In this lesson, we will recall all the previously studied methods of factoring a polynomial into factors and consider examples of their application, in addition, we will study a new method - the method of allocating a complete square and learn how to apply it to solving various problems.
Topic:Factoring Polynomials
Lesson:Factorization of polynomials. Full square selection method. Combination of methods
Let us recall the main methods of factoring polynomials that were studied earlier:
The method of taking the common factor out of the brackets, that is, such a factor that is present in all terms of the polynomial. Let's consider an example:
Recall that a monomial is a product of degrees and numbers. In our example, both members have some common, identical elements.
So, let's take the common factor out of the brackets:
;
Recall that by multiplying the multiplier by the parenthesis, you can check the correctness of the subtraction.
Grouping method. It is not always possible to take out a common factor in a polynomial. In this case, it is necessary to divide its members into groups so that in each group it is possible to take out the common factor and try to divide so that after the factors are taken out in the groups, a common factor appears for the whole expression, and the expansion can be continued. Let's consider an example:
Let's group the first term with the fourth, the second with the fifth, and the third, respectively, with the sixth:
Let's take out common factors in groups:
The expression has a common factor. Let's take it out:
Applying abbreviated multiplication formulas. Let's consider an example:
;
Let's write down the expression in detail:
Obviously, we have before us the formula for the square of the difference, since there is the sum of the squares of two expressions and their double product is subtracted from it. Let's collapse by the formula:
Today we will learn another method - the method of selecting a full square. It is based on the formulas for the square of the sum and the square of the difference. Let's remind them:
The formula for the square of the sum (difference);
The peculiarity of these formulas is that they contain the squares of two expressions and their doubled product. Let's consider an example:
Let's write the expression:
So the first expression is this and the second.
In order to compose the formula for the square of the sum or difference, the double product of expressions is not enough. It needs to be added and subtracted:
Let's fold the full square of the sum:
Let's transform the resulting expression:
We apply the formula for the difference of squares, recall that the difference between the squares of two expressions is the product and the sum by their difference:
So, this method consists, first of all, in the fact that it is necessary to identify the expressions a and b that are in the square, that is, determine which squares of which expressions are in this example. After that, you need to check for the presence of a doubled product and if it is not there, then add and subtract it, the meaning of the example will not change from this, but the polynomial can be factorized using the formulas for the square of the sum or the difference and difference of squares, if possible.
Let's move on to solving examples.
Example 1 - factorize:
Let's find expressions that are squared:
Let's write down what their doubled product should be:
Let's add and subtract twice the product:
Let's collapse the full square of the sum and give similar ones:
Let's write the difference of squares by the formula:
Example 2 - Solve the equation:
;
There is a trinomial on the left side of the equation. You need to factor it out. We use the formula for the square of the difference:
We have the square of the first expression and the doubled product, the square of the second expression is missing, add and subtract it:
Let's fold a full square and give similar terms:
Let's apply the formula for the difference of squares:
So, we have the equation 
We know that the product is zero only if at least one of the factors is zero. On this basis, we compose the equations:
Let's solve the first equation:
Let's solve the second equation:
Answer: or
;
We proceed similarly to the previous example - select the square of the difference.
What factorization? This is a way to turn an awkward and complex example into a simple and cute one.) Very powerful trick! It is found 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 an expression in another form while preserving its essence.
Meaning factorization extremely simple and straightforward. Straight from the name itself. You can forget (or not know) what a multiplier is, but the fact that this word comes from the word "multiply" can you figure it out?) Factoring means: represent an expression as multiplying something by something. Forgive me mathematics and the Russian language ...) And that's it.
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 very well that 12 and 3 4 same. The essence of the number 12 from conversion has not changed.
Is it possible to decompose 12 differently? Easy!
12 \u003d 3 4 \u003d 2 6 \u003d 3 2 2 \u003d 0.5 24 \u003d ........
Expansion options are endless.
Factoring numbers is a useful thing. It helps a lot, for example, when dealing with roots. But factoring algebraic expressions is not a thing that is useful, it is - necessary! Just for example:
Simplify:
Those who do not know how to factor an expression rests on the sidelines. Whoever knows how - simplifies and gets:
The effect is amazing, right?) By the way, the solution is quite simple. See for yourself below. Or, for example, a task like this:
Solve the equation:
x 5 - x 4 \u003d 0
Decided in the mind, by the way. Using factorization. Below we will solve this example. Answer: x 1 \u003d 0; x 2 \u003d 1.
Or, the same thing, but for the older ones):
Solve the equation:
With these examples I have shown main purpose factorization: simplify fractional expressions and solve some types of equations. I recommend remembering a rule of thumb:
If we are faced with a terrible fractional expression, you can try to factor the numerator and denominator into factors. Very often the fraction is shortened and simplified.
If we have an equation in front of us, where on the right is zero, and on the left - do not understand what, you can try to factor the left side into factors. Sometimes it helps).
Basic methods of factoring.
Here are the most popular ways:
4. Decomposition of a square trinomial.
These methods must be remembered. In that order. Complex examples are checked into all possible ways of decomposition. 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 the brackets.
A simple and reliable way. It never hurts! It happens either good or not.) Therefore, he is the first. Understanding.
Everyone knows (I believe!)) The rule:
a (b + c) \u003d ab + ac
Or, more generally:
a (b + c + d + .....) \u003d ab + ac + ad + ....
All equalities work from left to right, and vice versa, from right to left. You can write:
ab + ac \u003d a (b + c)
ab + ac + ad + .... = a (b + c + d + .....)
That's the whole point of taking the common factor out of parentheses.
On the left side and - common factor for all terms. Multiplied by everything that is). On the right is the most and is already outside the brackets.
We will consider the practical application of the method by examples. At first the option is simple, even primitive.) But on this option I will mark (in green) very important points for any factorization.
Factorize:
ah + 9x
What general the multiplier sits in both terms? X, of course! We will take it out of the brackets. We do this. We immediately write the X outside the brackets:
ax + 9x \u003d x (
And in brackets we write the result of division each term on this very x. In order:
That's all. Of course, there is no need to describe 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 down the results of dividing all terms by this very common factor. In order.
So we expanded the expression ah + 9x by factors. Turned it into multiplying x by (a + 9). Note that the original expression also contained 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) there is nothing except 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 as 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 but multiplication. This is the whole point of factoring.
By the way, is it possible to somehow 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 initial expression? If it works, everything is tip-top!)
x (a + 9) \u003d ax + 9x
Happened.)
There is no problem with this primitive example. But if there are several terms, and even with different signs ... In short, every third student mumbles). 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, you can endure it. Is there more general factor? Yes! This is a three. You can write the expression like this:
3ax + 3 3x
Here you can immediately see that the common factor will be 3x... Here we take it out:
3ax + 3 3x \u003d 3x (a + 3)
They laid it out.
And what will happen if you endure only x? Nothing special:
3ax + 9x \u003d x (3a + 9)
This will also be a factorization. But in this fascinating process, it is customary to lay out everything until it stops, while there is an opportunity. Here, in brackets, there is an opportunity to take out a triple. It turns out:
3ax + 9x \u003d x (3a + 9) \u003d 3x (a + 3)
The same thing, with only one extra action.) Remember:
When taking the common factor out of the brackets, we try to take out maximum common factor.
Do we continue the fun?)
Factor expression:
3ax + 9x-8a-24
What are we going to endure? Three, X? Nope ... You can't. I remind you that you can only endure general factor that is in allterms of expression. That's why he general. There is no such multiplier here ... What, you can not expand !? Well, yes, we were delighted, of course ... Meet:
2. Grouping.
Actually, grouping can hardly be called an independent way of factoring. Rather, it is 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 by example. So, before us is the expression:
3ax + 9x-8a-24
It can be seen that there are some common letters and numbers. But... General there is no factor in all the terms. We do not lose heart and break the expression into pieces. Let's group. So that in each piece there was a common factor, there was something to take out. How do we break? Just put the brackets.
Let me remind you that parentheses can be placed anywhere and in any way. If only the essence of the example has not changed. For example, you can do this:
3ax + 9x-8a-24=(3ax + 9x) - (8a + 24)
Pay attention to the second brackets! There is a minus sign in front of them, and 8a and 24 become positive! If, for verification, open the brackets back, the signs will change, and we will receive initial expression. Those. the essence of the expression from parentheses has not changed.
But if you just stuck in parentheses without considering the sign change, for example, like this:
3ax + 9x-8a-24=(3ax + 9x) - (8a-24 )
it will be a mistake. Right - already other expression. Open the brackets and everything will become visible. You don't have to decide further, yes ...)
But back to factoring. We look at the first brackets (3ax + 9x) and we think, can we endure anything? Well, we solved this example above, you can take 3x:
(3ax + 9x) \u003d 3x (a + 3)
We study the second brackets, there you can take out the eight:
(8a + 24) \u003d 8 (a + 3)
Our whole expression will turn out:
(3ax + 9x) - (8a + 24) \u003d 3x (a + 3) -8 (a + 3)
Factorized? No. The decomposition should result in only multiplication, and our minus sign spoils everything. But ... Both terms have a common factor! it (a + 3)... It was not for nothing that I said that the whole brackets are, as it were, one letter. Hence, these brackets can be taken out of the brackets. Yes, that's exactly what it sounds like.)
We do as described above. We write the common factor (a + 3), in the second parentheses we write the results of dividing the terms by (a + 3):
3x (a + 3) -8 (a + 3) \u003d (a + 3) (3x-8)
All! On the right, there is nothing but multiplication! So the factorization is successful!) Here it is:
3ax + 9x-8a-24 \u003d (a + 3) (3x-8)
Let us briefly repeat the essence of the grouping.
If the expression does not contain common multiplier for of all terms, we break the expression with parentheses so that inside the parentheses the common factor was. We take it out and see what happened. If you're lucky, and there are exactly the same expressions in the brackets, move these brackets outside the brackets.
I will add that grouping is a creative process). It doesn't always work the first time. Nothing wrong. Sometimes you have to change the places of terms, consider different options for grouping until you find a successful one. The main thing here is not to lose heart!)
Examples.
Now, having enriched with knowledge, you can solve tricky examples.) There were three of these at the beginning of the lesson ...
Simplify:
In fact, we have already solved this example. Unbeknownst to myself.) Let me remind you: if we are given a terrible fraction, we try to factor out the numerator and denominator. Other simplification options simply no.
Well, the denominator here does not expand, but the numerator ... We have already expanded the numerator in the course of the lesson! Like this:
3ax + 9x-8a-24 \u003d (a + 3) (3x-8)
We write the result of the 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 ones. Final simplification result:
I would like to emphasize: 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 for simplification. Of course, if the expressions various, then nothing will be reduced. Of course. But factoring gives a chance. This chance without decay is simply not there.
Example with equation:
Solve the equation:
x 5 - x 4 \u003d 0
We take out the common factor x 4 outside the brackets. We get:
x 4 (x-1) \u003d 0
We figure that the product of the factors is 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 this equality, the second factor does not bother us. Anyone can be, all the same in the end it will turn out to be zero. And what number in the fourth power of zero will give? Only zero! And nothing else ... So:
We sorted out the first factor, found one root. Let's deal with the second factor. Now we don't care about the first factor.):
So we found a solution: x 1 \u003d 0; x 2 \u003d 1... Any of these roots fit our equation.
A very important note. Please note that we solved the equation piece by piece! Every factor was set equal to zero, ignoring the rest of the 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 solve similar. Piece by piece. For instance:
(x-1) (x + 5) (x-3) (x + 2) \u003d 0
The one who opens the brackets, multiplies everything, he will forever hang on this equation.) A correct student will immediately see that there is nothing on the left except multiplication, on the right - zero. And it will begin (in the mind!) To equate to zero all the parentheses in order. And he will receive (in 10 seconds!) The correct solution: x 1 \u003d 1; x 2 \u003d -5; x 3 \u003d 3; x 4 \u003d -2.
Great, isn't it?) Such an elegant solution is possible if the left side of the equation decomposed into factors. 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, letters can hide sines, logarithms, and whatever you like! Factoring works in all mathematics.
We take out the common factor lg 4 x outside the brackets. We get:
lg 4 x \u003d 0
This is one root. Let's deal with the second factor.
Here's the final answer: x 1 \u003d 1; x 2 \u003d 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 deal with the formulas for abbreviated multiplication and the square trinomial.
If you like this site ...
By the way, I have a couple more interesting sites for you.)
You can practice solving examples and find out your level. Testing with instant verification. Learning - with interest!)
you can get acquainted with functions and derivatives.
We already know how to partially use the factorization of the difference of degrees - while studying the topic "Difference of squares" and "Difference of cubes" we learned to represent the difference of expressions as a product, which can be represented as squares or as cubes of some expressions or numbers.
Abbreviated multiplication formulas
According to the abbreviated multiplication formulas:
the difference of squares can be represented as the product of the difference of two numbers or expressions by their sum
The difference between cubes can be represented as the product of the difference of two numbers by the incomplete square of the sum
Transition to the difference of expressions to the 4th degree
Based on the formula for the difference of squares, let's try to factor the expression $ a ^ 4-b ^ 4 $
Let's remember how a power is raised to a power - for this the base remains the same, and the exponents are multiplied, that is, $ ((a ^ n)) ^ m \u003d a ^ (n * m) $
Then you can imagine:
$ a ^ 4 \u003d (((a) ^ 2)) ^ 2 $
$ b ^ 4 \u003d (((b) ^ 2)) ^ 2 $
So, our expression can be represented as $ a ^ 4-b ^ 4 \u003d (((a) ^ 2)) ^ 2 $ - $ (((b) ^ 2)) ^ 2 $
Now, in the first parenthesis, we again received the difference of numbers, which means that we can again factorize as the product of the difference of two numbers or expressions by their sum: $ a ^ 2-b ^ 2 \u003d \\ left (a-b \\ right) (a + b) $.
Now we calculate the product of the second and third brackets using the rule of the product of polynomials, - multiply each term of the first polynomial by each term of the second polynomial and add the result. To do this, first the first term of the first polynomial - $ a $ - is multiplied by the first and second terms of the second (by $ a ^ 2 $ and $ b ^ 2 $), i.e. we get $ a \\ cdot a ^ 2 + a \\ cdot b ^ 2 $, then the second term of the first polynomial - $ b $ - we multiply by the first and second terms of the second polynomial (by $ a ^ 2 $ and $ b ^ 2 $), those. we get $ b \\ cdot a ^ 2 + b \\ cdot b ^ 2 $ and compose the sum of the resulting expressions
$ \\ left (a + b \\ right) \\ left (a ^ 2 + b ^ 2 \\ right) \u003d a \\ cdot a ^ 2 + a \\ cdot b ^ 2 + b \\ cdot a ^ 2 + b \\ cdot b ^ 2 \u003d a ^ 3 + ab ^ 2 + a ^ 2b + b ^ 3 $
Let us write the difference of monomials of degree 4, taking into account the calculated product:
$ a ^ 4-b ^ 4 \u003d (((a) ^ 2)) ^ 2 $ - $ (((b) ^ 2)) ^ 2 \u003d ((a) ^ 2-b ^ 2) (a ^ 2 + b ^ 2) $ \u003d $ \\ \\ left (ab \\ right) (a + b) (a ^ 2 + b ^ 2) \\ $ \u003d
Go to the difference of expressions in the 6th degree
Based on the formula for the difference of squares, let's try to factorize the expression $ a ^ 6-b ^ 6 $
Let's remember how a power is raised to a power - for this the base remains the same, and the exponents are multiplied, that is, $ ((a ^ n)) ^ m \u003d a ^ (n \\ cdot m) $
Then you can imagine:
$ a ^ 6 \u003d (((a) ^ 3)) ^ 2 $
$ b ^ 6 \u003d (((b) ^ 3)) ^ 2 $
Hence, our expression can be represented as $ a ^ 6-b ^ 6 \u003d (((a) ^ 3)) ^ 2 - (((b) ^ 3)) ^ 2 $
In the first bracket we got the difference of cubes of monomials, in the second the sum of cubes of monomials, now we can again factorize the difference of cubes of monomials as the product of the difference of two numbers by the incomplete square of the sum $ a ^ 3-b ^ 3 \u003d \\ left (ab \\ right) ( a ^ 2 + ab + b ^ 2) $
The original expression takes the form
$ a ^ 6-b ^ 6 \u003d ((a) ^ 3-b ^ 3) \\ left (a ^ 3 + b ^ 3 \\ right) \u003d \\ left (ab \\ right) (a ^ 2 + ab + b ^ 2) (a ^ 3 + b ^ 3) $
We calculate the product of the second and third brackets using the rule of the product of polynomials, - multiply each term of the first polynomial by each term of the second polynomial and add the result.
$ (a ^ 2 + ab + b ^ 2) (a ^ 3 + b ^ 3) \u003d a ^ 5 + a ^ 4b + a ^ 3b ^ 2 + a ^ 2b ^ 3 + ab ^ 4 + b ^ 5 $
Let us write the difference of the monomials of the 6th degree, taking into account the calculated product:
$ a ^ 6-b ^ 6 \u003d ((a) ^ 3-b ^ 3) \\ left (a ^ 3 + b ^ 3 \\ right) \u003d \\ left (ab \\ right) (a ^ 2 + ab + b ^ 2) (a ^ 3 + b ^ 3) \u003d (ab) (a ^ 5 + a ^ 4b + a ^ 3b ^ 2 + a ^ 2b ^ 3 + ab ^ 4 + b ^ 5) $
Degree Difference Factoring
Let us analyze the formulas for the difference of cubes, difference of $ 4 $ degrees, difference of $ 6 $ degrees
We see that in each of these expansions there is some analogy, generalizing which we get:
Example 1
Factor $ (32x) ^ (10) - (243y) ^ (15) $
Decision: First, we represent each monomial as some 5th degree monomial:
\\ [(32x) ^ (10) \u003d ((2x ^ 2)) ^ 5 \\] \\ [(243y) ^ (15) \u003d ((3y ^ 3)) ^ 5 \\]
We use the power difference formula
Picture 1.
- 1. Factoring out the common factor and grouping method... In some cases, it is advisable to replace some terms with the sum (difference) of similar terms or introduce mutually annihilating terms.
- 2. Using abbreviated multiplication formulas.Sometimes you have to put factors out of brackets, group members, select a complete square and only then represent the sum of cubes, the difference of squares or the difference of cubes as a product.
- 3. Use of Bezout's theorem and the method of undefined coefficients.
Example ... Factorize:
P 3 (x) \u003d x 3 + 4x 2 + 5x + 2;
Since P 3 (-1) \u003d 0, the polynomial P 3 (x) is divisible by x + 1. Using the method of undefined coefficients, we find the quotient of the division of the polynomial
P 3 (x) \u003d x 3 + 4x 2 + 5x + 2 per binomial x + 1.
Let the quotient be the polynomial x 2 +. Since x 3 + 4x 2 + 5x + 2 \u003d (x + 1) (x 2 +) \u003d
X 3 + (+ 1) x 2 + () x +, we get the system:
Where from. Therefore, P 3 (x) \u003d (x + 1) (x 2 + 3x + 2).
Since x 2 + 3x + 2 \u003d x 2 + x + 2x + 2 \u003d x (x + 1) + 2 (x + 1) \u003d (x + 1) (x + 2), then P 3 (x ) \u003d (x + 1) 2 (x + 2).
4. Use of Bezout's theorem and column division.
Example ... Factor
P 4 (x) \u003d 5 x 4 + 9 x 3 -2 x 2 -4 x -8.
Decision ... Since P 4 (1) \u003d 5 + 9-2-4-8 \u003d 0, then P 4 (x) is divisible by (x-1). Division by "column" find the quotient
Consequently,
P 4 (x) \u003d (x-) (5 x 3 + 14x 2 + 12x + 8) \u003d
\u003d (x-1) P 3 (x).
Since P 3 (-2) \u003d -40 + 56-24 + 8 \u003d 0, the polynomial P 3 (x) \u003d 5 x 3 + 14x 2 + 12x + 8 is divisible by x + 2.
Let us find the quotient by dividing by "column":
Consequently,
P 3 (x) \u003d (x + 2) (5 x 2 + 4x + 4).
Since the discriminant of the square trinomial 5 x 2 + 4x + 4 is D \u003d -24<0, то этот
the square trinomial cannot be decomposed into linear factors.
So, P 4 (x) \u003d (x-1) (x + 2) (5 x 2 + 4x + 4)
5. Using Bezout's theorem and Horner's scheme. The quotient obtained by these methods can be factorized in any other way or in the same way.
Example ... Factorize:
P 3 (x) \u003d 2 x 3 -5 x 2 -196 x + 99;
Decision .
If a given polynomial has rational roots, then they can only be among the numbers 1/2, 1, 3/2, 3, 9/2, 11/2, 9, 33, 99, 11.
To find the root of a given polynomial, we use the following statement:
If at the ends of a certain segment the values \u200b\u200bof the polynomial have different signs, then on the interval (a; b) there is at least one root of this polynomial.
For a given polynomial P 3 (0) \u003d 99, P 3 (1) \u003d - 100. Therefore, on the interval (0; 1) there is at least one root of this polynomial. Therefore, among the 24 numbers written out above, it is advisable to first check those numbers that belong to the interval
(0; 1). Of these numbers, only the number belongs to this interval.
The value of P 3 (x) at x \u003d 1/2 can be found not only by direct substitution, but also in other ways, for example, according to Horner's scheme, since P () is equal to the remainder of dividing the polynomial P (x) by x-. Moreover, in many examples this method is preferable, since the coefficients of the quotient are found simultaneously.
According to Horner's scheme for this example, we get:
Since P 3 (1/2) \u003d 0, then x \u003d 1/2 is a root of the polynomial P 3 (x), and the polynomial P 3 (x) is divisible by x-1/2, i.e. 2 x 3 -5 x 2 -196 x + 99 \u003d (x-1/2) (2 x 2 -4 x-198).
Since 2 x 2 -4 x-198 \u003d 2 (x 2 -2 x + 1-100) \u003d 2 ((x-1) 2 -10 2) \u003d 2 (x + 9) x-11), then
P 3 (x) \u003d 2 x 3 -5 x 2 -196 x + 99 \u003d 2 (x-1/2) (x + 9) (x-11).
Polynomial ring concept
Let be TOand L commutative rings
Definition 1 : Ring TO called a simple ring extension K using elements x and write:
L \u003d K [x]if the conditions are met:
subring rings
The main set K [x]indicate somvol L, K [x].
Definition 2 : Simple extension L \u003d K [x] rings K through x - simple transcendental ring expansion K through xif the conditions are met:
subring rings
If, then
Definition 3 : Element x called transcendental over the ring K, if the condition is satisfied: if, then
Sentence... Let be K [x] simple transcendental expansion. If and, where, then
Evidence ... By condition, subtract the second from the first expression, we get: since the element x transcendental over K, then from (3) we get :.
Output. Any element of a simple nonzero transcendental extension of a commutative ring K using the element x admits a unique representation as a linear combination of integer non-negative powers of an element x
Definition: By the polynomial ring of the unknown x over non-zero ring K is called a simple transcendental extension of a nonzero commutative ring K using the element x.
Theorem ... For any nonzero commutative ring K, there is its simple transcendental extension with the element x, k [x]
Operations on polynomials
Let k [x] be a polynomial ring of a nonzero commutative ring K
Definition 1: The polynomials f and g belonging to k [x] are called equal and write f \u003d g if all coefficients of the polynomials f and g that are at the same degrees of the unknown are equal x.
Consequence ... In the notation of a polynomial, the order of the terms is not essential. Assigning and excluding from the notation of the polynomial terms with zero coefficient will not change the polynomial.
Definition 2. The sum of the polynomials f and g is the polynomial f + g defined by the equality:
Definition 3 : - the product of polynomials, denoted, which is determined by the rule:
Degree of polynomials
Let the ring be commutative. k [x] a ring of polynomials over a field K : ,
Definition : Let - any polynomial. If, then a non-negative integer n is the degree of the polynomials f... In this case, they write n \u003d deg f.
The numbers are the coefficients of the polynomial, where is the leading coefficient.
If, f - normalized. The degree of the zero polynomial is undefined.
Polynomial degree properties
K - integrity area
Evidence :
Since and. TO - the area of \u200b\u200bintegrity.
Corollary 1 : k [x] over the field TO (area of \u200b\u200bintegrity) is in turn an area of \u200b\u200bintegrity. For any area of \u200b\u200bintegrity, there is a particular area.
Corollary 2 : For any k [x] over the domain of integrity TO there is a private field.
Division into binomials and roots of a polynomial.
Let, the element is called the value of the polynomial f from the argument.
Bezout's theorem : For any polynomial and element, there is an element:.
Evidence : Let be any polynomial
Consequence : Remainder of dividing a polynomial by, equals.
Definition : The element is called the root of the polynomial f, if.
Theorem : Let, element is a root f if and only if divides f
Evidence:
Necessities. Let, from Bezout's theorem it follows that, from the divisibility properties it follows that
Sufficiency. Let that. ch.d.
The maximum number of roots of a polynomial over the domain of integrity.
Theorem : Let k be the domain of integrity. Number of roots of a polynomial f in integrity k no more than a degree n polynomial f.
Evidence :
By induction on the degree of a polynomial. Let the polynomial f has zero roots, and their number does not exceed.
Let the theorem be proved for any.
Let us show that item 2 implies the truth of the statement of the theorem for polynomials.
Let and, two cases are possible:
- A) Polynomial f has no roots, therefore, the statement of the theorem is true.
- B) Polynomial f has at least a root by Bezout's theorem, since k is the domain of integrity, then by property 3 (degree of the polynomial), it follows that
Because, k - integrity area.
Thus, all roots of the polynomial are the roots of the polynomial g since, then by the induction hypothesis, the number of all roots of the polynomial g not more n, Consequently, f has no more ( n +1) root.
Consequence : Let be k - domain of integrity, if the number of roots of the polynomial f more numbers n,where, then f is a zero polynomial.
Algebraic and functional equality of polynomials
Let, be some polynomial, it defines some function
in general, any polynomial can define one function.
Theorem : Let be k- domain of integrity, thus, for equality of polynomials and equality (identical equality ()) defined by and.
Evidence :
Necessities. Let and be the domain of integrity,.
Let, that is
Sufficiency. Let's pretend that. Consider since k integrity domain, then polynomial h has the number of roots, it follows from the corollary that h zero polynomial. Thus, ch.t.d.
Divisibility theorem with remainder
Definition : Euclidean ring K such an area of \u200b\u200bintegrity is called k,that a function is defined on the set h,taking non-negative integer values \u200b\u200band satisfies the condition
In the process of finding elements for these elements is called division with a remainder, - incomplete quotient, - remainder of division.
Let be a ring of polynomials over a field.
Theorem (on division with remainder) : Let be a ring of polynomials over a field and a polynomial there is a unique pair of polynomials such that the condition or is satisfied. or
Evidence : Existence of a polynomial. Let it be. The theorem is true, obviously, if is zero or, since or. Let us prove the theorem when. We carry out the proof by induction of the degree of the polynomial, suppose that the theorem is proved (except for uniqueness) for the polynomial. Let us show that in this case the assertion of the theorem holds for. Indeed, let is the leading coefficient of the polynomial, therefore, the polynomial will have the same leading coefficient and the same degree as the polynomial, therefore, the polynomial will have or is a zero polynomial. If, then, therefore, for and we get. If, then by the inductive hypothesis, therefore, that is, for we obtain or. The existence of the polynomial is proved.
Let us show that such a pair of polynomials is unique.
Let there exist or, subtract:. There are two possible cases or.
On the other hand. By the condition of the degree or, or.
If. A contradiction is thus obtained. The uniqueness has been proven.
Corollary 1 : The ring of polynomials over the field is Euclidean space.
Corollary 2 : The polynomial ring over, is the principal ideal ring (any ideal has a unique generator)
Any Euclidean ring is factorial: A polynomial ring over is called a factorial ring.
Euclid's algorithm. GCD of two polynomials
Let the ring of polynomials be over.
Definition 1 : Let and, if there is a polynomial, then the remainder of the division is equal to zero, then it is called the divisor of the polynomial and is denoted by: ().
Definition 2 : The greatest common divisor of polynomials is called a polynomial:
and (is a common factor of and).
(by any common divisor and).
The greatest common divisor of polynomials is denoted by gcd (;). Among the common divisors of any polynomials are all polynomials of degree zero from, that is, not a zero field. It may happen that two given polynomials and do not have common divisors, which are not zero polynomials.
Definition : If polynomials and do not have common divisors that are not polynomials of degree zero, then they are called coprime.
Lemma : If polynomials from over the field holds, then the greatest common divisor of the polynomials and the associated gcd. ~
Recording ( a ~ b) means that (and) by definition.
Evidence : Let u
and, from this it follows that we teach that is the common divisor of the polynomial and.
common divisor and, we obtain
Euclid's Algorithm
In general, this task involves a creative approach, since there is no universal method for solving it. But let's try to give a few tips.
In the overwhelming majority of cases, the factorization of a polynomial is based on a corollary from Bezout's theorem, that is, a root is found or selected and the degree of the polynomial is reduced by one by dividing by. A root is sought for the resulting polynomial, and the process is repeated until the expansion is complete.
If the root cannot be found, then specific decomposition methods are used: from grouping to the introduction of additional mutually exclusive terms.
The further presentation is based on the skills of solving equations of higher degrees with integer coefficients.
Factor out 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 more than factoring out the common factor.
Example.
Factor a third-degree polynomial.
Decision.
Obviously, it is the root of the polynomial, that is x can be taken outside the brackets:
Find the roots of the square trinomial 
In this way, 
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Factoring a polynomial with rational roots.
First, consider a method for decomposing a polynomial with integer coefficients of the form, the coefficient at the highest power is equal to one.
In this case, if the polynomial has integer roots, then they are divisors of the free term.
Example.
Decision.
Let's check if there are whole 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 written numbers. Let's check these numbers one by one according to Horner's scheme. Its convenience also lies in the fact that, as a result, we obtain the coefficients of the expansion of the polynomial: 
I.e, x \u003d 2 and x \u003d -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, therefore, it has no real roots.
Answer:
Comment:
instead of Horner's scheme, one could use the selection of a root and subsequent division of the polynomial by a polynomial.
Now consider the decomposition 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.
Factor expression.
Decision.
By performing variable replacement y \u003d 2x, we pass to a polynomial with a coefficient equal to one at the highest degree. To do this, first we 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:
Let us calculate successively the values \u200b\u200bof the function g (y) at these points until zero is obtained. 
I.e, y \u003d -5 is the root 
therefore is the root of the original function. We will divide the polynomial by a column (corner) by a binomial. 
In this way, 
It is impractical to continue checking the remaining divisors, since it is easier to factorize the resulting square trinomial 
Consequently, 
Unreleased polynomials. The theorem about the distribution of a polynomial for dobutok ignorant. Canonical distribution of the polynomial.