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 whole 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 zero only if at least one of the factors 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.

The ability to perform such a procedure is extremely necessary in many topics of mathematics related to square trinomialax 2 + bx + c . The most common:

1) Drawing parabolas y= ax 2 + bx+ c;

2) Solving many tasks for a square trinomial ( quadratic equations and inequalities, problems with parameters, etc.);

3) Working with some functions containing a square trinomial, as well as working with curves of the second order (for students).

Useful thing, in short! Are you up for a five? Then let's learn!)

What does it mean to select the full square of a binomial in a square trinomial?

This task means that the original square trinomial must be converted with the help to this form:

Number a what's on the left, what's on the right same. X-squared coefficient. That's why it's marked one letter. Multiplies on the right by square brackets. In the brackets themselves sits the same binomial, which is discussed in this topic. The sum of a pure x and some number m. Yes, please pay attention pure x! It is important.

And here are the letters m and n right - some new numbers. What will be obtained as a result of our transformations. They can turn out to be positive, negative, whole, fractional - all sorts! You will see for yourself in the examples below. These numbers depend from coefficientsa, bandc. They have their own special general formulas. Quite bulky, with fractions. Therefore, I will not give them right here and now. Why do your bright minds need extra garbage? Yes, and it's not interesting. Let's get creative.)

What do you need to know and understand?

First of all, you need to know by heart. At least two of them sum squared and difference squared.

These ones:

Without this couple of formulas - nowhere. Not only in this lesson, but in almost all other mathematics in general. Is the hint clear?)

But mere memorized formulas are not enough here. Need more smart be able to apply these formulas. And not so much directly, from left to right, but vice versa, from right to left. Those. by the original square trinomial, be able to decipher the square of the sum / difference. This means that you should easily, automatically, recognize type equalities:

x 2 +4 x+4 = (x+2) 2

x 2 -10 x+25 = (x-5) 2

x 2 + x+0,25 = (x+0,5) 2

Without this useful skill, there is no way either ... So if there are problems with these simple things, then close this page. It's too early for you here.) First, go to the link above. She is for you!

Oh, how long have you been on the subject? Excellent! Then read on.)

So:

How to select the full square of a binomial in a square trinomial?

Let's start, of course, with a simple one.

Level 1. Coefficient at x2 equals 1

This is the simplest situation requiring a minimum of additional transformations.

For example, given a square trinomial:

X 2 +4x+6

Externally, the expression is very similar to the square of the sum. We know that the square of the sum contains the pure squares of the first and second expressions ( a 2 and b 2 ), as well as the double product 2 ab these same expressions.

Well, we already have the square of the first expression in its pure form. it X 2 . Actually, this is precisely the simplicity of the examples of this level. Need to get the square of the second expression b 2 . Those. find b. And will serve as a clue expression with x in the first degree, i.e. 4x. After all 4x can be represented as double product xx for a deuce. Like this:

4 x = 2 ́ x 2

So if 2 ab=2x2 and a= x, then b=2 . You can write:

X 2 +4x+6 = x 2 +2 ́ x 2+2 2 ….

So us I want to. But! Mathematics I want our actions to be the essence of the original expression hasn't changed. That's how she's made. We have added to the double product 2 2 , thus changing the original expression. So, in order not to offend mathematics, this is the most 2 2 need it right now take away. Like this:

…= x 2 +2 ́ x 2+ 2 2 -2 2 ….

Almost all. It remains only to add 6, in accordance with the original trinomial. The six hasn't gone anywhere! We write:

= X 2 +2 ́ x 2+2 2 - 2 2 +6 = …

Now the first three terms give net (or - full) binomial square x+2 . Or (x+2) 2 . This is what we are trying to achieve.) I will not even be lazy and put brackets:

… = (x 2 +2 ́ x 2+2 2 ) - 2 2 +6 =…

Parentheses do not change the essence of the expression, but they clearly suggest what, how and why. It remains to collapse these three terms into a full square according to the formula, count the remaining tail in numbers -2 2 +6 (that would be 2) and write:

X 2 +4x+6 = (x+2) 2 +2

Everything. We singled out bracket square (x+2) 2 from the original square trinomial X 2 +4x+6. Turned it into a sum full square binomial (x+2) 2 and some constant number (two). And now I will write the entire chain of our transformations in a compact form. For clarity.

And that's all.) That's the whole point of the procedure for selecting a full square.

By the way, what are the numbers here m and n? Yes. Each of them is equal to two: m=2, n=2 . So it happened during the selection.

Another example:

Select the full square of the binomial:

X 2 -6x+8

And again, the first look is at the term with x. We turn 6x into twice the product of x and three. Before double - minus. So we single out difference squared. We add (to get a full square) and immediately subtract (to compensate) the triple in the square, i.e. 9. Well, do not forget about the eight. We get:

Here m=-3 and n=-1 . Both are negative.

Do you get the principle? Then it was time to master and general algorithm. Everything is the same, but through letters. So, we have a square trinomial x 2 + bx+ c (a=1) . What are we doing:

bx b /2 :

b With.

Clearly? The first two examples were very simple, with integers. For acquaintance. Worse, when in the course of transformations fractions get out. The main thing here is not to be afraid! And in order not to be afraid, everyone needs to know the actions with fractions, yes ...) But here is the five level, isn't it? We complicate the task.

Let's say the following trinomial is given:

X 2 +x+1

How to organize the square of the sum in this trinomial? No problem! Similar. We work on points.

1. We look at the term with x in the first degree ( bx) and turn it into twice the product of x byb /2 .

Our term with x is just x. So what? How can we turn lonely X into double product? Yes, very easy! Directly according to the instructions. Like this:

Number b in the original trinomial - one. That is, b/2 turns out to be fractional. A half. 1/2. Well, okay. Not small already.)

2. We add to the double product and immediately subtract the square of the number b/2. We add - to complement to a full square. We take away - for compensation. At the very end we add a free term With.

We continue:

3. We turn the first three terms into the square of the sum / difference according to the corresponding formula. The expression remaining outside is carefully calculated in numbers.

The first three terms are separated by brackets. You can not separate, of course. This is done purely for convenience and clarity of our transformations. Now you can clearly see that the full square of the sum is in brackets (x+1/2) 2 . And everything remaining outside the square of the sum (if you count) gives +3/4. Finish line:

Answer:

Here m=1/2 , a n=3/4 . Fractional numbers. It happens. Such a trinomial got caught ...

Such is the technology. Got it? Can you move to the next level?

Level 2. The coefficient at x 2 is not equal to 1 - what to do?

This is a more general case than the case a=1. The volume of calculations, of course, increases. It upsets, yes ... But overall solution generally remains the same. Just one new step is added to it. This makes me happy.)

For now, consider a harmless case, without any fractions and other pitfalls. For example:

2 x 2 -4 x+6

There is a minus in the middle. So, we will fit the square of the difference. But the coefficient at the square of x is a deuce. And it's easier to work with one. With pure x. What to do? And let's put this deuce out of brackets! So as not to interfere. We have the right! We get:

2(x 2 -2 x+3)

Like this. Now the trinomial in brackets - already with clean X squared! As required by the level 1 algorithm. And now it is already possible to work with this new trinomial according to the old well-established scheme. Here we are acting. Let's write it separately and transform it:

x 2 -2 x+3 = x 2 -2x1+1 2 -1 2 +3 = (x 2 -2x1+1 2 ) -1 2 +3 = (x-1) 2 +2

Half done. It remains to insert the resulting expression inside the brackets, and expand them back. Get:

2(x 2 -2 x+3) = 2((x-1) 2 +2) = 2(x-1) 2 +4

Ready!

Answer:

2 x 2 -4 x+6 = 2( x -1) 2 +4

We fix in the head:

If the coefficient at the square of x is not equal to one, then we take this coefficient out of brackets. With the trinomial remaining inside the brackets, we work according to the usual algorithm for a=1. Having selected a full square in it, paste the result in place, and open the outer brackets back.

But what if the coefficients b and c are not divisible by a? This is the most common and at the same time the worst case. Then only fractions, yes... There's nothing to be done. For example:

3 x 2 +2 x-5

Everything is the same, we send the three out of brackets, we get:

Unfortunately, neither two nor five are completely divisible by three, so the coefficients of the new (reduced) trinomial are fractional. Well, no big deal. Working directly with fractions: two thirds x turn into double product of x by one third, add the square of one third (i.e. 1/9), subtract it, subtract 5/3...

In general, you understand!

Decide what is already there. It should end up like this:

And one more rake. Many students famously crack down on positive integer and even fractional odds, but hang on negative ones. For example:

- x 2 +2 x-3

What to do with minus beforex 2 ? In the formula for the square of the sum / difference, any plus is needed ... Not a question! All the same. We take out this minus for brackets. Those. minus one. Like this:

- x 2 +2 x-3 = -(x 2 -2 x+3) = (-1) (x 2 -2 x+3)

And all things. And with the trinomial in brackets - again along the knurled track.

x 2 -2 x+3 = (x 2 -2 x+1) -1+3 = (x-1) 2 +2

So, minus:

- x 2 +2 x-3 = -((x-1) 2 +2) = -(x-1) 2 -2

That's all. What? Don't know how to put minus out of brackets? Well, this is a question for elementary algebra of the seventh grade, not for square trinomials ...

Remember: work with a negative coefficient a nothing inherently different from working with the positive. Bringing out the negative a out of brackets, and then - according to all the rules.

Why do you need to be able to select a full square?

The first useful thing is to draw parabolas quickly and without errors!

For example, such a task:

Plot the function:y=- x 2 +2 x+3

What we are going to do? Build by points? Of course it is possible. Little steps along the long road. Pretty dull and uninteresting...

First of all, I remind you that when building any parabolas, we always present her with a standard set of questions. There are two of them. Namely:

1) Where are the branches of the parabola directed?

2) Where is the top?

With the direction of the branches, everything is clear right from the original expression. Branches will be directed way down, because the coefficient beforex 2 - negative. Minus one. Minus before the x-square always flips the parabola.

But with the location of the top, everything is not so obvious. There is, of course, a general formula for calculating its abscissa through the coefficients a and b.

This one:

But not everyone remembers this formula, oh, not everyone ... And 50% of those who still remember stumble out of the blue and mess up in banal arithmetic (usually when counting a game). It's a shame, right?)

Now you will learn how to find the coordinates of the vertex of any parabola in my mind in one minute! Both x and y. In one fell swoop and without any formulas. How? By selecting a full square!

So, we select the full square in our expression. We get:

y=-x 2 +2 x+3 = -(x-1) 2 +4

Who is well versed in general information about the functions and mastered the topic well" function graph transformations ", he will easily figure out that our desired parabola is obtained from the usual parabola y= x 2 with the help of three transformations. It:

1) Change the direction of the branches.

This is indicated by the minus sign in front of the square brackets ( a=-1). It was y= x 2 , became y=- x 2 .

Conversion: f ( x ) -> - f ( x ) .

2) Parallel translation of the parabola y=- x 2 X 1 unit to the RIGHT.

This is how the intermediate schedule is obtained y=-(x-1 ) 2 .

Conversion: - f ( x ) -> - f ( x + m ) (m=-1).

Why is the shift to the right and not to the left, although there is a minus in brackets? This is the theory of graph transformations. This is a separate issue.

And finally,

3) Parallel transfer parabolas y=-( x -1) 2 by 4 units UP.

This is how the final parabola is obtained. y=-(x-1) 2 +4 .

Conversion: - f ( x + m ) -> - f ( x + m )+ n (n=+4)

And now we look at our chain of transformations and think: Where does the vertex of the parabola move?y=x 2 ? It was at the point (0; 0), after the first transformation the vertex did not move anywhere (the parabola simply turned over), after the second one it moved down by x by +1, and after the third one by y by +4. Total top hit the point (1; 4) . That's the whole secret!

The picture will be as follows:

Actually, it is for this reason that I have drawn your attention to numbers with such persistence. m and n obtained in the process of selecting a full square. Didn't guess why? Yes. The point is that the point with coordinates (- m ; n ) - it's always top of a parabola y = a ( x + m ) 2 + n . We just look at the numbers in the converted trinomial and in my mind we give the correct answer, where is the top. Convenient, right?)

Drawing parabolas is the first useful thing. Let's move on to the second.

The second useful thing is the solution of quadratic equations and inequalities.

Yes Yes! Selection of the full square in many cases turns out to be much faster and more efficient traditional methods of solving such problems. Doubt? Please! Here's a task for you:

Solve the inequality:

x 2 +4 x+5 > 0

Learned? Yes! It's classic square inequality . All such inequalities are solved by the standard algorithm. For this we need:

1) Make an equation of the standard form from the inequality and solve it, find the roots.

2) Draw the X axis and mark the roots of the equation with dots.

3) Schematically depict a parabola according to the original expression.

4) Determine the +/- areas in the figure. Select the desired areas according to the original inequality and write down the answer.

Actually, this whole process is annoying, yes ...) And, moreover, it does not always save from errors in non-standard situations like this example. Let's try the pattern first, shall we?

So let's do the first point. We make an equation from the inequality:

x 2 +4 x+5 = 0

Standard quadratic equation, no tricks. We decide! We consider the discriminant:

D = b 2 -4 ac = 4 2 - 4∙1∙5 = -4

That's it! And the discriminant is negative! The equation has no roots! And there is nothing to draw on the axis ... What should I do?

Here some may conclude that the original inequality also has no solutions.. This is a fatal delusion, yes ... But by highlighting the full square, the correct answer to this inequality can be given in half a minute! Doubt? Well, you can time it.

So, we select the full square in our expression. We get:

x 2 +4 x+5 = (x+2) 2 +1

The original inequality began to look like this:

(x+2) 2 +1 > 0

And now, without solving or transforming anything further, we simply turn on elementary logic and think: if to the square of some expression (the value is obviously non-negative!) add another one, then what number will we end up with? Yes! Strictly positive!

Now let's look at the inequality:

(x+2) 2 +1 > 0

We translate the entry from the mathematical language into Russian: for which x is strictly positive expression will be strictly more zero? Didn't guess? Yes! With any!

Here is your answer: x is any number.

Now let's get back to the algorithm. Still, understanding the essence and simple rote memorization are two different things.)

The essence of the algorithm is that we make a parabola from the left side of the standard inequality, and look where it is above the X axis, and where it is below. Those. where are positive values ​​of the left side, where are negative.

If we make a parabola from our left side:

y=x 2 +4 x+5

And draw its graph, we will see that all whole parabola passes above the x-axis. The picture will look like this:

The parabola is crooked, yes ... That's why it is schematic. But at the same time, everything that we need is visible in the picture. The parabola has no points of intersection with the X axis, there are no zero values ​​of the game. And, of course, there are no negative values ​​either. This is shown by shading the entire X-axis. By the way, the Y-axis and the coordinates of the vertex I have depicted here for good reason. Compare the parabola vertex coordinates (-2; 1) and our transformed expression!

y=x 2 +4 x+5 = ( x +2) 2 +1

And how do you? Yes! In our case m=2 and n=1 . Therefore, the vertex of the parabola has coordinates: (- m; n) = (-2; 1) . It's all logical.)

Another task:

Solve the equation:

x 2 +4 x+3 = 0

Simple quadratic equation. You can decide the old fashioned way. It is possible through . As you wish. Math doesn't mind.)

Let's get the roots: x 1 =-3 x 2 =-1

And if neither one nor the other way of that ... do not remember? Well, a deuce shines for you, in a good way, but ... So be it, I'll save you! I will show you how you can solve some quadratic equations using only the methods of the seventh grade. Again select a full square!)

x 2 +4 x+3 = (x+2) 2 -1

And now we write the resulting expression as ... difference of squares! Yes, yes, there is one in the seventh grade:

a 2 -b 2 = (a-b)(a+b)

Cast a brackets protrude(x+2) , and in the role b- one. We get:

(x+2) 2 -1 = (x+2) 2 -1 2 = ((x+2)-1)((x+2)+1) = (x+1)(x+3)

We insert this expansion into the equation instead of the square trinomial:

(x+1)(x+3)=0

It remains to figure out that the product of the factors is equal to zero then and only then when any of them is equal to zero. So we equate (in the mind!) To zero each bracket.

We get: x 1 =-3 x 2 =-1

That's all. The same two roots. Such is the skillful receiver. In addition to the discriminant.)

By the way, about the discriminant and about general formula roots of the quadratic equation:

In the lesson I omitted the derivation of this cumbersome formula. For uselessness. But here is the place for him.) Would you like to know how get this formula? Where does the expression for the discriminant come from and why exactlyb 2 -4ac, but not in some other way? Still, a complete understanding of the essence of what is happening is much more useful than thoughtless scribbling of all sorts of letters and symbols, right?)

The third useful thing is the derivation of the formula for the roots of a quadratic equation.

Here we go! We take the square trinomial in general view ax 2 + bx+ c and… we start to select a full square! Yes, straight through letters! There was arithmetic, it became algebra.) First, as usual, we take out the letter a outside the brackets, and divide all other coefficients by a:

Like this. This is a perfectly legal conversion: a not equal to zero, and can be divided by it. And we again work with brackets according to the usual algorithm: from the term with x we ​​make a double product, add / subtract the square of the second number ...

Everything is the same, but with letters.) Try to finish it yourself! Healthy!)

After all the transformations, you should get this:

And why do we need to build such heaps out of a harmless trinomial - you ask? Nothing, now it will be interesting! And now, of course, we equate this thing to zero:

We solve it like a normal equation, we work according to all the rules, only with letters. We do elementary:

1) Move the larger fraction to the right. When moving plus, we change to minus. In order not to draw a minus in front of the fraction itself, I will simply change all the signs in the numerator. On the left in the numerator was4ac-b 2 , and after the transfer becomes -( 4ac-b 2 ) , i.e. b 2 -4 ac. Something familiar, don't you think? Yes! Discriminant, he is the most ...) It will be like this:

2) We clear the square of brackets from the coefficient. We divide both parts by " a". On the left, before the brackets, the letter a disappears, and on the right goes into the denominator of a large fraction, turning it into 4 a 2 .

It turns out this equality:

Didn't it work out for you? Then the theme "" is for you. Get there immediately!

next step extract the root. We are interested in X, right? And the X sits under the square ... We extract according to the rules for extracting roots, of course. After extraction, this is what happens:

On the left is the square of the sum disappears and it remains just the sum itself. Which is required.) But on the right appears plus/minus. For our hefty fraction, despite its awesome appearance, is just some number. Fractional number. Coefficient dependent a, b, c. At the same time, the root from the numerator of this fraction is not beautifully extracted, there is a difference of two expressions. And here is the root of the denominator 4 a 2 quite extractable! It will turn out easy 2 a.

"Tricky" question for filling: did I have the right, extracting the root from the expression 4 a2, give an answer just 2a? After all, the extraction rule square root obliges to put the sign of the module, i.e.2|a| !

Think about why I still omitted the module sign. Very useful. Hint: the answer lies in the sign plus/minus before the fraction.)

There are empty spaces left. We provide a clean x on the left. To do this, move the small fraction to the right. With a change of sign, the pepper is clear. I remind you that the sign in a fraction can be changed anywhere and any way. We want to change before the fraction, we want in the denominator, we want in the numerator. I will change sign in the numerator. It was + b, became – b. I hope there are no objections?) After the transfer, it will become like this:

We add two fractions with the same denominators and get (finally!):

Well? What can I say? Wow!)

The fourth useful thing is for students to take note!

Now let's move smoothly from school to university. You will not believe it, but the selection of a full square in higher mathematics is also necessary!

For example, such a task:

Find the indefinite integral:

Where to start? Direct application does not roll. Only selecting a full square saves, yes ...)

Those who do not know how to select a full square will forever hang on this simple example. And who knows how, he allocates and receives:

x 2 +4 x+8 = (x+2) 2 +4

And now the integral (for those who know) is taken with one left!

It's great, right? And it's not just integrals! I'm already silent about analytic geometry, with its curves of the second order – ellipse, hyperbola, parabola and circle.

For example:

Determine the type of curve, given by the equation:

x 2 + y 2 -6 x-8 y+16 = 0

Without the ability to select a full square, the task cannot be solved, yes ... But the example could not be easier! For those in the know, of course.

We group the terms with x and with y into heaps and select full squares for each variable. Get:

(x 2 -6x) + (y 2 -8 y) = -16

(x 2 -6x+9)-9 + (y 2 -8 y+16)-16 = -16

(x-3) 2 + (y-4) 2 = 9

(x-3) 2 + (y-4) 2 = 3 2

So how is it? Did you find out what kind of animal?) Well, of course! A circle of radius three with center at the point (3; 4).

And that's all.) A useful thing is to select a full square!)

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 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 expressions in the denominator or , and then convert them, respectively, to or .

Example 9

Find the indefinite integral

it the simplest example, wherein 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, in the numerator we have linear function. How to solve such integrals?

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` the value of `11/4` is added to positive number, so we get a number greater than `11/4`. In this way, smallest value 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 . So the number 7 is highest value 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=`