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

After learning a few tricks factorization solutions can be classified as follows:

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

In order to factorize, it is necessary to simplify the expressions. This is necessary in order to be able to further reduce. The decomposition of a polynomial makes sense when its degree is not lower than the second. A polynomial with the first degree is called linear.

The article will reveal all the concepts of decomposition, theoretical basis and methods for factoring a polynomial.

Theory

Theorem 1

When 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 an and n linear factors (x - xi) , i = 1 , 2 , ... , n , then P n (x) = an (x - xn) (x - xn - 1) . . . · (x - x 1) , where x i , i = 1 , 2 , … , n - these 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 complex roots will occur in conjugate pairs. For example, the 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, hence we get 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. Consider the proof of the theorem of algebra, the consequences of Bezout's theorem.

Fundamental theorem of algebra

Theorem 2

Any 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 the 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 from 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.

Factorization of a square trinomial

A square trinomial of the form a x 2 + b x + c can be factored into linear factors. then we get that a x 2 + b x + c \u003d a (x - x 1) (x - x 2) , where x 1 and x 2 are roots (complex or real).

This shows that the decomposition itself reduces to solving the quadratic equation later.

Example 1

Factorize a square 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 according to the formula, then we get D \u003d (- 5) 2 - 4 4 1 \u003d 9. Hence we have that

x 1 = 5 - 9 2 4 = 1 4 x 2 = 5 + 9 2 4 = 1

From here we get that 4 x 2 - 5 x + 1 = 4 x - 1 4 x - 1.

To perform the check, you need to open the brackets. 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 verification, we arrive at the original expression. That is, we can conclude that the expansion is correct.

Example 2

Factorize a square trinomial of the form 3 x 2 - 7 x - 11 .

Solution

We get 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 - 1816

From here we get that 3 x 2 - 7 x - 11 = 3 x - 7 + 181 6 x - 7 - 181 6 .

Example 3

Factorize the polynomial 2 x 2 + 1.

Solution

Now you 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 that the decomposition itself can be represented as 2 x 2 + 1 = 2 x - 1 2 · i x + 1 2 · i.

Example 4

Expand the square 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 ix 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 value of the discriminant is negative, then the polynomials will remain second-order polynomials. Hence it follows that we will not decompose them into linear factors.

Methods for factoring a polynomial of degree higher than the second

The decomposition assumes a universal method. 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 lower 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 get a complete decomposition.

If the root is not found, then other methods of factorization are used: grouping, additional terms. This topic assumes the solution of equations with higher degrees 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 \u003d 0, then you can represent the polynomial in the form of an expression P n (x) \u003d 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

Factorize the third degree polynomial 4 x 3 + 8 x 2 - x.

Solution

We see that x 1 \u003d 0 is the root of the given polynomial, then we can bracket x out 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 the 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 xx - - 1 + 5 2 x - - 1 - 5 2 = = 4 xx + 1 - 5 2 x + 1 + 5 2

To begin with, let's take for 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 power is 1 .

When the polynomial has integer roots, then they are considered divisors of the free term.

Example 6

Expand the expression f (x) = x 4 + 3 x 3 - x 2 - 9 x - 18.

Solution

Consider whether there are integer roots. It is necessary to write out 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 according to the Horner scheme. It is very convenient and allows you to quickly obtain the expansion coefficients of a polynomial:

It follows that x \u003d 2 and x \u003d - 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 turn to the decomposition of a square trinomial of the form x 2 + 2 x + 3 .

Since the discriminant is negative, it means that there are no real roots.

Answer: f (x) \u003d x 4 + 3 x 3 - x 2 - 9 x - 18 \u003d (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 us proceed to consider 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 does not equal one.

This case takes place for fractional rational fractions.

Example 7

Factorize f (x) = 2 x 3 + 19 x 2 + 41 x + 15 .

Solution

It is necessary to change the variable y = 2 x , one should pass 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) \u003d y 3 + 19 y 2 + 82 y + 60 has integer roots, then their finding 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 proceed to the calculation of 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 get that y \u003d - 5 is the root of the equation of the form y 3 + 19 y 2 + 82 y + 60, which means that x \u003d y 2 \u003d - 5 2 is the root of the original function.

Example 8

It is necessary to divide by a column 2 x 3 + 19 x 2 + 41 x + 15 by x + 5 2.

Solution

We write 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 take the factorization of the resulting square 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

Hence 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 tricks when 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 decomposed 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 take it out of brackets.

Example 9

Factorize 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, we 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 a different method of decomposition and solution.

Grouping is required:

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, it is necessary to represent it as a 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 it is easy enough to choose terms. There is no definite way to solve it, therefore it is necessary to use special theorems and rules.

Example 10

Factorize 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 factoring, 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 and Newton's binomial formulas to factorize a polynomial

Appearance often does not always make it clear which way to use 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

Factorize 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 brackets is indicated by the expression x + 1 4 .

So 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 parenthesis. It is clear that there are no horses there, so the formula for the difference of squares should be applied again. We get an expression like

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 change 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 changing a variable, the degree is reduced and the polynomial is factorized.

Example 13

Factorize a polynomial of the form x 6 + 5 x 3 + 6 .

Solution

By the condition, it is clear that it is necessary to make a 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 the 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 have obtained the desired expansion.

The cases discussed above will help in considering and factoring a polynomial in various ways.

If you notice a mistake in the text, please highlight it and press Ctrl+Enter

Class: 9

Lesson type: a lesson in consolidating and systematizing knowledge.

Type of lesson: Verification, assessment and correction of knowledge and methods of action.

Goals:

• Educational:

- to develop in students the ability to decompose a square trinomial into factors;
- consolidation of knowledge in the process of solving various tasks on a specified topic;
– formation of mathematical thinking;
- increase interest in the subject in the process of repeating the material covered.

Educational:

- education of organization, concentration;
- fostering a positive attitude towards learning;
- cultivating 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 test knowledge, cards with homework, algebra textbook Yu.N. Makarychev.

Lesson plan.

Lesson stages	Time, min	Techniques and methods
I. Stage of updating knowledge. Motivation for learning problem	2	Teacher's conversation
II. The main content of the lesson Formation and consolidation of students' ideas about the formula for factoring a square trinomial into factors.	10	Teacher's explanation. Heuristic conversation
III. Formation of skills and abilities. Consolidation of the studied material	25	Problem solving. Answers to students' questions
IV. Checking the assimilation of knowledge. Reflection	5	Teacher's message. Student message
V. Homework	3	Task on cards

During the classes

I. Stage of updating knowledge. Motivation of the educational problem.

Organizing time.

Today in the lesson we will generalize and systematize knowledge on the topic: “Factorization of a square trinomial”. By doing various exercises, you should note for yourself the points that you need to pay special attention to when solving equations and practical problems. This is very important when preparing for the exam.
Write down the topic of the lesson: “Factorization of a square trinomial. Solving Examples.

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

oral work.

– To successfully factorize a square trinomial, you need to remember both the formulas for finding the discriminant and the formulas for finding the roots of a quadratic equation, the formula for factoring a square trinomial and put them into practice.

1. Look at the “Continue or complete the 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;

Define a square trinomial. Define the 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 a formula for calculating the roots of a quadratic equation

x2 + px + q= 0 by Vieta's theorem?

1) x 1 + x 2 =p,
x one · x 2 = q.

2) x 1 + x 2 = –p ,
x one · x 2 = q.

3)x 1 + x 2 = –p ,
x one · x 2 = – q .

5. Expand the square trinomial X 2 – 11x + 18 for multipliers.

Answer: ( X – 2)(X – 9)

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

Answer: ( X – 4)(X – 5)

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

1. Factorize 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 when reducing fractions.

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

4. Make a square trinomial whose roots are numbers:
but) x 1 = 4; x 2 = 2;
b) x 1 = 3; x 2 = -6;

Independent work.

Independently complete the task according to the options, followed by verification. The first two tasks must be answered "Yes" or "no". One student from each option is called (they work on the lapels of the board). After independent work is done 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. The number 2 is the root of the equation x 2 + 3x - 10 = 0.

3. Factorize the square trinomial into factors 6 x 2 – 5x + 1;

2nd option:

1.D>0. The equation has 2 roots.

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

3. Decompose the square trinomial into factors 2 X 2 – 5x + 3

IV. Checking the assimilation of knowledge. Reflection.

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

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

Solve

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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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The world is immersed in a huge number of numbers. Any calculations occur with their help.

People learn numbers in order not to fall for deception in later life. It is necessary to devote a huge amount of time to be educated and calculate your own budget.

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Mathematics is an exact science that plays a big role in life. At school, children learn numbers, and then, actions on them.

Actions on numbers are completely different: multiplication, expansion, addition, and others. In addition to simple formulas, more complex actions are also used in the study of mathematics. There are a huge number of formulas by which any values ​​\u200b\u200bare known.

At school, as soon as algebra appears, simplification formulas are added to the life of a student. There are equations when there are two unknown numbers, but find in a simple way will not work. A trinomial is a combination of three monomials, using a simple subtraction and addition method. The trinomial is solved using the Vieta theorem and the discriminant.

The formula for factoring a square trinomial into factors

There are two correct and simple solutions example:

• discriminant;
• Vieta's theorem.

A square trinomial has an unknown squared, as well as a number without a square. The first option for solving the problem uses the Vieta formula. This simple formula if the digits that come before unknown will be the minimum value.

For other equations, where the number is in front of the unknown, the equation must be solved through the discriminant. This is a more complicated solution, but the discriminant is used much more often than Vieta's theorem.

Initially, to find all the variables of the equation, it is necessary to raise the example to 0. The solution of the example can be checked and find out whether the numbers are adjusted correctly.

Discriminant

1. It is necessary to equate the equation to 0.

2. Each number before x will be called numbers a, b, c. Since there is no number before the first square x, it equates to 1.

3. Now the solution of the equation begins through the discriminant:

4. Now we have found the discriminant and find two x. The difference is that in one case b will be preceded by a plus, and in the other by a minus:

5. By solving two numbers, it turned out -2 and -1. Substitute under the original equation:

6. In this example, there are two correct options. If both solutions are correct, then each of them is true.

Solve through the discriminant and more complex equation. But if the value of the discriminant itself is less than 0, then the example is wrong. The discriminant in the search is always under the root, and a negative value cannot be in the root.

Vieta's theorem

It is used to solve easy problems, where the first x is not preceded by a number, that is, a=1. If the option matches, then the calculation is carried out through the Vieta theorem.

To solve any trinomial it is necessary to raise the equation to 0. The first steps for the discriminant and the Vieta theorem are the same.

2. Now there are differences between the two methods. Vieta's theorem uses not only "dry" calculation, but also logic and intuition. Each number has its own letter a, b, c. The theorem uses the sum and product of two numbers.

Remember! The number b is always added with the opposite sign, and the number c remains unchanged!

Substituting data values ​​in the example , we get:

3. Using the logic method, we substitute the most suitable numbers. Consider all possible solutions:

1. The numbers are 1 and 2. When added, we get 3, but if we multiply, we don’t get 4. Not suitable.
2. Value 2 and -2. When multiplied, it will be -4, but when added, it turns out 0. Not suitable.
3. Numbers 4 and -1. Since the multiplication contains a negative value, it means that one of the numbers will be with a minus. Suitable for addition and multiplication. Correct option.

4. It remains only to check, laying out the numbers, and see if the chosen option is correct.

5. Thanks to an online check, we found out that -1 does not match the condition of the example, which means it is the wrong solution.

When adding a negative value in the example, the number must be enclosed in brackets.

In mathematics, there will always be simple problems and difficult ones. Science itself includes a variety of problems, theorems and formulas. If you understand and correctly apply knowledge, then any difficulties with calculations will be trifling.

Mathematics does not need constant memorization. You need to learn to understand the solution and learn a few formulas. Gradually, according to logical conclusions, it is possible to solve similar problems, equations. Such a science may seem very difficult at first glance, but if one plunges into the world of numbers and tasks, then the view will change dramatically for the better.

Technical specialties always remain the most sought after in the world. Now, in the world modern technologies Mathematics has become an indispensable attribute of any field. You must always remember about useful properties mathematics.

Decomposition of a trinomial with brackets

In addition to solving in the usual ways, there is another one - decomposition into brackets. Used with Vieta's formula.

1. Equate the equation to 0.

ax 2 + bx+ c= 0

2. The roots of the equation remain the same, but instead of zero, they now use bracket expansion formulas.

ax 2 + bx + c = a (x-x 1) (x-x 2)

2 x 2 – 4 x – 6 = 2 (x + 1) (x – 3)

4. Solution x=-1, x=3