Topic 6 Arithmetic polynomials. Polynomials in one variable

– = 1

Solution:

(+) * 8 = 1*8,

* 8 – ,

2 x - (x – 3) =8,

2 x – 4 x + 3=8,

x = 8 – 3,

x=5.

Answer: 5.

Solve the equation:

+ = 2

2. Solve the problem:

The master produces 8 more parts per hour than the apprentice. The apprentice worked for 6 hours, and the master for 8 hours, and together they made 232 parts. How many parts did the student produce per hour?

Directions for solution:

a) fill out the table;

8 more parts

b) write an equation;

c) solve the equation;

d) check and write down the answer.

Option 3

(For strong students, given without a sample)

1. Solve the equation:

– = 2

2. Solve the problem:

Potatoes were brought to the dining room, packed in 3 kg bags. If it were packaged in 5 kg bags, then 8 bags less would be needed. How many kilograms of potatoes were brought to the canteen?

Independent work is carried out at the end of the lesson. After completing the work, a self-test using the key is used.

As homework, students are offered creative independent work:

Think of a problem that can be solved using the equation

30 x = 60(x– 4) and solve it.

Independent work No. 8

(carried out with the aim of developing skills and abilities to take the common factor out of brackets)

Option 1

A)mx + my; e)x 5 – x 4 ;

b) 5ab – 5 b; e) 4x 3 – 8 x 2 ;

V) – 4mn + n; *and) 2c 3 + 4c 2 + c ;

G) 7ab – 14a 2 ; * h)ax 2 +a 2 .

2. Additional task.

2 – 2 18 divisible by 14.

Option 2

1. Take the common factor out of brackets (check your actions by multiplying):

A) 10x + 10y;d)a 4 +a 3 ;

b) 4x + 20y;e) 2x 6 – 4x 3 ;

V) 9 ab + 3b; *and)y 5 + 3y 6 + 4y 2 ;

G) 5xy 2 + 15y; *h) 5bc 2 +bc.

2. Additional task.

Prove that the value of the expression is 8 5 – 2 11 divisible by 17.

Option 3

1. Take the common factor out of brackets (check your actions by multiplying):

A) 18ay + 8ax;d)m 6 +m 5 ;

b) 4ab - 16a;e) 5z 4 – 10z 2 ;

at 4mn + 5 n; * g) 3x 4 – 6 x 3 + 9 x 2 ;

d) 3x 2 y– 9 x; * h)xy 2 +4 xy.

2. Additional task.

Prove that the value of the expression is 79 2 + 79 * 11 is divisible by 30.

Option 4

1. Take the common factor out of brackets (check your actions by multiplying):

a) – 7xy + 7 y; e)y 7 - y 5 ;

b) 8mn + 4 n; e) 16z 5 – 8 z 3 ;

in 20a 2 + 4 ax; * g) 4x 2 – 6 x 3 + 8 x 4 ;

d) 5x 2 y 2 + 10 x; * h)xy +2 xy 2 .

2. Additional task.

Prove that the value of the expression is 313 * 299 – 313 2 divisible by 7.

CIndependent work is carried out at the beginning of the lesson. After the work is completed, a key check is used.

Correspondence school 7th grade. Task No. 2.

Methodological manual No. 2.

Themes:

Polynomials. Sum, difference and product of polynomials;

Solving equations and problems;

Factoring polynomials;

Abbreviated multiplication formulas;

Problems for independent solution.

Polynomials. Sum, difference and product of polynomials.

Definition. Polynomial is called the sum of monomials.

Definition. The monomials from which a polynomial is composed are called members of the polynomial.

Multiplying a monomial by a polynomial .

To multiply a monomial by a polynomial, you need to multiply this monomial by each term of the polynomial and add the resulting products.

Multiplying a polynomial by a polynomial .

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 problem solving:

Simplify the expression:

Solution.

Solution:

Since, by condition, the coefficient at must be equal to zero, then

Answer: -1.

Solving equations and problems.

Definition . An equality containing a variable is called equation with one variable or equation with one unknown.

Definition . Root of an equation (solution of an equation) is the value of a variable at which the equation becomes a true equality.

Solving an equation means finding many roots.

Definition. Equation of the form
, Where X variable, a And b – some numbers are called linear equations with one variable.

Definition.

A bunch of roots of a linear equation can:

Examples of problem solving:

Is the given number 7 the root of the equation:

Solution:

Thus, x=7 is the root of the equation.

Answer: Yes.

Solve the equations:

Solution:
Answer: -12		Answer: -0.4

A boat departed from the pier to the city at a speed of 12 km/h, and half an hour later a steamboat departed in this direction at a speed of 20 km/h. What is the distance from the pier to the city if the steamer arrived in the city 1.5 hours before the boat?

Solution:

Let us denote by x the distance from the pier to the city.

According to the conditions of the problem, the boat spent 2 hours more time than the steamer (since the ship left the pier half an hour later and arrived in the city 1.5 hours before the boat).

Let's create and solve the equation:

60 km – distance from the pier to the city.

Answer: 60 km.

The length of the rectangle was reduced by 4 cm and a square was obtained, the area of ​​which was 12 cm² less than the area of ​​the rectangle.

Solution:

Find the area of ​​the rectangle.

According to the conditions of the problem, the area of ​​a square is 12 cm² less than the area of ​​a rectangle.

Let's create and solve the equation:

7 cm is the length of the rectangle.

(cm²) – area of ​​the rectangle.

Answer: 21 cm².

The tourists covered the planned route in three days.

Solution:

On the first day they covered 35% of the planned route, on the second - 3 km more than on the first, and on the third - the remaining 21 km. How long is the route?

The total length of the path was x km.

Thus, we create and solve the equation:

0.35x+0.35x+21=x

0.7x+21=x

0.3x=21

70 km length of the entire route.

Answer: 70 km.

Definition Factoring polynomials.

. Representing a polynomial as a product of two or more polynomials is called factorization. .

Taking the common factor out of brackets :

Example .

Grouping method

Taking the common factor out of brackets :

The grouping must be done so that each group has a common factor; in addition, after taking the common factor out of brackets in each group, the resulting expressions must also have a common factor.

Abbreviated multiplication formulas.

The product of the difference of two expressions and their sum is equal to the difference of the squares of these expressions. solutions The square of the sum of two expressions is equal to the square of the first expression plus twice the product of the first and second expressions, plus the square of the second expression. . 1. Find the remainder of division polynomial x6 – 4x4 + x3 ... does not have solutions , A decisions the second is the pairs (1; 2) and (2; 1). Answer: (1; 2) , (2; 1). Tasks For solutions independent

• . Solve the system...

  Approximate curriculum for algebra and elementary analysis for grades 10-11 (profile level) Explanatory note

  Program Each paragraph gives the required amount Tasks For solutions tasks . 1. Find the remainder of division in order of increasing difficulty. ...decomposition algorithm by powers of binomial; polynomials by powers of binomial; with complex coefficients;

• with valid...

  Elective course “Solving non-standard problems. 9th grade" Completed by a mathematics teacher

  Elective course by powers of binomial; The equation is equivalent to the equation P(x) = Q(X), where P(x) and Q(x) are some with one variable x. Transferring Q(x) to the left side... = . ANSWER: x1=2, x2=-3, xs=, x4=. TASKS FOR INDEPENDENT SOLUTIONS

• . Solve the following equations: x4 – 8x...

  Approximate curriculum for algebra and elementary analysis for grades 10-11 (profile level) Explanatory note

  Elective program in mathematics for 8th grade Tasks Algebra theorem, Vieta's theorem Tasks . 1. Find the remainder of division quadratic trinomial and Each paragraph gives the required amount Tasks For solutions arbitrary degree, theorem on rational... material. It's not just a list

, but also the task of making a development model... is an expression that is a product of numbers, variables and powers with a natural exponent.

For example, each of the expressions,
,
is a monomial.

They say that the monomial has standard view , if it contains only one numerical factor in the first place, and each product of identical variables in it is represented by a degree. The numerical factor of a monomial written in standard form is called coefficient of the monomial . By the power of the monomial is called the sum of the exponents of all its variables.

Definition 3.4. Polynomial called the sum of monomials. The monomials that make up a polynomial are calledmembers of the polynomial .

Similar terms - monomials in a polynomial - are called similar terms of the polynomial .

Definition 3.5. Polynomial of standard form called a polynomial in which all terms are written in standard form and similar terms are given.Degree of a polynomial of standard form is called the greatest of the powers of the monomials included in it.

For example, is a polynomial of standard form of the fourth degree.

Actions on monomials and polynomials

The sum and difference of polynomials can be converted into a polynomial of standard form. When adding two polynomials, all their terms are written down and similar terms are given. When subtracting, the signs of all terms of the polynomial being subtracted are reversed.

For example:

The terms of a polynomial can be divided into groups and enclosed in parentheses. Since this is an identical transformation inverse to the opening of parentheses, the following is established bracketing rule: if a plus sign is placed before the brackets, then all terms enclosed in brackets are written with their signs; If a minus sign is placed in front of the brackets, then all terms enclosed in brackets are written with opposite signs.

For example,

Rule for multiplying a polynomial by a polynomial: To multiply a polynomial by a polynomial, it is enough to multiply each term of one polynomial by each term of another polynomial and add the resulting products.

For example,

Definition 3.6. Polynomial in one variable degrees called an expression of the form

Where
- any numbers that are called polynomial coefficients , and
,– non-negative integer.

If
, then the coefficient called leading coefficient of the polynomial
, monomial
- his senior member , coefficient – free member .

If instead of a variable to a polynomial
substitute real number , then the result will be a real number
which is called the value of the polynomial
at
.

Definition 3.7. Number calledroot of the polynomial
, If
.

Consider dividing a polynomial by a polynomial, where
And - integers. Division is possible if the degree of the polynomial dividend is
not less than the degree of the divisor polynomial
, that is
.

Divide a polynomial
to a polynomial
,
, means finding two such polynomials
And
, to

In this case, the polynomial
degrees
called polynomial-quotient ,
– the remainder ,
.

Remark 3.2. If the divisor
–is not a zero polynomial, then division
on
,
, is always feasible, and the quotient and remainder are uniquely determined.

Remark 3.3. In case
in front of everyone , that is

they say that it is a polynomial
completely divided(or shares)to a polynomial
.

The division of polynomials is carried out similarly to the division of multi-digit numbers: first, the leading term of the dividend polynomial is divided by the leading term of the divisor polynomial, then the quotient from the division of these terms, which will be the leading term of the quotient polynomial, is multiplied by the divisor polynomial and the resulting product is subtracted from the dividend polynomial . As a result, a polynomial is obtained - the first remainder, which is divided by the divisor polynomial in a similar way and the second term of the quotient polynomial is found. This process is continued until a zero remainder is obtained or the degree of the remainder polynomial is less than the degree of the divisor polynomial.

When dividing a polynomial by a binomial, you can use Horner's scheme.

Horner scheme

Suppose we want to divide a polynomial

by binomial
. Let us denote the quotient of division as a polynomial

and the remainder is . Meaning , polynomial coefficients
,
and the remainder Let's write it in the following form:

In this scheme, each of the coefficients
,
,
, …,obtained from the previous number in the bottom line by multiplying by the number and adding to the resulting result the corresponding number in the top line above the desired coefficient. If any degree is absent in the polynomial, then the corresponding coefficient is zero. Having determined the coefficients according to the given scheme, we write the quotient

and the result of division if
,

or ,

If
,

Theorem 3.1. In order for an irreducible fraction (

,

)was the root of the polynomial
with integer coefficients, it is necessary that the number was a divisor of the free term , and the number - divisor of the leading coefficient .

Theorem 3.2. (Bezout's theorem ) Remainder from dividing a polynomial
by binomial
equal to the value of the polynomial
at
, that is
.

When dividing a polynomial
by binomial
we have equality

This is true, in particular, when
, that is
.

Example 3.2. Divide by
.

Solution. Let's apply Horner's scheme:

Hence,

Example 3.3. Divide by
.

Solution. Let's apply Horner's scheme:

Hence,

,

Example 3.4. Divide by
.

Solution.

As a result we get

Example 3.5. Divide
on
.

Solution. Let's divide the polynomials by column:

Then we get

.

Sometimes it is useful to represent a polynomial as an equal product of two or more polynomials. Such an identity transformation is called factoring a polynomial . Let us consider the main methods of such decomposition.

Taking the common factor out of brackets. In order to factor a polynomial by taking the common factor out of brackets, you must:

1) find the common factor. To do this, if all the coefficients of the polynomial are integers, the largest modulo common divisor of all coefficients of the polynomial is considered as the coefficient of the common factor, and each variable included in all terms of the polynomial is taken with the largest exponent it has in this polynomial;

2) find the quotient of dividing a given polynomial by a common factor;

3) write down the product of the general factor and the resulting quotient.

Grouping of members. When factoring a polynomial using the grouping method, its terms are divided into two or more groups so that each of them can be converted into a product, and the resulting products would have a common factor. After this, the method of bracketing the common factor of the newly transformed terms is used.

Application of abbreviated multiplication formulas. In cases where the polynomial to be expanded into factors, has the form of the right side of any abbreviated multiplication formula; its factorization is achieved by using the corresponding formula written in a different order.

Let

, then the following are true abbreviated multiplication formulas:

Introduction of new auxiliary members. This method consists in replacing a polynomial with another polynomial that is identically equal to it, but containing a different number of terms, by introducing two opposite terms or replacing any term with an identically equal sum of similar monomials. The replacement is made in such a way that the method of grouping terms can be applied to the resulting polynomial.

Example 3.6..

Solution. All terms of a polynomial contain a common factor
. Hence,.

Answer: .

Example 3.7.

Solution. We group separately the terms containing the coefficient , and terms containing . Taking the common factors of groups out of brackets, we get:

.

Answer:
.

Example 3.8. Factor a polynomial
.

Solution. Using the appropriate abbreviated multiplication formula, we get:

Answer: .

Example 3.9. Factor a polynomial
.

Solution. Using the grouping method and the corresponding abbreviated multiplication formula, we obtain:

.

Answer: .

Example 3.10. Factor a polynomial
.

Solution. We will replace on
, group the terms, apply the abbreviated multiplication formulas:

.

Answer:
.

Example 3.11. Factor a polynomial

Solution. Because ,
,
, That

In this part of Algebra 7th grade you can study school lessons on the topic “Polynomials. Arithmetic operations on polynomials."

Educational video lessons on Algebra 7th grade “Polynomials. Arithmetic operations on polynomials" is taught by Valentin Alekseevich Tarasov, teacher of the Logos LV school. You can also study other topics in algebra

Degree as a special case of a polynomial

In this lesson, basic concepts and definitions will be discussed, the basis will be prepared for studying a complex and voluminous topic, namely: we will recall the theoretical material regarding degrees - definitions, properties, theorems, and solve several examples to consolidate the technique.

Reducing polynomials to standard form. Typical tasks

In this lesson, we will recall the basic definitions of this topic and consider some typical problems, namely, reducing a polynomial to a standard form and calculating a numerical value for given values ​​of variables. We will solve several examples in which reduction to a standard form will be used to solve various types of problems.

Addition and subtraction of polynomials. Typical tasks

In this lesson, the operations of addition and subtraction of polynomials will be studied, and the rules for addition and subtraction will be formulated. Examples are considered and some typical problems and equations are solved, and the skills of performing these operations are consolidated.

Multiplying a polynomial by a monomial. Typical tasks

In this lesson we 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 distributive law of multiplication and formulate the rule for multiplying any polynomial by a monomial. Let us also recall some properties of degrees. In addition, typical errors will be formulated when performing various examples.

Multiplying binomials. Typical tasks

In this lesson we will get acquainted with the operation of multiplying the simplest polynomials - binomials, and formulate the rule for their multiplication. Let us derive some formulas for abbreviated multiplication using this operation. In addition, we will solve a large number of examples and typical problems, namely the problem of simplifying an expression, a computational problem and equations.

Multiplying trinomials. Typical tasks

In this lesson, we will look at the operation of multiplying trinomials, deduce the rule for multiplying trinomials, and, in fact, formulate the rule for multiplying polynomials in general. Let's solve a few examples related to this topic in order to move on to the multiplication of polynomials in more detail.

Multiplying a polynomial by a polynomial

In this lesson we will remember everything that we have already learned about multiplying polynomials, sum up some results and formulate a general rule. After this, we will perform a series of examples to reinforce the technique of multiplying polynomials.

Multiplying polynomials in word problems

In this lesson we will recall the method of mathematical modeling and solve problems with its help. We will learn to compose polynomials and expressions with them from the conditions of a text problem and solve these problems, which means applying the acquired knowledge about polynomials in more complex types of work.

Multiplying polynomials in problems with geometry elements

In this lesson we will learn how to solve word problems with elements of geometry using the method of mathematical modeling. To do this, first recall the basic geometric facts and stages of problem solving.

Abbreviated multiplication formulas. Squared sum and squared difference

In this lesson we will get acquainted with the formulas for the square of the sum and the square of the difference and derive them. Let us prove the formula for the square of the sum geometrically. In addition, we will solve many different examples using these formulas.

Abbreviated multiplication formulas. Difference of squares

In this lesson, we will recall the abbreviated multiplication formulas we learned earlier, namely the square of the sum and the square of the difference. Let us derive the formula for the difference of squares and solve many different typical problems using this formula. In addition, we will solve problems involving the complex application of several formulas.

Abbreviated multiplication formulas. Difference of cubes and sum of cubes

In this lesson we will continue to study abbreviated multiplication formulas, namely, we will look at the difference and sum of cubes formulas. In addition, we will solve various typical problems using these formulas.

Shared use of abbreviated multiplication formulas

This video lesson will be useful to all those who want to independently study the topic “Combined application of abbreviated multiplication formulas.” With this video lecture you will be able to summarize, deepen and systematize the knowledge gained in previous lessons. The teacher will teach you how to use abbreviated multiplication formulas together.

Formulas for abbreviated multiplication in problems of increased complexity. Part 1

In this lesson we will apply our knowledge of polynomials and abbreviated multiplication formulas to solve a fairly complex geometric problem. This will allow us to strengthen our skills in working with polynomials.

Formulas for abbreviated multiplication in problems of increased complexity. Part 2

In this lesson, we will look at complicated problems using abbreviated multiplication formulas and perform many different examples to reinforce the technique.

Geometric problem on a parallelepiped using the abbreviated multiplication formula

In this video lesson, everyone will be able to study the topic “Geometric problem on a parallelepiped using the abbreviated multiplication formula.” In this activity, students will practice using the abbreviated multiplication formula for a parallelepiped. In particular, the teacher will give a geometric problem on a parallelepiped, which must be disassembled and solved.

Dividing a polynomial by a monomial

In this lesson, we will recall the rule for dividing a monomial by a monomial and formulate the basic supporting facts. Let's add some theoretical information to what is already known and derive the rule for dividing a polynomial by a monomial. After this, we will perform a number of examples of varying complexity to master the technique of dividing a polynomial by a monomial.

Goals: generalization and consolidation of the material covered: repeat the concept of a polynomial, the rule of multiplying a polynomial by a polynomial and consolidate this rule during the test work, consolidate the skills of solving equations and problems using equations.

Equipment: poster “Whoever does and thinks for himself from a young age later becomes more reliable, stronger, smarter” (V. Shukshin). Overhead projector, magnetic board, crossword puzzle, test cards.

Lesson plan.

1. Organizational moment.
2. Checking homework.
3. Oral exercises (crossword puzzle).
4. Solving exercises on the topic.
5. Test on the topic: “Polynomials and operations on them” (4 options).
6. Lesson summary.
7. Homework.

During the classes

I. Organizational moment

Students in the class are divided into groups of 4-5 people, the eldest in the group is selected.

II. Checking homework.

Students prepare their homework on a card at home. Each student checks their work through an overhead projector. The teacher offers to evaluate the homework for the student himself and puts a grade on the report sheet, indicating the evaluation criterion: “5” ─ the task was completed correctly and independently; “4” ─ the task was completed correctly and completely, but with the help of parents or classmates; “3” ─ in all other cases, if the task is completed. If the task is not completed, you can put a dash.

III. Oral exercises.

1) To review theoretical questions, students are offered a crossword puzzle. The crossword puzzle is solved orally by the group, and the answers are given by students from different groups. We give ratings: “5” ─ 7 correct words, “4” ─ 5.6 correct words, “3” ─ 4 correct words.

Questions for the crossword: (see Annex 1)

1. The property of multiplication used when multiplying a monomial by a polynomial;
2. method of factoring a polynomial;
3. an equality that is true for any value of the variable;
4. an expression representing the sum of monomials;
5. terms that have the same letter part;
6. the value of the variable at which the equation turns into a true equality;
7. numerical factor of monomials.

2) Follow these steps:

3. If the length of the rectangle is reduced by 4 cm and its width is increased by 7 cm, then you will get a square whose area will be 100 cm 2 greater than the area of ​​the rectangle. Determine the side of the square. (The side of the square is 24 cm).

Students solve tasks in groups, discussing and helping each other. When the groups have completed the task, they are checked against the solutions written on the board. After the check, grades are assigned: for this work, students receive two grades: self-assessment and group assessment. Evaluation criterion: “5” ─ solved everything correctly and helped his comrades, “4” ─ made mistakes when solving, but corrected them with the help of his comrades, “3” ─ was interested in the solution and solved everything with the help of classmates.

V. Test work.

Option I

1. Present in standard form the polynomial 3a – 5a∙a – 5 + 2a 2 – 5a +3.

3. Find the difference of the polynomials 2x 2 – x + 2 and ─ 3x 2 ─2x + 1.

5. Present the expression as a polynomial: 2 – (3a – 1)(a + 5).

Option II

1. Present in standard form the polynomial 5x 2 – 5 + 4x ─ 3x∙x + 2 – 2x.

3. Find the difference of the polynomials 4y 2 – 2y + 3 and - 2y 2 + 3y +2.

5. Solve the equation: ─3x 2 + 5x = 0.

6. Present as a product: 5a 3 – 3a 2 – 10a + 6.

Option III

1. Find the value of the polynomial ─ 6а 2 – 5аb + b 2 – (─3а 2 – 5аb + b 2) with а = ─, b=─3.

2. Simplify the expression: ─8x – (5x – (3x – 7)).

4. Multiply: ─3x∙(─ 2x 2 + x – 3)

6. Present it as a product: 3x 3 – 2x 2 – 6x + 4.

7. Present the expression as a product: a(x – y) ─2b(y – x)

IV option

1. Find the value of the polynomial ─ 8a 2 – 2ax – x 2 – (─4a 2 – 2ax – x 2) with a= ─, x= ─ 2.

2. Simplify the expression: ─ 5a – (2a – (3a – 5)).

4. Perform multiplication: ─4а ∙ (─5а 2 + 2а – 1).

6. Express it as a polynomial: (3x – 2)(─x 2 + x – 4).

7. Present the expression as a product: 2c(b – a) – d(a – b)

VI. Lesson summary

During the lesson, each student receives several grades. The student himself evaluates his knowledge by comparing it with the knowledge of others. Group evaluation is more effective because the evaluation is discussed by all group members. The guys point out shortcomings and shortcomings in the work of group members. All grades are entered into the work card by the group leader.

The teacher gives the final grade and communicates it to the whole class.

VII. Homework:

1. Follow these steps:

a) (a 2 + 3аb─b 2)(2а – b);
b) (x 2 + 2xy – 5y 2)(2x 2 – 3y).

2. Solve the equation:

a) (3x – 1)(2x + 7) ─ (x + 1)(6x – 5) = 16;
b) (x – 4)(2x2 – 3x + 5) + (x2 – 5x + 4)(1 – 2x) = 20.

3. If one side of the square is reduced by 1.2 m and the other by 1.5 m, then the area of ​​the resulting rectangle will be 14.4 m 2 less than the area of ​​the given square. Determine the side of the square.