HP MOBU "Poikovskaya secondary school No. 2"
Open algebra lesson in 7th grade
on this topic:
"Multiplication of a monomial by a polynomial"
math teachers
Limar T. A.
town of Poikovsky, 2014
Methodical information
Lesson type
Lesson of "discovery" of new knowledge
The objectives of the lesson (educational, developing, educational)
Activity goal of the lesson : the formation of students' abilities to independently build new ways of action on the topic "Multiplication of a monomial by a polynomial" based on the method of reflexive self-organization.
educational goal : 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:
The development of attention, memory, the ability to reason and argue their actions through the solution of a problematic task;
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 communication 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,
Method of reflective self-organization (activity method),
Formation of personal UUD.
Didactic support of the lesson:
computer presentation,
task cards,
class assessment cards,
Cards with practical tasks on a new topic.
Stages lesson
Teacher activity
Student activities
organizational stage. (1 min)
Objectives: updating the knowledge of students, determining the objectives of the lesson, dividing the class into groups (different levels), choosing a group leader.
Psychological attitude, greeting students.
Greets students, calls the epigraph of the lesson. Offers to take places in pre-distributed groups and gives a preliminary briefing.
Hello, have a seat. Guys, thousands of years before our birth, Aristotle said that "... mathematics ... reveals order, symmetry and certainty, and these are the most important types of beauty." And after each lesson in the world of mathematics, there is less uncertainty. I hope that today we will discover something new for ourselves.
During the lesson, you will complete the evaluation sheet, which lies on your tables, after completing each task.
Students are seated in pre-divided groups. Familiarize yourself with the score sheet.
Verbal counting.
Purpose: to check the assimilation of theoretical material on the topic: “Multiplying a monomial by a monomial. Exponentiation" and the ability to apply it in practice, the development of students' thinking skills, awareness of the value of joint activities, the struggle for the success of the group.
a) math dictation.
Give similar monomials.
a) 2x+4y+6x=
b) -4a + c-3a \u003d
c) 3c+2d+5d=
d) -2d + 4a-3a =
2. Multiply a monomial by a monomial
a) -2x 3x
b) (-4av) (-2c)
d) (-5av) (2z)
e) 2z (x + y)
The teacher offers to perform a mathematical dictation written on the blackboard. Controls the correct execution, leads to the study of new material.
Together with students, formulates the purpose and topic of the lesson
- Which of the dictation numbers caused you the most difficulty?
Let's try to find out Where there was a difficulty and Why?
- The purpose of our lesson: to learn how to multiply a monomial by a polynomial (the validity of your decision).
Lesson topic: « U multiplication of a monomial by a polynomial.
Students complete assignments. Together with the teacher, formulate the purpose and topic of the lesson. Write down the topic of the lesson in notebooks.
(the expected answer of students is e)
Develop (formulate) a rule for multiplying a monomial by a polynomial.
Introduction to a new topic
Purpose: to prepare students for learning new material .
Group work.
Group #1.
Calculate.
15 80+15 20= 1200+300=1500
15 (80+20)=15 100=1500
Group #2
Calculate.
20 40+20 100=800+2000=2800
20 (40+100)=20 140=2800
Group #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. Controls execution.
Each group needs to find the meaning of two expressions. Compare them and write the conclusion as an equality or inequality.
Students solve examples in groups, draw a conclusion.
1 member from each group writes the conclusion on the board.
On the board is written:
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 grade themselves on a score sheet. If the conclusion is formulated and written correctly, then put 5.
"Discovery" of new material by students.
Target: the formation of students' abilities to independently build new ways of action on the topic "Multiplication of a monomial by a polynomial" based on the method of reflexive self-organization.
Completing the task "Fill in the blanks"
Slide 2.
2z ∙(x + y )=2z ∙ +2z ∙
3x(a+b)=a+b
After a minute, the correct solution is displayed on the board.
The teacher gives instructions.
Conducts a survey. Makes a conclusion.
Using the equations written on the board, fill in the gaps in the following expressions
Pay attention to what comes before the bracket?
What is in brackets?
What happens in the answer?
And so, let's conclude how to multiply a monomial by a polynomial. After three minutes, they present their material to the class (a white sheet and felt-tip pens are used).
Generalizes
Let's check if you formulated the rule correctly. To do this, open the textbook on p.
The students work in groups, each group discusses how to fill in the gaps.
Check if the blanks are filled in correctly.
Each group puts forward its hypothesis and presents it to the class, there is a general discussion and a conclusion is made.
Read aloud the rule from the textbook.
Monomial
Polynomial
New polynomial
Primary fastening.
Purpose: practicing the skills of multiplying a monomial by a polynomial, developing the mental skills of students, understanding the value of joint activities, fighting for the success of the group, increasing the motivation for learning activities.
Group work.
Group #1, 3
x∙(
m∙(n+3)=__________________ ; 7a ∙ (2b -3c ) = _______________ ;
Group #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.
Perform a mutual check, group 1 exchanges cards with group 3, and group 2 with group 4. Assign marks to the groups on the score sheet:
5 correctly completed tasks - grade "5"; 4 - "4"; 3- "3"; less than 3 - "2".
They complete the task on the cards, conduct a mutual check.
The responsible member of group #1 asks any member of group #3. Puts a score on the score sheet.
the responsible member of group #2 asks any member of group #4. Puts a mark 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 instructs, shows cards on which monomials, polynomials and expressions that are neither monomials nor polynomials are written.
Students do the exercises in order
"Single member" - hands raised up; "Polynomial" - hands in front of you; "Another expression" - hands to the sides;
We closed our eyes, silently counted to 30, opened our eyes.
Math lotto
Purpose: to fix the algorithm for multiplying a monomial by a polynomial and to stimulate interest in mathematics
Group No. 1,3
c(3a-4c)=3ac-12s;
3) 3c(x-3y)=3cx-9cy;
4) -n(x-m)=-nx+nm;
5) 3z(x-y)= 3zx-3zy .
Answer cards:
3as-12sun; 3ac+12sun; 3ac-4c
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 #2, 4
Multiply a monomial by a polynomial
5a(b+3d)=5ab+15ad
A (3v + s) \u003d -3av-ac;
4x (5c -s )=20cx -4xs ;
a(3c+2b)=3ac+2ba
Answer cards:
3av-ac; 3ab + ac; you;
20cx -4xs ; 20cx+4xs ; 5c-4xs;
3ac+2ba; 3ac+6ba; 3ac-2ba;
cp-5cm; avg-5m; p-5cm.
5ab+ad; 5ab+5b; 5ab+15ad
Handing out envelopes. Tells the rules of the game. One envelope contains 5 examples of multiplying a monomial by a polynomial and 15 cards with answers.
I explain how to evaluate the work done.
The group receives a mark of "5" if the first to complete 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 a score of "4", the third - "3", the last - "2".
Receive envelopes with assignments.
Multiply a monomial by a monomial.
Choose the correct answer from all the given cards.
Self-test.
Get a self-check card. Put the score on the score sheet.
8 . Reflection of educational activity in the lesson (the result of the lesson).
Purpose: self-assessment by students of the results of their educational activities, awareness of the method of building boundaries and applying a new mode 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 evaluation sheets (their results are visible on the slide)
Returns to the motto of the lesson, draws a parallel between the epigraph and the algorithm derived in the lesson.
Hand over evaluation sheets, which clearly show the result of your activities.
Once again, let's return 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 derived today in the lesson will help us to make new discoveries in the future: the multiplication of a polynomial by a polynomial will help to learn the formulas for abbreviated multiplication, which are talked about a lot in algebra. A lot of interesting and important things await us in front of us.
Thank you for the lesson!!!
Students do self-analysis of their work, remember the algorithm studied in the lesson, answer questions.
APPLICATION.
CARD #1.
Group #1.
Calculate.
15 80+15 20= ______________________________
15 (80+20)= _______________________________
CARD #1.
Group #2
Calculate.
20 40+20 100 =_________________________________
20 (40+100)= __________________________________
CARD #1.
Group #3.
Calculate.
6 (2a + 3a) \u003d _____________________________________
6 2a+6 3a=_____________________________________
CARD #1
Group No. 4
Calculate
7 (4x + 2x) = _____________________________________
7 4x+7 2x= _____________________________________
CARD #2.
Group #3
x∙( z + y ) = __________________ ; a ∙ (c + d) \u003d ___________________;
5x∙(3a-6a)= _______ -________= _______.
CARD №4.
Group #2
7x ∙ (5d -8d )= ______ - ________= _______.CARD #2.
Group #1
x∙( z + y ) = __________________ ; a ∙ (c + d) \u003d ___________________;
m∙(n+3)=__________________ ; 7a∙(2b-3c) = _______________ ;
5x∙(3a-6a)= _______ -________= _______.
CARD №2.
Group #2
a ∙ (c-y ) = __________________ ; c ∙ (c + d) \u003d ___________________;
m∙(y+5)=__________________ ; 6m ∙ (2n -3k ) = ______________ ;
7x ∙ (5d -8d )= ______ - ________= _______.Math Lotto (two copies)
c(3a-4c)
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)
Answers to the lotto (two copies)
3as-12sun
3as+12sun
3ac-4c
zx+2zy;
zx-2zy
zx+2y
3sh-9su
3cx-3cy
3sh+3su
Nx+nm
nx+nm
nx-nm
zx-zy
3zx-y
3zx-3zy
3av-ac
3ab + ac;
you
20cx-4xs
20cx+4xs
5c-4xs
3ac+2ba
3ac+6ba
3ac-2ba
cp-5cm
Wed -5m
p-5cm.
5ab+ad
5ab+5b
>>Math: Multiplication of a polynomial by a monomial
Multiplication of a polynomial by a monomial
You may have noticed that so far Chapter 4 has been structured according to the same plan as Chapter 3. In both chapters, the basic concepts were first introduced: in Chapter 3, these were the monomial, the standard form of the monomial, the coefficient of the monomial; in chapter 4 - polynomial, the standard form of a polynomial. Then in Chapter 3 we looked at addition and subtraction of monomials; similarly, in chapter 4 - addition and subtraction of polynomials.
What happened next in chapter 3? Then we talked about the multiplication of monomials. So, by analogy, what should we talk about now? On the multiplication of polynomials. But here we have to proceed slowly: first (in this paragraph) we consider the multiplication of a polynomial by monomial(or a monomial by a polynomial, it doesn't matter), and then (in the next paragraph) - the multiplication of any polynomials. When you learned to multiply numbers in elementary school, you also acted gradually: first you learned to multiply a multi-digit number by a single-digit number and only then did you multiply a multi-digit number by a multi-digit one.
(a + b)c \u003d ac + bc.
Example 1 Perform multiplication 2a 2 - Zab) (-5a).
Solution. Let's introduce new variables:
x \u003d 2a 2, y \u003d Zab, z \u003d - 5a.
Then this product will be rewritten in the form (x + y)z, which, according to the distribution law, is equal to xr + yz. Now back to the old variables:
xz + yz - 2a 2 (- 5a) + (- Zab) (- 5a).
It only remains for us to find products of monomials. We get:
- 10a 3 + 15a 2 b
We give a brief notation of the solution (this is how we will write it in the future without introducing new variables):
(2a 2 - Zab) (- 5a) \u003d 2a 2 (- 5a) + (- Zab) (- 5a) \u003d -10a 3 + 15a 2 b.
Now we can formulate the corresponding rule for multiplying a polynomial by a monomial.
The same rule applies when multiplying a monomial by a polynomial:
- 5a (2a 2 - Zab) \u003d (- 5a) 2a 2 + (- 5a) (- Zab) \u003d 10a 3 + 15a 2 b
(we took example 1, but swapped the factors).
Example 2 Express a polynomial as a product of a polynomial and a monomial if:
a) p1(x, y) - 2x 2 y + 4a:;
b) p 2 (x, y) \u003d x 2 + Zu 2.
Solution.
a) Note that 2x 2 y \u003d 2x xy, and 4a: \u003d 2x 2. Hence,
2x 2 y + 4x = xy 2x + 2 2x = (xy + 2) 2x
b) In example a), we succeeded in the composition of each member of the polynomial p 1 (x, y) \u003d 2x 2 y + 4a: select the same part (the same factor) 2x. Here, there is no such common part. This means that the polynomial p 2 (x, y) \u003d x 2 + Zy 2 cannot be represented as a product of a polynomial and a monomial.
In fact, the polynomial p 2 (x, y) can also be represented as a product, for example, like this:
x2 + 3y2 = (2x2 + 6y2) 0.5
or like this:
x 2 + 3y 2 = (x 2 + 3y 2) 1
- the product of a number by a polynomial, but this is an artificial transformation and is not used without great necessity.
By the way, the requirement to represent a given polynomial as a product of a monomial and a polynomial is quite common in mathematics, so this procedure has been given a special name: taking the common factor out of brackets.
The task to take the common factor out of brackets may be correct (as in example 2a), or it may not be entirely correct (as in example 26). In the next chapter, we will deal specifically with this issue.
At the end of the section, we will solve problems that will show how, in practice, to work with mathematical models real situations, one has to make up an algebraic sum of polynomials, and multiply a polynomial by a monomial. So we study these operations not in vain.
Example 3 Points A, B and C are located on the highway as shown in Figure 3. The distance between A and B is 16 km. A pedestrian left B towards C. 2 hours later, a cyclist left A towards C, whose speed is 6 km/h more than the speed of a pedestrian. 4 hours after his departure, the cyclist caught up with the pedestrian at point C. What is the distance from B to C?
Solution.
First stage. Drawing up a mathematical model. Let x km/h be the speed of a pedestrian, then (x + 6) km/h is the speed of a cyclist.
The cyclist traveled the distance from A to C in 4 hours, which means that this distance is expressed by the formula 4 (x + 6) km; in other words, AC = 4 (x + 6).
The distance from B to C was covered by the pedestrian in 6 hours (after all, before the cyclist left, he had already been on the road for 2 hours), therefore, this distance is expressed by the formula 6x km; in other words, BC = 6x
And now pay attention to Figure 3: AC - BC = AB, i.e. AC - BC = 16. This is the basis for compiling a mathematical model of the problem. Recall that AC = 4 (x + 6), BC = 6x:; hence,
4(x + 6) -6x = 16.
A. V. Pogorelov, Geometry for grades 7-11, Textbook for educational institutions
Lesson content lesson summary support frame lesson presentation accelerative methods interactive technologies Practice tasks and exercises self-examination workshops, trainings, cases, quests homework discussion questions rhetorical questions from students Illustrations audio, video clips and multimedia photographs, pictures graphics, tables, schemes humor, anecdotes, jokes, comics parables, sayings, crossword puzzles, quotes Add-ons abstracts articles chips for inquisitive cheat sheets textbooks basic and additional glossary of terms other Improving textbooks and lessonscorrecting errors in the textbook updating a fragment in the textbook elements of innovation in the lesson replacing obsolete knowledge with new ones Only for teachers perfect lessons calendar plan for the year methodological recommendations of the discussion program Integrated LessonsA special case of multiplying a polynomial by a polynomial is the 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 is based 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 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 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, a short solution can be formulated 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 an 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 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 work has been recorded. Now we perform the next step, multiplying each term of the polynomial by the given monomial. In this case, it is convenient to first translate the decimal fractions into common 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 the 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.
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 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
I.To multiply a monomial by a polynomial, it is necessary to multiply each term of the polynomial by this monomial and add the resulting products.
Example 1 Multiply a monomial by a polynomial: 2a (4a 2 -0.5ab+5a 3).
Solution. Monomial 2a we will multiply by each monomial of the polynomial:
2a (4a 2 -0.5ab+5a 3)=2a∙4a 2 +2a∙(-0.5ab)+2a∙5a 3=8a 3 -a 2 b+10a 4 . We write the resulting polynomial in standard form:
10a 4 +8a 3 -a 2 b.
Example 2 Multiply a polynomial by a monomial: (3xyz 5 -4.5x 2 y+6xy 3 +2.5y 2 z)∙(-0.4x 3).
Solution. Each term in brackets is multiplied by a monomial (-0.4x3).
(3xyz 5 -4.5x 2y+6xy 3 +2.5y 2z)∙(-0.4x 3)=
3xyz 5 ∙(-0.4x 3) -4.5x 2 y∙(-0.4x 3)+6xy 3 ∙(-0.4x 3)+2.5y 2 z∙(-0.4x 3)=
=-1.2x 4 yz 5 +1.8x 5 y-2.4x 4 y 3 -x 3 y 2 z.
II.Representing a polynomial as a product of two or more polynomials is called factoring a polynomial.
III.Taking the common factor out of brackets is the simplest way to factorize a polynomial.
Example 3 Factorize the polynomial: 5a 3 +25ab-30a 2 .
Solution. We take out the common factor of all members of the polynomial in brackets. This is a monomial 5a, because on 5a each of the terms of this polynomial is divisible. So, 5a we write before the brackets, and in brackets we write the quotients from dividing each monomial by 5a.
5a 3 + 25ab-30a 2 \u003d 5a (a 2 + 5b-6a). Checking ourselves: if we multiply 5a to polynomial in brackets a 2 +5b-6a, then we get this polynomial 5a 3 +25ab-30a 2.
Example 4 Take the common factor out of brackets: (x+2y) 2 -4 (x+2y).
Solution.(x+2y) 2 -4 (x+2y)= (x+2y)(x+2y-4).
The common factor here is the binomial (x + 2y). We took it out of brackets, and in brackets we wrote the private members of the division of these members (x+2y) 2 And -4 (x+2y) to their common divisor
(x + 2y). As a result, we presented this polynomial as a product of two polynomials (x+2y) And (x+2y-4), in other words, we have expanded the polynomial (x+2y) 2 -4 (x+2y) for multipliers. Answer: (x+2y)(x+2y-4).
IV.To multiply a polynomial by a polynomial, you need to multiply each term of one polynomial by each term of the other polynomial and write the resulting products as a sum of monomials. If necessary, add like terms.
Example 5 Perform polynomial multiplication: (4x2 -6xy+9y2)(2x+3y).
Solution. As a rule, we must multiply each term of the first polynomial (4x 2 -6xy + 9y 2) by each term of the second polynomial (2x + 3y). In order not to get confused, always do this: first, multiply each term of the first polynomial by 2x, then again multiply each term of the first polynomial by 3y.
(4x 2 -6xy+9y 2)( 2x+3y)=4x 2 ∙ 2x-6xy∙ 2x+9y 2 ∙ 2x+4x 2 ∙ 3y-6xy∙ 3y+9y 2 ∙ 3y=
8x 3 -12x 2 y+18xy 2 +12x 2 y-18xy 2 +27y 3 =8x 3 +27y 3 .
Similar terms -12x 2 y and 12x 2 y, as well as 18xy 2 and -18xy 2 turned out to be opposite, their sums are equal to zero.
Answer: 8x 3 +27y 3 .
Page 1 of 1 1
For a single? How to correctly place signs when multiplying?
Rule.
To multiply a polynomial by, you need to multiply each term of the polynomial by a monomial and add the results.
It is convenient to write the monomial before the brackets.
In order to correctly place the signs during multiplication, it is better to use the rule for opening brackets, which are preceded by a plus sign or a minus sign.
Multiplications of a polynomial by a monomial can be represented using a diagram.
We multiply the monomial by each term of the polynomial in brackets (“fountain”).
If there is a "+" sign in front of the brackets, the characters in the brackets do not change:
If there is a "-" sign in front of the brackets, each character in the brackets is reversed:
Consider how to multiply a polynomial by a monomial, using specific examples.
Examples.
Multiply a polynomial by a monomial:
Solution:
We multiply the monomial by each term of the polynomial in parentheses. Since there is a plus sign in front of the parentheses, the characters in the parentheses do not change:
We multiply the numbers separately, separately - with the same bases:
We multiply the monomial by each term of the polynomial. Since there is a multiplier in front of the brackets, we change the sign of each term in brackets to the opposite:
Usually they write shorter, the multiplication of powers and numbers (with the exception of ordinary fractions and mixed numbers) is performed orally.
If the coefficients are ordinary fractions, then we multiply them according to the rule for multiplying ordinary fractions: the numerator by the numerator, the denominator by the denominator, and immediately writing them under one fractional line. If the coefficients are mixed numbers, we translate them into improper fractions:
Attention!
We do not reduce fractions until we have written down all the actions to the end. As practice shows, if you immediately start with the reduction of fractions, then the rest of the terms do not reach - they are simply forgotten.
