Definition
Expressions of the form 2 x 2 + 3 x + 5 are called quadratic trinomials. IN general case a square trinomial is an expression of the form a x 2 + b x + c, where a, b, c a, b, c - arbitrary numbers, and a ≠ 0.
Consider the quadratic 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 "allocation full square from a square trinomial".
Determine the perfect square from the quadratic 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 `(1/2)^2` to the resulting expression, we get
`((3x)^2+2*1/2*3x+(1/2)^2)+1-(1/2)^2=(3x+1/2)^2+3/4`.
We will show how the method of isolating a perfect square from a quadratic trinomial is used to factorize a square trinomial.
Factor the quadratic trinomial 4 x 2 - 12 x + 5.
We select the perfect square from the quadratic 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 we 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 the quadratic trinomial - 9 x 2 + 12 x + 5.
9 x 2 + 12 x + 5 = - 9 x 2 - 12 x + 5. Now we 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 quadratic trinomial 3 x 2 - 14 x - 5 .
We cannot represent the expression 3 x 2 as the square of some expression, because we have not yet studied this in school. You will go through this later, and in Task No. 4 we will study square roots. Let's show how you can factor a given quadratic 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'll show you how to use the perfect square method to find the largest or smallest value of a quadratic trinomial.
Consider the quadratic trinomial x 2 - x + 3. Select a complete 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 quadratic trinomial is `11/4`, and when `x!=1/2` the value of `11/4` is added positive number, so we get a number greater than `11/4`. Thus, smallest value quadratic trinomial is `11/4` and it is obtained when `x=1/2`.
Find the largest value of the quadratic trinomial - 16 2 + 8 x + 6.
We select a perfect square from a quadratic 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 .
When `x=1/4` the value of the quadratic trinomial is 7, and when `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 quadratic trinomial, and it is obtained when `x=1/4`.
Factor the numerator and denominator of the fraction `(x^2+2x-15)/(x^2-6x+9)` and reduce the fraction.
Note that the denominator of the fraction x 2 - 6 x + 9 = x - 3 2. Let's factorize the numerator of the fraction using the method of isolating a complete square from a 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 led 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 method of isolating a complete square 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 sophisticated the fraction, the more difficult it is to find its integral. In this regard, you have to resort to various tricks, which I will now tell you about. Prepared readers can immediately take advantage of table of contents:
- Method of subsuming the differential sign for simple fractions
Artificial numerator conversion method
Example 1
By the way, the considered integral can also be solved by the change of variable method, denoting , but writing the solution will be much longer.
Example 2
Find indefinite integral. Perform check.
This is an example for independent decision. It should be noted that the variable replacement method will no longer work here.
Attention, important! Examples No. 1, 2 are typical and occur frequently. In particular, such integrals often arise during the solution of other integrals, in particular, when integrating irrational functions (roots).
The considered technique also works in the case if the highest degree of the numerator is greater senior degree denominator.
Example 3
Find the indefinite integral. Perform check.
We begin to select the numerator.
The algorithm for selecting the numerator is something like this:
1) In the numerator I need to organize , but there . What to do? I put it in brackets and multiply by: .
2) Now I try to open these brackets, what happens? . Hmm... that’s better, but there’s no two in the numerator initially. What to do? You need to multiply by:
3) I open the brackets again: . And here is the first success! It turned out just right! But the problem is that an extra term has appeared. What to do? To prevent the expression from changing, I must add the same to my construction: 
. Life has become easier. Is it possible to organize again in the numerator?
4) It is possible. Let's try: 
. Open the brackets of the second term: 
. Sorry, but in the previous step I actually had , not . What to do? You need to multiply the second term by: 
5) Again, to check, I open the brackets in the second term: 
. Now it's normal: derived from the final construction of point 3! But again there is a small “but”, an extra term has appeared, which means I must add to my expression: 
If everything is done correctly, then when we open all the brackets we should get the original numerator of the integrand. We check:
Hood.
Thus: 
Ready. In the last term, I used the method of subsuming a function under a differential.
If we find the derivative of the answer and reduce the expression to common denominator, then we get exactly the original integrand function. The considered method of decomposition into a sum is nothing more than reverse action to reduce an expression to a common denominator.
Algorithm for selecting the numerator in similar examples It's better to do it in draft form. With some skills it will work mentally. I remember a record-breaking case when I was performing a selection for the 11th power, and the expansion of the numerator took up almost two lines of Verd.
Example 4
Find the indefinite integral. Perform check.
This is an example for you to solve on your own.
Method of subsuming the differential sign for simple fractions
Let's move on to consider next type fractions.
, , , (coefficients and are not equal to zero).
In fact, a couple of cases with arcsine and arctangent have already been mentioned in the lesson Variable change method in indefinite integral. Such examples are solved by subsuming the function under the differential sign and further integrating using a table. Here's another typical examples with long and high logarithm:
Example 5
Example 6
Here it is advisable to pick up a table of integrals and see what formulas and How transformation takes place. Note, how and why The squares in these examples are highlighted. In particular, in Example 6 we first need to represent the denominator in the form 
, then bring it under the differential sign. And all this needs to be done in order to use the standard tabular formula 
.
Why look, try to solve examples No. 7, 8 yourself, especially since they are quite short:
Example 7
Example 8
Find the indefinite integral:
If you also manage to check these examples, then great respect - your differentiation skills are excellent.
Full square selection method
Integrals of the form 
(coefficients and are not equal to zero) are solved complete square extraction method, which has already appeared in the lesson Geometric transformations of graphs.
In fact, such integrals reduce to one of the four tabular integrals we just looked at. And this is achieved using familiar abbreviated multiplication formulas:
The formulas are applied precisely in this direction, that is, the idea of the method is to artificially organize the expressions either in the denominator, and then convert them accordingly to either.
Example 9
Find the indefinite integral
This simplest example, in which with the term – unit coefficient(and not some number or minus).
Let's look at the denominator, here the whole matter clearly comes down to chance. Let's start converting the denominator:
Obviously, you need to add 4. And, so that the expression does not change, subtract the same four:
Now you can apply the formula:
After the conversion is completed ALWAYS it is advisable to perform reverse stroke: , everything is fine, there are no errors.
The final design of the example in question should look something like this: 
Ready. Summarizing "freebie" complex function under the differential sign: , in principle, could be neglected
Example 10
Find the indefinite integral:
This is an example for you to solve on your own, 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, we need to take the minus out of brackets and arrange the terms in the order we need: . Constant(“two” in in this case) don't touch!
Now we add one in parentheses. Analyzing the expression, we come to the conclusion that we need to add one outside the brackets:
Here we get the formula, apply:
ALWAYS We check the draft:
, which was what needed to be checked.
The clean example looks something like this: 
Making the task more difficult
Example 12
Find the indefinite integral: 
Here the term is no longer a unit coefficient, but a “five”.
(1) If there is a constant at, then we immediately take it out of brackets.
(2) In general, it is always better to move this constant outside the integral so that it does not get in the way.
(3) Obviously, everything will come down to the formula. We need to understand the term, namely, get the “two”
(4) Yeah, . This means that we add to the expression and subtract the same fraction.
(5) Now select a complete square. In the general case, we also need to calculate , but here we have the formula for a long logarithm 
, and there is no point in performing the action; why will become clear below.
(6) Actually, we can apply the formula 
, only instead of “X” we have , which does not negate the validity of the table integral. Strictly speaking, one step was missed - before integration, the function should have been subsumed under the differential sign: 
, but, as I have repeatedly noted, this is often neglected.
(7) In the answer under the root, it is advisable to expand all the brackets back:
Difficult? This is not the most difficult part of integral calculus. Although, the examples under consideration are not so much complex as they require good computing techniques.
Example 13
Find the indefinite integral:
This is an example for you to solve on your own. The answer is at the end of the lesson.
There are integrals with roots in the denominator, which, using a substitution, are reduced to integrals of the type considered; you can read about them in the article Complex integrals, but it is designed for very prepared students.
Subsuming the numerator under the differential sign
This final part lesson, however, integrals of this type occur quite often! If you're tired, maybe it's better to read tomorrow? ;)
The integrals that we will consider are similar to the integrals of the previous paragraph, they have the form: or 
(coefficients , and are not equal to zero).
That is, in our numerator we have linear function. How to solve such integrals?
Online calculator.
Squaring a binomial and factoring it quadratic trinomial.
This math program distinguishes the square binomial from the square trinomial, i.e. does a transformation like: \(ax^2+bx+c \rightarrow a(x+p)^2+q \) and factorizes a quadratic trinomial: \(ax^2+bx+c \rightarrow a(x+n)(x+m) \)
Those. the problems boil down 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 may be useful for high school students secondary schools in preparation for tests and exams, when testing knowledge before the Unified State Exam, 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 it done as quickly as possible? homework in mathematics or algebra? In this case, you can also use our programs with detailed solutions.
This way you can conduct your own training and/or training of yours. younger brothers or sisters, while the level of education in the field of problems being solved increases.
If you are not familiar with the rules for entering a quadratic trinomial, we recommend that you familiarize yourself with them.
Rules for entering a quadratic 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 whole or fractional numbers.
Moreover, fractional numbers can be entered not only as a decimal, but also as an ordinary fraction.
Rules for entering decimal fractions.
In decimals fraction can be separated from the whole by either a period 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 numerical fraction The numerator is separated from the denominator by a division sign: /
The whole part is separated from the fraction by the ampersand sign: &
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 parentheses. In this case, when solving, the introduced expression is first simplified.
For example: 1/2(x-1)(x+1)-(5x-10&1/2)
Example detailed solution
Isolating the square of a 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 little theory.
Isolating the square of a 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.
From the trinomial 2x 2 +12x+14 we extract the square of the binomial.
\(2x^2+12x+14 = 2(x^2+6x+7) \)
To do this, imagine 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 extract the square binomial from the square trinomial, and showed that:
$$ 2x^2+12x+14 = 2(x+3)^2-4 $$
Factoring a quadratic trinomial
If the square trinomial ax 2 +bx+c is represented in the form a(x+n)(x+m), where n and m are real numbers, then the operation is said to have been performed factorization of a quadratic trinomial.
Let us show with an example how this transformation is done.
Let's factor the quadratic 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, imagine 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 factored the quadratic trinomial, and showed that:
$$ 2x^2+4x-6 = 2(x-1)(x+3) $$
Note that factoring a quadratic trinomial is possible only when, quadratic equation, corresponding to this trinomial has roots.
Those. in our case, it is possible to factor the trinomial 2x 2 +4x-6 if the quadratic equation 2x 2 +4x-6 =0 has roots. In the process of factorization, we established 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 previously studied methods of factoring a polynomial and consider examples of their application, in addition, we will study new method- method of identifying a complete square and learn how to apply it in solving various problems.
Subject:Factoring polynomials
Lesson:Factoring polynomials. Method for selecting a complete square. Combination of methods
Let us recall the basic methods of factoring a polynomial that were studied earlier:
The method of putting a common factor out of brackets, that is, a factor that is present in all terms of the polynomial. Let's look at an example:
Recall that a monomial is the product of powers and numbers. In our example, both terms have some common, identical elements.
So let's take it out common multiplier outside of brackets:
;
Let us remind you that by multiplying the factor taken out by a bracket, you can check the correctness of the factor taken out.
Grouping method. It is not always possible to extract 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 in the entire expression, and you can continue the decomposition. Let's look at an example:
Let's group the first term with the fourth, the second with the fifth, and the third with the sixth:
Let's take out the common factors in the groups:
The expression now has a common factor. Let's take it out:
Application of abbreviated multiplication formulas. Let's look at an example:
;
Let's write the expression in detail:
Obviously, we have before us the formula for the squared difference, since it is the sum of the squares of two expressions and their double product is subtracted from it. Let's use the formula:
Today we will learn another method - the method of selecting a complete square. It is based on the formulas of the square of the sum and the square of the difference. Let's remind them:
Formula for the square of the sum (difference);
The peculiarity of these formulas is that they contain the squares of two expressions and their double product. Let's look at an example:
Let's write down the expression:
So, the first expression is , and the second is .
In order to create a formula for the square of a sum or difference, twice the product of expressions is not enough. It needs to be added and subtracted:
Let's complete the square of the sum:
Let's transform the resulting expression:
Let's apply the formula for the difference of squares, recall that the difference of the squares of two expressions is the product of and the sum of 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 in the square, that is, to determine which expressions’ squares are in in this example. After this, 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 factorized 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:
Let's find expressions that are squared:
Let us write down what their double product should be:
Let's add and subtract double the product:
Let's complete the square of the sum and give similar ones:
Let's write it using the difference of squares formula:
Example 2 - solve the equation:
;
On the left side of the equation is a trinomial. You need to factor it into factors. We use the squared difference formula:
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's fold a complete square and give similar 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 equal to zero. Let's create the following equations based on this:
Let's solve the first equation:
Let's solve the second equation:
Answer: or
;
We proceed similarly to the previous example - select the square of the difference.