Multiplication of a polynomial by a monomial. Typical tasks
NR MOBU "Poikovskaya average comprehensive school No. 2 "
Public lesson in algebra in grade 7
on this topic:
"Multiplication of a monomial by a polynomial"
Math teachers
Limar T.A.
Poikovsky settlement, 2014
Methodical information
Lesson type
The lesson of "discovering" new knowledge
Lesson objectives (educational, developmental, educational)
Activity purpose of the lesson : the formation of students' abilities for independent construction of new ways of action on the topic "Multiplication of a monomial by a polynomial" based on the method of reflexive self-organization.
Educational purpose : expansion of the conceptual base on the topic "Polynomials" by including new elements in it: multiplication of monomials by a polynomial.
Lesson Objectives
educational:
Develop an algorithm for multiplying a monomial by a polynomial, consider examples of its application.
developing:
Development of attention, memory, the ability to reason and argue their actions through solving a problem problem;
Development of cognitive interest in the subject;
Formation of an emotionally positive attitude among students through the use of active forms of lesson management and the use of ICT;
The development of reflective skills through the analysis of the results of the lesson and introspection of their own achievements.
educational:
Development of communicative skills of students through the organization of group, pair and frontal work in the lesson.
Methods used
Verbal methods(conversation, reading),
Visual (demonstration of the presentation),
Problem search engine,
Reflexive self-organization method (activity method),
Formation of personal UUD.
Didactic support of the lesson:
Computer presentation,
Quest cards,
Assessment cards for the work in the lesson,
Cards with practical assignments on new topic.
Stages of a lesson
Teacher activity
Student activities
Organizational stage. (1 minute)
Objectives: updating students' knowledge, defining the objectives of the lesson, dividing the class into groups (different levels), choosing a group leader.
Psychological attitude, greeting students.
Greets the students, names the epigraph of the lesson. He offers to take places in pre-assigned groups and gives preliminary instructions.
Hello, have a seat. Guys, thousands of years before our birth, Aristotle said that “... mathematics ... reveals order, symmetry and certainty, and this - the most important species beautiful ". And after each lesson in the world of mathematics, the uncertainty becomes less. I hope that today we will discover something new for ourselves.
During the lesson, you will fill out the assessment sheet that lies on your desks after completing each assignment.
Students are seated in pre-divided groups. Acquainted with the score sheet.
Verbal counting.
Purpose: check assimilation theoretical material on the topic: “Multiplication of a monomial by a monomial. Exponentiation "and the ability to apply it in practice, the development of students' thinking skills, awareness of the value joint activities fighting for the success of the group.
a) mathematical dictation.
Bring similar monomials.
a) 2x + 4y + 6x =
b) -4a + c-3a =
c) 3c + 2d + 5d =
d) -2d + 4a-3a =
2. Multiply a monomial by a monomial
a) -2xy 3x
b) (-4av) (-2c)
d) (-5av) (2z)
e) 2z (x + y)
The teacher offers to complete the math dictation written on the board. Controls the correctness of implementation, leads to the study of new material.
Together with the students, formulates the purpose and topic of the lesson
- Which of the numbers of the dictation caused you the most difficulties?
Let's try to find out where it was the difficulty that arose and why?
- The purpose of our lesson: to learn how to perform multiplication of a monomial by a polynomial (the fairness of your decision).
Lesson topic: "U multiplication of a monomial by a polynomial ”.
Students complete assignments. Together with the teacher, formulates the purpose and topic of the lesson. Write down the topic of the lesson in notebooks.
(expected student answer d)
Develop (formulate) a rule for multiplying a monomial by a polynomial.
Leading up to a new topic
Objective: to prepare students to learn new material .
Group work.
Group # 1.
Calculate.
15 80+15 20= 1200+300=1500
15 (80+20)=15 100=1500
Group no. 2
Calculate.
20 40+20 100=800+2000=2800
20 (40+100)=20 140=2800
Group number 3.
Calculate.
6 (2a + 3a) = 6 5a = 30a
6 2a + 6 3a = 12a + 18a = 30
Group No. 4
Calculate
7 (4x + 2x) = 7 6x = 42
7 4x + 7 2x = 28x + 14x = 42x
The teacher gives instructions. Monitors execution.
Each group needs to find the meaning of two expressions. Compare them and write the output as equality or inequality.
Students solve examples in groups, make a conclusion.
1 member from each group writes the conclusion on the board.
The chalkboard reads:
15 80+15 20=15 (80+20)
20 40+20 100=20 (40+100)
6 (2a + 3a) = 6 2a + 6 3
7 (4x + 2x) = 7 4x + 7 2x
Students put themselves on a grade sheet. If the conclusion is formulated and written correctly, then put 5.
"Discovery" of new material by students.
Target: the formation of students' abilities for the independent construction of new methods of action on the topic "Multiplication of a monomial by a polynomial" based on the method of reflexive self-organization.
Completion of the task "Fill in the blanks"
Slide 2.
2z ∙ (x + y) = 2z ∙ + 2z ∙
3x (a + b) = a + b
A minute later, the board displays the right decision.
The teacher gives instructions.
Conducts a survey. Concludes.
Using the equations on the chalkboard, fill in the blanks in the following expressions
Notice what comes before the parenthesis?
What's in brackets?
What is the answer?
And so, let's conclude how to multiply a monomial by a polynomial. After three minutes, present their material to the class (used White list and felt-tip pens).
Summarizes
Let's check if you have formulated the rule correctly. To do this, open the tutorial on p.
Students work in groups, with each group discussing how to fill in the blanks.
Check the correctness of filling in the gaps.
Each group comes up with a hypothesis and presents it to the class, there is a general discussion and a conclusion is made.
Read aloud a rule from a textbook.
Monomial
Polynomial
New polynomial
Primary anchoring.
Goal: working out the skills of multiplying a monomial by a polynomial, developing the thinking skills of students, realizing the value of joint activities, fighting for the success of the group, increasing motivation learning activities.
Group work.
Group No. 1, 3
x ∙ (
m ∙ (n +3) = __________________; 7a ∙ (2b -3c) = _______________;
Group No. 2, 4
a ∙ (c-y) = __________________; c ∙ (c + d) = ___________________;
m ∙ (y + 5) = __________________; 6m ∙ (2n-3k) = ______________;
7
The teacher gives instructions.
Take on the desk card number 2 A prerequisite is when deciding to pronounce the rule to each other.
Do a cross-check, group 1 exchanges cards with group 3, and group 2 with group 4. Put the marks to the groups on the score sheet:
5 correctly completed tasks - score "5"; 4 - "4"; 3- "3"; less than 3- "2".
They carry out the task on the cards, carry out a mutual check.
The responsible member of group # 1 asks any member of group # 3. Puts the grade on the score sheet.
the responsible member of group # 2 asks any member of group # 4. Puts the grade on the score sheet
6. Mathematical exercises.
Purpose: to increase or maintain the mental performance of children in the classroom;
provide short-term active rest for students during the lesson.
The teacher gives instructions, shows cards on which monomials, polynomials and expressions are written that are neither monomials nor polynomials.
Students complete the exercises in commands
"Monomial" - hands raised up; "Polynomial" - hands in front of you; "Another expression" - arms to the sides;
They closed their eyes, counted to themselves to 30, opened their eyes.
Mathematical Lotto
Objective: to consolidate the algorithm for multiplying a monomial by a polynomial and to stimulate interest in mathematics
Group No. 1.3
s (3a-4b) = 3ac-12vs;
3) 3c (x-3y) = 3cx-9cy;
4) -n (x-m) = - nx + nm;
5) 3z (x-y) = 3zx-3zy .
Answer cards:
3ac-12vs; 3ac + 12vs; 3ac-4v
zx + 2zy; zx-2zy; zx + 2y;
3cx-9cy; 3cx + 9cy; 3cx-3cy;
Nx + nm; nx + nm; nx-nm;
3zx-3zy; 3zx-y; zx-zy.
Group No. 2, 4
Multiply a monomial by a polynomial
5a (b + 3d) = 5ab + 15ad
A (3b + c) = - 3av-ac;
4x (5c -s) = 20cx -4xs;
a (3c + 2b) = 3ac + 2ba
Answer cards:
3av-as; 3av + ac; you;
20cx -4xs; 20cx + 4xs; 5c -4xs;
3ac + 2ba; 3ac + 6ba; 3ac-2ba;
cp-5cm; Wed-5m; p-5cm.
5ab + ad; 5ab + 5b; 5ab + 15ad
Distributes envelopes. Tells the rules of the game. One envelope contains 5 examples of multiplication of a monomial by a polynomial and 15 cards with answers.
I am explaining how to evaluate the work performed.
The group receives a grade of "5" if the first one completed all the tasks correctly, 4 tasks - "4"; 3 tasks - "3", less than three - "2", the group that completes the game in the lotto second, while completing all the tasks, correctly receives the mark "4", the third - "3", the last - "2".
Receive envelopes with assignments.
The multiplication of a monomial by a monomial is carried out.
Choose the correct answers from all the cards offered.
Self-test.
Get a self-test card. Put the mark on the score sheet.
8 . Reflection of educational activities in the lesson (lesson summary).
Purpose: self-assessment by students of the results of their educational activities, awareness of the method of constructing boundaries and the use of a new method of action.
Frontal conversation on the questions on the slide:
What algorithm for multiplying a monomial by a polynomial exists in mathematics?
What is the result of your activity?
The teacher analyzes the grade sheets (their results are visible on the slide)
Returns to the motto of the lesson, draws a parallel between the epigraph and the algorithm deduced in the lesson.
Pass in scorecards that clearly show the result of your activity.
Let's go back to the motto of our lesson: "... mathematics ... reveals order, symmetry and certainty, and these are the most important types of beauty." The algorithm that we brought out today in the lesson will help us make new discoveries in the future: multiplying a polynomial by a polynomial will help to learn the abbreviated multiplication formulas, which are much talked about in algebra. A lot of interesting and important things await us in the front.
Thank you for the lesson!!!
Students make an introspection of their work, recall the algorithm learned in the lesson, answer questions.
APPENDIX.
CARD # 1.
Group # 1.
Calculate.
15 80+15 20= ______________________________
15 (80+20)= _______________________________
CARD # 1.
Group no. 2
Calculate.
20 40+20 100 =_________________________________
20 (40+100)= __________________________________
CARD # 1.
Group number 3.
Calculate.
6 (2a + 3a) = _____________________________________
6 2a + 6 3a = _____________________________________
CARD # 1
Group No. 4
Calculate
7 (4x + 2x) = _____________________________________
7 4x + 7 2x = _____________________________________
CARD # 2.
Group No. 3
x ∙ ( z + y) = __________________; a ∙ (c + d) = ___________________;
5x ∙ (3a-6a) = _______ -________ = _______.
CARD №4.
Group no. 2
7x ∙ (5d -8d) = ______ - ________ = _______.CARD # 2.
Group # 1
x ∙ ( z + y) = __________________; a ∙ (c + d) = ___________________;
m ∙ (n + 3) = __________________; 7a ∙ (2b-3c) = _______________;
5x ∙ (3a-6a) = _______ -________ = _______.
CARD №2.
Group no. 2
a ∙ (c -y) = __________________; c ∙ (c + d) = ___________________;
m ∙ (y +5) = __________________; 6m ∙ (2n -3k) = ______________;
7x ∙ (5d -8d) = ______ - ________ = _______.Mathematical Lotto (two copies)
s (3a-4b)
z (x + 2y)
3c (x-3y)
-n (x-m)
3z (x-y)
-a (3v + s)
4x (5c -s)
a (3c + 2b)
c (p-5m)
5a (b + 3d)
Lotto Answers (two copies)
3ac-12vs
3ac + 12vs
3ac-4v
zx + 2zy;
zx-2zy
zx + 2y
3sx-9su
3cx-3cy
3sх + 3su
Nx + nm
nx + nm
nx-nm
zx-zy
3zx-y
3zx-3zy
3av-as
3av + ac;
you
20cx -4xs
20cx + 4xs
5c -4xs
3ac + 2ba
3ac + 6ba
3ac-2ba
cp-5cm
Wed -5m
p-5cm.
5ab + ad
5ab + 5b
This lesson will study the operation of multiplying a polynomial by a monomial, which is the basis for studying the multiplication of polynomials. Let us recall the distribution law of multiplication and formulate the rule for multiplying any polynomial by a monomial. Let us also recall some properties of the degrees. In addition, will be formulated typical errors when performing various examples.
Topic: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 distribution law of multiplication. Let's remind him:
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, each term of the polynomial must be multiplied by this monomial. Add the algebraically obtained products, and then perform over the polynomial necessary actions- namely, bring it to standard view.
Let's consider an example:
A comment: given example is solved by exactly following the rule: each term of the polynomial is multiplied by a monomial. In order to understand and master 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 substitution of the initial values was performed. 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:
Commentary: this example is solved according to the distribution law and is similar to the previous example - each member of the polynomial is multiplied by a monomial, the resulting polynomial is already written in a standard form, so it cannot be simplified.
Example 2 - perform actions and get a polynomial in standard form:
Commentary: to solve this example, first we will multiply for the first and second binomials according to the distribution law, then we will bring the resulting polynomial to the standard form - we will bring similar terms.
Now we will formulate the main problems associated with the operation of multiplying a polynomial by a monomial, and give examples of their solution.
Objective1 is to simplify the expression:
Commentary: this example is solved similarly to the previous one, namely, first, the multiplication of polynomials by the corresponding monomials is performed, after which similar ones are reduced.
Task 2 is to simplify and calculate:
Example 1 :;
Commentary: this example is solved similarly to the previous one, with the only addition that after bringing similar members, you need to substitute its specific value instead of a variable and calculate the value of the polynomial. Recall to easily multiply decimal by ten, you need to move the comma one place to the right.
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 1It 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 (3 + a - 2 a 2 + 3 a - 2) = a 2 b + 4 a 3 b - 2 a 4 b.
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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 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 power of 1 less, that is, x to the 1st power, 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.
