Lesson "multiplication of a monomial by a polynomial". Multiplication of a polynomial by a monomial

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

1. Multiplication of powers with the same bases... Let, for example, require a 3 ∙ a 5. Knowing the meaning of exponentiation, we will write the same in more detail:

a ∙ a ∙ a ∙ a ∙ a ∙ a ∙ a ∙ a

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

Let b 42 ∙ b 28 be required. One 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 a factor of 70 times. i.e. b 70. So, b 42 ∙ b 28 = b 70. From this it is already clear that when multiplying degrees 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 in 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, the exponents of which are indicated by letters, for example, xn (x to the power n). Expressions like these take some getting used to. Here are some examples:

Let us explain some of these examples: bn - 3 ∙ b 5 base b should be left unchanged, and the indicators should be added, that is, (n - 3) + (+5) = n - 3 + 5 = n + 2. Of course, such additions must be learned to do quickly in the mind.

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

You can above the found order, how to perform multiplication of powers with the same bases, can now be expressed by equality:

a m ∙ a n = a m + n

2. Multiplication of a monomial by a monomial. Suppose, for example, you want 3a²b³c ∙ 4ab²d². We see that here one multiplication is indicated by a dot, but we know that the same multiplication sign is meant 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 can multiply them by any groups in any order. Let us rearrange them so that the coefficients and degrees with the same bases are next to each other, i.e.

3 ∙ 4 ∙ a² ∙ a ∙ b³ ∙ b² ∙ c ∙ d².

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

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

More examples:

3. Multiplication of a polynomial by a monomial. Suppose you must first some polynomial, for example, a - b - c + d multiplied by a positive integer, for example, +3. Because positive numbers are considered to be the same as arithmetic, then it is the same as (a - b - c + d) ∙ 3, i.e. a - b - c + d is taken 3 times as a term, or

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

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

Hence follows:

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

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

etc.

Now let it be necessary to multiply (a - b - c + d) by a positive fraction, for example, by +. It's like multiplying by an arithmetic fraction, which means taking the parts of (a - b - c + d). Taking one fifth of this polynomial is easy: 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 +.

Now let it be necessary to multiply (a - b - c + d) by a negative number, whole 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 term of the polynomial must be multiplied by this monomial.

4. Multiplying a polynomial by a polynomial... Let it be necessary (a + b + c) ∙ (d + e). Since d and e stand for numbers, so (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 temporarily take d + e 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,

that is, to multiply a polynomial by a polynomial, you have to multiply each term of one polynomial by each term of the other. It is convenient (for this, 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 the casting of similar members.

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

From the multiplication of the leading term of the first binomial by the senior term of the second, ie 4x² by 3x, we obtain 12x³ the senior term in the product - there will obviously not be any similar to it. Next, we are looking for, from the multiplication of which terms, we get terms with less than 1 power of the letter x, that is, with x². We can easily see that such terms are obtained from multiplying the 2nd term of the first factor by the 1st term of the second and from multiplying the 1st term of the first factor by the 2nd term of the second (the brackets at the bottom of the example indicate this). It is not difficult to perform these multiplications in the mind and also to perform the reduction of these two similar terms (after which we get the term –19x²). Then we notice that the next term containing the letter x to a degree 1 less, that is, x to the 1st degree, will be obtained only from 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 do examples like the following:

The senior term is obtained by multiplying the senior term by the senior, there will be no similar terms to it, and it = 2a³. Then we look for which multiplications will result in terms with a² - 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 parentheses; 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 parentheses above. After completing them and combining the resulting terms into one, we get + 11a. Finally, note that the least significant term in the product (+10), which does not contain a at all, is obtained by multiplying the least significant term (–2) of one polynomial by the least significant term (–5) of the other.

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

From all the previous examples, we also get the general result: the leading term of the product is always obtained from the multiplication of the leading terms of the factors, and there can be no similar terms; also, the lowest term of the product is obtained from the multiplication of the lowest terms of the factors, and there cannot be similar terms.

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 are mutually annihilated, and only the older and younger ones 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 (write only the result)
(x 4 - x³ + x² - x + 1) (x + 1) = x 5 + 1 etc.

These results are noteworthy and useful to remember.

The following case of multiplication is especially important:

(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, when applied to arithmetic, we have the product of the sum of two numbers by their difference, and the result is the difference of the squares of these numbers.

If we see a similar case, then there is no need to perform the multiplication in detail, as was done above, but you 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 is 1, and the second factor is the difference of the same numbers; therefore, the result should be: the square of the first number (that is, 3a ∙ 3a = 9a²) minus the square of the second number (1 ∙ 1 = 1), that is.

(3a + 1) ∙ (3a - 1) = 9a² - 1.

Also

(ab - 5) ∙ (ab + 5) = a²b² - 25, etc.

So let's remember

(a + b) (a - b) = a² - b²

that is, the product of the sum of two numbers by their difference is equal to the difference of the squares of these numbers.

Target:

1. Ensure the assimilation of the initial knowledge on the topic "Multiplication of a monomial by a polynomial";
2. Develop analytical and synthesizing thinking;
3. Foster motives for learning and a positive attitude towards knowledge.

Team building of the class.

Tasks:

1. Get acquainted with the algorithm for multiplying a monomial by a polynomial;
2. Work out practical use algorithm.

Equipment: cards with tasks, 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 "Multiplication of 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 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 your tables in which we will enter your points scored throughout the lesson, and a grade will be given as a result. We will represent the points in the form of emoticons. ( Annex 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 you gained in previous lessons.

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

Frontal work:

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

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

2) What are the terms 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 notation in literal form.

III. The stage of assimilating new knowledge.

We have repeated the rule of multiplying 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 walks over to the screen, writes down the missing part of the expression, and explains their point of view.)

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

What expression is obtained as a result of performing these actions?

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

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

IV. Anchoring:

1. Physical education:

Guys, sit back, close your eyes, relax, now we are resting, the muscles are relaxed, we are studying the topic "Multiplication of 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 expressions obtained. We open our eyes.

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

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

(When executing the number, the most typical errors are analyzed)

3. Competition by options (decoding of the pictogram). (Appendix 2)

Option 1:		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) 10x 2 y (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 NS 2 y (NSy 3 - 3NS+ y 2) 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)

The tasks are presented on individual cards and on the screen. Each student completes his task, finds a letter and writes it down on the screen opposite the expression that he transformed. If you get the right answer, you get the word: well done! smart guys 7a

In this video tutorial, we will consider in detail the question of multiplying a polynomial by an expression that meets the definition of "monomial", or a monomial. Any free numerical value represented by natural number(to any extent, with any sign) or some variable (with similar attributes). It is worth remembering that a polynomial is a set of algebraic elements called terms of the polynomial. Sometimes some members can be similarized and abbreviated. It is strongly recommended to carry out the procedure of reduction of such terms after the operation of multiplication. The final answer, then, is the standardized form of the polynomial.

As it follows from our video, the process of multiplying a monomial by a polynomial can be viewed from two perspectives: linear algebra and geometry. Consider the operation of multiplying a polynomial on each side - this contributes to the universality of the application of the rules, especially in the case of complex problems.

In the algebraic sense, multiplication of a polynomial by a monomial corresponds to the standard rule for multiplication by a sum: each element of the sum must be multiplied by a given value, and the resulting value must be added algebraically. It should be understood that any polynomial is an expanded algebraic sum. After multiplying each term of the polynomial by some value, we get a new algebraic sum, which is usually reduced to standard view, if it is possible of course.

Consider the multiplication of a polynomial in this case:

3а * (2а 2 + 3с - 3)

It is easy to understand that here the expression (2a 2 + 3c - 3) is a polynomial, and 3a is a free factor. To solve this expression, it is enough to multiply each of the three terms of the polynomial by 3a:

It is worth remembering that the sign is an important attribute of the variable on the right, and it cannot be lost. The "+" sign is usually not written if an expression begins with it. When multiplying numerical-letter expressions, all coefficients for variables are elementarily multiplied. The same variables increase the degree. Different variables remain unchanged, and are written in one element: a * c = ac. Knowledge of these simple addition rules contributes to the correct and quick solution of any such exercises.

We got three values, which are, in fact, members of the final polynomial, which is the answer to the example. It is only necessary to add these values ​​algebraically:

6a 3 + 9ac + (- 9a) = 6a 3 + 9ac - 9a

We open the brackets, keeping the signs, since this is an algebraic addition, and by definition there is a plus sign between the monomials. The final standard form of the polynomial is the correct answer to the presented example.

The geometric form of multiplying a polynomial by a monomial is the process of finding the area of ​​a rectangle. Suppose we have a certain rectangle with sides a and c. The figure is divided by two segments into three rectangles of different areas, so that side c is common for all, or the same. And the sides a1, a2 and a3 add up to the initial a. As is known from the axiomatic definition of the area of ​​a rectangle, to find this parameter, it is necessary to multiply the sides: S = a * c. Or, S = (a1 + a2 + a3) * c. Let's multiply the polynomial (formed by the sides of the smaller rectangles) by the monomial - the main side of the figure, and we get the expression for S: a1 * c + a2 * c + a3 * c. But if you look closely, you will notice that this polynomial is the sum of the areas of three smaller rectangles that make up the initial figure. Indeed, for the first rectangle S = a1c (by the axiom), etc. Algebraically, the correctness of the reasoning when adding a polynomial is confirmed by calculations of linear algebra. And geometrically - by the rules for adding areas in a single simplest figure.

When carrying out manipulations with multiplying a polynomial by a monomial, it should be remembered that in this case the degrees of the monomial and the polynomial (general) add up - and the resulting value is the degree of the new polynomial (answer).

All of the above rules along with the basics algebraic addition are used in examples of the simplest simplification of expressions, where such terms are reduced and elements are multiplied to simplify the entire polynomial.

When multiplying a polynomial by a monomial, we will use one of the laws of multiplication. He received in mathematics the name of the distribution law of multiplication. Distributional multiplication law:

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 sufficient to multiply each of the members of the polynomial by a monomial. After that, add the resulting works. The following figure shows a scheme for multiplying a monomial by a polynomial.

The order of multiplication is not important, if, for example, you need to multiply a polynomial by a monomial, then you need to act in exactly the same way. So there is no difference between 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 monomial written above. And we will show on specific example how to do it correctly:

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

Using the distribution 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 by 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 distribution 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 is 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 distribution 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, each of the monomials is reduced to its standard form. We get:

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

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

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

The rule of 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 relative to addition. Literally, this property is written as follows: (a + b) c = a c + b c (a, b and c- some numbers). In this entry, the expression (a + b) c is precisely the product of the polynomial (a + b) and the monomial c... The right side of equality a c + b c is the sum of the products of monomials a and b on a monomial c.

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

Definition 1

To carry out the action of multiplying a polynomial by a monomial, it is necessary:

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

Let us additionally explain the above algorithm.

To compose the product of a polynomial by a monomial, the original polynomial is enclosed in parentheses; then the multiplication sign is placed between it and the given monomial. In the case when the record of a monomial begins with a minus sign, it must also be enclosed in parentheses. 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 a 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 to multiply 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. Suppose that after the first step of the algorithm we got the expression (2 x 2 + x + 3) 5 x, then in the second step we multiply each term of the polynomial 2 x 2 + x + 3 with 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 monomials 10 x 3, 5 x 2 and 15 x.

The last action according to the rule is the addition of the received works. From the proposed example, after completing this step of the 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 the polynomial 2 x 2 + x + 3 and monomial 5 x we 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, a short solution can be formulated as follows: (2 x 2 + x + 3) 5 x = 10 x 3 + 5 x 2 + 15 x.

The examples considered 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 multiplicable polynomial and monomial.

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

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.

Solution

The first step of the rule has already been completed - the piece has been recorded. Now we carry out the next step, multiplying each term of the polynomial by the given monomial. In this case, it is convenient to first translate decimal fractions of ordinary ones. 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 non-standard form, before finding their work, it is advisable to bring them to a standard form.

Example 2

Polynomial 3 + a - 2 a 2 + 3 a - 2 and monomial - 0.5 a b (- 2) a... It is necessary to find their work.

Solution

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 we carry out the multiplication of the monomial a 2 b for each term of the polynomial 1 + 4 a - 2 a 2

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 have brought the initial data to a standard form: the solution in this case would be more cumbersome. In this case, the last step would arise the need to bring such members. For understanding, we will give 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 a (3 + a - 2 a 2 + 3 a - 2) = a 2 b + 4 a 3 b - 2 a 4 b.

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