yourself the points to which you need to pay special attention when solving equations and practical problems. This is very important when preparing for the exam.
Record lesson topic: “Factoring a square trinomial. Solution of examples ”.

II. The main content of the lesson. Formation and consolidation of students' ideas about the formula for factoring a square trinomial.

Oral work.

- For a successful factorization of a quadratic trinomial, you need to remember both the formulas for finding the discriminant and the formula for finding the roots of the quadratic equation, the formula for factoring the quadratic trinomial into factors and apply them in practice.

1. Look at the “Continue or Complete Statement” cards.

2. Look at the board.

1. Which of the proposed polynomials is not square?

1) x 2 – 4x +3 = 0;
2) – 2x 2 +x– 3 = 0;
3) x 4 – 2x 3 + 2 = 0;
4) 2x 3 – 2x 2 + 2 = 0;

Give the definition of a square trinomial. Give the definition of a root of a square trinomial.

2. Which of the formulas is not a formula for calculating the roots of a quadratic equation?

1) x 1,2 = ;
2) x 1,2 = – b+ ;
3) x 1,2 = .

3. Find the coefficients a, b, c of the square trinomial - 2 x 2 + 5x +7

1) – 2; 5; 7;
2) 5; – 2; 7;
3) 2; 7; 5.

4. Which of the formulas is the formula for calculating the roots of a quadratic equation

x 2 + px + q\u003d 0 by Vieta's theorem?

1) x 1 + x 2 \u003d p,
x 1 · x 2 \u003d q.

2) x 1 + x 2 = – p,
x 1 · x 2 \u003d q.

3) x 1 + x 2 = – p,
x 1 · x 2 \u003d - q.

5. Expand the square trinomial x 2 – 11x +18 by factors.

Answer: ( x – 2)(x – 9)

6. Expand the square trinomial at 2 – 9y +20 multipliers

Answer: ( x – 4)(x – 5)

III. Formation of skills and abilities. Consolidation of the studied material.

1. Factor the square trinomial:
a) 3 x 2 – 8x + 2;
b) 6 x 2 – 5x + 1;
in 3 x 2 + 5x – 2;
d) -5 x 2 + 6x – 1.

2. Factoring helps us to reduce fractions.

3. Without using the root formula, find the roots of the square trinomial:
and) x 2 + 3x + 2 = 0;
b) x 2 – 9x + 20 = 0.

4. Make a square trinomial, the roots of which are numbers:
and) x 1 = 4; x 2 = 2;
b) x 1 = 3; x 2 = -6;

Independent work.

Perform the task independently according to the options with subsequent verification. The first two tasks must be answered “Yes” or “No”. One student from each option is called (they work on the tops of the board). After the independent work is completed on the board, a joint check of the solution is carried out. Students evaluate their work.

1st option:

1.D<0. Уравнение имеет 2 корня.

2. Number 2 is the root of the equation x 2 + 3x - 10 \u003d 0.

3. Factor the square trinomial 6 x 2 – 5x + 1;

2nd option:

1.D\u003e 0. The equation has 2 roots.

2. The number 3 is the root of the quadratic equation x 2 - x - 12 \u003d 0.

3. Factor the square trinomial 2 x 2 – 5x +3

IV. Testing the assimilation of knowledge. Reflection.

- The lesson showed that you know the basic theoretical material of this topic. We have generalized knowledge