What is the rule for multiplying a monomial by a polynomial. Multiplication of monomials and polynomials

special case multiplication of a polynomial by a polynomial - multiplication of a polynomial by a monomial. In this article, we formulate the rule for performing this action and analyze the theory with practical examples.

Rule for multiplying a polynomial by a monomial

Let's figure out what is the basis of multiplying a polynomial by a monomial. This action relies on the distributive property of multiplication with respect to addition. Literally, this property is written as follows: (a + b) c \u003d a c + b c (a, b and c are some numbers). In this entry, the expression (a + b) c is just the product of the polynomial (a + b) and the monomial c. The right side of the equality a c + b c is the sum of products of monomials a and b into a monomial c.

The above reasoning allows us to formulate the rule for multiplying a polynomial by a monomial:

Definition 1

To carry out the action of multiplying a polynomial by a monomial, you must:

• write down the product of a polynomial and a monomial, which must be multiplied;
• multiply each term of the polynomial by the given monomial;
• find the sum of the resulting products.

Let us further explain the above algorithm.

To compose the product of a polynomial by a monomial, the original polynomial is enclosed in brackets; further, a multiplication sign is placed between it and the given monomial. In the case when the entry of a monomial begins with a minus sign, it must also be enclosed in brackets. For example, the product of a polynomial − 4 x 2 + x − 2 and monomial 7 y write as (− 4 x 2 + x − 2) 7 y, and the product of the polynomial a 5 b − 6 a b and monomial − 3 a 2 compose in the form: (a 5 b − 6 a b) (− 3 a 2).

The next step of the algorithm is the multiplication of each term of the polynomial by a given monomial. The components of the polynomial are monomials, i.e. in fact, we need to perform the multiplication of a monomial by a monomial. Let us assume that after the first step of the algorithm we have obtained the expression (2 x 2 + x + 3) 5 x, then the second step is to multiply each term of the polynomial 2 x 2 + x + 3 with a monomial 5 x, thus obtaining: 2 x 2 5 x = 10 x 3 , x 5 x = 5 x 2 and 3 5 x = 15 x. The result will be the monomials 10 x 3, 5 x 2 and 15 x.

The last action according to the rule is the addition of the resulting products. From the given example, doing this step algorithm, we get: 10 x 3 + 5 x 2 + 15 x.

By default, all steps are written as a chain of equalities. For example, finding the product of a polynomial 2 x 2 + x + 3 and monomial 5 x let's write it like this: (2 x 2 + x + 3) 5 x = 2 x 2 5 x + x 5 x + 3 5 x = 10 x 3 + 5 x 2 + 15 x . Eliminating the intermediate calculation of the second step, short solution can be done as follows: (2 x 2 + x + 3) 5 x = 10 x 3 + 5 x 2 + 15 x.

The considered examples make it possible to notice important nuance: as a result of multiplying a polynomial and a monomial, a polynomial is obtained. This statement is true for any multiplying polynomial and monomial.

By analogy, a monomial is multiplied by a polynomial: a given monomial is multiplied with each member of the polynomial, and the resulting products are summed.

Examples of multiplying a polynomial by a monomial

Example 1

It is necessary to find the product: 1 , 4 · x 2 - 3 , 5 · y · - 2 7 · x .

Decision

The first step of the rule has already been completed - the work has been recorded. Now we perform the next step, multiplying each term of the polynomial by the given monomial. AT this case it is convenient to first translate decimal fractions into ordinary fractions. Then we get:

1 , 4 x 2 - 3 , 5 y - 2 7 x = 1 , 4 x 2 - 2 7 x - 3 , 5 y - 2 7 x = = - 1 , 4 2 7 x 2 x + 3 , 5 2 7 x y = - 7 5 2 7 x 3 + 7 5 2 7 x y = - 2 5 x 3 + x y

Answer: 1 , 4 x 2 - 3 , 5 y - 2 7 x = - 2 5 x 3 + x y .

Let us clarify that when the original polynomial and/or monomial are given in a non-standard form, before finding their product, it is desirable to reduce them to standard form.

Example 2

Given a polynomial 3 + a − 2 a 2 + 3 a − 2 and monomial − 0 , 5 a b (− 2) a. You need to find their work.

Decision

We see that the initial data is presented in a non-standard form, therefore, for the convenience of further calculations, we will bring them into a standard form:

− 0 , 5 a b (− 2) a = (− 0 , 5) (− 2) (a a) b = 1 a 2 b = a 2 b 3 + a − 2 a 2 + 3 a − 2 = (3 − 2) + (a + 3 a) − 2 a 2 = 1 + 4 a − 2 a 2

Now let's do the multiplication of the monomial a 2 b for each member of the polynomial 1 + 4 a − 2 a2

a 2 b (1 + 4 a − 2 a 2) = a 2 b 1 + a 2 b 4 a + a 2 b (− 2 a 2) = = a 2 b + 4 a 3 b − 2 a 4 b

We could not bring the initial data to the standard form: the solution would then turn out to be more cumbersome. In this case, the last step would be the need to reduce similar terms. For understanding, here is a solution according to this scheme:

− 0 .5 a b (− 2) a (3 + a − 2 a 2 + 3 a − 2) = = − 0 . 5 a b (− 2) a 3 − 0 . 5 a b (− 2) a a − 0 . 5 a b (− 2) a (− 2 a 2) − 0 . 5 a b (− 2) a 3 a − 0 , 5 a b (− 2) a (− 2) = = 3 a 2 b + a 3 b − 2 a 4 b + 3 a 3 b − 2 a 2 b = a 2 b + 4 a 3 b − 2 a 4 b

Answer: − 0 , 5 a b (− 2) a (3 + a − 2 a 2 + 3 a − 2) = a 2 b + 4 a 3 b − 2 a 4 b.

If you notice a mistake in the text, please highlight it and press Ctrl+Enter

On the this lesson the operation of multiplying a polynomial by a monomial will be studied, which is the basis for studying the multiplication of polynomials. Let us recall the distributive law of multiplication and formulate the rule for multiplying any polynomial by a monomial. We also recall some properties of degrees. In addition, typical errors will be formulated when performing various examples.

Subject:Polynomials. Arithmetic operations on monomials

Lesson:Multiplication of a polynomial by a monomial. Typical tasks

The operation of multiplying a polynomial by a monomial is the basis for considering the operation of multiplying a polynomial by a polynomial, and you must first learn how to multiply a polynomial by a monomial in order to understand the multiplication of polynomials.

The basis of this operation is the distributive law of multiplication. Recall it:

In essence, we see the rule for multiplying a polynomial, in this case a binomial, by a monomial, and this rule can be formulated as follows: in order to multiply a polynomial by a monomial, you need to multiply each term of the polynomial by this monomial. Add the algebraically obtained products, and then perform the necessary actions on the polynomial - namely, bring it to the standard form.

Consider an example:

Comment: given example is solved exactly following the rule: each term of a polynomial is multiplied by a monomial. In order to understand and assimilate the distribution law well, in this example, the terms of the polynomial were replaced by x and y, respectively, and the monomial by c, after which an elementary action was performed in accordance with the distribution law and the initial values ​​were substituted. You should be careful with the signs and correctly multiply by minus one.

Consider an example of multiplying a trinomial by a monomial and make sure that it is no different from the same operation with a binomial:

Let's move on to solving examples:

Comment: this example is solved according to the distribution law and similarly to the previous example - each term of the polynomial is multiplied by a monomial, the resulting polynomial is already written in standard form, so it cannot be simplified.

Example 2 - perform actions and get a polynomial in standard form:

Comment: to solve this example, first we will multiply for the first and second binomials according to the distribution law, after that we will bring the resulting polynomial to the standard form - we will bring like terms.

Now let us formulate the main problems associated with the operation of multiplying a polynomial by a monomial and give examples of their solution.

Task 1 - simplify the expression:

Comment: this example is solved similarly to the previous one, namely, first the polynomials are multiplied by the corresponding monomials, after that the similar ones are reduced.

Task 2 - simplify and calculate:

Example 1:;

Comment: this example is solved similarly to the previous one, with the only addition that after the reduction of such members, it is necessary to substitute its specific value instead of the variable and calculate the value of the polynomial. Recall that it is easy to multiply decimal by ten, you need to move the decimal point one place to the right.

If the numbers are denoted by different letters, then it is only possible to designate from the product; let, for example, the number a be multiplied by the number b, - we can denote this either a ∙ b or ab, but there can be no question of somehow performing this multiplication. However, when we are dealing with monomials, then, due to 1) the presence of coefficients and 2) the fact that these monomials can include factors denoted by the same letters, it is possible to talk about the multiplication of monomials; such a possibility is even wider for polynomials. Let's analyze a number of cases where it is possible to perform multiplication, starting from the simplest.

1. Multiplying powers with the same grounds . Let, for example, a 3 ∙ a 5 be required. Let's write, knowing the meaning of raising to a power, the same thing in more detail:

a ∙ a ∙ a ∙ a ∙ a ∙ a ∙ a ∙ a

Looking at this detailed entry, we see that we have written a multiplier of 8 times, or, in short, a 8 . So, a 3 ∙ a 5 = a 8 .

Let b 42 ∙ b 28 be required. We would have to write first the factor b 42 times, and then again the factor b 28 times - in general, we would get that b is taken by the factor 70 times. i.e. b 70 . So, b 42 ∙ b 28 \u003d b 70. From this it is already clear that when multiplying powers with the same bases, the base of the degree remains unchanged, and the exponents are added. If we have a 8 ∙ a, then we have to keep in mind that the factor a implies an exponent of 1 (“a to the first power”), therefore, a 8 ∙ a = a 9 .

Examples: x ∙ x 3 ∙ x 5 = x 9 ; a 11 ∙ a 22 ∙ a 33 = a 66; 3 5 ∙ 3 6 ∙ 3 = 3 12 ; (a + b) 3 ∙ (a + b) 4 = (a + b) 7 ; (3x – 1) 4 ∙ (3x – 1) = (3x – 1) 5 etc.

Sometimes you have to deal with degrees whose exponents are indicated by letters, for example, xn (x to the power of n). You have to get used to using these expressions. Here are some examples:

Let's explain some of these examples: b n - 3 ∙ b 5 you need to leave the base b unchanged, and add the indicators, i.e. (n - 3) + (+5) \u003d n - 3 + 5 \u003d n + 2. Of course, such additions must be learned to perform quickly in the mind.

Another example: x n + 2 ∙ x n - 2, - the base of x must be left unchanged, and the indicator should be added, i.e. (n + 2) + (n - 2) = n + 2 + n - 2 = 2n.

It is possible to express the order found above, how to perform the multiplication of powers with the same bases, now by the equality:

a m ∙ a n = a m + n

2. Multiplication of a monomial by a monomial. Let, for example, 3a²b³c ∙ 4ab²d² be required. We see that here one multiplication is indicated by a dot, but we know that the same multiplication sign is implied between 3 and a², between a² and b³, between b³ and c, between 4 and a, between a and b², between b² and d². Therefore, we can see the product of 8 factors here and we can multiply them by any groups in any order. Let's rearrange them so that the coefficients and powers with the same bases are close, i.e.

3 ∙ 4 ∙ a² ∙ a ∙ b³ ∙ b² ∙ c ∙ d².

Then we can multiply 1) coefficients and 2) powers with the same base and get 12a³b5cd².

So, when multiplying a monomial by a monomial, we can multiply the coefficients and powers with the same bases, and the remaining factors have to be rewritten without change.

More examples:

3. Multiplication of a polynomial by a monomial. Suppose we first need to multiply some polynomial, for example, a - b - c + d, by a positive integer, for example, +3. As positive numbers are considered to coincide with arithmetic, then it is the same as (a - b - c + d) ∙ 3, i.e. take a - b - c + d as a summand 3 times, or

(a - b - c + d) ∙ (+3) = a - b - c + d + a - b - c + d + a - b - c + d = 3a - 3b - 3c + 3d,

i.e., as a result, each term of the polynomial had to be multiplied by 3 (or by +3).

It follows from this:

(a - b - c + d) ÷ (+3) = a - b - c + d,

i.e., each term of the polynomial had to be divided by (+3). Also, summarizing, we get:

etc.

Let now it is necessary to multiply (a - b - c + d) by positive fraction, for example, to +. It's the same as multiplying by arithmetic fraction, which means to take parts from (a - b - c + d). It is easy to take one-fifth of this polynomial: you need to divide (a - b - c + d) by 5, and we already know how to do this - we get . It remains to repeat the result obtained 3 times or multiply by 3, i.e.

As a result, we see that we had to multiply each term of the polynomial by or by +.

Let now it is necessary to multiply (a - b - c + d) by a negative number, integer or fractional,

i.e., in this case, each term of the polynomial had to be multiplied by -.

Thus, whatever the number m, always (a - b - c + d) ∙ m = am - bm - cm + dm.

Since each monomial is a number, here we see an indication of how to multiply a polynomial by a monomial - each member of the polynomial must be multiplied by this monomial.

4. Multiplication of a polynomial by a polynomial. Let it be (a + b + c) ∙ (d + e). Since d and e mean numbers, then (d + e) ​​expresses any one number.

(a + b + c) ∙ (d + e) ​​= a(d + e) ​​+ b(d + e) ​​+ c(d + e)

(we can explain it this way: we have the right to take d + e temporarily for a monomial).

Ad + ae + bd + be + cd + ce

As a result, you can change the order of the members.

(a + b + c) ∙ (d + e) ​​= ad + bd + ed + ae + be + ce,

i.e., to multiply a polynomial by a polynomial, one has to multiply each term of one polynomial by each term of the other. It is convenient (for this purpose the order of the obtained terms was changed above) to multiply each term of the first polynomial first by the first term of the second (by + d), then by the second term of the second (by + e), then, if it were, by the third, etc. d.; after that, you should make a reduction of similar terms.

In these examples, the binomial is multiplied by the binomial; in each binomial, the terms are arranged in descending powers of the letter common to both binomials. Such multiplications are easy to perform in your head and immediately write the final result.

From multiplying the senior term of the first binomial by the senior term of the second, i.e. 4x² by 3x, we get 12x³ the senior term of the product - obviously there will be no similar ones. Next, we look for the terms from the multiplication of which terms will be obtained with the power of the letter x less by 1, i.e. with x². It is easy to see that such terms are obtained by multiplying the 2nd term of the first factor by the 1st term of the second and by multiplying the 1st term of the first factor by the 2nd term of the second one (the brackets at the bottom of the example indicate this). Doing these multiplications in your head and also doing the reduction of these two similar terms (after which we get the -19x² term) is not difficult. Then we notice that the next term, containing the letter x to the power 1 less, i.e. x to the 1st power, will only be obtained by multiplying the second term by the second, and there will be no similar ones.

Another example: (x² + 3x)(2x - 7) = 2x³ - x² - 21x.

It is also easy to mentally execute examples like the following:

The senior term is obtained by multiplying the senior term by the senior one, there will be no similar terms for it, and it = 2a³. Then we look for from which multiplications the terms with a² will be obtained - from the multiplication of the 1st term (a²) by the 2nd (-5) and from the multiplication of the second term (-3a) by the 1st (2a) - this is indicated below in brackets; after performing these multiplications and combining the resulting terms into one, we get -11a². Then we look for which multiplications will result in terms with a in the first degree - these multiplications are marked with brackets from above. After completing them and combining the resulting members into one, we get + 11a. Finally, we notice that the low term of the product (+10), which does not contain a at all, is obtained by multiplying the low term (–2) of one polynomial by the low term (–5) of another.

Another example: (4a 3 + 3a 2 - 2a) ∙ (3a 2 - 5a) \u003d 12a 5 - 11a 4 - 21a 3 + 10a 2.

From all the previous examples, we also get overall result: the highest term of the product is always obtained from the multiplication of the highest terms of the factors, and there cannot be members similar to it; also, the lowest term of the product is obtained by multiplying the lowest terms of the factors, and there can be no similar terms either.

The rest of the terms obtained by multiplying a polynomial by a polynomial may be similar, and it may even happen that all these terms cancel each other out, and only the older and younger remain.

Here are some examples:

(a² + ab + b²) (a - b) = a³ + a²b + ab² - a²b - ab² - b³ = a³ - b³
(a² - ab + b²) (a - b) = a³ - a²b + ab² + a²b - ab² + b³ = a³ + b³
(a³ + a²b + ab² + b³) (a - b) = a 4 - b 4 (we only write the result)
(x 4 - x³ + x² - x + 1) (x + 1) = x 5 + 1, etc.

These results are noteworthy and useful to remember.

Especially important next case multiplications:

(a + b) (a - b) = a² + ab - ab - b² = a² - b²
or (x + y) (x - y) = x² + xy - xy - y² = x² - y²
or (x + 3) (x - 3) = x² + 3x - 3x - 9 = x² - 9 etc.

In all these examples, applied to arithmetic, we have the product of the sum of two numbers and their difference, and the result is the difference of the squares of these numbers.

If we see such a case, then there is no need to carry out the multiplication in detail, as was done above, but we can immediately write the result.

For example, (3a + 1) ∙ (3a – 1). Here the first factor, from the point of view of arithmetic, is the sum of two numbers: the first number is 3a and the second 1, and the second factor is the difference of the same numbers; therefore, the result should be: the square of the first number (i.e. 3a ∙ 3a = 9a²) minus the square of the second number (1 ∙ 1 = 1), i.e.

(3a + 1) ∙ (3a - 1) = 9a² - 1.

Also

(ab - 5) ∙ (ab + 5) = a²b² - 25, etc.

So let's remember

(a + b) (a - b) = a² - b²

i.e., the product of the sum of two numbers and their difference is equal to the difference of the squares of these numbers.

When multiplying a polynomial by a monomial, we will use one of the laws of multiplication. It received in mathematics the name of the distributive law of multiplication. Distributive law of multiplication:

1. (a + b)*c = a*c + b*c

2. (a - b)*c = a*c - b*c

In order to multiply a monomial by a polynomial, it is enough to multiply each of the terms of the polynomial by a monomial. After that, add the resulting products. The following figure shows a scheme for multiplying a monomial by a polynomial.

The order of multiplication is unimportant, if, for example, you need to multiply a polynomial by a monomial, then you need to do exactly the same. Thus, there is no difference between the entries 4*x * (5*x^2*y - 4*x*y) and (5*x^2*y - 4*x*y)* 4*x.

Let's multiply the polynomial and the monomial written above. And we'll show specific example how to do it right:

4*x * (5*x^2*y - 4*x*y)

Using the distributive law of multiplication, we compose the product:

4*x*5*x^2*y - 4*x*4*x*y.

In the resulting sum, we bring each of the monomials to the standard form and get:

20*x^3*y - 16*x^2*y.

This will be the product of a monomial and a polynomial: (4*x) * (5*x^2*y - 4*x*y) = 20*x^3*y - 16*x^2*y.

Examples:

1. Multiply the monomial 4*x^2 by the polynomial (5*x^2+4*x+3). Using the distributive law of multiplication, we compose the product. We have
(4*x^2*5*x^2) +(4*x^2* 4*x) +(4*x^2*3).

20*x^4 +16*x^3 +12*x^2.

This will be the product of a monomial and a polynomial: (4*x^2)*(5*x^2+4*x+3)= 20*x^4 +16*x^3 +12*x^2.

2. Multiply the monomial (-3*x^2) by the polynomial (2*x^3-5*x+7).

Using the distributive law of multiplication, we will compose the product. We have:

(-3*x^2 * 2*x^3) +(-3*x^2 * -5*x) +(-3*x^2 *7).

In the resulting sum, we reduce each of the monomials to its standard form. We get:

6*x^5 +15*x^3 -21*x^2.

This will be the product of a monomial and a polynomial: (-3*x^2) * (2*x^3-5*x+7)= -6*x^5 +15*x^3 -21*x^2.

Target:

1. Ensure the assimilation of initial knowledge on the topic "Multiplying a monomial by a polynomial";
2. Develop analytical and synthesizing thinking;
3. To cultivate the motives of teaching and a positive attitude towards knowledge.

Team building of the class.

Tasks:

1. To get acquainted with the algorithm for multiplying a monomial by a polynomial;
2. work out practical use algorithm.

Equipment: task cards, computer, interactive projector.

Lesson type: combined.

During the classes

I. Organizational moment:

Hello guys, sit down.

Today we continue to study the section "Polynomials" and the topic of our lesson is "Multiplying a monomial by a polynomial." Open your notebooks and write down the number and the topic of the lesson "Multiplying a monomial by a polynomial."

The task of our lesson is to derive the rule for multiplying a monomial by a polynomial and learn how to apply it in practice. The knowledge gained today is necessary for you throughout the study of the entire course of algebra.

You have forms on the tables in which we will enter your points scored throughout the lesson, and the results will be graded. We will display points in the form of emoticons. ( Appendix 1)

II. The stage of preparing students for the active and conscious assimilation of new material.

When studying new topic we will need the knowledge that you received in the previous lessons.

Students perform tasks on cards on the topic "Degree and its properties." (5-7 minutes)

Front work:

1) Two monomials are given: 12p 3 and 4p 3

a) the amount;
b) difference;
c) a work;
e) private;
e) the square of each monomial.

2) Name the members of the polynomial and determine the degree of the polynomial:

a)5 ab – 7a 2 + 2b – 2,6
b)6 xy 5 + x 2 y - 2

3) Today we need the distributive property of multiplication.

Let's formulate this property and record in literal form.

III. Stage of assimilation of new knowledge.

We have repeated the rule of multiplication of a monomial by a monomial, the distributive property of multiplication. Now let's complicate the task.

Divide into 4 groups. Each group has 4 expressions on the cards. Try to restore the missing link in the chain and explain your point of view.

• 8x 3 (6x 2 – 4x + 3) = ………………….……= 48x 5 – 32x 4 + 24x 3
• 5a 2 (2a 2 + 3a - 7) = …………………...…..= 10a 4 + 15a 3 - 35a 2
• 3y(9y 3 - 4y 2 - 6) = ………………………. =27y 4 – 12y 3 – 18y
• 6b 4 (6b 2 + 4b - 5) = ………….……………= 36b 6 + 24b 5 - 30b 4

(One representative from each group comes to the screen, writes down the missing part of the expression and explains his point of view.)

Try to formulate a rule (algorithm) for multiplying a polynomial by a monomial.

What expression is obtained as a result of these actions?

To test yourself, open the textbook on page 126 and read the rule (1 person reads aloud).

Do our conclusions match the rule in the textbook? Write down the rule for multiplying a monomial by a polynomial in a notebook.

IV. Fixing:

1. Physical education minute:

Guys, sit back, close your eyes, relax, now we are resting, the muscles are relaxed, we are studying the topic "Multiplying a monomial by a polynomial."

And so we remember the rule and repeat after me: to multiply a monomial by a polynomial, you need to multiply the monomial by each term of the polynomial and write down the sum of the resulting expressions. We open our eyes.

2. Work according to textbook No. 614 at the blackboard and in notebooks;

a) 2x (x 2 - 7x - 3) \u003d 2x 3 - 14x 2 - 6x
b) -4v 2 (5v 2 - 3v - 2) = -20v 4 + 12v 3 + 8v 2
c) (3a 3 - a 2 + a) (- 5a 3) \u003d -15a 6 + 5a 5 - 5a 4
d) (y 2 - 2.4y + 6) 1.5y \u003d 1.5y 3 - 3.6y 2 + 9y
e) -0.5x 2 (-2x 2 - 3x + 4) \u003d x 4 + 1.5x 3 - 2x 2
e) (-3y 2 + 0.6y) (- 1.5y 3) \u003d 4.5y 5 - 0.9y 4

(When performing the number, the most typical mistakes are analyzed)

3. Competition by variants (deciphering the pictogram). (Annex 2)

1 option:		Option 2:
1) -3x 2 (- x 3 + x - 5) 2) 14 x(3 xy 2 – x 2 y + 5) 3) -0,2 m 2 n(10 mn 2 – 11 m 3 – 6) 4) (3a 3 - a 2 + 0.1a) (-5a 2) 5) 1/2 with(6 with 3 d – 10c 2 d 2) 6) 1.4p 3 (3q - pq + 5p) 7) 10x2y(5.4xy - 7.8y - 0.4) 8) 3 ab(a 2 – 2ab + b 2)		1) 3a 4 x (a 2 - 2ax + x 3 - 1) 2) -11a(2a 2 b - a 3 + 5b 2) 3) -0,5 X 2 y(Xy 3 - 3X+y2) 4) (6b 4 - b 2 + 0.01) (-7b 3) 5) 1/3m 2 (9m 3 n 2 - 15mn) 6) 1.6c 4 (2c 2 d - cd + 5d) 7) 10p 4 (0.7pq - 6.1q - 3.6) 8) 5xy(x 2 - 3xy + x 3)

Tasks are presented on individual cards and on the screen. Each student completes his task, finds a letter and writes it on the screen opposite the expression that he transformed. If the correct answer is received, then the word will turn out: well done! smarties 7a