Multiplication of a polynomial by a monomial. Typical tasks
>>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 the specified procedure is assigned special name: parenthesize the common factor.
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 faster than that 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:; Consequently,
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 photos, 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 guidelines discussion programs Integrated LessonsIn the presented video tutorial, we will consider in detail the issue of multiplying a polynomial by any expression that meets the definition of "mono", or monomial. A monomial can be any free numeric value represented by natural number(to any degree, with any sign) or some variable (with similar attributes). It is worth remembering that a polynomial is a set of algebraic elements called members of a polynomial. Sometimes some terms can be given with similarity and abbreviated. It is strongly recommended to carry out the procedure of reducing like terms after the multiplication operation. The final answer, in this case, will be the standardized form of the polynomial.
As follows from our video, the process of multiplying a monomial by a polynomial can be considered from two positions: 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.
Algebraically speaking, multiplying a polynomial by a monomial follows the standard rule for multiplying by a sum: each element of the sum must be multiplied by a given value, and the resulting value must be algebraically added. It should be understood that any polynomial is an expanded algebraic sum. After multiplying each term of the polynomial by a certain 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:
3a * (2a 2 + 3c - 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, as a rule, is not written if the 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 algebraically add these values:
6a 3 + 9ac + (- 9a) \u003d 6a 3 + 9ac - 9a
We open the brackets, preserving the signs, since this is algebraic addition, and by definition there is a plus sign between monomials. The resulting standard form of the polynomial is the correct answer to the presented example.
The geometric view of multiplying a polynomial by a monomial is the process of finding the area of a rectangle. Suppose we have a 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 to 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) * s. We will multiply the polynomial (formed by the sides of smaller rectangles) by the monomial - the main side of the figure, and we will obtain an expression for S: a1 * c + a2 * c + a3 * c. But if you look closely, you can see that this polynomial is the sum of the areas of three smaller rectangles, which make up the initial figure. Indeed, for the first rectangle S = a1c (according to the axiom), and so on. Algebraically, the correctness of reasoning when adding a polynomial is confirmed by linear algebra calculations. And geometrically - the rules for adding areas in a single simple figure.
When carrying out manipulations with the multiplication of a polynomial by a monomial, it should be remembered that in this case the degrees of the monomial and the polynomial (general) are added - 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 similar terms are reduced and elements are multiplied to simplify the entire polynomial.
Target:
- Ensure the assimilation of initial knowledge on the topic "Multiplying a monomial by a polynomial";
- Develop analytical and synthesizing thinking;
- To cultivate the motives of teaching and a positive attitude towards knowledge.
Team building of the class.
Tasks:
- To get acquainted with the algorithm for multiplying a monomial by a polynomial;
- 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. ( Attachment 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
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.
