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 initial 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 initial 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! it (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 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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By the way, I have a couple more interesting sites for you.)
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What to do if, in the process of solving a problem from the Unified State Examination or at the entrance exam in mathematics, you received a polynomial that cannot be factored by the standard methods that you learned at school? In this article, a math tutor will talk about one effective way, the study of which is outside the scope of the school curriculum, but with which it will not be difficult to factor a polynomial. Read this article to the end and watch the attached video tutorial. The knowledge you gain will help you in the exam.
Factoring a polynomial by the division method
In the event that you received a polynomial greater than the second degree and were able to guess the value of a variable at which this polynomial becomes equal to zero (for example, this value is equal to), know! This polynomial can be divided without remainder by .
For example, it is easy to see that a fourth degree polynomial vanishes at . This means that it can be divided by without a remainder, thus obtaining a polynomial of the third degree (less than one). That is, put it in the form:
where A, B, C and D- some numbers. Let's expand the brackets:
Since the coefficients at the same powers must be the same, we get:
So we got:
Move on. It is enough to sort through several small integers to see that the polynomial of the third degree is again divisible by . This results in a polynomial of the second degree (less than one). Then we move on to a new record:
where E, F and G- some numbers. Opening the brackets again, we arrive at the following expression:
Again, from the condition of equality of the coefficients at the same powers, we obtain:
Then we get:
That is, the original polynomial can be factored as follows:
In principle, if desired, using the difference of squares formula, the result can also be represented in the following form:
Here is such a simple and effective way to factorize polynomials. Remember it, it may come in handy in an exam or math olympiad. Check if you have learned how to use this method. Try to solve the following problem yourself.
Factorize a polynomial:
Write your answers in the comments.
Prepared by Sergey Valerievich
In order to factorize, it is necessary to simplify the expressions. This is necessary in order to be able to further reduce. The decomposition of a polynomial makes sense when its degree is not lower than the second. A polynomial with the first degree is called linear.
Yandex.RTB R-A-339285-1
The article will reveal all the concepts of decomposition, theoretical foundations and methods for factoring a polynomial.
Theory
Theorem 1When any polynomial with degree n having the form P n x = a n x n + a n - 1 x n - 1 + . . . + a 1 x + a 0 , are represented as a product with a constant factor with the highest degree a n and n linear factors (x - x i) , i = 1 , 2 , … , n , then P n (x) = a n (x - x n) (x - x n - 1) . . . · (x - x 1) , where x i , i = 1 , 2 , … , n - these are the roots of the polynomial.
The theorem is intended for roots of complex type x i , i = 1 , 2 , … , n and for complex coefficients a k , k = 0 , 1 , 2 , … , n . This is the basis of any decomposition.
When coefficients of the form a k , k = 0 , 1 , 2 , … , n are real numbers, then complex roots will occur in conjugate pairs. For example, the roots x 1 and x 2 related to a polynomial of the form P n x = a n x n + a n - 1 x n - 1 + . . . + a 1 x + a 0 are considered complex conjugate, then the other roots are real, hence we get that the polynomial takes the form P n (x) = a n (x - x n) (x - x n - 1) · . . . (x - x 3) x 2 + p x + q, where x 2 + p x + q = (x - x 1) (x - x 2) .
Comment
The roots of a polynomial can be repeated. Consider the proof of the theorem of algebra, the consequences of Bezout's theorem.
Fundamental theorem of algebra
Theorem 2Any polynomial with degree n has at least one root.
Bezout's theorem
After dividing a polynomial of the form P n x = a n x n + a n - 1 x n - 1 + . . . + a 1 x + a 0 on (x - s) , then we get the remainder, which is equal to the polynomial at the point s , then we get
P n x = a n x n + a n - 1 x n - 1 + . . . + a 1 x + a 0 = (x - s) Q n - 1 (x) + P n (s) , where Q n - 1 (x) is a polynomial with degree n - 1 .
Corollary from Bezout's theorem
When the root of the polynomial P n (x) is considered to be s , then P n x = a n x n + a n - 1 x n - 1 + . . . + a 1 x + a 0 = (x - s) Q n - 1 (x) . This corollary is sufficient when used to describe the solution.
Factorization of a square trinomial
A square trinomial of the form a x 2 + b x + c can be factored into linear factors. then we get that a x 2 + b x + c \u003d a (x - x 1) (x - x 2) , where x 1 and x 2 are roots (complex or real).
This shows that the decomposition itself reduces to solving the quadratic equation later.
Example 1
Factorize a square trinomial.
Solution
It is necessary to find the roots of the equation 4 x 2 - 5 x + 1 = 0. To do this, you need to find the value of the discriminant according to the formula, then we get D \u003d (- 5) 2 - 4 4 1 \u003d 9. Hence we have that
x 1 = 5 - 9 2 4 = 1 4 x 2 = 5 + 9 2 4 = 1
From here we get that 4 x 2 - 5 x + 1 = 4 x - 1 4 x - 1.
To perform the check, you need to open the brackets. Then we get an expression of the form:
4 x - 1 4 x - 1 = 4 x 2 - x - 1 4 x + 1 4 = 4 x 2 - 5 x + 1
After verification, we arrive at the original expression. That is, we can conclude that the expansion is correct.
Example 2
Factorize a square trinomial of the form 3 x 2 - 7 x - 11 .
Solution
We get that it is necessary to calculate the resulting quadratic equation of the form 3 x 2 - 7 x - 11 = 0.
To find the roots, you need to determine the value of the discriminant. We get that
3 x 2 - 7 x - 11 = 0 D = (- 7) 2 - 4 3 (- 11) = 181 x 1 = 7 + D 2 3 = 7 + 181 6 x 2 = 7 - D 2 3 = 7 - 1816
From here we get that 3 x 2 - 7 x - 11 = 3 x - 7 + 181 6 x - 7 - 181 6 .
Example 3
Factorize the polynomial 2 x 2 + 1.
Solution
Now you need to solve the quadratic equation 2 x 2 + 1 = 0 and find its roots. We get that
2 x 2 + 1 = 0 x 2 = - 1 2 x 1 = - 1 2 = 1 2 i x 2 = - 1 2 = - 1 2 i
These roots are called complex conjugate, which means that the decomposition itself can be represented as 2 x 2 + 1 = 2 x - 1 2 · i x + 1 2 · i.
Example 4
Expand the square trinomial x 2 + 1 3 x + 1 .
Solution
First you need to solve a quadratic equation of the form x 2 + 1 3 x + 1 = 0 and find its roots.
x 2 + 1 3 x + 1 = 0 D = 1 3 2 - 4 1 1 = - 35 9 x 1 = - 1 3 + D 2 1 = - 1 3 + 35 3 i 2 = - 1 + 35 i 6 = - 1 6 + 35 6 i x 2 = - 1 3 - D 2 1 = - 1 3 - 35 3 i 2 = - 1 - 35 i 6 = - 1 6 - 35 6 i
Having obtained the roots, we write
x 2 + 1 3 x + 1 = x - - 1 6 + 35 6 i x - - 1 6 - 35 6 i = = x + 1 6 - 35 6 i x + 1 6 + 35 6 i
Comment
If the value of the discriminant is negative, then the polynomials will remain second-order polynomials. Hence it follows that we will not decompose them into linear factors.
Methods for factoring a polynomial of degree higher than the second
The decomposition assumes a universal method. Most of all cases are based on a corollary of Bezout's theorem. To do this, you need to select the value of the root x 1 and lower its degree by dividing by a polynomial by 1 by dividing by (x - x 1) . The resulting polynomial needs to find the root x 2 , and the search process is cyclical until we get a complete decomposition.
If the root is not found, then other methods of factorization are used: grouping, additional terms. This topic assumes the solution of equations with higher powers and integer coefficients.
Taking the common factor out of brackets
Consider the case when the free term is equal to zero, then the form of the polynomial becomes P n (x) = a n x n + a n - 1 x n - 1 + . . . + a 1 x .
It can be seen that the root of such a polynomial will be equal to x 1 \u003d 0, then you can represent the polynomial in the form of an expression P n (x) \u003d a n x n + a n - 1 x n - 1 +. . . + a 1 x = = x (a n x n - 1 + a n - 1 x n - 2 + . . . + a 1)
This method is considered to be taking the common factor out of brackets.
Example 5
Factorize the third degree polynomial 4 x 3 + 8 x 2 - x.
Solution
We see that x 1 \u003d 0 is the root of the given polynomial, then we can bracket x out of the entire expression. We get:
4 x 3 + 8 x 2 - x = x (4 x 2 + 8 x - 1)
Let's move on to finding the roots of the square trinomial 4 x 2 + 8 x - 1. Let's find the discriminant and the roots:
D = 8 2 - 4 4 (- 1) = 80 x 1 = - 8 + D 2 4 = - 1 + 5 2 x 2 = - 8 - D 2 4 = - 1 - 5 2
Then it follows that
4 x 3 + 8 x 2 - x = x 4 x 2 + 8 x - 1 = = 4 x x - - 1 + 5 2 x - - 1 - 5 2 = = 4 x x + 1 - 5 2 x + 1 + 5 2
To begin with, let's take for consideration a decomposition method containing integer coefficients of the form P n (x) = x n + a n - 1 x n - 1 + . . . + a 1 x + a 0 , where the coefficient of the highest power is 1 .
When the polynomial has integer roots, then they are considered divisors of the free term.
Example 6
Expand the expression f (x) = x 4 + 3 x 3 - x 2 - 9 x - 18.
Solution
Consider whether there are integer roots. It is necessary to write out the divisors of the number - 18. We get that ± 1 , ± 2 , ± 3 , ± 6 , ± 9 , ± 18 . It follows that this polynomial has integer roots. You can check according to the Horner scheme. It is very convenient and allows you to quickly obtain the expansion coefficients of a polynomial:
It follows that x \u003d 2 and x \u003d - 3 are the roots of the original polynomial, which can be represented as a product of the form:
f (x) = x 4 + 3 x 3 - x 2 - 9 x - 18 = (x - 2) (x 3 + 5 x 2 + 9 x + 9) = = (x - 2) (x + 3) (x 2 + 2 x + 3)
We turn to the decomposition of a square trinomial of the form x 2 + 2 x + 3 .
Since the discriminant is negative, it means that there are no real roots.
Answer: f (x) \u003d x 4 + 3 x 3 - x 2 - 9 x - 18 \u003d (x - 2) (x + 3) (x 2 + 2 x + 3)
Comment
It is allowed to use root selection and division of a polynomial by a polynomial instead of Horner's scheme. Let us proceed to consider the expansion of a polynomial containing integer coefficients of the form P n (x) = x n + a n - 1 x n - 1 + . . . + a 1 x + a 0 , the highest of which does not equal one.
This case takes place for fractional rational fractions.
Example 7
Factorize f (x) = 2 x 3 + 19 x 2 + 41 x + 15 .
Solution
It is necessary to change the variable y = 2 x , one should pass to a polynomial with coefficients equal to 1 at the highest degree. You need to start by multiplying the expression by 4 . We get that
4 f (x) = 2 3 x 3 + 19 2 2 x 2 + 82 2 x + 60 = = y 3 + 19 y 2 + 82 y + 60 = g (y)
When the resulting function of the form g (y) \u003d y 3 + 19 y 2 + 82 y + 60 has integer roots, then their finding is among the divisors of the free term. The entry will look like:
± 1 , ± 2 , ± 3 , ± 4 , ± 5 , ± 6 , ± 10 , ± 12 , ± 15 , ± 20 , ± 30 , ± 60
Let's proceed to the calculation of the function g (y) at these points in order to get zero as a result. We get that
g (1) = 1 3 + 19 1 2 + 82 1 + 60 = 162 g (- 1) = (- 1) 3 + 19 (- 1) 2 + 82 (- 1) + 60 = - 4 g (2) = 2 3 + 19 2 2 + 82 2 + 60 = 308 g (- 2) = (- 2) 3 + 19 (- 2) 2 + 82 (- 2) + 60 = - 36 g (3) = 3 3 + 19 3 2 + 82 3 + 60 = 504 g (- 3) = (- 3) 3 + 19 (- 3) 2 + 82 (- 3) + 60 = - 42 g (4) = 4 3 + 19 4 2 + 82 4 + 60 = 756 g (- 4) = (- 4) 3 + 19 (- 4) 2 + 82 (- 4) + 60 = - 28 g (5) = 5 3 + 19 5 2 + 82 5 + 60 = 1070 g (- 5) = (- 5) 3 + 19 (- 5) 2 + 82 (- 5) + 60
We get that y \u003d - 5 is the root of the equation of the form y 3 + 19 y 2 + 82 y + 60, which means that x \u003d y 2 \u003d - 5 2 is the root of the original function.
Example 8
It is necessary to divide by a column 2 x 3 + 19 x 2 + 41 x + 15 by x + 5 2.
Solution
We write and get:
2 x 3 + 19 x 2 + 41 x + 15 = x + 5 2 (2 x 2 + 14 x + 6) = = 2 x + 5 2 (x 2 + 7 x + 3)
Checking the divisors will take a lot of time, so it is more profitable to take the factorization of the resulting square trinomial of the form x 2 + 7 x + 3. By equating to zero, we find the discriminant.
x 2 + 7 x + 3 = 0 D = 7 2 - 4 1 3 = 37 x 1 = - 7 + 37 2 x 2 = - 7 - 37 2 ⇒ x 2 + 7 x + 3 = x + 7 2 - 37 2 x + 7 2 + 37 2
Hence it follows that
2 x 3 + 19 x 2 + 41 x + 15 = 2 x + 5 2 x 2 + 7 x + 3 = = 2 x + 5 2 x + 7 2 - 37 2 x + 7 2 + 37 2
Artificial tricks when factoring a polynomial
Rational roots are not inherent in all polynomials. To do this, you need to use special methods to find factors. But not all polynomials can be decomposed or represented as a product.
Grouping method
There are cases when you can group the terms of a polynomial to find a common factor and take it out of brackets.
Example 9
Factorize the polynomial x 4 + 4 x 3 - x 2 - 8 x - 2.
Solution
Because the coefficients are integers, then the roots can presumably also be integers. To check, we take the values 1 , - 1 , 2 and - 2 in order to calculate the value of the polynomial at these points. We get that
1 4 + 4 1 3 - 1 2 - 8 1 - 2 = - 6 ≠ 0 (- 1) 4 + 4 (- 1) 3 - (- 1) 2 - 8 (- 1) - 2 = 2 ≠ 0 2 4 + 4 2 3 - 2 2 - 8 2 - 2 = 26 ≠ 0 (- 2) 4 + 4 (- 2) 3 - (- 2) 2 - 8 (- 2) - 2 = - 6 ≠ 0
This shows that there are no roots, it is necessary to use a different method of decomposition and solution.
Grouping is required:
x 4 + 4 x 3 - x 2 - 8 x - 2 = x 4 + 4 x 3 - 2 x 2 + x 2 - 8 x - 2 = = (x 4 - 2 x 2) + (4 x 3 - 8 x) + x 2 - 2 = = x 2 (x 2 - 2) + 4 x (x 2 - 2) + x 2 - 2 = = (x 2 - 2) (x 2 + 4 x + 1)
After grouping the original polynomial, it is necessary to represent it as a product of two square trinomials. To do this, we need to factorize. we get that
x 2 - 2 = 0 x 2 = 2 x 1 = 2 x 2 = - 2 ⇒ x 2 - 2 = x - 2 x + 2 x 2 + 4 x + 1 = 0 D = 4 2 - 4 1 1 = 12 x 1 = - 4 - D 2 1 = - 2 - 3 x 2 = - 4 - D 2 1 = - 2 - 3 ⇒ x 2 + 4 x + 1 = x + 2 - 3 x + 2 + 3
x 4 + 4 x 3 - x 2 - 8 x - 2 = x 2 - 2 x 2 + 4 x + 1 = = x - 2 x + 2 x + 2 - 3 x + 2 + 3
Comment
The simplicity of grouping does not mean that it is easy enough to choose terms. There is no definite way to solve it, therefore it is necessary to use special theorems and rules.
Example 10
Factorize the polynomial x 4 + 3 x 3 - x 2 - 4 x + 2.
Solution
The given polynomial has no integer roots. The terms should be grouped. We get that
x 4 + 3 x 3 - x 2 - 4 x + 2 = = (x 4 + x 3) + (2 x 3 + 2 x 2) + (- 2 x 2 - 2 x) - x 2 - 2 x + 2 = = x 2 (x 2 + x) + 2 x (x 2 + x) - 2 (x 2 + x) - (x 2 + 2 x - 2) = = (x 2 + x) (x 2 + 2 x - 2) - (x 2 + 2 x - 2) = (x 2 + x - 1) (x 2 + 2 x - 2)
After factoring, we get that
x 4 + 3 x 3 - x 2 - 4 x + 2 = x 2 + x - 1 x 2 + 2 x - 2 = = x + 1 + 3 x + 1 - 3 x + 1 2 + 5 2 x + 1 2 - 5 2
Using abbreviated multiplication and Newton's binomial formulas to factorize a polynomial
Appearance often does not always make it clear which way to use during decomposition. After the transformations have been made, you can build a line consisting of Pascal's triangle, otherwise they are called Newton's binomial.
Example 11
Factorize the polynomial x 4 + 4 x 3 + 6 x 2 + 4 x - 2.
Solution
It is necessary to convert the expression to the form
x 4 + 4 x 3 + 6 x 2 + 4 x - 2 = x 4 + 4 x 3 + 6 x 2 + 4 x + 1 - 3
The sequence of coefficients of the sum in brackets is indicated by the expression x + 1 4 .
So we have x 4 + 4 x 3 + 6 x 2 + 4 x - 2 = x 4 + 4 x 3 + 6 x 2 + 4 x + 1 - 3 = x + 1 4 - 3 .
After applying the difference of squares, we get
x 4 + 4 x 3 + 6 x 2 + 4 x - 2 = x 4 + 4 x 3 + 6 x 2 + 4 x + 1 - 3 = x + 1 4 - 3 = = x + 1 4 - 3 = x + 1 2 - 3 x + 1 2 + 3
Consider the expression that is in the second parenthesis. It is clear that there are no horses there, so the formula for the difference of squares should be applied again. We get an expression like
x 4 + 4 x 3 + 6 x 2 + 4 x - 2 = x 4 + 4 x 3 + 6 x 2 + 4 x + 1 - 3 = x + 1 4 - 3 = = x + 1 4 - 3 = x + 1 2 - 3 x + 1 2 + 3 = = x + 1 - 3 4 x + 1 + 3 4 x 2 + 2 x + 1 + 3
Example 12
Factorize x 3 + 6 x 2 + 12 x + 6 .
Solution
Let's change the expression. We get that
x 3 + 6 x 2 + 12 x + 6 = x 3 + 3 2 x 2 + 3 2 2 x + 2 3 - 2 = (x + 2) 3 - 2
It is necessary to apply the formula for abbreviated multiplication of the difference of cubes. We get:
x 3 + 6 x 2 + 12 x + 6 = = (x + 2) 3 - 2 = = x + 2 - 2 3 x + 2 2 + 2 3 x + 2 + 4 3 = = x + 2 - 2 3 x 2 + x 2 + 2 3 + 4 + 2 2 3 + 4 3
A method for replacing a variable when factoring a polynomial
When changing a variable, the degree is reduced and the polynomial is factorized.
Example 13
Factorize a polynomial of the form x 6 + 5 x 3 + 6 .
Solution
By the condition, it is clear that it is necessary to make a replacement y = x 3 . We get:
x 6 + 5 x 3 + 6 = y = x 3 = y 2 + 5 y + 6
The roots of the resulting quadratic equation are y = - 2 and y = - 3, then
x 6 + 5 x 3 + 6 = y = x 3 = y 2 + 5 y + 6 = = y + 2 y + 3 = x 3 + 2 x 3 + 3
It is necessary to apply the formula for the abbreviated multiplication of the sum of cubes. We get expressions of the form:
x 6 + 5 x 3 + 6 = y = x 3 = y 2 + 5 y + 6 = = y + 2 y + 3 = x 3 + 2 x 3 + 3 = = x + 2 3 x 2 - 2 3 x + 4 3 x + 3 3 x 2 - 3 3 x + 9 3
That is, we have obtained the desired expansion.
The cases discussed above will help in considering and factoring a polynomial in various ways.
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In the general case, this task involves a creative approach, since there is no universal method for solving it. However, let's try to give a few hints.
In the vast majority of cases, the decomposition of a polynomial into factors is based on the consequence of the Bezout theorem, that is, the root is found or selected and the degree of the polynomial is reduced by one by dividing by. The resulting polynomial is searched for a root and the process is repeated until complete expansion.
If the root cannot be found, then specific decomposition methods are used: from grouping to introducing additional mutually exclusive terms.
Further presentation is based on the skills of solving equations of higher degrees with integer coefficients.
Bracketing the common factor.
Let's start with the simplest case, when the free term is equal to zero, that is, the polynomial has the form .
Obviously, the root of such a polynomial is , that is, the polynomial can be represented as .
This method is nothing but taking the common factor out of brackets.
Example.
Decompose a polynomial of the third degree into factors.
Solution.
It is obvious that is the root of the polynomial, that is, X can be bracketed:
Find the roots of a square trinomial 
In this way, 
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Factorization of a polynomial with rational roots.
First, consider the method of expanding a polynomial with integer coefficients of the form , the coefficient at the highest degree is equal to one.
In this case, if the polynomial has integer roots, then they are divisors of the free term.
Example.
Solution.
Let's check if there are integer roots. To do this, we write out the divisors of the number -18
: . That is, if the polynomial has integer roots, then they are among the numbers written out. Let's check these numbers sequentially according to Horner's scheme. Its convenience also lies in the fact that in the end we will also obtain the expansion coefficients of the polynomial: 
That is, x=2 and x=-3 are the roots of the original polynomial and it can be represented as a product:
It remains to expand the square trinomial.
The discriminant of this trinomial is negative, hence it has no real roots.
Answer:
Comment:
instead of Horner's scheme, one could use the selection of a root and the subsequent division of a polynomial by a polynomial.
Now consider the expansion of a polynomial with integer coefficients of the form , and the coefficient at the highest degree is not equal to one.
In this case, the polynomial can have fractionally rational roots.
Example.
Factorize the expression.
Solution.
By changing the variable y=2x, we pass to a polynomial with a coefficient equal to one at the highest degree. To do this, we first multiply the expression by 4
.
If the resulting function has integer roots, then they are among the divisors of the free term. Let's write them down:
Calculate sequentially the values of the function g(y) at these points until reaching zero. 
Factoring an equation is the process of finding terms or expressions that, when multiplied, lead to the initial equation. Factoring is a useful skill for solving basic algebraic problems, and becomes a practical necessity 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 rule out certain possible answers faster than you can by manually solving the equation.
Steps
Factorization of numbers and basic algebraic expressions
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Factorization of numbers. The concept of factoring is simple, but in practice factoring can be tricky (given a complex equation). So let's start with the concept of factoring using numbers as an example, continue with simple equations, and then move on to complex equations. The factors of a given number are the 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.
- Similarly, you can think of the factors of a number as its divisors, that is, the numbers that the given 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 a fairly large number of factors.
- 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 factored. To do this, find the multipliers of the coefficient at the variable. Knowing how to factorize the terms of the 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. by factoring 12 into the factors that work best for you.
- You can lay out 12x multiple times in a row. In other words, you shouldn't stop at 3(4x) or 2(6x); continue 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. by factoring 12 into the factors that work best for you.
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Apply the distributive property of multiplication to factorize algebraic equations. Knowing how to factorize numbers and terms of an expression (coefficients with variables), you can simplify simple algebraic equations by finding the common factor of a number and a term of an expression. Usually, to simplify the equation, you need to find the greatest common divisor (gcd). Such a 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 that 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 decomposed into 1/2(x+8); for example, -7x+(-21) can be decomposed into -7(x+3).
Factorization of quadratic equations
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Make sure the equation is in quadratic form (ax 2 + bx + c = 0). Quadratic equations are: 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 this equation has 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 equate it to zero.
- For example, given the equation: 5x 2 + 7x - 9 = 4x 2 + x - 18. It can be converted to the equation x 2 + 6x + 9 = 0, which is a quadratic equation.
- Equations with a 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 (only if such equations cannot be simplified to quadratic equations with the variable x to the power of 2).
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Quadratic equations, where a \u003d 1, are decomposed into (x + d) (x + e), where d * e \u003d c and d + e \u003d b. If the quadratic equation given to you has the form: x 2 + bx + c \u003d 0 (that is, the coefficient at x 2 is equal to 1), then such an equation can (but not guaranteed) be decomposed 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 the brackets are opened, leads to the original equation.
- For example, given the quadratic equation x 2 + 5x + 6 = 0. 3*2=6 and 3+2=5, so you can expand the equation into (x+3)(x+2).
- For negative terms, make the following minor changes to the factorization process:
- If the quadratic equation has the form x 2 -bx + c, then it decomposes into: (x-_) (x-_).
- If the quadratic equation has the form x 2 -bx-c, then it decomposes into: (x + _) (x-_).
- Note: spaces can be replaced with fractions or decimals. For example, the equation x 2 + (21/2)x + 5 = 0 is decomposed into (x + 10) (x + 1/2).
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Factorization by trial and error. Simple quadratic equations can be factored by simply substituting numbers into possible solutions until you find the correct solution. If the equation has the form ax 2 +bx+c, where a>1, the possible solutions are written as (dx +/- _)(ex +/- _), where d and e are numerical coefficients other than zero, 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 the 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.
