Isolating the common factor out of brackets. Taking the common factor out of brackets

\(5x+xy\) can be represented as \(x(5+y)\). These are indeed identical expressions, we can verify this if we open the brackets: \(x(5+y)=x \cdot 5+x \cdot y=5x+xy\). As you can see, as a result we get the original expression. This means that \(5x+xy\) is indeed equal to \(x(5+y)\). By the way, this is a reliable way to check the correctness of the common factors - open the resulting bracket and compare the result with the original expression.

The main rule for bracketing:

For example, in the expression \(3ab+5bc-abc\) only \(b\) can be taken out of the bracket, because it is the only one that is present in all three terms. The process of taking common factors out of brackets is shown in the diagram below:

Bracketing rules

In mathematics, it is customary to take out all common factors at once.

Example:\(3xy-3xz=3x(y-z)\)
Please note that here we could expand like this: \(3(xy-xz)\) or like this: \(x(3y-3z)\). However, these would be incomplete decompositions. Both the C and the X must be taken out.

Sometimes the common members are not immediately visible.

Example:\(10x-15y=2·5·x-3·5·y=5(2x-3y)\)
In this case, the common term (five) was hidden. However, having expanded \(10\) as \(2\) multiplied by \(5\), and \(15\) as \(3\) multiplied by \(5\) - we “pulled the five into the light of God”, after which they were easily able to take it out of the bracket.

If a monomial is removed completely, one remains from it.

Example: \(5xy+axy-x=x(5y+ay-1)\)
We put \(x\) out of brackets, and the third monomial consists only of x. Why does one remain from it? Because if any expression is multiplied by one, it will not change. That is, this same \(x\) can be represented as \(1\cdot x\). Then we have the following chain of transformations:

\(5xy+axy-\)\(x\) \(=5xy+axy-\)\(1 \cdot x\) \(=\)\(x\) \((5y+ay-\)\ (1\)\()\)

Moreover, this is the only correct way to extract it, because if we do not leave one, then when opening the brackets we will not return to the original expression. Indeed, if we do the extraction like this \(5xy+axy-x=x(5y+ay)\), then when expanded we will get \(x(5y+ay)=5xy+axy\). The third member is missing. This means that such a statement is incorrect.

You can place a minus sign outside the bracket, and the signs of the terms in the bracket are reversed.

Example:\(x-y=-(-x+y)=-(y-x)\)
Essentially, here we are putting out the “minus one”, which can be “selected” in front of any monomial, even if there was no minus in front of it. We use here the fact that one can be written as \((-1) \cdot (-1)\). Here is the same example, described in detail:

\(x-y=\)
\(=1·x+(-1)·y=\)
\(=(-1)·(-1)·x+(-1)·y=\)
\(=(-1)·((-1)·x+y)=\)
\(=-(-x+y)=\)
\(-(y-x)\)

A parenthesis can also be a common factor.

Example:\(3m(n-5)+2(n-5)=(n-5)(3m+2)\)
We most often encounter this situation (removing brackets from brackets) when factoring using the grouping method or

Algebra lesson in 7th grade "Bracketing a common factor"

Komarova Galina Aleksandrovna

Target: improving the practical skills of students in factoring a polynomial by taking the common factor out of brackets and using it in solving equations. Conduct diagnostics of the assimilation of a system of knowledge and skills and its application to perform practical tasks at a standard level with a transition to a higher level. Develop skills: apply rules, analyze, compare, generalize, highlight the main thing.

Tasks:

create a situation of success in the lesson, conditions for independent activity of students in the lesson;

promote understanding of the lesson material;

cultivate communication and tolerance in student relationships.

Lesson type: combined.

Methods: stimulating, search, visual, practical, verbal, gaming, differentiated work.

Forms of implementation: individual, collective, group.

Knowledge is assessed using a 5-point system.

Lesson type: generalization and systematization of knowledge with didactic games.

Learning outcomes: Be able to put the common factor out of brackets, be able to use this method when factoring, be able to use put the common factor out of brackets when solving equations.

Lesson progress

1. Organizational moment.

Greeting students.

When Pythagoras' disciples woke up, they had to recite the following verses:

“Before you rise from the sweet dreams evoked at night,

Think through what the day has in store for you.”

2. Warm-up - graphic test of theoretical material.

Is the statement, definition, property true?

1. A monomial is called amount numerical and alphabetic factors. (No -)

2. Numerical the factor of a monomial written in standard form is called the coefficient of the monomial. (yes Λ)

3. Identical or differing from each other only in coefficients are called similar terms. (yes Λ)

4. The algebraic sum of several monomials is called monomial. (No -)

5. When any number or expression is multiplied by zero, the result is zero. (yes Λ)

6. Multiplying a monomial by a polynomial results in a polynomial. (yes Λ)

7. When we open brackets preceded by the sign “-”, we omit the brackets and the signs of the members that were enclosed in brackets, don't change to the opposite. (No-)

8.The common numerical factor is the greatest common divisor of the coefficients of monomials. (yes Λ)

9. Of the identical letter factors of monomials, we take it out of bracketsthe smallest degree . (yes Λ)

Examination: ––ΛΛ- ΛΛ-ΛΛ

Give yourself ratings:

“5” - no errors “4” - two errors “3” - four errors “2” - more than four errors

3. Updating of basic knowledge.

Individual work on cards No. 1, No. 2, No. 3 (3 students).

Frontal work with the class:

Task 1 . Continue the sentence:

One way to factor a polynomial is... (putting the common factor out of brackets );

When taking the common factor out of brackets,... (distributive property );

If all terms of a polynomial contain a common factor, then...(this factor can be taken out of brackets )

Task 2 .

What numerical factor will be common in the following expressions: 12 y 3 -8 y 2 ; 15x 2 - 75x. (4у 2 ; 15x)

What degree of multipliers A And X can be taken out of brackets

a 2 x - a 5 x 3 + 3a 3 x 2 ( A 2 X )

Formulate an algorithm for removing the common factor.

Algorithm:

Find the gcd for all coefficients of the monomials and take it out of brackets:

2) the smallest degree:

divide :

4. Studying new material.

Determine the common factor in these expressions and take it out of brackets:

2a+6=

3 xy-3y=

18m-9nm=

x 2 -x 3 +x 6 =

3y+3xy=

(Work in pairs, peer review )

Using the cipher key, decipher the word.

3y(x-1) or

-3у(-х+1)

9m(2-n)

2(a+3)

X 2 (1-x +x 4)

3(7c 2 -5a 3)

Answer: Galois.

Evariste Galois (1811-1832)

Galois is the pride of French science. While still a child, he read Legendre's geometry as a fascinating book. By the age of 16, Galois's talents had manifested themselves to such an extent that they placed him among the greatest mathematicians of that time . Galois's scientific works on the theory of algebraic equations of higher degrees laid the foundation for the development of modern algebra.

The brilliant mathematician, the pride of world science, lived only 20 years, five of which he devoted to mathematics. 2011 marks the 200th anniversary of his birth.

I suggest you solve an equation on the left side of which is a polynomial of the second degree.
12x 2 +6 x =0. Let's take 3x out of brackets. We'll get it.

6x(2x+1)=0 The product is zero when at least 6x=0 or 2x+1=0. one of the factors is zero.

x=0:6 2x=-1

x=0 x = -1:2

x=-0.5

and we find x=0 or x= -0.5

Answer: x 1 =0, x 2 = -0,5

5. Physical education minute.

Statements are read to students. If the statement is true, then students should raise their hands up, and if it is false, then sit down and clap.

7 2 =49 (Yes).

30 = 3 (No).

The greatest common factor of the polynomial 5a-15b is 5 (Da).

5 2 =10 (No).

There are 10 fingers on the hands. There are 100 fingers on 10 hands (No).

5 0 =1 ( Yes)

0 is divisible by all numbers without a remainder ( Yes).

question for filling 5:0=0

6. Homework.

Group I, II

Rule in notebook, No. 709(e,f), 718(g,)719(g),

III group:

Rule in notebook, No. 710 (a, b), 715 (c, d)

Additional task (optional)

It is known that for some values a andb expression value A-b equals 3. What is the value of the expression for the same a and b?

a) 5a-5 b ; b) 12b - 12a; V) (A -b ) 2 ; G) (b -a) 2 ;

7. Consolidation.

,Group II decide number 710(a,c)

Group III decides number 709(a,c)

Come up with a second degree equation yourself

Students work on the assignment of card No. 5-6 at the board and in notebooks. (diff)

Find the mistake

5. Independent work.

Students are asked to complete independent educational work in the form of a test, followed by self-test; the correct answers can be placed on the back of the board.

6. Summing up the lesson.

Reflection: Who did the best job in our lesson today?

What rating will we give them?

I worked well

Understood how to solve equations by taking out

Common multiplier in brackets

Happy with the lesson

I need help from a teacher or consultant

WE A How did we work together today?

Examples of cards.

Card No. 1.

2x-2y

5ab+10a

2a 3 -a 5

a(x-2)+b(x-2)

-7xy+y

Card No. 2.

Take the common factor out of brackets:

5ab-10ac

4xy-16x2

a 2 -4a+3a 5

0.3a 2 b+0.6ab 2

x 2 (y-6)-x(y-6)

Card number 3.

Take out the total factor

outside of brackets:

-3x 2 y-12y 2

5a 2 -10a 3 +15a 5

6c 2 x 3 -4c 3 x 3 +2x 2 c

7a 2 b 3 -1.4a 3 b 4 +2.1a 2 b 5

3a(x-5)+7(5-x)

Card No. 5- 1

Take the common factor out of brackets:

3x + 3y;

5a – 15b;

8x+12y;

Solve the equation

1) 2x ² + 5x = 0

Card No. 5-2

1) 10 a – 10 v

2) 3 xy – x 2 y 2

3) 5 for 2 + 15 for 3

2.Solve the equation

2x² - 9x = 0

Card No. 6

1. Take the common factor out of brackets:

1) 8 a + 8 c.

2) 4 x y + x 3 y 3

3) 3 in y – 6 in.

2. Solve the equation

2x² +7x = 0

Additional tasks

1.Find the error:

3x (x-3)=3x 2 -6x; 2x+3xy=x(2+y);

2.Insert the missing expression:

5x(2x 2 -x)=10x 3 -…; -3ау-12у=-3у (а+...);

3. Take the common factor out of brackets:

5a - 5b; 3x + 6 y; 15a – 25b; 2.4x + 7.2y.

7a + 7b; 8x – 32a; 21a + 28b; 1.25x – 1.75a.

8x – 8y; 7a + 14b; 24x – 32a; 0.01a + 0.03y.

4.Replace “M” with a monomial so that the resulting equality is true:

a) M × (a – b) = 4 ac – 4 bc;

b) M × (3a – 1) = 12a 3 – 4a 2;

c) M × (2a – b) = 10a 2 – 5a b.

VIII. Frontal work (on attentiveness, on learning new rules).

Expressions are written on the board. Find errors in these equalities, if any, and correct them.

2 x 3 – 3 x 2 – x = x (2 x 2 – 3 x).

2 x + 6 = 2 (x + 3).

8 x + 12 y = 4 (2 x - 3y).

a 6 – a 2 = a 2 (a 2 – 1).

4 -2a = – 2 (2 – a).

Algorithm:

Find the gcd for all coefficients of monomials and take it out of brackets

2) Of the identical letter factors of monomials, remove it from bracketsthe smallest degree

3) Each monomial of a polynomialdivide by the common factor and the result of division written in brackets

Student's knowledge control sheet of class 7A __________________________________________

dictation

2. encryption

3.Individual Working with cards

4.test

5.Total points

6.Teacher's mark

answer

Test

1.What power of the factor a can be taken out of brackets for the polynomial

a²x - ax³

a) a b) a² c) a³

2 x³ -8x²

a) 4 b) 8 c) 2

a²+ab – ac +a

A ) a(a+b-c+1) b) a (a+b-c)

V) a 2 (a+b-c+1)

7m³ + 49m²

a) 7 m ² (m +7m 2) b) 7m ² (m +7)

c) 7 m² (7m +7)

5.Factorize:

x(x – y) + a(x – y)

A ) (x-y)(x+a) b) (y-x)(x+a)

V ) (x+a)(x+y)

6. Solve the equation

6y-(y-1)=2(2y-4)

a) -9 b) 8 c) 9

d) another answer

7.Add the common factor

x(x – y) + a(y- x)

A ) (x-y)(x- a) b) (y-x)(x+a)

V ) (x+a)(x+y)

Answers

Test

1.What power of the factor b can be taken out of brackets for the polynomial

b² - a³b³

A) b b) b ² c) b ³

2.What numerical factor can be taken out of brackets for a polynomial

15a³ - 25a

A) 15 b) 5 c) 25

3. Take out the common factor of all terms of the polynomial

x² - xy + xp – x

A) x (x -y +p -1) b) x (x -y +p )

V) x 2 (x-y+p-1)

4. Present the polynomial as a product

9b² - 81b

A) 9b(b-81) b) 9b 2 (b-9)

V) 9b(b-9)

5.Factorize:

a(a + 3) – 2(a +3)

A ) (a+3)(a+2) b) (a+3)(a-2)

V ) (a-2)(a-3)

6. Solve the equation

3x-(12x-x)=4(5-x)

a) -4 b) 4 c) 2

d) another answer

7.Add the common factor

a (a - 3) – 2(3-a)

A ) (a -3)(a+2) b) (a+3)(a-2)

V ) (a-2)(a-3)

Answers

Option I

Perform action:

(3x+10y) – (6x+3y)

a) 9x+7y; b) 7u-3x; c) 3x-7y; d) 9x-7y

6x 2 -3x

A ) 3x(2x-1); b) 3x(2x-x); c) 3x 2 (2-x); d)3x(2x+1)

3. Reduce polynomial to standard form:

X+5x 2 +4x-x 2

a) 6x 2 +3x; b) 4x 2 +3x; c)4x 2 +5x; G) 6x 2 -3x

4. Perform action:

3x 2 (2x-0.5y)

a) 6x 2 -1.5x 2 y; b) 6x 2 -1.5xy; V) 6x 3 -1.5x 2 at; d) 6x 3 -0.5x 2 y;

5. Solve the equation:

8x+5(2-x)=13

a) x=3; b) x=-7; c)x=-1; G) x=1;

6. Take the common factor out of brackets:

x(x-y)-6y(x-y)

A) (x-y)(x-6y)) ; b) (x-y)(x+6y);

c) (x+y)(x-6y); d) (x-y)(6y-x);

7. Solve the equation:

X 2 +8x=0

a) 0 and -8 b) 0 and 8; c) 8 and -8

Option II

Perform action:

(2a-1)+(3+6a)

a) 8a+3; b) 8a+4; V) 8a+2; d) 6a+2

Take the common factor out of brackets:

7a-7b

A) 7(a-c); b) 7(a+c); c)7(c-a); d) a(7-c);

Reduce the polynomial to standard form:

4x 2 +3x-5x 2

A) -X 2 +3x; b) 9x 2 +3x; c) 2x 2; d) –x 2 -3x;

Perform multiplication:

4a 2 (a-c)

a) 4a 3 -c; b) 4a 3 -4av; V) 4a 3 -4a 2 V; d) 4a 2 -4a 2 c;

Factorize:

a(v-1)-3(v-1)

A) (c-1)(a-3); b) (c-1)(a+3) ; c) (c+1)(a-3) ; d) (c-3)(a-1) ;

Solve the equation:

4(a-5)+a=5

a) a=1; b) a=-5; c) a=3; G) a=5;

7. Solve the equation:

6x 2 -30x=0

a) 0 and 5 b) 0 and -5 c) 5 and -5

Galois

A boy came in in a poor frock coat,

To buy tobacco and Madeira in the shop.

She invited me kindly, like a younger brother,

Broken mistress and continue to come.

She walked me to the door, sighing tiredly,

After him, she threw up her hands: “Eccentric!

I cheated by four centimes again,

And four centimes is no small thing now!

Someone told me, like a prominent scientist,

Some mathematician, Monsieur Galois.

How can the laws of the world be revealed?

This, if I may say so, is the head?!”

But he went up to the attic, deceived by her,

I took the treasured sketch in the attic dust

And he proved again with all mercilessness,

That the owners of full stomachs are zeros. (A. Markov

Option 1

1 . 4-2x

A. 2(2 + x).B. 4(1 - x).

B. 2(2).G. 4(1 + x).

2. A 3 V 2 - A 4 V

A. a 4 c(c - a).B. a 3 in (in - a).

B. a 3 in 2 (1 - a). D. a 3 in (1 - a).

3. 15x y 2 + 5x y - 20x 2 y

A. 5x y (3y + 1 - 4x).B. 5xy (3y - 4x).

B. 5x(3 y 2 + y - 2x).G. 5x(3y 2 + y - 4x).

4. A( b +3) +( b + 3).

A. ( b + 3) (a + 1).B. (b + 3)a.

B. (3 + b ) (a - 1).G. (3 + b )(1-a).

5. X(y - z ) - (z - y ).

A. (x - 1) ( y - z).B. (x - 1) (z - y).

B.(x + 1)(y- z).T.(x + 1)(z -y).

6. Solve the equation

3y - 12 y 2 =0

Factoring polynomials

Option 2

1. 6a-3.

A. 3(2a-1).B. 6(a-1).

B. 3(2a+1).G. 3(a-1).

2. A 2 b 3 – a 3 b 4

A. a 2 b 3 (1 - ab).B. a 3 (b 3 – b 4).

B. a b 3 (1 - a 2 b).G. b 3 (x 2 - x 3).

3. 12x 2 y - 6xy - 24xy 2 .

A. 6xy(2x - 1 - 4y).B. 6xy (2x - 4y).

B. 6xy (6x - 1 - 4y). D. 6xy(2x + 4y + 1).

4. X( y + 5) + ( y +5).

A. (x - 1) (y + 5).B. (x + 1) (y + 5).

B.(y + 5)x.G. (x - 1) (5 - y).

5. a(c-b )- (b -With).

A. (a - 1) ( b + c).B. (a - 1) (b - c).

B. (a + 1) (c - b).G. (a + 1) (b - c).

6. Solve the equation

Within the framework of the study of identity transformations, the topic of taking the common factor out of brackets is very important. In this article we will explain what exactly such a transformation is, derive the basic rule and analyze typical examples of problems.

Yandex.RTB R-A-339285-1

The concept of taking a factor out of brackets

To successfully apply this transformation, you need to know what expressions it is used for and what result you need to get in the end. Let us clarify these points.

You can take the common factor out of brackets in expressions that represent sums in which each term is a product, and in each product there is one factor that is common (the same) for everyone. This is called the common factor. It is this that we will take out of the brackets. So, if we have works 5 3 And 5 4, then we can take the common factor 5 out of brackets.

What does this transformation consist of? During it, we represent the original expression as the product of a common factor and an expression in parentheses containing the sum of all original terms except the common factor.

Let's take the example given above. Let's add a common factor of 5 to 5 3 And 5 4 and we get 5 (3 + 4) . The final expression is the product of the common factor 5 by the expression in brackets, which is the sum of the original terms without 5.

This transformation is based on the distributive property of multiplication, which we have already studied before. In literal form it can be written as a (b + c) = a b + a c. By changing the right side with the left, we will see a scheme for taking the common factor out of brackets.

The rule for taking the common factor out of brackets

Using everything said above, we derive the basic rule for such a transformation:

Definition 1

To remove the common factor from brackets, you need to write the original expression as the product of the common factor and brackets that include the original sum without the common factor.

Example 1

Let's take a simple example of rendering. We have a numeric expression 3 7 + 3 2 − 3 5, which is the sum of three terms 3 · 7, 3 · 2 and a common factor 3. Taking the rule we derived as a basis, we write the product as 3 (7 + 2 − 5). This is the result of our transformation. The entire solution looks like this: 3 7 + 3 2 − 3 5 = 3 (7 + 2 − 5).

We can put the factor out of brackets not only in numerical, but also in literal expressions. For example, in 3 x − 7 x + 2 you can take out the variable x and get 3 x − 7 x + 2 = x (3 − 7) + 2, in the expression (x 2 + y) x y − (x 2 + y) x 3– common factor (x2+y) and get in the end (x 2 + y) · (x · y − x 3).

It is not always possible to immediately determine which factor is the common one. Sometimes an expression must first be transformed by replacing numbers and expressions with identically equal products.

Example 2

So, for example, in the expression 6 x + 4 y it is possible to derive a common factor 2 that is not written down explicitly. To find it, we need to transform the original expression, representing six as 2 · 3 and four as 2 · 2. That is 6 x + 4 y = 2 3 x + 2 2 y = 2 (3 x + 2 y). Or in expression x 3 + x 2 + 3 x we can take out of brackets the common factor x, which is revealed after the replacement x 3 on x · x 2 . This transformation is possible due to the basic properties of the degree. As a result, we get the expression x (x 2 + x + 3).

Another case that should be discussed separately is the removal of a minus from brackets. Then we take out not the sign itself, but minus one. For example, let us transform the expression in this way − 5 − 12 x + 4 x y. Let's rewrite the expression as (− 1) 5 + (− 1) 12 x − (− 1) 4 x y, so that the overall multiplier is more clearly visible. Let's take it out of brackets and get − (5 + 12 · x − 4 · x · y) . This example shows that in brackets the same amount is obtained, but with opposite signs.

In conclusions, we note that transformation by placing the common factor out of brackets is very often used in practice, for example, to calculate the value of rational expressions. This method is also useful when you need to represent an expression as a product, for example, to factor a polynomial into individual factors.

If you notice an error in the text, please highlight it and press Ctrl+Enter

§ 10. Factoring polynomials using the method putting the common factor out of brackets

In 6th grade, we factored composite numbers into prime factors, that is, we presented natural numbers as a product. For example, 12 = 2 2 ∙ 3; 105 = 3 ∙ 5 ∙ 7 dr.

Some polynomials can also be represented as a product. This means that these polynomials can be factorized. For example, 5a: - 5y - 5(x - y); a 3 and 3a 2 = a 2 (a + 3) and the like.

Let's consider one of the ways to factor polynomials - taking the common factor out of brackets. One of the examples of such an expansion known to us is the distributive property of multiplication a(b + c) = ab + ac, if written in reverse order: ab + ac - a(b + c). This means that the polynomial ab + ac was decomposed into two factors a and b + c.

When factoring polynomials with integer coefficients, the factor that is taken out of brackets is chosen so that the terms of the polynomial that remains in brackets do not have a common letter factor, and the modules of their coefficients do not have common divisors.

Let's look at a few examples.

Example 1. Factor the expression:

3) 15a 3 b - 10a 2 b 2.

R a s i z a n i .

1) The common factor is the number 4, so

8m + 4 = 4 . 2m+ 4 ∙ 1 = 4(2m + 1).

2) The common factor is the variable a, therefore

at + 7ap = a(t + 7p).

3) In this case, the common numerical factor is the greatest common divisor of the numbers 10 and 15 - the number 5, and the common letter factor is the monomial a 2 b. So,

15a 3 b - 10a 2 b 2 = 5a 2 b ∙ 3a - 5a 2 b ∙ b = 5a 2 b(3a - 2b).

Example 2. Factor into:

1) 2m(b - s) + 3р(b - s);

2) x(y - t) + c(t - c).

R az v ’ i z a n n i.

1) In this case, the common factor is the binomial b = c.

Therefore, 2m( b - With) + 3р( b - c) = (b - с)(2m + 3р).

2) The terms have factors in - t and t - in, which are opposite expressions. Therefore, in the second term we take the factor -1 out of brackets, we get: c(t - в) = -с(у - t).

Therefore, x(y - t) + c(t - b) = x(y - t) - c(y - t) = (y - t) (x - c).

To check the correctness of the factorization, you should multiply the resulting factors. The result must equal the given polynomial.

Factoring polynomials often simplifies the process of solving an equation.

Example 3. Find the roots of the equation 5x 2 - 7x = 0.

R az v ’ i z a n n i. Let's factorize the left side of the equation by taking the common factor out of brackets: x(5x - 7) = 0. Considering that the product is equal to zero if and only if at least one of the factors is equal to zero, we will have: x = 0 or 5x - 7 = 0, whence x = 0 or x = 1.4.

Answer: 0; 1.4.

What transformation is called the factorization of a polynomial? Using the example of the polynomial ab + ac, explain how factorization is performed by placing the common factor out of brackets.

(Oral) Find the common factor in the expression:

(Oral) Factor into:

Take the common factor out of brackets:

(Orally) correctly performed the factorizations:

1) 7a + 7 = 7a;

2) 5m - 5 = 5(m - 5);

3) 2a - 2 = 2(a - 1);

4) 7xy - 14x = 7x - (y - 2);

5) 5mn + bn = 5m(n + 3);

6) 7ab + 8cb = 15b(a + c)?

Write the amount as a product:

Factor it out:

Factor it out:

4) 7a + 21ау;

5) 9x 2 - 27x;

6) 3a - 9a 2;

8) 12ax - 4a 2;

9) -18xy + 24v 2;

10) a 2 b - ab 2 ;

11) rm - p 2 m;

12) -x 2 y 2 - xy.

Take the common factor out of brackets:

4) 15xy + 5x;

6) 15m - 30m 2 ;

7) 9xy + 6x 2;

9) -p 2 q - pq 2.

Factor it out:

5) 3b 2 - 9b 3;

7) 4y 2 + 12y 4 ;

8) 5m 5 + 15m 2 ;

9) -16a 4 - 20a.

Factor it out:

4) 18p 3 - 12p 2 ;

5) 14b 3 + 7b 4 ;

6) -25m 3 - 20m.

Write the sum 6x 2 in + 15x as a product and find its value if x = -0.5, y = 5.
Write the expression 12a 2 b - 8a as a product and find its value if a = 2, 6 = .
Take the common factor out of brackets:

1) a 4 + a 3 - a 2;

2) m 9 - m 2 + m 7;

3) b 6 + b 5 - b 9 ;

4) - at 7 - at 12 - at 3.

Present it as a product:

1) p 7 + p 3 - p 4;

2) a 10 - a 5 + a 8;

3) b 7 - b 5 - b 2;

4) -m 8 - m 2 - m 4.

Calculate in a convenient way:

1) 132 ∙ 27 + 132 ∙ 73;

2) 119 ∙ 37 - 19 ∙ 37.

Solve the equation:

1) x 2 - 2x = 0;

2) x 2 + 4x = 0.

Find the roots of the equation:

1) x 2 + 3x = 0;

2) x 2 -7x = 0.

1) 4a 3 + 2a 2 - 8a;

2) 9b 3 - 3b 2 - 27b 6;

3) 16m 2 - 24m 6 - 22m 3;

4) -5b 3 - 20b 2 - 25b 5.

Take the common factor out of brackets:

1) 5s 8 - 5s 7 + 10s 4;

2) 9m 4 + 27m 3 - 81m;

3) 8r 7 - 4r 5 + 10r 3;

4) 21b - 28b 4 - 14b 3.

Take the common factor out of brackets:

1) 7m 4 - 21m 2 n 2 + 14m 3 ;

2) 12a 2 b - 18ab 2 + 30ab 3;

3) 8x 2 y 2 - 4x 3 in 5 + 12x 4 in 3;

4) 5p 4 q 2 - 10p 2 q 4 + 15pq 3.

Factor the polynomial:

1) 12a - 6a 2 x 2 - 9a 3;

2) 12b 2 in - 18b 3 - 30b 4 in;

3) 16bx 2 - 8b 2 x 3 + 24b 3 x;

4) 60m 4 n 3 - 45m 2 n 4 + 30m 3 n 5.

Calculate in a convenient way:

1) 843 ∙ 743 - 743 2 ;

2) 1103 2 - 1103 ∙ 100 - 1103 ∙ 3.

Find the meaning of the expression:

1) 4.23 a - a 2, if a = 5.23;

2) x 2 y + x 3, if x = 2.51, b = -2.51;

3) am 5 - m 6, if = -1, a = -5;

4) -xy - x 2, if x = 2.7, b = 7.3.

Find the meaning of the expression:

1) 9.11 a + a 2, if a = -10.11;

2) 5x 2 + 5a 2 x, if a = ; x = .

Factor the polynomial:

1) 2p(x - y) + q(x - y);

2) a(x + y) - (x + y);

3) (a - 7) - b(a - 7);

4) 5(a + 1) + (a + 1) 2;

5) (x + 2) 2 - x(x + 2);

6) -5m(m - 2) + 4(m - 2) 2 .

Express the expression as a product:

1) a(x - y) + b(y - x);

2) g(b - 5) - n(5 - b);

3) 7x - (2b - 3) + 5y(3 - 2b);

4) (x - y) 2 - a(y - x);

5) 5(x - 3) 2 - (3 - x);

6) (a + 1)(2b - 3) - (a + 3)(3 - 2b).

Factor it out:

1) 3x(b - 2) + y(b - 2);

2) (m 2 - 3) - x(m 2 - 3);

3) a(b - 9) + c(9 - b);

4) 7(a + 2) + (a + 2) 2;

5) (s - m) 2 - 5(m - s);

6) -(x + 2y) - 5(x + 2y) 2.

Find the roots of the equation:

1) 4x 2 - x = 0;

2) 7x 2 + 28x = 0;

3) x 2 + x = 0;

4)x 2 - x = 0.

Solve the equation:

1) 12x 2 + x = 0;

2) 0.2 x 2 - 2x = 0;

3) x 2 - x = 0;

4) 1 - x 2 + - x = 0.

Solve the equation:

1) x(3x + 2) - 5(3x + 2) = 0;

2) 2x(x - 2) - 5(2 - x) = 0.

Solve the equation:

1) x(4x + 5) - 7(4x + 5) = 0;

2) 7(x - 3) - 2x(3 - x) = 0.

1) 17 3 + 17 2 is a multiple of 18;

2) 9 14 - 81 6 is a multiple of 80.

Prove that the meaning of the expression is:

1) 39 9 - 39 8 is divided by 38;

2) 49 5 - 7 8 is divided by 48.

Take the common factor out of brackets:

1) (5m - 10) 2 ;

2) (18a + 27b) 2 .

Find the roots of the equation:

1) x(x - 3) = 7x - 21;

2) 2x(x - 5) = 20 - 4x.

Solve the equation:

1) x(x - 2) = 4x - 8;

2) 3x(x - 4) = 28 - 7x.

Prove that the number:

1) 10 4 + 5 3 is divisible by 9;

2) 4 15 - 4 14 + 4 13 is divided by 13;

3) 27 3 - 3 7 + 9 3 is divided by 25;

4) 21 3 + 14 a - 7 3 is divided by 34.

Exercises to repeat

Simplify the expression and find its meaning:

1) -3x 2 + 7x 3 – 4x 2 + 3x 2, if x = 0.1;

2) 8m + 5n - 7m + 15n, if m = 7, n = -1.

Write the following monomial coefficients instead of asterisks so that the equality turns into an identity:

1) 2m 2 - 4mn + n 2 + (*m 2 - *m - *n 2) = 3m 2 - 9mn - 5n 2 ;

2) 7x 2 - 10y 2 - xy - (*x 2 - *xy + * 2) = -x 2 + 3y 2 + xy.

The length of a rectangle is three times its width. If the length of a rectangle is reduced by 5 cm, then its area will decrease by 40 cm 2. Find the length and width of the rectangle.

Interesting tasks for lazy students

It is known that a< b < с. Могут ли одновременно выполняться неравенства |а| >|s| and |b|< |с|?

Among the various expressions that are considered in algebra, sums of monomials occupy an important place. Here are examples of such expressions:
\(5a^4 - 2a^3 + 0.3a^2 - 4.6a + 8\)
\(xy^3 - 5x^2y + 9x^3 - 7y^2 + 6x + 5y - 2\)

The sum of monomials is called a polynomial. The terms in a polynomial are called terms of the polynomial. Monomials are also classified as polynomials, considering a monomial to be a polynomial consisting of one member.

For example, a polynomial
\(8b^5 - 2b \cdot 7b^4 + 3b^2 - 8b + 0.25b \cdot (-12)b + 16 \)
can be simplified.

Let us represent all the terms in the form of monomials of the standard form:
\(8b^5 - 2b \cdot 7b^4 + 3b^2 - 8b + 0.25b \cdot (-12)b + 16 = \)
\(= 8b^5 - 14b^5 + 3b^2 -8b -3b^2 + 16\)

Let us present similar terms in the resulting polynomial:
\(8b^5 -14b^5 +3b^2 -8b -3b^2 + 16 = -6b^5 -8b + 16 \)
The result is a polynomial, all terms of which are monomials of the standard form, and among them there are no similar ones. Such polynomials are called polynomials of standard form.

For degree of polynomial of a standard form take the highest of the powers of its members. Thus, the binomial \(12a^2b - 7b\) has the third degree, and the trinomial \(2b^2 -7b + 6\) has the second.

Typically, the terms of polynomials of standard form containing one variable are arranged in descending order of exponents of its degree. For example:
\(5x - 18x^3 + 1 + x^5 = x^5 - 18x^3 + 5x + 1\)

The sum of several polynomials can be transformed (simplified) into a polynomial of standard form.

Sometimes the terms of a polynomial need to be divided into groups, enclosing each group in parentheses. Since bracketing is the inverse transformation of opening brackets, it is easy to formulate rules for opening brackets:

If a “+” sign is placed before the brackets, then the terms enclosed in brackets are written with the same signs.

If a “-” sign is placed before the brackets, then the terms enclosed in the brackets are written with opposite signs.

Transformation (simplification) of the product of a monomial and a polynomial

Using the distributive property of multiplication, you can transform (simplify) the product of a monomial and a polynomial into a polynomial. For example:
\(9a^2b(7a^2 - 5ab - 4b^2) = \)
\(= 9a^2b \cdot 7a^2 + 9a^2b \cdot (-5ab) + 9a^2b \cdot (-4b^2) = \)
\(= 63a^4b - 45a^3b^2 - 36a^2b^3 \)

The product of a monomial and a polynomial is identically equal to the sum of the products of this monomial and each of the terms of the polynomial.

This result is usually formulated as a rule.

To multiply a monomial by a polynomial, you must multiply that monomial by each of the terms of the polynomial.

We have already used this rule several times to multiply by a sum.

Product of polynomials. Transformation (simplification) of the product of two polynomials

In general, the product of two polynomials is identically equal to the sum of the product of each term of one polynomial and each term of the other.

Usually the following rule is used.

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

Abbreviated multiplication formulas. Sum squares, differences and difference of squares

You have to deal with some expressions in algebraic transformations more often than others. Perhaps the most common expressions are \((a + b)^2, \; (a - b)^2 \) and \(a^2 - b^2 \), i.e. the square of the sum, the square of the difference and difference of squares. You noticed that the names of these expressions seem to be incomplete, for example, \((a + b)^2 \) is, of course, not just the square of the sum, but the square of the sum of a and b. However, the square of the sum of a and b does not occur very often; as a rule, instead of the letters a and b, it contains various, sometimes quite complex, expressions.

The expressions \((a + b)^2, \; (a - b)^2 \) can be easily converted (simplified) into polynomials of the standard form; in fact, you have already encountered such a task when multiplying polynomials:
\((a + b)^2 = (a + b)(a + b) = a^2 + ab + ba + b^2 = \)
\(= a^2 + 2ab + b^2 \)

It is useful to remember the resulting identities and apply them without intermediate calculations. Brief verbal formulations help this.

\((a + b)^2 = a^2 + b^2 + 2ab \) - the square of the sum is equal to the sum of the squares and the double product.

\((a - b)^2 = a^2 + b^2 - 2ab \) - the square of the difference is equal to the sum of squares without the doubled product.

\(a^2 - b^2 = (a - b)(a + b) \) - the difference of squares is equal to the product of the difference and the sum.

These three identities allow one to replace its left-hand parts with right-hand ones in transformations and vice versa - right-hand parts with left-hand ones. The most difficult thing is to see the corresponding expressions and understand how the variables a and b are replaced in them. Let's look at several examples of using abbreviated multiplication formulas.