Very often, the numerator and denominator of a fraction are algebraic expressions that must first be factored, and then, having found identical ones among them, divide both the numerator and denominator by them, that is, reduce the fraction. An entire chapter of the 7th grade algebra textbook is devoted to the task of factoring a polynomial. Factorization can be done 3 ways, as well as a combination of these methods.
1. Application of abbreviated multiplication formulas
As is known, to multiply a polynomial by a polynomial, you need to multiply each term of one polynomial by each term of the other polynomial and add the resulting products. There are at least 7 (seven) frequently occurring cases of multiplying polynomials that are included in the concept. For example,
Table 1. Factorization in the 1st way
2. Taking the common factor out of brackets
This method is based on the application of the distributive multiplication law. For example,
We divide each term of the original expression by the factor that we take out, and we get an expression in parentheses (that is, the result of dividing what was by what we take out remains in parentheses). First of all you need determine the multiplier correctly, which must be taken out of the bracket.
The common factor can also be a polynomial in brackets:
When performing the “factorize” task, you need to be especially careful with the signs when putting the total factor out of brackets. To change the sign of each term in a parenthesis (b - a), let’s take the common factor out of brackets -1 , and each term in the bracket will be divided by -1: (b - a) = - (a - b) .
If the expression in brackets is squared (or to any even power), then numbers inside brackets can be swapped completely freely, since the minuses taken out of brackets will still turn into a plus when multiplied: (b - a) 2 = (a - b) 2, (b - a) 4 = (a - b) 4 and so on…
3. Grouping method
Sometimes not all terms in an expression have a common factor, but only some. Then you can try group terms in brackets so that some factor can be taken out of each one. Grouping method- this is a double removal of common factors from brackets.
4. Using several methods at once
Sometimes you need to apply not one, but several methods of factoring a polynomial at once.
This is a summary of the topic "Factorization". Choose what to do next:
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What's happened factorization? This is a way to turn an inconvenient and complex example into a simple and cute one.) A very powerful technique! It is found at every step in both elementary and higher mathematics.
Such transformations in mathematical language are called identical transformations of expressions. For those who are not in the know, take a look at the link. There is very little there, simple and useful.) The meaning of any identity transformation is the recording of the expression in another form while maintaining its essence.
Meaning factorization extremely simple and clear. Right from the name itself. You may forget (or not know) what a multiplier is, but you can figure out that this word comes from the word “multiply”?) Factoring means: represent an expression in the form of multiplying something by something. May mathematics and the Russian language forgive me...) That's all.
For example, you need to expand 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 understand perfectly well that 12 and 3 4 the same thing. The essence of the number 12 from transformation hasn't changed.
Is it possible to decompose 12 differently? Easily!
12=3·4=2·6=3·2·2=0.5·24=........
The decomposition options are endless.
Factoring numbers is a useful thing. It helps a lot, for example, when working with roots. But factoring algebraic expressions is not only useful, it is necessary! Just for example:
Simplify:
Those who do not know how to factor an expression rest on the sidelines. Those who know how - simplify and get:
The effect is amazing, right?) By the way, the solution is quite simple. You'll see for yourself below. Or, for example, this task:
Solve the equation:
x 5 - x 4 = 0
It is decided in the mind, by the way. Using factorization. We will solve this example below. Answer: x 1 = 0; x 2 = 1.
Or, the same thing, but for the older ones):
Solve the equation:
In these examples I showed main purpose factorization: simplifying fractional expressions and solving some types of equations. Here's a rule of thumb to remember:
If we have a scary fractional expression in front of us, we can try factoring 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 there is zero, and on the left - I don’t understand what, we can try to factorize the left side. Sometimes it helps).
Basic methods of factorization.
Here they are, the most popular methods:
4. Expansion of a quadratic trinomial.
These methods must be remembered. Exactly 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... So let’s start in order.)
1. Taking the common factor out of brackets.
A simple and reliable way. Nothing bad comes from him! It happens either well or not at all.) That’s why he comes first. Let's figure it out.
Everyone knows (I believe!) the rule:
a(b+c) = ab+ac
Or, more general view:
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 taking the common factor out of brackets.
On the left side A - common multiplier for all terms. Multiplied by everything that exists). On the right is the most A is already located outside the brackets.
Practical Application Let's look at the method using examples. At first the option is simple, even primitive.) But on this option I will note ( green) Very important points for any factorization.
Factorize:
ah+9x
Which general does the multiplier appear in both terms? X, of course! We will take it out of the brackets. Let's do this. We immediately write X outside the brackets:
ax+9x=x(
And in parentheses we write the result of division each term on this very X. In order:
That's it. Of course, there is no need to describe it in such detail, this is done in the mind. But it is advisable to understand what’s what). We record in memory:
We write the common factor outside the brackets. In parentheses we write the results of dividing all terms by this common factor. In order.
So we have expanded the expression ah+9x by 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 was not factorized! Because in addition to multiplication, this expression also contained addition, the “+” sign! And in expression x(a+9) There is nothing but multiplication!
How so!? - I hear the indignant voice of the people - And in brackets!?)
Yes, there is addition inside the parentheses. 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 entirely, as with one letter. In this sense, in the expression x(a+9) There is nothing except multiplication. This is the whole point of factorization.
By the way, is it possible to somehow check whether we did everything correctly? Easily! It’s enough to multiply back what you put out (x) by brackets and see if it worked original expression? If it works, everything is great!)
x(a+9)=ax+9x
It worked.)
There are no problems 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.
Factorize:
3ax+9x
We are looking for a common factor. Well, everything is clear with X, it can be taken out. Is there more general factor? Yes! This is a three. You can write the expression like this:
3ax+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.
What happens if you take it out 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 out everything to the limit while there is an opportunity. Here in brackets there is an opportunity to put out a three. It will turn out:
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 factor.
Shall we continue the fun?)
Factor the expression:
3akh+9х-8а-24
What will we take away? Three, X? Nope... You can't. I remind you that you can only take out general multiplier that is in all terms of the expression. That's why he general. There is no such multiplier here... What, you don’t have to expand it!? Well, yes, we were so happy... Meet:
2. Grouping.
Actually, it’s hard to name the group in an independent way factorization. It's more of a way to get out complex example.) We need to group the terms so that everything works out. This can only be shown by example. So, we have the expression:
3akh+9х-8а-24
It can be seen that there are some common letters and numbers. But... General there is no multiplier to be in all terms. Let's not lose heart and break the expression into pieces. Let's group. So that each piece has a common factor, there is something to take away. How do we break it? Yes, we just put parentheses.
Let me remind you that parentheses can be placed anywhere and however you want. Just the essence of the example hasn't changed. For example, you can do this:
3akh+9х-8а-24=(3ах+9х)-(8а+24)
Please pay attention to the second brackets! They are preceded by a minus sign, and 8a And 24 turned positive! If, to check, we open the brackets back, the signs will change, and we get original expression. Those. the essence of the expression from the brackets has not changed.
But if you just inserted parentheses without taking into account the change of sign, for example, like this:
3akh+9х-8а-24=(3ax+9x) -(8a-24 )
it would be a mistake. On the right - already other expression. Open the brackets and everything will become visible. You don’t have to decide further, yes...)
But let's return to factorization. Let's look at the first brackets (3ax+9x) and we think, is there anything we can take out? Well, we solved this example above, we can take it 3x:
(3ax+9x)=3x(a+3)
Let's study the second brackets, we can add an eight there:
(8a+24)=8(a+3)
Our entire expression will be:
(3ax+9x)-(8a+24)=3x(a+3)-8(a+3)
Factored? No. The result of decomposition should be only multiplication but with us the minus sign spoils everything. But... Both terms have a common factor! This (a+3). It was not for nothing that I said that the entire brackets are, as it were, one letter. This means that these brackets can be taken out of brackets. Yes, that's exactly what it sounds like.)
We do as described above. We 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)
All! There is nothing on the right except multiplication! This means that factorization has been completed successfully!) Here it is:
3ax+9x-8a-24=(a+3)(3x-8)
Let us briefly repeat the essence of the group.
If the expression does not general multiplier for everyone terms, we break the expression into brackets so that inside the brackets the common factor was. We take it out and see what happens. If you are lucky and there are absolutely identical expressions left in the brackets, we move these brackets out of brackets.
I will add that grouping is a creative process). It doesn't always work out the first time. It's OK. Sometimes you have to swap terms and consider different options groups until a successful one is found. The main thing here is not to lose heart!)
Examples.
Now, having enriched yourself with knowledge, you can tricky examples decide.) At the beginning of the lesson there were three of these...
Simplify:
In essence, we have already solved this example. Unbeknownst to ourselves.) I remind you: if we are given a terrible fraction, we try to factor the numerator and denominator. Other simplification options just no.
Well, the denominator here is not expanded, but the numerator... We have already expanded the numerator during the lesson! Like this:
3ax+9x-8a-24=(a+3)(3x-8)
We write the result of the expansion into the numerator of the fraction:
According to the rule of reducing fractions (the main property of a fraction), we can divide (at the same time!) the numerator and denominator by the same number, or expression. Fraction from this doesn't change. So we divide the numerator and denominator by the expression (3x-8). And here and there we will get ones. The final result of the simplification:
I would like to especially emphasize: reducing 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 for simplification. Of course, if the expressions different, then nothing will be reduced. It will happen. But factorization gives a chance. This chance without decomposition is simply not there.
Example with equation:
Solve the equation:
x 5 - x 4 = 0
We take out the common factor x 4 out of brackets. We get:
x 4 (x-1)=0
We realize that the product of factors is equal to zero then and only then, when any of them is 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 such an equality, the second factor does not concern us. Anyone can be, but in the end it will still be zero. What number to the fourth power does zero give? Only zero! And no other... Therefore:
We figured out the first factor and found one root. Let's look at the second factor. Now we don’t care about the first factor anymore.):
Here we found a solution: x 1 = 0; x 2 = 1. Any of these roots fits our equation.
Very important note. Please note that we solved the equation piece by piece! Each factor was equal to zero, regardless of other factors. By the way, if in such an equation there are not two factors, like ours, but three, five, as many as you like, we will solve exactly the same. Piece by piece. For example:
(x-1)(x+5)(x-3)(x+2)=0
Anyone who opens the brackets and multiplies everything will be stuck on this equation forever.) A correct student will immediately see that there is nothing on the left except multiplication, and zero on the right. And he will begin (in his mind!) to equate all brackets in order to zero. And he will get (in 10 seconds!) the correct solution: x 1 = 1; x 2 = -5; x 3 = 3; x 4 = -2.
Cool, right?) Such an elegant solution is possible if the left side of the equation factorized. Got the hint?)
Well, one last example, for the older ones):
Solve the equation:
It’s 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 the letters! Factoring works throughout mathematics.
We take out the common factor lg 4 x out of brackets. We get:
log 4 x=0
This is one root. Let's look at the second factor.
Here is the final answer: x 1 = 1; x 2 = 10.
I hope you've realized the power of factoring in simplifying fractions and solving equations.)
In this lesson we learned about common factoring and grouping. It remains to understand the formulas for abbreviated multiplication and the quadratic trinomial.
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You can get acquainted with functions and derivatives.
Factoring an equation is the process of finding those terms or expressions that, when multiplied, lead to the initial equation. Factoring is a useful skill for solving basic algebraic problems, and becomes almost essential when working with quadratic equations and other polynomials. Factoring is used to simplify algebraic equations to make them easier to solve. Factoring can help you eliminate certain possible answers faster than you would by solving an equation by hand.
Steps
Factoring numbers and basic algebraic expressions
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Factoring numbers. The concept of factoring is simple, but in practice, factoring can be challenging (if a complex equation is given). Therefore, first, let's look at the concept of factorization using numbers as an example, and continue with simple equations, and then move on to complex equations. Multipliers given number- These are numbers that, when multiplied, give the original number. For example, the factors of the number 12 are the numbers: 1, 12, 2, 6, 3, 4, since 1*12=12, 2*6=12, 3*4=12.
- Likewise, you can think of the factors of a number as its divisors, that is, the numbers that the number is divisible by.
- Find all the factors of the number 60. We often use the number 60 (for example, 60 minutes in an hour, 60 seconds in a minute, etc.) and this number has quite large number multipliers.
- 60 multipliers: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30 and 60.
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Remember: terms of an expression containing a coefficient (number) and a variable can also be factorized. To do this, find the coefficient factors of the variable. Knowing how to factor the terms of equations, you can easily simplify this equation.
- For example, the term 12x can be written as the product of 12 and x. You can also write 12x as 3(4x), 2(6x), etc., breaking down 12 into the factors that work best for you.
- You can deal 12x multiple times in a row. In other words, you shouldn't stop at 3(4x) or 2(6x); continue the expansion: 3(2(2x)) or 2(3(2x)) (obviously 3(4x)=3(2(2x)), etc.)
- For example, the term 12x can be written as the product of 12 and x. You can also write 12x as 3(4x), 2(6x), etc., breaking down 12 into the factors that work best for you.
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Apply the distributive property of multiplication to factor algebraic equations. Knowing how to factor numbers and terms of expressions (coefficients with variables), you can simplify simple algebraic equations, finding the common factor of the number and term of the expression. Typically, to simplify an equation, you need to find the greatest common factor (GCD). This simplification is possible due to the distributive property of multiplication: for any numbers a, b, c, the equality a(b+c) = ab+ac is true.
- Example. Factor the equation 12x + 6. First, find the gcd of 12x and 6. 6 is the largest number, which divides both 12x and 6, so you can factor this equation into: 6(2x+1).
- This process is also true for equations that have negative and fractional terms. For example, x/2+4 can be factored into 1/2(x+8); for example, -7x+(-21) can be factored into -7(x+3).
Factoring Quadratic Equations
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Make sure the equation is given in quadratic form (ax 2 + bx + c = 0). Quadratic equations have the form: ax 2 + bx + c = 0, where a, b, c are numerical coefficients other than 0. If you are given an equation with one variable (x) and in this equation there are one or more terms with a second-order variable , you can move all the terms of the equation to one side of the equation and set it equal to zero.
- For example, given the equation: 5x 2 + 7x - 9 = 4x 2 + x – 18. This can be converted into the equation x 2 + 6x + 9 = 0, which is a quadratic equation.
- Equations with variable x of large orders, for example, x 3, x 4, etc. are not quadratic equations. These are cubic equations, fourth-order equations, and so on (unless such equations can be simplified to quadratic equations with the variable x raised to the power of 2).
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Quadratic equations, where a = 1, are expanded into (x+d)(x+e), where d*e=c and d+e=b. If the quadratic equation given to you has the form: x 2 + bx + c = 0 (that is, the coefficient of x 2 is 1), then such an equation can (but is not guaranteed) be expanded into the above factors. To do this, you need to find two numbers that, when multiplied, give “c”, and when added, “b”. Once you find these two numbers (d and e), substitute them into the following expression: (x+d)(x+e), which, when opening the parentheses, leads to the original equation.
- For example, given a quadratic equation x 2 + 5x + 6 = 0. 3*2=6 and 3+2=5, so you can factor this equation into (x+3)(x+2).
- For negative terms, make the following minor changes to the factorization process:
- If a quadratic equation has the form x 2 -bx+c, then it expands into: (x-_)(x-_).
- If a quadratic equation has the form x 2 -bx-c, then it expands into: (x+_)(x-_).
- Note: spaces can be replaced with fractions or decimal numbers. For example, the equation x 2 + (21/2)x + 5 = 0 is expanded into (x+10)(x+1/2).
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Factorization by trial and error. Simple quadratic equations can be factored by simply plugging the numbers into the possible solutions until you find the right decision. If the equation has the form ax 2 +bx+c, where a>1, possible solutions are written in the form (dx +/- _)(ex +/- _), where d and e are non-zero numerical coefficients, which when multiplied give a. Either d or e (or both coefficients) can be equal to 1. If both coefficients are equal to 1, then use the method described above.
- For example, given the equation 3x 2 - 8x + 4. Here 3 has only two factors (3 and 1), so possible solutions are written as (3x +/- _)(x +/- _). In this case, substituting -2 for spaces, you will find the correct answer: -2*3x=-6x and -2*x=-2x; - 6x+(-2x)=-8x and -2*-2=4, that is, such an expansion when opening the brackets will lead to the terms of the original equation.
Considering the multiplication of polynomials, we remembered several formulas, namely: formulas for (a + b)², for (a – b)², for (a + b) (a – b), for (a + b)³ and for (a – b)³.
If a given polynomial turns out to coincide with one of these formulas, then it will be possible to factorize it. For example, the polynomial a² – 2ab + b², we know, is equal to (a – b)² [or (a – b) · (a – b), i.e. we managed to factor a² – 2ab + b² into 2 factors]; Also
Let's look at the second of these examples. We see that the polynomial given here fits the formula obtained by squaring the difference of two numbers (the square of the first number, minus the product of two by the first number and the second, plus the square of the second number): x 6 is the square of the first number, and therefore , the first number itself is x 3 , the square of the second number is the last term of the given polynomial, i.e. 1, the second number itself is, therefore, also 1; the product of two by the first number and the second is the term –2x 3, because 2x 3 = 2 x 3 1. Therefore, our polynomial was obtained by squaring the difference of the numbers x 3 and 1, i.e. it is equal to (x 3 – 1) 2. Let's look at another 4th example. We see that this polynomial a 2 b 2 – 25 can be considered as the difference of the squares of two numbers, namely the square of the first number is a 2 b 2, therefore, the first number itself is ab, the square of the second number is 25, why is the second number itself is 5. Therefore, our polynomial can be considered as obtained from multiplying the sum of two numbers by their difference, i.e.
(ab + 5) (ab – 5).
Sometimes it happens that in a given polynomial the terms are not arranged in the order to which we are accustomed, for example.
9a 2 + b 2 + 6ab – mentally we can rearrange the second and third terms, and then it will become clear to us that our trinomial = (3a + b) 2.
... (we mentally rearrange the first and second terms).
25a 6 + 1 – 10x 3 = (5x 3 – 1) 2, etc.
Let's consider another polynomial
a 2 + 2ab + 4b 2 .
We see that its first term is the square of the number a and the third term is the square of the number 2b, but the second term is not the product of two by the first number and the second - such a product would be equal to 2 a 2b = 4ab. Therefore, it is impossible to apply the formula for the square of the sum of two numbers to this polynomial. If someone wrote that a 2 + 2ab + 4b 2 = (a + 2b) 2, then this would be incorrect - one must carefully consider all the terms of the polynomial before applying factorization to it using formulas.
40. A combination of both techniques. Sometimes, when factoring polynomials, you have to combine both the technique of taking the common factor out of brackets and the technique of using formulas. Here are examples:
1. 2a 3 – 2ab 2. Let’s first take the common factor 2a out of brackets, and we get 2a (a 2 – b 2). The factor a 2 – b 2, in turn, is decomposed according to the formula into factors (a + b) and (a – b).
Sometimes you have to use the formula decomposition technique multiple times:
1. a 4 – b 4 = (a 2 + b 2) (a 2 – b 2)
We see that the first factor a 2 + b 2 does not fit any of the familiar formulas; Moreover, recalling special cases of division (item 37), we will establish that a 2 + b 2 (the sum of the squares of two numbers) cannot be factorized at all. The second of the resulting factors a 2 – b 2 (the difference by the square of two numbers) is decomposed into factors (a + b) and (a – b). So,
41. Application of special cases of division. Based on paragraph 37, we can immediately write that, for example,
Factoring a polynomial. Part 1
Factorization is a universal technique that helps solve complex equations and inequalities. The first thought that should come to mind when solving equations and inequalities in which there is a zero on the right side is to try to factor the left side.
Let's list the main ways to factor a polynomial:
- putting the common factor out of brackets
- using abbreviated multiplication formulas
- using the formula for factoring a quadratic trinomial
- grouping method
- dividing a polynomial by a binomial
- method of uncertain coefficients
In this article we will dwell in detail on the first three methods, and we will consider the rest in subsequent articles.
1. Taking the common factor out of brackets.
To take the common factor out of brackets, you must first find it. Common multiplier factor equal to the greatest common divisor of all coefficients.
Letter part the common factor is equal to the product of the expressions included in each term with the smallest exponent.
The scheme for assigning a common multiplier looks like this:
Attention!
The number of terms in brackets is equal to the number of terms in the original expression. If one of the terms coincides with the common factor, then when dividing it by the common factor, we get one.
Example 1.
Factor the polynomial:
Let's take the common factor out of brackets. To do this, we will first find it.
1. Find the greatest common divisor of all coefficients of the polynomial, i.e. numbers 20, 35 and 15. It is equal to 5.
2. We establish that the variable is contained in all terms, and the smallest of its exponents is equal to 2. The variable is contained in all terms, and the smallest of its exponents is 3.
The variable is contained only in the second term, so it is not part of the common factor.
So the total factor is
3. We take the multiplier out of brackets using the diagram given above:
Example 2. Solve the equation:
Solution. Let's factorize the left side of the equation. Let's take the factor out of brackets:
So we get the equation 
Let's equate each factor to zero:
We get - the root of the first equation.
Roots:
Answer: -1, 2, 4
2. Factorization using abbreviated multiplication formulas.
If the number of terms in the polynomial we are going to factor is less than or equal to three, then we try to apply the abbreviated multiplication formulas.
1. If the polynomial isdifference of two terms, then we try to apply square difference formula:
or difference of cubes formula:
Here are the letters and denote a number or algebraic expression.
2. If a polynomial is the sum of two terms, then perhaps it can be factored using sum of cubes formulas:
3. If a polynomial consists of three terms, then we try to apply square sum formula:
or squared difference formula:
Or we try to factorize by formula for factoring a quadratic trinomial:
Here and are the roots of the quadratic equation 
Example 3.Factor the expression:
Solution. We have before us the sum of two terms. Let's try to apply the formula for the sum of cubes. To do this, you first need to represent each term as a cube of some expression, and then apply the formula for the sum of the cubes:
Example 4. Factor the expression: 
Decision. Here we have the difference of the squares of two expressions. First expression: , second expression:
Let's apply the formula for the difference of squares:
Let's open the brackets and add similar terms, we get: