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
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- 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

$11x$

$2y - 3$

$2x + y$

$2xy + 11z+1$

$x^2 +2+x$

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.

$11x^2$

$6xy$

$-2z$

$10y$

$-9$

$82mnq$

$x + xy$

$2p + 11q$

$y + 1$

$11 - 3xy$

$x + xy+1$

$2p + 11q+pq$

$y + 2+y^2$

$11 -3xy+x$

$x + y + z + 1$

$3pq$

$11xyz - 10x$

$4x + 2y + z$

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

$2x$, $3x$ are like term

$5x^2$ and $-9x^2$ are like terms.

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

$2x$, $3y$ are unlike term

$5x^2$ and $-9zx^2$ are unlike terms.

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 $4x - 7xy + 3y + 12$ from $12x - 9xy + 5y - 3$

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

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

Arranging the like term together

$=8x-2xy+2y-15$

General steps for Multiplication

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

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

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

- group like terms

- Multiplication of monomial to monomial

- Multiplication of monomial to binomial, trinomial or more terms polynomials

- Multiplication of binomial, trinomial or more terms polynomials to monomial

- Multiplication of binomial to binomial, trinomial or more terms polynomials

- Multiplication of trinomial to trinomial or more terms polynomials

(i) $a^2 \times (2b^{22}) \times (4a^{26})$

(ii $(\frac {-10pq^3}{3} \times (\frac {6p^3}{5})$

We will use the below property extensively in above questions

$x^m \times x^n \times x^o=x^{m+n+o}$

1. Multiplying the constant and using the exponent property given above

we get

$a^2 \times (2b^{22}) \times (4a^{26})=8a^{28}b^{22}$

2. $(\frac {-10pq^3}{3} \times (\frac {6p^3}{5})$

$=(-12p^4q^3)$

(i) $(2a + 6b)$ and $(4a - 3b)$

(ii) $(x - 1)$ and $(3x - 2)$

Let first understand how to multiply the terms $(a+b)$ and $(c+d)$

Multiplication can be done by distributive law

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

$=(a \times c)+(a \times d)+(b \times c)+(b \times d)$

We will use the same concept in all the question below

(i) (2a + 6b) and (4a - 3b)

$=2a \times 4a - 2a \times 3b + 6b \times 4a - 6b \times 3b$

= 8a² - 6ab + 24ab -18b

= 8a² + 18ab -18b

(ii) (x - 1) and (3x - 2)

= x × 3x - 2x - 1 × 3x + 2

= 3x

= 3x

(3x + 1)(4x

=3x(4x

=12x

=12x

$(a + b + c)(a + b - c)$

$=a(a + b -c) + b(a + b -c)+c(a + b -c)$

$=a^2 +ab-ac +ab +b^2 -bc +ac+bc-c^2$

$=a^2 + b^2 -c^2 +2ab$

- $(2x-1)(x-2)$
- $(2y+3+z)(y+3)$
- $(2z+x)(z+x)$
- $(x+9)(x+y+1)$
- $(x+11)(11-y)$
- $x^2 \times (xy)^10 \times y^10 \times zx^2 \times 11x \times 2z$
- $(2x+1) -(1-x+y)$
- $(x^2 +y^2 +1) + (1 -z^2)$

$(a + b)^2 = a^2 + 2ab + b^2$

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

$x^2=1$

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

- $(a + b)^2 = a^2 + 2ab + b^2$

- $(a - b)^2 = a^2 - 2ab + b^2$

- $(a + b) (a - b) = a^2- b^2$

- $(x + a) (x + b) = x^2 + (a + b) x + ab$

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) $(y + 1) (y + 1)$

$=(y+1)^2$

Now $(a + b)^2 = a^2 + 2ab + b^2$

$=y^2 +2y+1$

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

Now $(a + b) (a - b) = a^2- b^2$

$=4x^2 -1$

(iii) $(2z - 3) (2z - 3)$

$=(2z-3)^2$

Now as $(a - b)^2 = a^2 - 2ab + b^2$

$=4z^2 -12z+9$

- $(x-1)(x-1)$
- $(y+3)(y+3)$
- $(2z+1)(2z-1)$
- $(x-9)(x+10)$
- $(x+7)(x+7)$
- $(x-5)(x-7)$

**Notes****Worksheets**-
**Ncert Solutions**- Algebraic Expressions and Identities NCERT solutions Exercise 9.1
- Algebraic Expressions and Identities NCERT solutions Exercise 9.2
- Algebraic Expressions and Identities NCERT solutions Exercise 9.3
- Algebraic Expressions and Identities NCERT solutions Exercise 9.4
- Algebraic Expressions and Identities NCERT solutions Exercise 9.5

Class 8 Maths Class 8 Science

Given below are the links of some of the reference books for class 8 Math.

- Mathematics Foundation Course for JEE/Olympiad : Class 8 This book can take students maths skills further. Only buy if child is interested in Olympiad/JEE foundation courses.
- Mathematics for Class 8 by R S Aggarwal Detailed Mathematics book to clear basics and concepts. I would say it is a must have book for class 8 student.
- Pearson Foundation Series (IIT -JEE / NEET) Physics, Chemistry, Maths & Biology for Class 8 (Main Books) | PCMB Combo : These set of books could help your child if he aims to get extra knowledge of science and maths. These would be helpful if child wants to prepare for competitive exams like JEE/NEET. Only buy if you can provide help to the child while studying.
- Reasoning Olympiad Workbook - Class 8 :- Reasoning helps sharpen the mind of child. I would recommend students practicing reasoning even though they are not appearing for Olympiad.

You can use above books for extra knowledge and practicing different questions.

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