- What is algebraic expression
- |
- Terms, Factors and Coefficients
- |
- Monomials, Binomials, trinomial and Polynomials
- |
- Like and Unlike Terms
- |
- Addition and Subtraction of Algebraic Expressions
- |
- Multiplication of Algebraic expression
- |
- What is an Identity

In this page we will explain the topics for the chapter 9 of Algebraic Expressions and Identities Class 8 Maths.We have given quality notes and video to explain various things so that students can benefits from it and learn maths in a fun and easy manner, Hope you like them and do not forget to like , social share
and comment at the end of the page.

11

2

2x +

2

x

11x + 1

1) This expression is made up of two terms, 11x and 1.

2) Terms are added to form algebraic expressions.

3) Terms themselves can be formed as the product of factors. The term 11x is the product of its factors 11 and x. The term 1 is made up of just one factor, i.e., 1.

4) The numerical factor of a term is called its numerical coefficient or simply coefficient. The coefficient in the term 11x is 11 and the coefficient in the constant term 1 is 1.

11x

6

–2

10

–9

82mnq

2

11 –3

2

11 –3

3

11

4

When the variable part of the terms is same, they are called like terms

2x, 3x are like term

5

Like term can be add or subtracted. Basically We need to add or subtract the coefficent

2x, 3y are unlike term

5

1) We write each expression to be added in a separate row. While doing so we write like terms one below the other

Or

We add the expression together on the same line and arrange the like term together

2) Add the like terms

3) Write the Final algebraic expression

Add the following expression x – y + xy, y – z + yz, z – x + zx

= (x – y + xy)+( y – z + yz)+( z – x + zx)

Arranging the like term together

= x-x -y+y-z+z+xy+yz+zx

=xy+yz+zx as x-x=y-y=z-z=0

1) We write each expression to be subtracted in a separate row. While doing so we write like terms one below the other and then we change the sign of the expression which is to be subtracted i.e. + becomes – and – becomes +

Or

We subtract the expression together on the same line, change the sign of all the term which is to be subtracted and then arrange the like term together

2) Add the like terms

3) Write the Final algebraic expression

Subtract

=(

=12x-9xy+5y-3 -4x+7xy-3y-12

Arranging the like term together

=8x-2xy+2y-15

General steps for Multiplication

1) We have to use distributive law and distribute each term of the first polynomial to every term of the second polynomial.

2) when you multiply two terms together you must multiply the coefficient (numbers) and add the exponents

3) Also as we already know ++ equals =, +- or -+ equals - and -- equals +

4) group like terms

Multiplication may involve

a) Multiplication of monomial to monomial

b) Multiplication of monomial to binomial, trinomial or more terms polynomials

c) Multiplication of binomial, trinomial or more terms polynomials to monomial

d) Multiplication of binomial to binomial, trinomial or more terms polynomials

e) Multiplication of trinomial to trinomial or more terms polynomials

- a
^{2}× (2b^{22}) × (4a^{26})

x

1. As you know

So, we get

a

2. (10pq

=(-12p

(i) (2a + 6b) and (4a – 3b)

(ii) (x – 1) and (3x – 2)

then by distributive law

( a+b) (c+d)= a(c+d) + b( c+d)

=(a × c)+(a × d)+(b × c)+(b × d)

We will use the same concept in all the question below

1) (2a + 6b) and (4a – 3b)

= 2a × 4a – 2a × 3b + 6 b × 4a – 6b × 3b

= 8a² - 6ab + 24ab -18b

= 8a² + 18ab -18b

2) (x – 1) and (3x – 2)

= x × 3x – 2x - 1 × 3x + 2

= 3x

= 3x

(3x + 1)(4x

=3x(4x

=12x

=12x

(

=a(

=a

=a

(

It is true for all the values of a and b

On the other hand, an equation is true only for certain values of its variables. An equation is not an identity

x

The below four identities are useful in carrying out squares and products of algebraic expressions.

(

(

(

These allow easy alternative methods to calculate products of numbers.

Use a suitable identity to get each of the following products.

(i) (y + 1) (y + 1)

(ii) (2x + 1) (2x -1)

(iii) (2z – 3) (2z – 3)

i) (y + 1) (y + 1)

=(y+1)

Now (

=y

ii) (2x + 1) (2x -1)

Now (

=4x

iii) (2z – 3) (2z – 3)

=(2z-3)

Now as (

=4z

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