The concept of a polynomial
Definition of polynomial: A polynomial is the sum of monomials. Polynomial example:
here we see the sum of two monomials, and this is a polynomial, i.e. sum of monomials.
The terms that make up a polynomial are called terms of the polynomial.
Is the difference of monomials a polynomial? Yes, it is, because the difference is easily reduced to a sum, example: 5a – 2b = 5a + (-2b).
Monomials are also considered polynomials. But a monomial has no sum, then why is it considered a polynomial? And you can add zero to it and get its sum with a zero monomial. So, a monomial is a special case of a polynomial; it consists of one term.
The number zero is the zero polynomial.
Standard form of polynomial
What is a polynomial of standard form? A polynomial is the sum of monomials, and if all these monomials that make up the polynomial are written in standard form, and there should be no similar ones among them, then the polynomial is written in standard form.
An example of a polynomial in standard form:
here the polynomial consists of 2 monomials, each of which has a standard form; among the monomials there are no similar ones.
Now an example of a polynomial that does not have a standard form:
here two monomials: 2a and 4a are similar. You need to add them up, then the polynomial will take the standard form:
Another example:
This polynomial is reduced to standard view? No, his second term is not written in standard form. Writing it in standard form, we obtain a polynomial of standard form:
Polynomial degree
What is the degree of a polynomial?
Polynomial degree definition:
The degree of a polynomial is the highest degree that the monomials that make up a given polynomial of standard form have.
Example. What is the degree of the polynomial 5h? The degree of the polynomial 5h is equal to one, because this polynomial contains only one monomial and its degree is equal to one.
Another example. What is the degree of the polynomial 5a 2 h 3 s 4 +1? The degree of the polynomial 5a 2 h 3 s 4 + 1 is equal to nine, because this polynomial includes two monomials, the first monomial 5a 2 h 3 s 4 has the highest degree, and its degree is 9.
Another example. What is the degree of the polynomial 5? The degree of a polynomial 5 is zero. So, the degree of a polynomial consisting only of a number, i.e. without letters, equals zero.
The last example. What is the degree of the zero polynomial, i.e. zero? The degree of the zero polynomial is not defined.
In this lesson, we will recall the basic definitions of this topic and consider some typical problems, namely, reducing a polynomial to a standard form and calculating a numerical value for given values of variables. We will solve several examples in which reduction to a standard form will be used to solve various kinds of problems.
Subject:Polynomials. Arithmetic operations on monomials
Lesson:Reducing a polynomial to standard form. Typical tasks
Let us recall the basic definition: a polynomial is the sum of monomials. Each monomial that is part of a polynomial as a term is called its member. For example:
Binomial;
Polynomial;
Binomial;
Since a polynomial consists of monomials, the first action with a polynomial follows from here - you need to bring all monomials to a standard form. Let us remind you that to do this you need to multiply all the numerical factors - get a numerical coefficient, and multiply the corresponding powers - get the letter part. In addition, let us pay attention to the theorem about the product of powers: when multiplying powers, their exponents add up.
Let's consider important operation- bringing the polynomial to standard form. Example:
Comment: to bring a polynomial to a standard form, you need to bring all the monomials included in its composition to a standard form, after which, if there are similar monomials - and these are monomials with the same letter part - perform actions with them.
So, we looked at the first typical problem - bringing a polynomial to a standard form.
Next typical task- calculation of a specific value of a polynomial for given numerical values of the variables included in it. Let's continue to look at the previous example and set the values of the variables:
Comment: let us recall that one to any natural power is equal to one, and zero to any natural power is equal to zero, in addition, we recall that when multiplying any number by zero, we get zero.
Let's look at a number of examples of typical operations of bringing a polynomial to a standard form and calculating its value:
Example 1 - bring to standard form:
Comment: the first step is to bring the monomials to standard form, you need to bring the first, second and sixth; second action - we bring similar terms, that is, we perform the given tasks on them arithmetic operations: we add the first with the fifth, the second with the third, the rest are rewritten without changes, since they have no similar ones.
Example 2 - calculate the value of the polynomial from example 1 given the values of the variables:
Comment: when calculating, you should remember that a unit to any natural power is one; if it is difficult to calculate powers of two, you can use the table of powers.
Example 3 - instead of an asterisk, put a monomial such that the result does not contain a variable:
Comment: regardless of the task, the first action is always the same - bring the polynomial to a standard form. In our example, this action comes down to bringing similar terms. After this, you should carefully read the condition again and think about how we can get rid of the monomial. Obviously, for this you need to add the same monomial to it, but with opposite sign- . Next, we replace the asterisk with this monomial and make sure that our solution is correct.
In studying the topic of polynomials, it is worth mentioning separately that polynomials occur in both standard and non-standard forms. In this case, the polynomial non-standard type can be reduced to a standard form. Actually, this question will be discussed in this article. Let's reinforce the explanations with examples with a detailed step-by-step description.
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The meaning of reducing a polynomial to standard form
Let's delve a little deeper into the concept itself, the action - “bringing a polynomial to a standard form.”
Polynomials, like any other expressions, can be transformed identically. As a result, in this case we obtain expressions that are identically equal to the original expression.
Definition 1
Reduce the polynomial to standard form– means replacing the original polynomial with an equal polynomial of standard form, obtained from the original polynomial using identical transformations.
A method for reducing a polynomial to standard form
Let's speculate on the topic of exactly what identity transformations will lead the polynomial to the standard form.
Definition 2
According to the definition, each polynomial of a standard form consists of monomials of a standard form and does not contain similar terms. A polynomial of a non-standard form may include monomials of a non-standard form and similar terms. From the above, a rule is naturally deduced about how to reduce a polynomial to a standard form:
- first of all, the monomials that make up a given polynomial are reduced to standard form;
- then the reduction of similar members is carried out.
Examples and solutions
Let us examine in detail examples in which we reduce the polynomial to standard form. We will follow the rule derived above.
Note that sometimes the terms of a polynomial in the initial state already have a standard form, and all that remains is to bring similar terms. It happens that after the first step of actions there are no such terms, then we skip the second step. IN general cases it is necessary to perform both actions from the rule above.
Example 1
Polynomials are given:
5 x 2 y + 2 y 3 − x y + 1 ,
0, 8 + 2 a 3 0, 6 − b a b 4 b 5,
2 3 7 · x 2 + 1 2 · y · x · (- 2) - 1 6 7 · x · x + 9 - 4 7 · x 2 - 8 .
It is necessary to bring them to a standard form.
Solution
Let's first consider the polynomial 5 x 2 y + 2 y 3 − x y + 1 : its members have a standard form, there are no similar terms, which means the polynomial is specified in a standard form, and no additional actions are required.
Now let's look at the polynomial 0, 8 + 2 · a 3 · 0, 6 − b · a · b 4 · b 5. It includes non-standard monomials: 2 · a 3 · 0, 6 and − b · a · b 4 · b 5, i.e. we need to bring the polynomial to standard form, for which the first step is to transform the monomials into standard form:
2 · a 3 · 0, 6 = 1, 2 · a 3;
− b · a · b 4 · b 5 = − a · b 1 + 4 + 5 = − a · b 10 , thus we obtain the following polynomial:
0, 8 + 2 · a 3 · 0, 6 − b · a · b 4 · b 5 = 0, 8 + 1, 2 · a 3 − a · b 10.
In the resulting polynomial, all terms are standard, there are no similar terms, which means our actions to bring the polynomial to standard form are completed.
Consider the third given polynomial: 2 3 7 x 2 + 1 2 y x (- 2) - 1 6 7 x x + 9 - 4 7 x 2 - 8
Let's bring its members to standard form and get:
2 3 7 · x 2 - x · y - 1 6 7 · x 2 + 9 - 4 7 · x 2 - 8 .
We see that the polynomial contains similar members, let’s bring similar members:
2 3 7 x 2 - x y - 1 6 7 x 2 + 9 - 4 7 x 2 - 8 = = 2 3 7 x 2 - 1 6 7 x 2 - 4 7 x 2 - x · y + (9 - 8) = = x 2 · 2 3 7 - 1 6 7 - 4 7 - x · y + 1 = = x 2 · 17 7 - 13 7 - 4 7 - x · y + 1 = = x 2 0 - x y + 1 = x y + 1
Thus, the given polynomial 2 3 7 x 2 + 1 2 y x (- 2) - 1 6 7 x x + 9 - 4 7 x 2 - 8 takes the standard form − x y + 1 .
Answer:
5 x 2 y + 2 y 3 − x y + 1- the polynomial is set as standard;
0, 8 + 2 a 3 0, 6 − b a b 4 b 5 = 0, 8 + 1, 2 a 3 − a b 10;
2 3 7 · x 2 + 1 2 · y · x · (- 2) - 1 6 7 · x · x + 9 - 4 7 · x 2 - 8 = - x · y + 1 .
In many problems, the action of reducing a polynomial to a standard form is intermediate when searching for an answer to asked question. Let's consider this example.
Example 2
The polynomial 11 - 2 3 z 2 · z + 1 3 · z 5 · 3 - 0 is given. 5 · z 2 + z 3 . It is necessary to bring it to a standard form, indicate its degree and arrange the terms of a given polynomial in descending degrees of the variable.
Solution
Let us reduce the terms of the given polynomial to the standard form:
11 - 2 3 z 3 + z 5 - 0 . 5 · z 2 + z 3 .
Next step Here are some similar terms:
11 - 2 3 z 3 + z 5 - 0 . 5 z 2 + z 3 = 11 + - 2 3 z 3 + z 3 + z 5 - 0, 5 z 2 = 11 + 1 3 z 3 + z 5 - 0, 5 z 2
We have obtained a polynomial of standard form, which allows us to designate the degree of the polynomial (equal to the highest degree of its constituent monomials). Obviously, the required degree is 5.
All that remains is to arrange the terms in decreasing powers of the variables. For this purpose, we simply rearrange the terms in the resulting polynomial of standard form, taking into account the requirement. Thus, we get:
z 5 + 1 3 · z 3 - 0 , 5 · z 2 + 11 .
Answer:
11 - 2 3 · z 2 · z + 1 3 · z 5 · 3 - 0, 5 · z 2 + z 3 = 11 + 1 3 · z 3 + z 5 - 0, 5 · z 2, while the degree of the polynomial - 5 ; as a result of arranging the terms of the polynomial in decreasing powers of the variables, the polynomial will take the form: z 5 + 1 3 · z 3 - 0, 5 · z 2 + 11.
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Any decimal fraction can be written as a ,bc ... · 10 k . Such records are often found in scientific calculations. It is believed that working with them is even more convenient than with ordinary decimal notation.
Today we will learn how to convert any decimal fraction to this form. At the same time, we will make sure that such an entry is already “overkill”, and in most cases it does not provide any advantages.
First, a little repetition. As you know, decimal fractions can be multiplied not only among themselves, but also by ordinary integers (see lesson “”). Of particular interest is multiplication by powers of ten. Take a look:
Task. Find the value of the expression: 25.81 10; 0.00005 1000; 8.0034 100.
Multiplication is performed according to the standard scheme, with the significant part being allocated for each factor. Let's briefly describe these steps:
For the first expression: 25.81 10.
- Significant parts: 25.81 → 2581 (shift right by 2 digits); 10 → 1 (shift left by 1 digit);
- Multiply: 2581 · 1 = 2581;
- Total shift: right by 2 − 1 = 1 digit. We perform a reverse shift: 2581 → 258.1.
For the second expression: 0.00005 1000.
- Significant parts: 0.00005 → 5 (shift right by 5 digits); 1000 → 1 (shift left by 3 digits);
- Multiply: 5 · 1 = 5;
- Total shift: right by 5 − 3 = 2 digits. We perform the reverse shift: 5 → .05 = 0.05.
Last expression: 8.0034 100.
- Significant parts: 8.0034 → 80034 (shift right by 4 digits); 100 → 1 (shift left by 2 digits);
- Multiply: 80,034 · 1 = 80,034;
- Total shift: right by 4 − 2 = 2 digits. We perform a reverse shift: 80,034 → 800.34.
Let's rewrite the original examples a little and compare them with the answers:
- 25.81 · 10 1 = 258.1;
- 0.00005 10 3 = 0.05;
- 8.0034 · 10 2 = 800.34.
What's happening? It turns out that multiplying a decimal fraction by the number 10 k (where k > 0) is equivalent to shifting the decimal point to the right by k places. To the right - because the number is increasing.
Likewise, multiplying by 10 −k (where k > 0) is equivalent to dividing by 10 k, i.e. shift by k digits to the left, which leads to a decrease in number. Take a look at the examples:
Task. Find the value of the expression: 2.73 10; 25.008:10; 1.447: 100;
In all expressions, the second number is a power of ten, so we have:
- 2.73 · 10 = 2.73 · 10 1 = 27.3;
- 25.008: 10 = 25.008: 10 1 = 25.008 · 10 −1 = 2.5008;
- 1.447: 100 = 1.447: 10 2 = 1.447 10 −2 = .01447 = 0.01447.
It follows that the same decimal fraction can be written infinite number ways. For example: 137.25 = 13.725 10 1 = 1.3725 10 2 = 0.13725 10 3 = ...
The standard form of a number is expressions of the form a ,bc ... · 10 k , where a , b , c , ... are ordinary numbers, and a ≠ 0. The number k is an integer.
- 8.25 · 10 4 = 82,500;
- 3.6 10−2 = 0.036;
- 1.075 · 10 6 = 1,075,000;
- 9.8 10−6 = 0.0000098.
For each number written in standard form, the corresponding decimal fraction is indicated next to it.
Switch to standard view
The algorithm for transitioning from an ordinary decimal fraction to a standard form is very simple. But before you use it, be sure to review what the significant part of a number is (see the lesson “Multiplying and dividing decimals”). So, the algorithm:
- Write out the significant part of the original number and put a decimal point after the first significant digit;
- Find the resulting shift, i.e. How many places has the decimal point moved compared to the original fraction? Let this be the number k;
- Compare the significant part that we wrote down in the first step with the original number. If the significant part (including the decimal point) is less than the original number, add a factor of 10 k. If more, add a factor of 10 −k. This expression will be the standard view.
Task. Write the number in standard form:
- 9280;
- 125,05;
- 0,0081;
- 17 000 000;
- 1,00005.
- 9280 → 9.28. Shift the decimal point 3 places to the left, the number decreased (obviously 9.28< 9280). Результат: 9,28 · 10 3 ;
- 125.05 → 1.2505. Shift - 2 digits to the left, the number has decreased (1.2505< 125,05). Результат: 1,2505 · 10 2 ;
- 0.0081 → 8.1. This time the shift was to the right by 3 digits, so the number increased (8.1 > 0.0081). Result: 8.1 · 10 −3 ;
- 17000000 → 1.7. The shift is 7 digits to the left, the number has decreased. Result: 1.7 · 10 7 ;
- 1.00005 → 1.00005. There is no shift, so k = 0. Result: 1.00005 · 10 0 (this happens too!).
As you can see, not only decimal fractions are represented in standard form, but also ordinary integers. For example: 812,000 = 8.12 · 10 5 ; 6,500,000 = 6.5 10 6.
When to use standard notation
In theory, standard number notation should make fractional calculations even easier. But in practice, a noticeable gain is obtained only when performing a comparison operation. Because comparing numbers written in standard form is done like this:
- Compare powers of ten. The largest number will be the one with this degree greater;
- If the degrees are the same, we begin to compare significant figures- as in ordinary decimal fractions. The comparison goes from left to right, from the most significant to the least significant. The largest number will be the one in which the next digit is larger;
- If the powers of ten are equal, and all the digits are the same, then the fractions themselves are also equal.
Of course, all this is true only for positive numbers. For negative numbers, all signs are reversed.
A remarkable property of fractions written in standard form is that any number of zeros can be assigned to their significant part - both on the left and on the right. A similar rule exists for other decimal fractions (see lesson “ Decimals”), but they have their own limitations.
Task. Compare the numbers:
- 8.0382 10 6 and 1.099 10 25;
- 1.76 · 10 3 and 2.5 · 10 −4 ;
- 2.215 · 10 11 and 2.64 · 10 11 ;
- −1.3975 · 10 3 and −3.28 · 10 4 ;
- −1.0015 · 10 −8 and −1.001498 · 10 −8 .
- 8.0382 10 6 and 1.099 10 25. Both numbers are positive, and the first has a lower degree of ten than the second (6< 25). Значит, 8,0382 · 10 6 < 1,099 · 10 25 ;
- 1.76 · 10 3 and 2.5 · 10 −4. The numbers are again positive, and the degree of ten for the first of them is greater than for the second (3 > −4). Therefore, 1.76 · 10 3 > 2.5 · 10 −4 ;
- 2.215 10 11 and 2.64 10 11. The numbers are positive, the powers of ten are the same. We look at the significant part: the first digits also coincide (2 = 2). The difference starts at the second digit: 2< 6, поэтому 2,215 · 10 11 < 2,64 · 10 11 ;
- −1.3975 · 10 3 and −3.28 · 10 4 . This negative numbers. The first has a degree of ten less (3< 4), поэтому (в силу отрицательности) само число будет больше: −1,3975 · 10 3 >−3.28 · 10 4 ;
- −1.0015 · 10 −8 and −1.001498 · 10 −8 . Negative numbers again, and the powers of ten are the same. The first 4 digits of the significant part are also the same (1001 = 1001). At the 5th digit the difference begins, namely: 5 > 4. Since the original numbers are negative, we conclude: −1.0015 10 −8< −1,001498 · 10 −8 .
We noted that any monomial can be bring to standard form. In this article we will understand what is called bringing a monomial to standard form, what actions allow this process to be carried out, and consider solutions to examples with detailed explanations.
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What does it mean to reduce a monomial to standard form?
It is convenient to work with monomials when they are written in standard form. However, quite often monomials are specified in a form different from the standard one. In these cases, you can always go from the original monomial to a monomial of the standard form by performing identity transformations. The process of carrying out such transformations is called reducing a monomial to a standard form.
Let us summarize the above arguments. Reduce the monomial to standard form- this means performing identical transformations with it so that it takes on a standard form.
How to bring a monomial to standard form?
It's time to figure out how to reduce monomials to standard form.
As is known from the definition, monomials of non-standard form are products of numbers, variables and their powers, and possibly repeating ones. And a monomial of the standard form can contain in its notation only one number and non-repeating variables or their powers. Now it remains to understand how to bring products of the first type to the type of the second?
To do this you need to use the following the rule for reducing a monomial to standard form consisting of two steps:
- First, a grouping of numerical factors is performed, as well as identical variables and their powers;
- Secondly, the product of the numbers is calculated and applied.
As a result of applying the stated rule, any monomial will be reduced to a standard form.
Examples, solutions
All that remains is to learn how to apply the rule from the previous paragraph when solving examples.
Example.
Reduce the monomial 3 x 2 x 2 to standard form.
Solution.
Let's group numerical factors and factors with a variable x. After grouping, the original monomial will take the form (3·2)·(x·x 2) . The product of numbers in the first brackets is equal to 6, and the rule for multiplying powers with the same bases allows the expression in the second brackets to be represented as x 1 +2=x 3. As a result, we obtain a polynomial of the standard form 6 x 3.
Here is a short summary of the solution: 3 x 2 x 2 =(3 2) (x x 2)=6 x 3.
Answer:
3 x 2 x 2 =6 x 3.
So, to bring a monomial to a standard form, you need to be able to group factors, multiply numbers, and work with powers.
To consolidate the material, let's solve one more example.
Example.
Present the monomial in standard form and indicate its coefficient.
Solution.
The original monomial has a single numerical factor in its notation −1, let's move it to the beginning. After this, we will separately group the factors with the variable a, separately with the variable b, and there is nothing to group the variable m with, we will leave it as is, we have 
. After performing operations with powers in brackets, the monomial will take the standard form we need, from which we can see the coefficient of the monomial equal to −1. Minus one can be replaced with a minus sign: .
