X 3 3x 2 10x 24

x 32 x 2

24 10 x

x2 x12

ax 2 bx c , where a ,b and c are given numbers, and a 0 .

We transform the square trinomial ax 2 bx c as follows.

x2 :

coefficient

x ):a x

which is the square of a number

As a result, we get:

Now noticing that

Get

4a 2

Example. Select a full square.

2 x 12

2x2 4x5 2x2 2x5

2x2 2x1 15

a r 1a r 2a r 1r 2,

3. a r 1r 2 a r 1r 2,

4. abr 1 ar 1 br 1,

a r 1

ar 1

br 1

1. Multiply 8

x3 12x7.

24x23.

8 x 3 12 x 7 x 8 x 12 x 8 12 x 24

2. Factorize

a2x3

3 y 3 ;

4 3 85

16 6

2 520 9 519

16 0,25

16 0,25

Definition

Expressions like 2 x 2 + 3 x + 5 are called the square trinomial. In the general case, a square trinomial is an expression of the form a x 2 + b x + c, where a, b, c a, b, c are arbitrary numbers, and a ≠ 0.

Consider the square trinomial x 2 - 4 x + 5 . Let's write it in this form: x 2 - 2 2 x + 5. Let's add 2 2 to this expression and subtract 2 2 , we get: x 2 - 2 2 x + 2 2 - 2 2 + 5. Note that x 2 - 2 2 x + 2 2 = (x - 2) 2, so x 2 - 4 x + 5 = (x - 2) 2 - 4 + 5 = (x - 2) 2 + 1 . The transformation we made is called "selection of a full square from a square trinomial".

Select the perfect square from the square trinomial 9 x 2 + 3 x + 1 .

Note that 9 x 2 = (3 x) 2 , `3x=2*1/2*3x`. Then `9x^2+3x+1=(3x)^2+2*1/2*3x+1`. Add and subtract to the resulting expression `(1/2)^2`, we get

`((3x)^2+2*1/2*3x+(1/2)^2)+1-(1/2)^2=(3x+1/2)^2+3/4`.

Let us show how the method of extracting a full square from a square trinomial is used to factorize a square trinomial.

Factor the square trinomial 4 x 2 - 12 x + 5 .

We select the full square from the square trinomial: 2 x 2 - 2 2 x 3 + 3 2 - 3 2 + 5 = 2 x - 3 2 - 4 = (2 x - 3) 2 - 2 2 . Now apply the formula a 2 - b 2 = (a - b) (a + b) , we get: (2 x - 3 - 2) (2 x - 3 + 2) = (2 x - 5) (2 x - 1 ) .

Factor out the square trinomial - 9 x 2 + 12 x + 5 .

9 x 2 + 12 x + 5 = - 9 x 2 - 12 x + 5 . Now notice that 9 x 2 = 3 x 2 , - 12 x = - 2 3 x 2 .

We add the term 2 2 to the expression 9 x 2 - 12 x, we get:

3 x 2 - 2 3 x 2 + 2 2 - 2 2 + 5 = - 3 x - 2 2 - 4 + 5 = 3 x - 2 2 + 4 + 5 = - 3 x - 2 2 + 9 = 3 2 - 3 x - 2 2 .

We apply the formula for the difference of squares, we have:

9 x 2 + 12 x + 5 = 3 - 3 x - 2 3 + (3 x - 2) = (5 - 3 x) (3 x + 1) .

Factor the square trinomial 3 x 2 - 14 x - 5 .

We can't represent the expression 3 x 2 as the square of some expression because we haven't learned that in school yet. You will go through this later, and already in Task No. 4 we will study square roots. Let us show how we can factorize a given square trinomial:

`3x^2-14x-5=3(x^2-14/3 x-5/3)=3(x^2-2*7/3 x+(7/3)^2-(7/3) ^2-5/3)=`

`=3((x-7/3)^2-49/9-5/3)=3((x-7/3)^2-64/9)=3((x-7/3)^ 2-8/3)^2)=`

`=3(x-7/3-8/3)(x-7/3+8/3)=3(x-5)(x+1/3)=(x-5)(3x+1) `.

We will show how the full square method is used to find the largest or smallest values ​​of a square trinomial.
Consider the square trinomial x 2 - x + 3 . Selecting a full square:

`(x)^2-2*x*1/2+(1/2)^2-(1/2)^2+3=(x-1/2)^2+11/4`. Note that when `x=1/2` the value of the square trinomial is `11/4`, and when `x!=1/2` a positive number is added to the value of `11/4`, so we get a number greater than `11/ 4`. Thus, the smallest value of the square trinomial is `11/4` and it is obtained with `x=1/2`.

Find the largest value of the square trinomial - 16 2 + 8 x + 6 .

We select the full square from the square trinomial: - 16 x 2 + 8 x + 6 = - 4 x 2 - 2 4 x 1 + 1 - 1 + 6 = - 4 x - 1 2 - 1 + 6 = - 4 x - 1 2 + 7 .

With `x=1/4` the value of the square trinomial is 7 , and with `x!=1/4` a positive number is subtracted from the number 7, that is, we get a number less than 7 . Thus, the number 7 is the largest value of the square trinomial, and it is obtained with `x=1/4`.

Factor the numerator and denominator of `(x^2+2x-15)/(x^2-6x+9)` and cancel the fraction.

Note that the denominator of the fraction x 2 - 6 x + 9 = x - 3 2 . We decompose the numerator of the fraction into factors using the method of extracting the full square from the square trinomial. x 2 + 2 x - 15 = x 2 + 2 x 1 + 1 - 1 - 15 = x + 1 2 - 16 = x + 1 2 - 4 2 = = (x + 1 + 4) (x + 1 - 4) = (x + 5) (x - 3) .

This fraction was reduced to the form `((x+5)(x-3))/(x-3)^2` after reduction by (x - 3) we get `(x+5)/(x-3)`.

Factor the polynomial x 4 - 13 x 2 + 36.

Let us apply the full square method to this polynomial. `x^4-13x^2+36=(x^2)^2-2*x^2*13/2+(13/2)^2-(13/2)^2+36=(x^ 2-13/2)^2-169/4+36=(x^2-13/2)^2-25/4=`

As I already noted, in integral calculus there is no convenient formula for integrating a fraction. And therefore, there is a sad trend: the more “fancy” the fraction, the more difficult it is to find the integral from it. In this regard, one has to resort to various tricks, which I will now discuss. Prepared readers can immediately use table of contents:

• The method of subsuming under the sign of the differential for simple fractions

Numerator Artificial Transformation Method

Example 1

By the way, the considered integral can also be solved by the change of variable method, denoting , but the solution will be much longer.

Example 2

Find the indefinite integral. Run a check.

This is a do-it-yourself example. It should be noted that here the variable replacement method will no longer work.

Attention important! Examples No. 1, 2 are typical and are common. In particular, such integrals often arise in the course of solving other integrals, in particular, when integrating irrational functions (roots).

The above method also works in the case if the highest power of the numerator is greater than the highest power of the denominator.

Example 3

Find the indefinite integral. Run a check.

Let's start with the numerator.

The numerator selection algorithm is something like this:

1) In the numerator I need to organize , but there . What to do? I enclose in brackets and multiply by: .

2) Now I try to open these brackets, what happens? . Hmm ... already better, but there is no deuce with initially in the numerator. What to do? You need to multiply by:

3) Opening the brackets again: . And here is the first success! Needed turned out! But the problem is that an extra term has appeared. What to do? In order for the expression not to change, I must add the same to my construction:
. Life has become easier. Is it possible to organize again in the numerator?

4) You can. We try: . Expand the brackets of the second term:
. Sorry, but I actually had in the previous step, and not . What to do? We need to multiply the second term by:

5) Again, for verification, I open the brackets in the second term:
. Now it's normal: obtained from the final construction of paragraph 3! But again there is a small “but”, an extra term has appeared, which means that I must add to my expression:

If everything is done correctly, then when opening all the brackets, we should get the original numerator of the integrand. We check:
Good.

In this way:

Ready. In the last term, I applied the method of bringing the function under the differential.

If we find the derivative of the answer and bring the expression to a common denominator, then we get exactly the original integrand. The considered method of expansion into a sum is nothing more than the reverse action to bring the expression to a common denominator.

The numerator selection algorithm in such examples is best performed on a draft. With some skills, it will also work mentally. I remember a record time when I did a selection for the 11th power, and the expansion of the numerator took almost two lines of Werd.

Example 4

Find the indefinite integral. Run a check.

This is a do-it-yourself example.

The method of subsuming under the sign of the differential for simple fractions

Let's move on to the next type of fractions.
, , , (the coefficients and are not equal to zero).

In fact, a couple of cases with arcsine and arctangent have already slipped in the lesson Variable change method in indefinite integral. Such examples are solved by bringing the function under the sign of the differential and then integrating using the table. Here are some more typical examples with a long and high logarithm:

Example 5

Example 6

Here it is advisable to pick up a table of integrals and follow what formulas and how transformation takes place. Note, how and why squares are highlighted in these examples. In particular, in Example 6, we first need to represent the denominator as , then bring under the sign of the differential. And you need to do all this in order to use the standard tabular formula .

But what to look at, try to solve examples No. 7,8 on your own, especially since they are quite short:

Example 7

Example 8

Find the indefinite integral:

If you can also check these examples, then great respect is your differentiation skills at their best.

Full square selection method

Integrals of the form , (coefficients and are not equal to zero) are solved full square selection method, which has already appeared in the lesson Geometric Plot Transformations.

In fact, such integrals reduce to one of the four table integrals that we have just considered. And this is achieved using the familiar abbreviated multiplication formulas:

Formulas are applied in this direction, that is, the idea of ​​the method is to artificially organize the expressions either in the denominator, and then convert them, respectively, to or .

Example 9

Find the indefinite integral

This is the simplest example where with the term - unit coefficient(and not some number or minus).

We look at the denominator, here the whole thing is clearly reduced to the case. Let's start converting the denominator:

Obviously, you need to add 4. And so that the expression does not change - the same four and subtract:

Now you can apply the formula:

After the conversion is finished ALWAYS it is desirable to perform a reverse move: everything is fine, there are no errors.

The clean design of the example in question should look something like this:

Ready. Bringing a "free" complex function under the differential sign: , in principle, could be neglected

Example 10

Find the indefinite integral:

This is an example for self-solving, the answer is at the end of the lesson.

Example 11

Find the indefinite integral:

What to do when there is a minus in front? In this case, you need to take the minus out of brackets and arrange the terms in the order we need:. Constant("double" in this case) do not touch!

Now we add one in parentheses. Analyzing the expression, we come to the conclusion that we need one behind the bracket - add:

Here is the formula, apply:

ALWAYS we perform a check on the draft:
, which was to be verified.

The clean design of the example looks something like this:

We complicate the task

Example 12

Find the indefinite integral:

Here, with the term, it is no longer a single coefficient, but a “five”.

(1) If a constant is found at, then we immediately take it out of brackets.

(2) In general, it is always better to take this constant out of the integral, so that it does not get in the way.

(3) It is obvious that everything will be reduced to the formula . It is necessary to understand the term, namely, to get a "two"

(4) Yep, . So, we add to the expression, and subtract the same fraction.

(5) Now select a full square. In the general case, it is also necessary to calculate , but here we have a long logarithm formula , and the action does not make sense to perform, why - it will become clear a little lower.

(6) Actually, we can apply the formula , only instead of "x" we have, which does not negate the validity of the tabular integral. Strictly speaking, one step is missing - before integration, the function should have been brought under the differential sign: , but, as I have repeatedly noted, this is often neglected.

(7) In the answer under the root, it is desirable to open all the brackets back:

Difficult? This is not the most difficult in integral calculus. Although, the examples under consideration are not so much complicated as they require good calculation technique.

Example 13

Find the indefinite integral:

This is a do-it-yourself example. Answer at the end of the lesson.

There are integrals with roots in the denominator, which, with the help of a replacement, are reduced to integrals of the considered type, you can read about them in the article Complex integrals, but it is designed for highly prepared students.

Bringing the numerator under the sign of the differential

This is the final part of the lesson, however, integrals of this type are quite common! If fatigue has accumulated, maybe it is better to read tomorrow? ;)

The integrals that we will consider are similar to the integrals of the previous paragraph, they have the form: or (the coefficients , and are not equal to zero).

That is, we have a linear function in the numerator. How to solve such integrals?

Online calculator.
Selection of the square of the binomial and factorization of the square trinomial.

Those. the problems are reduced to finding the numbers \(p, q \) and \(n, m \)

The program not only gives the answer to the problem, but also displays the solution process.

This program can be useful for high school students in preparation for tests and exams, when testing knowledge before the Unified State Examination, for parents to control the solution of many problems in mathematics and algebra. Or maybe it's too expensive for you to hire a tutor or buy new textbooks? Or do you just want to get your math or algebra homework done as quickly as possible? In this case, you can also use our programs with a detailed solution.

In this way, you can conduct your own training and/or the training of your younger brothers or sisters, while the level of education in the field of tasks to be solved is increased.

If you are not familiar with the rules for entering a square trinomial, we recommend that you familiarize yourself with them.

Rules for entering a square polynomial

Any Latin letter can act as a variable.
For example: \(x, y, z, a, b, c, o, p, q \) etc.

Numbers can be entered as integers or fractions.
Moreover, fractional numbers can be entered not only in the form of a decimal, but also in the form of an ordinary fraction.

Rules for entering decimal fractions.
In decimal fractions, the fractional part from the integer can be separated by either a dot or a comma.
For example, you can enter decimals like this: 2.5x - 3.5x^2

Rules for entering ordinary fractions.
Only a whole number can act as the numerator, denominator and integer part of a fraction.

The denominator cannot be negative.

When entering a numerical fraction, the numerator is separated from the denominator by a division sign: /
The integer part is separated from the fraction by an ampersand: &
Input: 3&1/3 - 5&6/5x +1/7x^2
Result: \(3\frac(1)(3) - 5\frac(6)(5) x + \frac(1)(7)x^2 \)

When entering an expression you can use brackets. In this case, when solving, the introduced expression is first simplified.
For example: 1/2(x-1)(x+1)-(5x-10&1/2)

Detailed Solution Example

Selection of the square of the binomial.$$ ax^2+bx+c \rightarrow a(x+p)^2+q $$ $$2x^2+2x-4 = $$ $$2x^2 +2 \cdot 2 \cdot\left( \frac(1)(2) \right)\cdot x+2 \cdot \left(\frac(1)(2) \right)^2-\frac(9)(2) = $$ $$2\left (x^2 + 2 \cdot\left(\frac(1)(2) \right)\cdot x + \left(\frac(1)(2) \right)^2 \right)-\frac(9 )(2) = $$ $$2\left(x+\frac(1)(2) \right)^2-\frac(9)(2) $$ Answer:$$2x^2+2x-4 = 2\left(x+\frac(1)(2) \right)^2-\frac(9)(2) $$ Factorization.$$ ax^2+bx+c \rightarrow a(x+n)(x+m) $$ $$2x^2+2x-4 = $$
$$ 2\left(x^2+x-2 \right) = $$
$$ 2 \left(x^2+2x-1x-1 \cdot 2 \right) = $$ $$ 2 \left(x \left(x +2 \right) -1 \left(x +2 \right ) \right) = $$ $$ 2 \left(x -1 \right) \left(x +2 \right) $$ Answer:$$2x^2+2x-4 = 2 \left(x -1 \right) \left(x +2 \right) $$

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A bit of theory.

Extraction of a square binomial from a square trinomial

If the square trinomial ax 2 + bx + c is represented as a (x + p) 2 + q, where p and q are real numbers, then they say that from square trinomial, the square of the binomial is highlighted.

Let us extract the square of the binomial from the trinomial 2x 2 +12x+14.

\(2x^2+12x+14 = 2(x^2+6x+7) \)

To do this, we represent 6x as a product of 2 * 3 * x, and then add and subtract 3 2 . We get:
$$ 2(x^2+2 \cdot 3 \cdot x + 3^2-3^2+7) = 2((x+3)^2-3^2+7) = $$ $$ = 2 ((x+3)^2-2) = 2(x+3)^2-4 $$

That. we selected the square of the binomial from the square trinomial, and showed that:
$$ 2x^2+12x+14 = 2(x+3)^2-4 $$

Factorization of a square trinomial

If the square trinomial ax 2 +bx+c is represented as a(x+n)(x+m), where n and m are real numbers, then the operation is said to be performed factorizations of a square trinomial.

Let's use an example to show how this transformation is done.

Let's factorize the square trinomial 2x 2 +4x-6.

Let us take the coefficient a out of brackets, i.e. 2:
\(2x^2+4x-6 = 2(x^2+2x-3) \)

Let's transform the expression in brackets.
To do this, we represent 2x as the difference 3x-1x, and -3 as -1*3. We get:
$$ = 2(x^2+3 \cdot x -1 \cdot x -1 \cdot 3) = 2(x(x+3)-1 \cdot (x+3)) = $$
$$ = 2(x-1)(x+3) $$

That. we factorize the square trinomial, and showed that:
$$ 2x^2+4x-6 = 2(x-1)(x+3) $$

Note that the factorization of a square trinomial is possible only when the quadratic equation corresponding to this trinomial has roots.
Those. in our case, factoring the trinomial 2x 2 +4x-6 is possible if the quadratic equation 2x 2 +4x-6 =0 has roots. In the process of factoring, we found that the equation 2x 2 +4x-6 =0 has two roots 1 and -3, because with these values, the equation 2(x-1)(x+3)=0 turns into a true equality.

In this lesson, we will recall all the previously studied methods of factoring a polynomial and consider examples of their application, in addition, we will study a new method - the full square method and learn how to apply it in solving various problems.

Topic:Factoring polynomials

Lesson:Factorization of polynomials. Full square selection method. Combination of methods

Recall the main methods for factoring a polynomial that were studied earlier:

The method of taking a common factor out of brackets, that is, a factor that is present in all members of the polynomial. Consider an example:

Recall that a monomial is a product of powers and numbers. In our example, both members have some common, identical elements.

So, let's take the common factor out of brackets:

;

Recall that by multiplying the rendered multiplier by the bracket, you can check the correctness of the rendering.

grouping method. It is not always possible to take out a common factor in a polynomial. In this case, you need to divide its members into groups in such a way that in each group you can take out a common factor and try to break it down so that after taking out the factors in the groups, a common factor appears for the entire expression, and expansion could be continued. Consider an example:

Group the first term with the fourth, the second with the fifth, and the third with the sixth respectively:

Let's take out the common factors in the groups:

The expression has a common factor. Let's take it out:

Application of abbreviated multiplication formulas. Consider an example:

;

Let's write the expression in detail:

Obviously, we have before us the formula for the square of the difference, since there is a sum of the squares of two expressions and their double product is subtracted from it. Let's roll by the formula:

Today we will learn another way - the full square selection method. It is based on the formulas of the square of the sum and the square of the difference. Recall them:

The formula for the square of the sum (difference);

The peculiarity of these formulas is that they contain squares of two expressions and their double product. Consider an example:

Let's write the expression:

So the first expression is , and the second .

In order to make a formula for the square of the sum or difference, the double product of the expressions is not enough. It needs to be added and subtracted:

Let's collapse the full square of the sum:

Let's transform the resulting expression:

We apply the difference of squares formula, recall that the difference of the squares of two expressions is the product and the sums by their difference:

So, this method consists, first of all, in the fact that it is necessary to identify the expressions a and b that are squared, that is, to determine which expressions are squared in this example. After that, you need to check for the presence of a double product and if it is not there, then add and subtract it, this will not change the meaning of the example, but the polynomial can be factored using the formulas for the square of the sum or difference and difference of squares, if possible.

Let's move on to solving examples.

Example 1 - factorize:

Find expressions that are squared:

Let's write down what their double product should be:

Let's add and subtract the double product:

Let's collapse the full square of the sum and give similar ones:

We will write according to the formula of the difference of squares:

Example 2 - solve the equation:

;

There is a trinomial on the left side of the equation. You need to factor it out. We use the formula of the square of the difference:

We have the square of the first expression and the double product, the square of the second expression is missing, let's add and subtract it:

Let us collapse the full square and give like terms:

Let's apply the difference of squares formula:

So we have the equation

We know that the product is equal to zero only if at least one of the factors is equal to zero. Based on this, we will write equations:

Let's solve the first equation:

Let's solve the second equation:

Answer: or

;

We act similarly to the previous example - select the square of the difference.