Rule for calculating the product of polynomials.
In order to consider the product of polynomials, first let’s remember how to multiply a monomial by a polynomial.
The product of a monomial and a polynomial is found as follows:
- the product of a monomial and a polynomial is composed.
- The parentheses open.
- numbers are grouped with numbers that are the same variables friend with a friend.
- numbers are multiplied and the powers of the corresponding identical variables are added.
Let us now consider the multiplication of two polynomials using an example:
Example 1
Let's multiply the polynomial $x-y+z$ by the polynomial $\(xy)^5+y^6-(xz)^5$.
First, let's write down the product of polynomials:
\[\left(x-y+z\right)((xy)^5+y^6-(xz)^5)\]
Let's make the following replacement. Let $x-y+z=t$, we get:
We obtained the product of a monomial and a polynomial. Let's find it using the rule stated above.
Let's expand the brackets:
Let's make a reverse substitution:
\[(\left(x-y+z\right)xy)^5+(\left(x-y+z\right)y)^6-(\left(x-y+z\right)xz) ^5\]
IN this expression we see the presence of three products of monomials and a polynomial. Let's find them separately using the above rule:
\[(\left(x-y+z\right)xy)^5=x(xy)^5-y(xy)^5+z(xy)^5=(x^2y)^5-(xy )^6+z(xy)^5\] \[(\left(x-y+z\right)y)^6=xy^6-yy^6+zy^6=xy^6-y^7 +zy^6\] \[(\left(x-y+z\right)xz)^5=x(xz)^5-y(xz)^5+z(xz)^5=x^2z^ 5-xyz^5+(xz)^6\]
Let's rewrite our expression:
\[\left((x^2y)^5-(xy)^6+z(xy)^5\right)+\left(xy^6-y^7+zy^6\right)-(x^ 2z^5-xyz^5+(xz)^6)\]
Let's open the brackets. Let us remind you that if there is a plus sign in front of the brackets, then the signs in the brackets remain unchanged, and if there is a minus sign in front of the brackets, then the signs in the brackets will change to the opposite. We get
\[(x^2y)^5-(xy)^6+z(xy)^5+xy^6-y^7+zy^6-x^2z^5+xyz^5-(xz)^6 \]
We got a polynomial. All that remains is to bring him to standard view. In total, the answer will be:
\[(x^2y)^5+xy^5z-y^7+zy^6-x^2z^5+xyz^5-(xz)^6\]
Taking a closer look at the result obtained, we get next rule multiplying a polynomial by a polynomial:
Rule: In order to multiply a polynomial by a polynomial, it is necessary to multiply each term of the first polynomial by each term of the second polynomial, add the resulting products and reduce the resulting polynomial to standard form.
Example 2
Multiply $2x+y$ and $x^2+2y+3$.
Let's write down the product:
\[\left(2x+y\right)(x^2+2y+3)\]
\[\left(2x+y\right)\left(x^2+2y+3\right)=2x^3+4xy+6x+x^2y+2y^2+3y\]
We see that the resulting polynomial has a standard form, which means the multiplication is complete.
Examples of problems involving the product of polynomials
Example 3
Multiply a polynomial by a polynomial:
a) $(2z+1)\ and\ (z^2-7z-3)$
b) $(1-4x^2)\ and\ (5y^2-3x-2)$
Solution:
a) $(2z+1)\ and\ (z^2-7z-3)$
Let's compose a piece:
\[(2z+1)\cdot (z^2-7z-3)\]
Let's open the brackets according to the rule of product of polynomials:
b) $(1-4x^2)\ and\ (5y^2-3x-2)$
Let's compose a piece:
\[(1-4x^2)\cdot (5y^2-3x-2)\]
Let's open the brackets according to the rule of product of polynomials:
We see that the resulting polynomial has a standard form, therefore:
Answer: $5y^2-3x-2-20x^2y^2+12x^3+8x^2$.
c) $(2n-5n^3)\ and\ (3n^2-n^3+n)$
Let's compose a piece:
\[(2n-5n^3)\cdot (3n^2-n^3+n)\]
Let's open the brackets according to the rule of product of polynomials:
Let's give given polynomial to standard form:
d) $(a^2+a+1)\ and\ (a^2-24a+6)$
Let's compose a piece:
\[(a^2+a+1)\cdot (a^2-24a+6)\]
Let's open the brackets according to the rule of product of polynomials:
Let us reduce this polynomial to standard form.
One of the operations with polynomials is to multiply a polynomial by a polynomial. In this article we will consider the rule of such multiplication and apply it to solve problems.
Rule for multiplying a polynomial by a polynomial
Let us define two polynomials a + b and c + d and perform their multiplication.
First of all, we write down the product of the original polynomials: we put a multiplication sign between them, having previously enclosed the polynomials in parentheses. We get: (a + b) (c + d). Now let's denote the factor (c+d) How x, then the expression will look like: (a + b) x, which is essentially the product of a polynomial and a monomial. Let's do the multiplication: (a + b) x = a x + b x, and then replace it back X on (c + d) : a · (c + d) + b · (c + d) . And again applying the rule of multiplying a polynomial by a monomial, we transform the expression into: a · c + a · d + b · c + b · d. To summarize: the product of given polynomials a+b And c + d corresponds to equality (a + b) · (c + d) = a · c + a · d + b · c + b · d.
The reasoning we presented above makes it possible to draw important conclusions:
- The result of multiplying a polynomial by a polynomial is a polynomial. This statement valid for any multiplyable polynomials.
- The product of polynomials is the sum of the products of each term of one polynomial by each term of another. Where can we conclude that when multiplying polynomials containing m And n members accordingly, the indicated sum of products of members consists of m n terms.
Now we can formulate the rule for multiplying polynomials:
Definition 1
To multiply a polynomial by a polynomial, you need to multiply each term of one polynomial by each term of another polynomial and find the sum of the resulting products.
Examples of multiplying a polynomial by a polynomial
IN practical solution problems of finding the product of polynomials is decomposed into several sequential actions:
- recording the product of multiplied polynomials (polynomials are enclosed in brackets and the multiplication sign is written between them);
- constructing the sum of the products of each term of the first polynomial by each term of the second. To this end, the first term of the first polynomial is multiplied by each term of the second polynomial, then the second term of the first polynomial is multiplied by each term of the second polynomial, and so on;
- if possible, the resulting sum is written as a polynomial of standard form.
Polynomials are given: 2 − 3 x And x 2 − 7 x + 1
Solution
Let's write down the product of the original polynomials. We get: (2 − 3 x) (x 2 − 7 x + 1).
The next step is to compile the sum of the products of each term of the polynomial 2 − 3 x for each term of the polynomial x 2 − 7 x + 1. Let's take a closer look: multiply the first term of the first polynomial (number 2) by each term of the second polynomial, we get: 2 x 2, 2 (− 7 x) and 2 1. Then we multiply the second term of the first polynomial by each term of the second polynomial and get: − 3 x x 2, − 3 x (− 7 x) and − 3 x 1. We collect all the resulting expressions into a sum: 2 x 2 + 2 (− 7 x) + 2 1 − 3 x x 2 − 3 x (− 7 x) − 3 x 1.
Let's check whether we have missed the product of any terms: to do this, we recalculate the number of terms in the written sum, we get 6. This is true because the original polynomials consist of 2 and 3 terms, giving a total of 6.
The last action Let's transform the written sum into a polynomial of the standard form: 2 x 2 + 2 (− 7 x) + 2 1 − 3 x x 2 − 3 x (− 7 x) − 3 x 1 = 2 x 2 − 14 x + 2 − 3 x 3 + 21 x 2 − 3 x = = (2 x 2 + 21 x 2) + (− 14 x − 3 x) + 2 − 3 x 3 = 23 · x 2 − 17 · x + 2 − 3 · x 3
Briefly without explanation, the solution will look like this:
(2 − 3 x) (x 2 − 7 x + 1) = 2 x 2 + 2 (− 7 x) + 2 1 − 3 x x 2 − 3 x (− 7 x) − 3 x 1 = = 2 x 2 − 14 x + 2 − 3 x 3 + 21 x 2 − 3 x = = (2 x 2 + 21 x 2) + (− 14 x − 3 x) + 2 − 3 x 3 = 23 x 2 − 17 x + 2 − 3 x 3
Answer: (2 − 3 x) (x 2 − 7 x + 1) = 23 x 2 − 17 x + 2 − 3 x 3.
Let us clarify that when the original polynomials are given in a non-standard form, before finding their product, it is advisable to reduce them to a standard form. The result, of course, will be the same, but the solution will be more convenient and shorter.
Example 2
Given polynomials 1 7 x 2 (- 3) y + 3 x - 2 7 x y x and x y − 1. You need to find their work.
Solution
One of the given polynomials is written in non-standard form. Let's fix this by bringing it to the standard form:
1 7 x 2 (- 3) y + 3 x - 2 7 x x y x = - 3 7 x 2 + 3 x - 2 7 x 2 y = = - 3 7 x 2 y - 2 7 x 2 y + 3 x = - 5 7 x 2 y + 3 x
Now let's find the required product:
5 7 x 2 y + 3 x x y - 1 = = - 5 7 x 2 y x y - 5 7 x 2 y (- 1) + 3 x x · y + 3 · x · (- 1) = = - 5 7 · x 3 · y 2 + 5 7 · x 2 · y + 3 · x 2 · y - 3 · x = - 5 7 · x 3 · y 2 + 3 5 7 x 2 y - 3 x
Answer:- 5 7 x 2 y + 3 x x y - 1 = - 5 7 x 3 y 2 + 3 5 7 x 2 y - 3 x
Finally, let us clarify the situation in which there is a need to multiply three or more polynomials. In this case, finding the product is reduced to sequential multiplication of polynomials by two: i.e. First, the first two polynomials are multiplied; the result obtained is multiplied by the third polynomial; the result of this multiplication is by the fourth polynomial and so on.
Example 3
The polynomials are given: x 2 + x · y − 1 , x + y and 2 · y − 3 . You need to find their work.
Solution
Let's record the work: (x 2 + x y − 1) (x + y) (2 y − 3).
Multiplying the first two polynomials, we get: (x 2 + x y − 1) (x + y) = x 2 x + x 2 y + x y x + x y y − 1 x − 1 · y = = x 3 + 2 · x 2 · y + x · y 2 − x − y .
The initial recording of the work takes the form: (x 2 + x y − 1) (x + y) (2 y − 3) = (x 3 + 2 x 2 y + x y 2 − x − y) (2 y − 3).
Let's find the result of this multiplication:
(x 3 + 2 x 2 y + x y 2 − x − y) (2 y − 3) = = x 3 2 y + x 3 (− 3) + 2 x 2 y 2 y + 2 x 2 y (− 3) + x y 2 2 y + + x y 2 (− 3) − x 2 y − x (− 3) − y · 2 · y − y · (− 3) = = 2 · x 3 · y − 3 · x 3 + 4 · x 2 · y 2 − 6 · x 2 · y + 2 · x · y 3 - − 3 x x y 2 − 2 x y + 3 x − 2 y 2 + 3 y
Answer:
(x 2 + x y − 1) (x + y) (2 y − 3) = 2 x 3 y − 3 x 3 + 4 x 2 y 2 − 6 x 2 y + + 2 x y 3 − 3 x y 2 − 2 x y + 3 x − 2 y 2 + 3 y
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Lesson objectives:(Presentation. Slide 2)
Educational:
- derive the rule for multiplying a polynomial by a polynomial;
- develop the ability to apply this rule.
Educational:
- development of attention;
- developing the ability to analyze and generalize knowledge on the topic;
- development of mental counting skills.
Educational:
- education of neatness;
- nurturing a sustainable interest in the subject.
Lesson type: Lesson on studying and initially consolidating new knowledge.
During the classes
I. Oral work(Presentation. Slide 3)
Do the multiplication.
a) a (x – y);
b) 2p (3 – q);
c) –2x (x – 4);
d) 4y(y 3 + 0.25);
e) – 0.5 s 2 (c 3 + 2);
e) –5x (3x 2 – 4);
g) 2a 4 (a 3 – 0.5);
h) –q 7 (q 3 – q 5).
II. Explanation of new material (Presentation. Slide 4)
The explanation is carried out in several stages according to the material in the textbook.
1. Derive the rule for multiplying a polynomial by a polynomial and visually present it on a slide (or board):
2. Formulate the resulting rule and ask several students to repeat it.
3. Analyze examples of application of the rule.
Because the this topic is new to students, it is advisable to give several simple examples of the direct application of the rule of multiplication of two polynomials. It is better to consider examples of using this rule in solving a number of problems in the following lessons.
Example 1.(Presentation. Slide 5) Multiply the polynomial (3a – 2b) by the polynomial (2a + 3b).
Solution: (3a – 2b)(2a + 3b) = 3a * 2a + 3a * 3b + (– 2b) * 2a + (– 2b) * 3b = 6a 2 + 9ab – 4 ab – 6b 2 = 6a 2 + 5ab – 6b 2 .
Example 2.(Presentation. Slide 6) Simplify the expression: (2x – 3)(5 – x) – 3x(4 – x).
Solution: (2x – 3)(5 – x) – 3x(4 – x) = 10x – 2x 2 – 15 + 3x – 12x + 3x 2 = x 2 + x – 15.
Example 3.(Presentation. Slide 7) Let us prove that for any natural value n the value of the expression (n + 1)(n + 2) – (3n – 1)(n + 3) + 5n(n + 2) + n +7 is a multiple of 3.
Solution: (p + 1)(p + 2) – (3p – 1)(p + 3) + 5p(p + 2) + p +7 = p 2 + 2p + p + 2 – 3p 2 – 9p + p + 3 + 5p 2 + 10p + p +7 = 3p 2 + 6p + 12 = 3 (p 2 + 2p + 4).
III. Formation of abilities and skills (Presentation. Slide 8)
During the lesson, you should survey as many students as possible to make sure that they have learned the rule for multiplying a polynomial by a polynomial. Therefore, three students can be called to the board at once to complete each task.
1. № 677, № 678.
In these polynomial multiplication problems, each of the factors is linear. It is important that students monitor the accuracy of application of the relevant rule and do not make mistakes in the signs.
2. № 680.
These tasks are somewhat more difficult because, in addition to applying the rules for multiplying polynomials, students must remember the properties of powers.
c) 12a 4 – a 2 b 2 – b 4;
e) 56p 3 – 51p 2 + 10p.
3. № 682 (a, c).
a) (x + 10) 2 = (x + 10) (x + 10) = x 2 + 10x + 10x + 100 = x 2 + 20x + 100;
c) (3a – 1) 2 = (3a – 1) (3a – 1) = 9a 2 – 3a – 3a – 1 = 9a 2 – 6a + 1.
IV. Lesson summary (Presentation. Slide 9)
– How to multiply a monomial by a polynomial?
– Formulate the rule for multiplying a polynomial by a polynomial.
– What signs will the terms obtained by multiplying polynomials have:
a) (x + y) (a – b);
b) (n – m) (p – q)?
V. Homework: (Presentation. Slide 10)
No. 679; No. 681; No. 682 (b, d).
Textbooks used and teaching aids: (Presentation. Slide 11)
- Textbook “Algebra 7”. Yu.N. Makarychev, N.G. Mindyuk, K.I. Neshkov, S.B. Suvorova, edited by S.A. Telyakovsky. Moscow “Enlightenment”. 2010.
- Rurukin A.N., Lupenko G.V., Maslennikova I.A. Lesson-based developments in algebra: 7th grade.
The design used.
We continue to study actions with polynomials. In this article we will look at multiplying a polynomial by a polynomial. Here we will obtain the multiplication rule, after which we will consider its application in solving examples of the multiplication of polynomials of various types.
Page navigation.
Rule
To approach the rule for multiplying a polynomial by a polynomial, consider an example. Let's take two polynomials a+b and c+d and multiply them.
First, let's compose their product; to do this, we enclose each of the polynomials in brackets, and put a multiplication sign between them, we have (a+b)·(c+d) . Now we denote (c+d) as x, after this replacement the written product will take the form (a+b) x. Let's perform the multiplication in the same way as multiplying a polynomial by a monomial: (a+b) x=a x+b x . At this stage, we will reversely replace x with c+d, which will lead us to the expression a·(c+d)+b·(c+d), which, using the rule of multiplying a monomial by a polynomial, is transformed into the form a·c+a· d+b·c+b·d . Thus, the multiplication of the original polynomials a+b and c+d corresponds to the equality (a+b)·(c+d)=a·c+a·d+b·c+b·d.
From the above reasoning, two conclusions can be made: important conclusions. First, the result of multiplying a polynomial by a polynomial is a polynomial. This statement is true for any multiplyable polynomials, not just those we took in the example. Secondly, the product of polynomials is equal to the sum of the products of each term of one polynomial by each term of the other. It follows that when multiplying polynomials containing m and n terms, respectively, the specified sum of products of terms will consist of m n terms.
Now the conclusions drawn allow us to formulate the rule for multiplying polynomials:
To multiply a polynomial by a polynomial, you need to multiply each term of one polynomial by each term of another polynomial and add the resulting products.
Examples of multiplying a polynomial by a polynomial
In practice, when solving examples, the rule for multiplying a polynomial by a polynomial, obtained in previous paragraph, is divided into successive steps:
- This is how the product of the polynomials being multiplied is first written. In this case, the polynomials to be multiplied are enclosed in brackets and the sign “·” is placed between them.
- Next, the sum of the products of each term of the first polynomial and each term of the second is constructed. To do this, take the first term of the first polynomial and multiply it by each term of the second polynomial. After this, the second term of the first polynomial is taken and also multiplied by each term of the second polynomial. And so on.
- Finally, if possible, it remains to transform the resulting sum into a polynomial of the standard form.
Let's look at this with a specific example.
Example.
Multiply the polynomials 2−3 x and x 2 −7 x+1 .
Solution.
We write the product: (2−3·x)·(x 2 −7·x+1) .
Now we compose the sum of the products of each term of the polynomial 2−3·x by each term of the polynomial x 2−7·x+1. To do this, we take the first term of the first polynomial, that is, 2, and multiply it by each term of the second polynomial, we have 2·x2, 2·(−7·x) and 2·1. Now we take the second term of the first polynomial −3 x and multiply it by each term of the second polynomial, we have −3 x x 2, −3 x (−7 x) and −3 x 1. From all the obtained expressions we make up the sum: 2 x 2 +2 (−7 x)+2 1− 3 x x 2 −3 x (−7 x)−3 x 1.
To make sure that we did everything correctly and did not forget about the product of any terms, let’s count the number of terms in the resulting sum. There are 6 of them. This is as it should be, since the original polynomials consist of 2 and 3 terms, and 2·3=6.
It remains to transform the resulting sum into a polynomial of the standard form:
2 x 2 +2 (−7 x)+2 1− 3 x x 2 −3 x (−7 x)−3 x 1= 23 x 2 −17 x+2−3 x 3 .
Thus, multiplying the original polynomials gives the polynomial 23 x 2 −17 x+2−3 x 3 .
It is convenient to write the solution in the form of a chain of equalities, which reflects all the actions performed. For our example short solution looks like that:
(2−3 x) (x 2 −7 x+1)= 2 x 2 +2 (−7 x)+2 1− 3 x x 2 −3 x (−7 x)−3 x 1= 2 x 2 −14 x+2−3 x 3 +21 x 2 −3 x= (2 x 2 +21 x 2)+(−14 x−3 x)+2−3 x 3 = 23 x 2 −17 x+2−3 x 3 .
Answer:
(2−3 x) (x 2 −7 x+1)=23 x 2 −17 x+2−3 x 3.
It is worth noting that if the polynomials to be multiplied are given in a form different from the standard one, then before multiplying it is advisable to reduce them to the standard form. The result will be the same as when multiplying polynomials in the original non-standard form, but the solution will be much shorter.
Example.
Perform multiplication of polynomials and x·y−1 .
Solution.
The polynomial is not given in standard form. Before performing the multiplication, let’s reduce the polynomial to its standard form: 
Now you can multiply polynomials: 
Answer:
In conclusion, sometimes you have to multiply three, four and more polynomials. It comes down to sequential multiplication of two polynomials. That is, first the first two polynomials are multiplied, the resulting result is multiplied by the third polynomial, this result is multiplied by the fourth polynomial, and so on.
Example.
Find the product of three polynomials x 2 +x·y−1, x+y and 2·y−3.
Bibliography.
- Algebra: textbook for 7th grade general education institutions / [Yu. N. Makarychev, N. G. Mindyuk, K. I. Neshkov, S. B. Suvorova]; edited by S. A. Telyakovsky. - 17th ed. - M.: Education, 2008. - 240 p. : ill. - ISBN 978-5-09-019315-3.
- Mordkovich A. G. Algebra. 7th grade. At 2 p.m. Part 1. Textbook for students educational institutions/ A. G. Mordkovich. - 17th ed., add. - M.: Mnemosyne, 2013. - 175 p.: ill. ISBN 978-5-346-02432-3.
- Algebra and started mathematical analysis. 10th grade: textbook. for general education institutions: basic and profile. levels / [Yu. M. Kolyagin, M. V. Tkacheva, N. E. Fedorova, M. I. Shabunin]; edited by A. B. Zhizhchenko. - 3rd ed. - M.: Education, 2010.- 368 p. : ill. - ISBN 978-5-09-022771-1.
- Gusev V. A., Mordkovich A. G. Mathematics (a manual for those entering technical schools): Proc. allowance.- M.; Higher school, 1984.-351 p., ill.