List the methods of factorization. How to factorize an algebraic equation
The factorization of polynomials is an identical transformation, as a result of which a polynomial is transformed into a product of several factors - polynomials or monomials.
There are several ways to factorize polynomials.
Method 1. Bracketing the common factor.
This transformation is based on the distributive law of multiplication: ac + bc = c(a + b). The essence of the transformation is to single out the common factor in the two components under consideration and “put it out” of the brackets.
Let us factorize the polynomial 28x 3 - 35x 4.
Decision.
1. We find a common divisor for elements 28x3 and 35x4. For 28 and 35 it will be 7; for x 3 and x 4 - x 3. In other words, our common factor is 7x3.
2. We represent each of the elements as a product of factors, one of which
7x 3: 28x 3 - 35x 4 \u003d 7x 3 ∙ 4 - 7x 3 ∙ 5x.
3. Bracketing the common factor
7x 3: 28x 3 - 35x 4 \u003d 7x 3 ∙ 4 - 7x 3 ∙ 5x \u003d 7x 3 (4 - 5x).
Method 2. Using abbreviated multiplication formulas. The "mastery" of mastering this method is to notice in the expression one of the formulas for abbreviated multiplication.
Let us factorize the polynomial x 6 - 1.
Decision.
1. We can apply the difference of squares formula to this expression. To do this, we represent x 6 as (x 3) 2, and 1 as 1 2, i.e. 1. The expression will take the form:
(x 3) 2 - 1 \u003d (x 3 + 1) ∙ (x 3 - 1).
2. To the resulting expression, we can apply the formula for the sum and difference of cubes:
(x 3 + 1) ∙ (x 3 - 1) \u003d (x + 1) ∙ (x 2 - x + 1) ∙ (x - 1) ∙ (x 2 + x + 1).
So,
x 6 - 1 = (x 3) 2 - 1 = (x 3 + 1) ∙ (x 3 - 1) = (x + 1) ∙ (x 2 - x + 1) ∙ (x - 1) ∙ (x 2 + x + 1).
Method 3. Grouping. The grouping method consists in combining the components of a polynomial in such a way that it is easy to perform operations on them (addition, subtraction, taking out a common factor).
We factorize the polynomial x 3 - 3x 2 + 5x - 15.
Decision.
1. Group the components in this way: the 1st with the 2nd, and the 3rd with the 4th
(x 3 - 3x 2) + (5x - 15).
2. In the resulting expression, we take the common factors out of brackets: x 2 in the first case and 5 in the second.
(x 3 - 3x 2) + (5x - 15) \u003d x 2 (x - 3) + 5 (x - 3).
3. We take out the common factor x - 3 and get:
x 2 (x - 3) + 5 (x - 3) \u003d (x - 3) (x 2 + 5).
So,
x 3 - 3x 2 + 5x - 15 \u003d (x 3 - 3x 2) + (5x - 15) \u003d x 2 (x - 3) + 5 (x - 3) \u003d (x - 3) ∙ (x 2 + 5 ).
Let's fix the material.
Factor the polynomial a 2 - 7ab + 12b 2 .
Decision.
1. We represent the monomial 7ab as the sum 3ab + 4ab. The expression will take the form:
a 2 - (3ab + 4ab) + 12b 2 . 
Let's open the brackets and get:
a 2 - 3ab - 4ab + 12b 2 .
2. Group the components of the polynomial in this way: the 1st with the 2nd and the 3rd with the 4th. We get:
(a 2 - 3ab) - (4ab - 12b 2).
3. Let's take out the common factors:
(a 2 - 3ab) - (4ab - 12b 2) \u003d a (a - 3b) - 4b (a - 3b).
4. Let's take out the common factor (a - 3b):
a(a – 3b) – 4b(a – 3b) = (a – 3b) ∙ (a – 4b).
So,
a 2 - 7ab + 12b 2 =
= a 2 - (3ab + 4ab) + 12b 2 =
= a 2 - 3ab - 4ab + 12b 2 =
= (a 2 - 3ab) - (4ab - 12b 2) =
= a(a - 3b) - 4b(a - 3b) =
= (а – 3 b) ∙ (а – 4b).
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The concepts of "polynomial" and "factorization of a polynomial" in algebra are very common, because you need to know them in order to easily perform calculations with large multi-valued numbers. This article will describe several decomposition methods. All of them are quite simple to use, you just have to choose the right one in each case.
The concept of a polynomial
A polynomial is the sum of monomials, that is, expressions containing only the multiplication operation.
For example, 2 * x * y is a monomial, but 2 * x * y + 25 is a polynomial, which consists of 2 monomials: 2 * x * y and 25. Such polynomials are called binomials.
Sometimes, for the convenience of solving examples with multivalued values, the expression must be transformed, for example, decomposed into a certain number of factors, that is, numbers or expressions between which the multiplication operation is performed. There are a number of ways to factorize a polynomial. It is worth considering them starting from the most primitive, which is used even in primary classes.
Grouping (general entry)
The formula for factoring a polynomial into factors by the grouping method in general looks like this:
ac + bd + bc + ad = (ac + bc) + (ad + bd)
It is necessary to group the monomials so that a common factor appears in each group. In the first parenthesis, this is the factor c, and in the second - d. This must be done in order to then take it out of the bracket, thereby simplifying the calculations.
Decomposition algorithm on a specific example
The simplest example of factoring a polynomial into factors using the grouping method is given below:
10ac + 14bc - 25a - 35b = (10ac - 25a) + (14bc - 35b)
In the first bracket, you need to take the terms with the factor a, which will be common, and in the second - with the factor b. Pay attention to the + and - signs in the finished expression. We put before the monomial the sign that was in the initial expression. That is, you need to work not with the expression 25a, but with the expression -25. The minus sign, as it were, is “glued” to the expression behind it and always take it into account in calculations.
At the next step, you need to take out the factor, which is common, out of the bracket. That's what grouping is for. To take it out of the bracket means to write out before the bracket (omitting the multiplication sign) all those factors that are repeated exactly in all the terms that are in the bracket. If there are not 2, but 3 or more terms in the bracket, the common factor must be contained in each of them, otherwise it cannot be taken out of the bracket.
In our case, only 2 terms in brackets. The overall multiplier is immediately visible. The first parenthesis is a, the second is b. Here you need to pay attention to the digital coefficients. In the first bracket, both coefficients (10 and 25) are multiples of 5. This means that not only a, but also 5a can be bracketed. Before the bracket, write out 5a, and then divide each of the terms in brackets by the common factor that was taken out, and also write down the quotient in brackets, not forgetting the + and - signs. Do the same with the second bracket, take out 7b, since 14 and 35 multiple of 7.
10ac + 14bc - 25a - 35b = (10ac - 25a) + (14bc - 35b) = 5a (2c - 5) + 7b (2c - 5).
It turned out 2 terms: 5a (2c - 5) and 7b (2c - 5). Each of them contains a common factor (the whole expression in brackets here is the same, which means it is a common factor): 2c - 5. It also needs to be taken out of the bracket, that is, the terms 5a and 7b remain in the second bracket:
5a(2c - 5) + 7b(2c - 5) = (2c - 5)*(5a + 7b).
So the full expression is:
10ac + 14bc - 25a - 35b \u003d (10ac - 25a) + (14bc - 35b) \u003d 5a (2c - 5) + 7b (2c - 5) \u003d (2c - 5) * (5a + 7b).
Thus, the polynomial 10ac + 14bc - 25a - 35b is decomposed into 2 factors: (2c - 5) and (5a + 7b). The multiplication sign between them can be omitted when writing
Sometimes there are expressions of this type: 5a 2 + 50a 3, here you can bracket not only a or 5a, but even 5a 2. You should always try to take the largest possible common factor out of the bracket. In our case, if we divide each term by a common factor, we get:
5a 2 / 5a 2 = 1; 50a 3 / 5a 2 = 10a(when calculating the quotient of several powers with equal bases, the base is preserved, and the exponent is subtracted). Thus, one remains in the bracket (in no case do not forget to write one if you take out one of the terms entirely from the bracket) and the quotient of division: 10a. It turns out that:
5a 2 + 50a 3 = 5a 2 (1 + 10a)
Square formulas
For the convenience of calculations, several formulas have been derived. They are called reduced multiplication formulas and are used quite often. These formulas help factorize polynomials containing powers. This is another powerful way to factorize. So here they are:
- a 2 + 2ab + b 2 = (a + b) 2 - the formula, called the "square of the sum", since as a result of the expansion into a square, the sum of the numbers enclosed in brackets is taken, that is, the value of this sum is multiplied by itself 2 times, which means it is a multiplier.
- a 2 + 2ab - b 2 = (a - b) 2 - the formula of the square of the difference, it is similar to the previous one. The result is a difference enclosed in brackets, contained in a square power.
- a 2 - b 2 \u003d (a + b) (a - b)- this is the formula for the difference of squares, since initially the polynomial consists of 2 squares of numbers or expressions between which subtraction is performed. It is perhaps the most commonly used of the three.
Examples for calculating by formulas of squares
Calculations on them are made quite simply. For example:
- 25x2 + 20xy + 4y 2 - use the formula "square of the sum".
- 25x 2 is the square of 5x. 20xy is twice the product of 2*(5x*2y), and 4y 2 is the square of 2y.
- So 25x 2 + 20xy + 4y 2 = (5x + 2y) 2 = (5x + 2y)(5x + 2y). This polynomial is decomposed into 2 factors (the factors are the same, therefore it is written as an expression with a square power).
Operations according to the formula of the square of the difference are performed similarly to these. What remains is the difference of squares formula. Examples for this formula are very easy to identify and find among other expressions. For example:
- 25a 2 - 400 \u003d (5a - 20) (5a + 20). Since 25a 2 \u003d (5a) 2, and 400 \u003d 20 2
- 36x 2 - 25y 2 \u003d (6x - 5y) (6x + 5y). Since 36x 2 \u003d (6x) 2, and 25y 2 \u003d (5y 2)
- c 2 - 169b 2 \u003d (c - 13b) (c + 13b). Since 169b 2 = (13b) 2
It is important that each of the terms is the square of some expression. Then this polynomial is to be factored by the difference of squares formula. For this, it is not necessary that the second power is above the number. There are polynomials containing large powers, but still suitable for these formulas.
a 8 +10a 4 +25 = (a 4) 2 + 2*a 4 *5 + 5 2 = (a 4 +5) 2
In this example, a 8 can be represented as (a 4) 2 , that is, the square of a certain expression. 25 is 5 2 and 10a is 4 - this is the double product of the terms 2*a 4 *5. That is, this expression, despite the presence of degrees with large exponents, can be decomposed into 2 factors in order to work with them later.
Cube formulas
The same formulas exist for factoring polynomials containing cubes. They are a little more complicated than those with squares:
- a 3 + b 3 \u003d (a + b) (a 2 - ab + b 2)- this formula is called the sum of cubes, since in its initial form the polynomial is the sum of two expressions or numbers enclosed in a cube.
- a 3 - b 3 \u003d (a - b) (a 2 + ab + b 2) - a formula identical to the previous one is denoted as the difference of cubes.
- a 3 + 3a 2 b + 3ab 2 + b 3 = (a + b) 3 - sum cube, as a result of calculations, the sum of numbers or expressions is obtained, enclosed in brackets and multiplied by itself 3 times, that is, located in the cube
- a 3 - 3a 2 b + 3ab 2 - b 3 = (a - b) 3 - the formula, compiled by analogy with the previous one with a change in only some signs of mathematical operations (plus and minus), is called the "difference cube".
The last two formulas are practically not used for the purpose of factoring a polynomial, since they are complex, and it is quite rare to find polynomials that completely correspond to just such a structure so that they can be decomposed according to these formulas. But you still need to know them, since they will be required for actions in the opposite direction - when opening brackets.
Examples for cube formulas
Consider an example: 64a 3 − 8b 3 = (4a) 3 − (2b) 3 = (4a − 2b)((4a) 2 + 4a*2b + (2b) 2) = (4a−2b)(16a 2 + 8ab + 4b 2 ).
We've taken fairly prime numbers here, so you can immediately see that 64a 3 is (4a) 3 and 8b 3 is (2b) 3 . Thus, this polynomial is expanded by the formula difference of cubes into 2 factors. Actions on the formula of the sum of cubes are performed by analogy.
It is important to understand that not all polynomials can be decomposed in at least one of the ways. But there are such expressions that contain larger powers than a square or a cube, but they can also be expanded into abbreviated multiplication forms. For example: x 12 + 125y 3 =(x 4) 3 +(5y) 3 =(x 4 +5y)*((x 4) 2 − x 4 *5y+(5y) 2)=(x 4 + 5y)( x 8 − 5x 4 y + 25y 2).
This example contains as many as 12 degrees. But even it can be factored using the sum of cubes formula. To do this, you need to represent x 12 as (x 4) 3, that is, as a cube of some expression. Now, instead of a, you need to substitute it in the formula. Well, the expression 125y 3 is the cube of 5y. The next step is to write the formula and do the calculations.
At first, or when in doubt, you can always check by inverse multiplication. You only need to open the brackets in the resulting expression and perform actions with similar terms. This method applies to all the listed methods of reduction: both to work with a common factor and grouping, and to operations on formulas of cubes and square powers.
What factorization? It's a way of turning an awkward and complicated example into a simple and cute one.) Very powerful trick! It occurs at every step both in elementary mathematics and in higher mathematics.
Such transformations in mathematical language are called identical transformations of expressions. Who is not in the subject - take a walk on the link. There is very little, simple and useful.) The meaning of any identical transformation is to write the expression in a different form while preserving its essence.
Meaning factorizations extremely simple and understandable. Right from the title itself. You can forget (or not know) what a multiplier is, but can you figure out that this word comes from the word "multiply"?) Factoring means: represent an expression as a multiplication of something by something. Forgive me mathematics and the Russian language ...) And that's it.
For example, you need to decompose the number 12. You can safely write:
So we presented the number 12 as a multiplication of 3 by 4. Please note that the numbers on the right (3 and 4) are completely different than on the left (1 and 2). But we are well aware that 12 and 3 4 same. The essence of the number 12 from the transformation hasn't changed.
Is it possible to decompose 12 in another way? Easily!
12=3 4=2 6=3 2 2=0.5 24=........
The decomposition options are endless.
Decomposing numbers into factors is a useful thing. It helps a lot, for example, when dealing with roots. But the factorization of algebraic expressions is not something that is useful, it is - necessary! Just for example:
Simplify:
Those who do not know how to factorize the expression, rest on the sidelines. Who knows how - simplifies and gets:
The effect is amazing, right?) By the way, the solution is quite simple. You will see for yourself below. Or, for example, such a task:
Solve the equation:
x 5 - x 4 = 0
Decided in the mind, by the way. With the help of factorization. Below we will solve this example. Answer: x 1 = 0; x2 = 1.
Or, the same thing, but for the older ones):
Solve the equation:
In these examples, I have shown main purpose factorizations: simplification of fractional expressions and solution of some types of equations. I recommend to remember the rule of thumb:
If we have a terrible fractional expression in front of us, we can try to factorize the numerator and denominator. Very often, the fraction is reduced and simplified.
If we have an equation in front of us, where on the right is zero, and on the left - don’t understand what, you can try to factorize the left side. Sometimes it helps.)
Basic methods of factorization.
Here are the most popular ways:
4. Decomposition of a square trinomial.
These methods must be remembered. It's in that order. Complex examples are checked for all possible decomposition methods. And it’s better to check in order, so as not to get confused ... Let’s start in order.)
1. Taking the common factor out of brackets.
Simple and reliable way. It doesn't get bad from him! It happens either well or not at all.) Therefore, he is the first. We understand.
Everyone knows (I believe!) the rule:
a(b+c) = ab+ac
Or, more generally:
a(b+c+d+.....) = ab+ac+ad+....
All equalities work both from left to right, and vice versa, from right to left. You can write:
ab+ac = a(b+c)
ab+ac+ad+.... = a(b+c+d+.....)
That's the whole point of putting the common factor out of brackets.
On the left side a - common factor for all terms. Multiplied by everything.) Right is the most a is already outside the brackets.
We will consider the practical application of the method with examples. At first, the variant is simple, even primitive.) But in this variant I will mark (in green) very important points for any factorization.
Multiply:
ah+9x
Which general is the multiplier in both terms? X, of course! We will take it out of brackets. We do so. We immediately write x outside the brackets:
ax+9x=x(
And in brackets we write the result of division each term on this very x. In order:
That's all. Of course, it is not necessary to paint in such detail, This is done in the mind. But to understand what's what, it is desirable). We fix in memory:
We write the common factor outside the brackets. In parentheses, we write the results of dividing all the terms by this very common factor. In order.
Here we have expanded the expression ah+9x for multipliers. Turned it into multiplying x by (a + 9). I note that in the original expression there was also a multiplication, even two: a x and 9 x. But it has not been factorized! Because in addition to multiplication, this expression also contained addition, the "+" sign! And in the expression x(a+9) nothing but multiplication!
How so!? - I hear the indignant voice of the people - And in brackets!?)
Yes, there is addition inside the brackets. But the trick is that while the brackets are not opened, we consider them like one letter. And we do all the actions with brackets in their entirety, like one letter. In this sense, in the expression x(a+9) nothing but multiplication. This is the whole point of factorization.
By the way, is there any way to check if we did everything right? Easy! It is enough to multiply back what was taken out (x) by brackets and see if it worked out original expression? If it worked out, everything is tip-top!)
x(a+9)=ax+9x
Happened.)
There is no problem in this primitive example. But if there are several terms, and even with different signs ... In short, every third student messes up). Therefore:
If necessary, check the factorization by inverse multiplication.
Multiply:
3ax+9x
We are looking for a common factor. Well, everything is clear with X, it can be endured. Is there any more general factor? Yes! This is a trio. You can also write the expression like this:
3x+3 3x
Here it is immediately clear that the common factor will be 3x. Here we take it out:
3ax+3 3x=3x(a+3)
Spread out.
And what happens if you take only x? Nothing special:
3ax+9x=x(3a+9)
This will also be a factorization. But in this fascinating process, it is customary to lay everything out until it stops, while there is an opportunity. Here in brackets there is an opportunity to take out a triple. Get:
3ax+9x=x(3a+9)=3x(a+3)
The same thing, only with one extra action.) Remember:
When taking the common factor out of brackets, we try to take out maximum common multiplier.
Let's continue the fun?
Factoring the expression:
3ax+9x-8a-24
What will we take out? Three, X? No-ee... You can't. I remind you that you can only take general multiplier that is in all terms of the expression. That's why he general. There is no such multiplier here ... What, you can not lay out!? Well, yes, we were delighted, how ... Meet:
2. Grouping.
Actually, grouping can hardly be called an independent way of factorization. This is rather a way to get out of a complex example.) You need to group the terms so that everything works out. This can only be shown with an example. So we have an expression:
3ax+9x-8a-24
It can be seen that there are some common letters and numbers. But... General there is no multiplier to be in all terms. Do not lose heart and we break the expression into pieces. We group. So that in each piece there was a common factor, there was something to take out. How do we break? Yes, just parentheses.
Let me remind you that brackets can be placed anywhere and any way. If only the essence of the example didn't change. For example, you can do this:
3ax+9x-8a-24=(3ax + 9x) - (8a + 24)
Please pay attention to the second brackets! They are preceded by a minus sign, and 8a and 24 become positive! If, for verification, we open the brackets back, the signs will change, and we get original expression. Those. the essence of the expression from brackets has not changed.
But if you just put in parentheses, not taking into account the sign change, for example, like this:
3ax+9x-8a-24=(3ax + 9x) -(8a-24 )
it will be a mistake. Right - already other expression. Expand the brackets and everything will become clear. You can decide no further, yes ...)
But back to factorization. Look at the first brackets (3ax + 9x) and think, is it possible to endure something? Well, we solved this example above, we can take it out 3x:
(3ax+9x)=3x(a+3)
We study the second brackets, there you can take out the eight:
(8a+24)=8(a+3)
Our entire expression will be:
(3ax + 9x) - (8a + 24) \u003d 3x (a + 3) -8 (a + 3)
Multiplied? No. The decomposition should result in only multiplication, and we have a minus sign spoils everything. But... Both terms have a common factor! This is (a+3). It was not in vain that I said that the brackets as a whole are, as it were, one letter. So these brackets can be taken out of the brackets. Yes, that's exactly what it sounds like.)
We do as described above. Write the common factor (a+3), in the second brackets we write the results of dividing the terms by (a+3):
3x(a+3)-8(a+3)=(a+3)(3x-8)
Everything! On the right, there is nothing but multiplication! So the factorization is completed successfully!) Here it is:
3ax + 9x-8a-24 \u003d (a + 3) (3x-8)
Let's recap the essence of the group.
If the expression does not general multiplier for all terms, we split the expression with brackets so that inside the brackets the common factor was. Let's take it out and see what happens. If we are lucky, and exactly the same expressions remain in the brackets, we take these brackets out of the brackets.
I will add that grouping is a creative process). It doesn't always work the first time. It's OK. Sometimes you have to swap terms, consider different grouping options until you find a good one. The main thing here is not to lose heart!)
Examples.
Now, having enriched with knowledge, you can also solve tricky examples.) At the beginning of the lesson, there were three of these ...
Simplify:
In fact, we have already solved this example. Imperceptibly to myself.) I remind you: if we are given a terrible fraction, we try to decompose the numerator and denominator into factors. Other simplification options simply no.
Well, the denominator is not decomposed here, but the numerator... We have already decomposed the numerator in the course of the lesson! Like this:
3ax + 9x-8a-24 \u003d (a + 3) (3x-8)
We write the result of expansion into the numerator of the fraction:
According to the rule of reduction of fractions (the main property of a fraction), we can divide (simultaneously!) The numerator and denominator by the same number, or expression. Fraction from this does not change. So we divide the numerator and denominator by the expression (3x-8). And here and there we get units. Final simplification result:
I emphasize in particular: reduction of a fraction is possible if and only if in the numerator and denominator, in addition to multiplying expressions there is nothing. That is why the transformation of the sum (difference) into multiplication so important to simplify. Of course, if the expressions various, then nothing will be reduced. Byvet. But the factorization gives a chance. This chance without decomposition - simply does not exist.
Equation example:
Solve the equation:
x 5 - x 4 = 0
Taking out the common factor x 4 for brackets. We get:
x 4 (x-1)=0
We assume that the product of the factors is equal to zero then and only then when any of them is equal to zero. If in doubt, find me a couple of non-zero numbers that, when multiplied, will give zero.) So we write, first the first factor:
With this equality, the second factor does not bother us. Anyone can be, anyway, in the end, zero will turn out. What is the number to the fourth power of zero? Only zero! And nothing else ... Therefore:
We figured out the first factor, we found one root. Let's deal with the second factor. Now we don't care about the first multiplier.):
Here we found a solution: x 1 = 0; x2 = 1. Any of these roots fit our equation.
A very important note. Note that we have solved the equation bit by bit! Each factor was set to zero. regardless of other factors. By the way, if in such an equation there are not two factors, as we have, but three, five, as many as you like, we will decide similar. Piece by piece. For example:
(x-1)(x+5)(x-3)(x+2)=0
The one who opens the brackets, multiplies everything, will forever hang on this equation.) The correct student will immediately see that there is nothing on the left except for multiplication, on the right - zero. And he will begin (in his mind!) To equate to zero all the brackets in order. And he will get (in 10 seconds!) the correct solution: x 1 = 1; x 2 \u003d -5; x 3 \u003d 3; x4 = -2.
Great, right?) Such an elegant solution is possible if the left side of the equation split into multiples. Is the hint clear?)
Well, the last example, for the older ones):
Solve the equation:
It is somewhat similar to the previous one, don't you think?) Of course. It's time to remember that in seventh grade algebra, sines, logarithms, and anything else can be hidden under letters! Factoring works in all mathematics.
Taking out the common factor lg4x for brackets. We get:
lg 4x=0
This is one root. Let's deal with the second factor.
Here is the final answer: x 1 = 1; x2 = 10.
I hope you've realized the power of factoring in simplifying fractions and solving equations.)
In this lesson, we got acquainted with the removal of the common factor and grouping. It remains to deal with the formulas for abbreviated multiplication and the square trinomial.
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A polynomial is an expression consisting of the sum of monomials. The latter are the product of a constant (number) and the root (or roots) of the expression to the power k. In this case, one speaks of a polynomial of degree k. The decomposition of a polynomial involves the transformation of the expression, in which the terms are replaced by factors. Let us consider the main ways of carrying out this kind of transformation.
Method for expanding a polynomial by extracting a common factor
This method is based on the laws of the distribution law. So, mn + mk = m * (n + k).
- Example: expand 7y 2 + 2uy and 2m 3 – 12m 2 + 4lm.
7y 2 + 2uy = y * (7y + 2u),
2m 3 - 12m 2 + 4lm = 2m(m 2 - 6m + 2l).
However, the factor that is necessarily present in each polynomial may not always be found, so this method is not universal.
Polynomial expansion method based on abbreviated multiplication formulas
Abbreviated multiplication formulas are valid for a polynomial of any degree. In general, the transformation expression looks like this:
u k – l k = (u – l)(u k-1 + u k-2 * l + u k-3 *l 2 + … u * l k-2 + l k-1), where k is a representative of natural numbers .
Most often in practice, formulas for polynomials of the second and third orders are used:
u 2 - l 2 \u003d (u - l) (u + l),
u 3 - l 3 \u003d (u - l) (u 2 + ul + l 2),
u 3 + l 3 = (u + l)(u 2 - ul + l 2).
- Example: expand 25p 2 - 144b 2 and 64m 3 - 8l 3 .
25p 2 - 144b 2 \u003d (5p - 12b) (5p + 12b),
64m 3 - 8l 3 = (4m) 3 - (2l) 3 = (4m - 2l)((4m) 2 + 4m * 2l + (2l) 2) = (4m - 2l)(16m 2 + 8ml + 4l 2 ).
Polynomial decomposition method - grouping terms of an expression
This method in some way echoes the technique of deriving a common factor, but has some differences. In particular, before isolating the common factor, one should group the monomials. Grouping is based on the rules of associative and commutative laws.
All monomials presented in the expression are divided into groups, in each of which a common value is taken out such that the second factor will be the same in all groups. In general, such a decomposition method can be represented as an expression:
pl + ks + kl + ps = (pl + ps) + (ks + kl) ⇒ pl + ks + kl + ps = p(l + s) + k(l + s),
pl + ks + kl + ps = (p + k)(l + s).
- Example: expand 14mn + 16ln - 49m - 56l.
14mn + 16ln - 49m - 56l = (14mn - 49m) + (16ln - 56l) = 7m * (2n - 7) + 8l * (2n - 7) = (7m + 8l)(2n - 7).
Polynomial Decomposition Method - Full Square Formation
This method is one of the most efficient in the course of polynomial decomposition. At the initial stage, it is necessary to determine the monomials that can be “folded” into the square of the difference or sum. For this, one of the following relations is used:
(p - b) 2 \u003d p 2 - 2pb + b 2,
- Example: expand the expression u 4 + 4u 2 – 1.
Among its monomials, we single out the terms that form a complete square: u 4 + 4u 2 - 1 = u 4 + 2 * 2u 2 + 4 - 4 - 1 =
\u003d (u 4 + 2 * 2u 2 + 4) - 4 - 1 \u003d (u 4 + 2 * 2u 2 + 4) - 5.
Complete the transformation using the rules of abbreviated multiplication: (u 2 + 2) 2 - 5 = (u 2 + 2 - √5) (u 2 + 2 + √5).
That. u 4 + 4u 2 - 1 = (u 2 + 2 - √5)(u 2 + 2 + √5).
