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NCERT Solutions Class 8 Maths chapter 8 Algebraic Expressions and Identities

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NCERT Solutions for Maths Class 8 Chapter 8 - FREE PDF Download

NCERT Solutions for Class 8 Maths chapter 8 Algebraic Expressions and Identities by Vedantu introduces you to the powerful tool of identities and various types of expressions such as monomials, binomials, and polynomials. Algebraic expressions are combinations of variables, constants, and operators that represent a value.

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Table of Content
1. NCERT Solutions for Maths Class 8 Chapter 8 - FREE PDF Download
2. Glance on Maths chapter 8 Class 8 - Algebraic Expressions and Identities
3. Access Exercise wise NCERT Solutions for chapter 8 Maths Class 8
4. Exercises Under NCERT Solutions for Class 8 Maths chapter 8 Algebraic Expressions and Identities
5. Access NCERT Solutions for Class 8 Maths chapter 8 –  Algebraic Expressions and Identities
    5.1Exercise - 8.1
    5.2Exercise - 8.2
    5.3Exercise - 8.3
    5.4Exercise - 8.4
6. NCERT Solutions for Class 8 Maths chapter 8 PDF
    6.1Overview of Deleted Syllabus for CBSE Class 8 Maths Algebraic Expressions and Identities
7. Class 8 Maths chapter 8: Exercises Breakdown
8. Conclusion 
9. Other Study Material for CBSE Class 8 Maths chapter 8
10. Chapter-Specific NCERT Solutions for Class 8 Maths
11. Important Related Links for CBSE Class 8 Maths
FAQs


In this chapter, you will learn how to simplify, and manipulate algebraic expressions. Additionally, you will delve into important algebraic identities that simplify complex expressions and solve equations efficiently. Vedantu’s Class 8 Maths NCERT Solutions provide step-by-step explanations to help you understand these concepts thoroughly. The clear and concise explanations make it easier to grasp the material and apply it to solve problems.


Glance on Maths chapter 8 Class 8 - Algebraic Expressions and Identities

  • This chapter explains that expressions are combinations of variables, numbers, and mathematical operations (like addition, subtraction, multiplication, division).

  • Expressions are of different types based on the number of terms they have:

  • Monomials (one term) - $4x,xy^{^{2}}$

  • Binomials (two terms) - $8x+9, y^{^{2}}-3$

  • Trinomials (three terms) - 3x + 6y - 6

  • Identities are equations that are always true, no matter what value you assign to the variable.

  • This article algebraic expressions and identities class 8 contains chapter notes, important questions, exemplar solutions, exercises and video links for Chapter  - Algebraic Expressions and Identities, which you can download as PDFs.

  • There are four exercises (15 fully solved questions) in class 8th Maths chapter 8 Algebraic Expressions and Identities.


Access Exercise wise NCERT Solutions for chapter 8 Maths Class 8

Exercises Under NCERT Solutions for Class 8 Maths chapter 8 Algebraic Expressions and Identities

  • Exercise 8.1: This exercise consists of 2 Questions and Solutions. This exercise deals with addition and subtraction of Algebraic Expressions.

  • Exercise 8.2: This exercise consists of 5 Questions and Solutions. This exercise deals with multiplication of Algebraic Expressions, multiplying a monomial by a monomial and multiplying three or more monomials.

  • Exercise 8.3: This exercise consists of 5 Questions and Solutions. This exercise deals with multiplying a monomial by a polynomial, multiplying a monomial by a binomial, multiplying a monomial by a trinomial.

  • Exercise 8.4: This exercise consists of 3 Questions and Solutions. This exercise deals with multiplying a polynomial by a polynomial, multiplying a binomial by a binomial, multiplying a binomial by a trinomial.


Access NCERT Solutions for Class 8 Maths chapter 8 –  Algebraic Expressions and Identities

Exercise - 8.1

1. Add the following:

(i) ${\text{ab - bc,bc - ca,ca - ab}}$

Ans:  

${\text{    12a - 9ab + 5b - 3}} $

Therefore, the sum of the given expressions is o.


(ii) ${\text{a - b + ab,b - c + bc,c - a + ac}}$

Ans: 

$  {\text{          }}a - b + ab $

$  {\text{             }} + b{\text{       }} - c + bc $

$  {\text{  }} + \quad  - a{\text{           }} + c{\text{         + ac}} $

$  \overline {{\text{                  ab           + bc + ac}}}  $

Thus the sum of given expressions is ${\text{ab + bc + ac}}$


(iii) ${\text{2}}{{\text{p}}^{\text{2}}}{{\text{q}}^{\text{2}}}{\text{ - 3pq + 4,5 + 7pq - 3}}{{\text{p}}^{\text{2}}}{{\text{q}}^{\text{2}}}$

Ans: 

$ {\text{     2}}{{\text{p}}^{\text{2}}}{{\text{q}}^{\text{2}}}{\text{  - 3pq  + 4}} $

$  {\text{ +    - 3}}{{\text{p}}^{\text{2}}}{{\text{q}}^{\text{2}}}{\text{ + 7pq + 5}}$

$  \overline {{\text{     - }}{{\text{p}}^{\text{2}}}{{\text{q}}^{\text{2}}}{\text{   + 4pq    + 9}}}   $

Therefore, the sum of given expressions is ${\text{ - }}{{\text{p}}^{\text{2}}}{{\text{q}}^{\text{2}}}{\text{ + 4pq  + 9}}$


(iv) ${{\text{l}}^{\text{2}}}{\text{ + }}{{\text{m}}^{\text{2}}}{\text{,}}{{\text{m}}^{\text{2}}}{\text{ + }}{{\text{n}}^{\text{2}}}{\text{,}}{{\text{n}}^{\text{2}}}{\text{ + }}{{\text{l}}^{\text{2}}}{\text{,2lm + 2mn + 2nl}}$

Ans:  

$  {\text{      }}{{\text{l}}^{\text{2}}}{\text{ + }}{{\text{m}}^{\text{2}}} $

$  {\text{  +           }}{{\text{m}}^{\text{2}}}{\text{ +  }}{{\text{n}}^{\text{2}}} $

$  {\text{ +    }}{{\text{l}}^{\text{2}}}{\text{           +  }}{{\text{n}}^{\text{2}}} $

$  {\text{ +                             2lm + 2mn + 2nl}} $

$  \overline {{\text{    2}}{{\text{l}}^{^{\text{2}}}}{\text{ + 2}}{{\text{m}}^{\text{2}}}{\text{ + 2}}{{\text{n}}^{\text{2}}}{\text{ + 2lm + 2mn + 2nl}}}  $

Therefore, the sum of the given expressions is ${\text{2}}{{\text{l}}^{^{\text{2}}}}{\text{ + 2}}{{\text{m}}^{\text{2}}}{\text{ + 2}}{{\text{n}}^{\text{2}}}{\text{ + 2lm + 2mn + 2nl}}$


2. Solve the following:

(i) Subtract ${\text{4a - 7ab + 3b + 12}}$ from ${\text{12a - 9ab + 5b - 3}}$

Ans:  

$  {12a - 9ab + 5b - 3} $

$  {4a - 7ab + 3b + 12} $

$  {( - )\quad ( + )\quad ( - )( - )} $ 

$  {\overline {8a - 2ab + 2b - 15} } $


(ii) Subtract ${\text{3xy + 5yz - 7zx}}$ from ${\text{5xy - 2yz - 2zx + 10xyz}}$

Ans:  

$  {\text{5xy - 2yz - 2zx + 10xyz}} $

$  {\text{3xy + 5yz - 7zx}} $

$  {\text{( - )( - )}}\quad {\text{( + )}} $

$  \overline {{\text{2xy - 7yz + 5zx + 10xyz}}} $


(iii) Subtract ${\text{4p 2q  -  3pq  +  5pq2  -  8p  +  7q  -  10}}$from ${\text{18  -  3p  -  11q  +  5pq  -  2pq2  +  5p 2q}}$

Ans: 

$ {\text{18 - 3p - 11q + 5pq - 2p}}{{\text{q}}^{\text{2}}}{\text{ + 5}}{{\text{p}}^{\text{2}}}{\text{q}} $

$  {\text{ - 10 - 8p + 7q - 3pq + 5p}}{{\text{q}}^{\text{2}}}{\text{ + 4}}{{\text{p}}^{\text{2}}}{\text{q}} $

$  \dfrac{{{\text{( + )( + )( - )( + )( - )}}\quad {\text{( - )}}}}{{{\text{28 + 5p - 18q + 8pq - 7p}}{{\text{q}}^{\text{2}}}{\text{ + }}{{\text{p}}^{\text{2}}}{\text{q}}}} $


Exercise - 8.2

1. Find the product of the following pairs of monomials.

(i) ${\text{4,7p}}$

Ans:  ${{4  \times  7p  =  4  \times  7  \times  p  =  28p}}$


(ii) $\dfrac{{{\text{first monomial}} \to }}{{{\text{second monomial}} \downarrow }}$

Ans:  ${{ -  4p  \times  7p  =   -  4  \times  p  \times  7  \times  p  =  }}\left( {{{ -  4  \times  7}}} \right){{  \times  }}\left( {{{p  \times  p}}} \right){\text{  =   -  28 }}{{\text{p}}^2}$


(iii) ${\text{ - 4p,7pq}}$

Ans:  ${{ -  4p  \times  7pq  =   -  4  \times  p  \times  7  \times  p  \times  q  =  }}\left( {{{ -  4  \times  7}}} \right){{  \times  }}\left( {{{p  \times  p  \times  q}}} \right){\text{  =   -  28}}{{\text{p}}^2}{\text{q }}$


(iv) ${\text{4}}{{\text{p}}^{\text{3}}}{\text{ ,  -  3p }}$

Ans:  ${\text{ 4}}{{\text{p}}^{\text{3}}}{{  \times   -  3p  =  4  \times  }}\left( {{\text{ -  3}}} \right){{  \times  p  \times  p  \times  p  \times  p  =   -  12 }}{{\text{p}}^{\text{4}}}$


(v) ${\text{4p, 0}}$

Ans:  ${{4p  \times  0  =  4  \times  p  \times  0  =  0 }}$


2. Find the areas of rectangles with the following pairs of monomials as their lengths and breadths respectively.

$\left( {{\text{p, q}}} \right){\text{; }}\left( {{\text{10m, 5n}}} \right){\text{; }}\left( {{\text{20}}{{\text{x}}^{\text{2}}}{\text{ , 5}}{{\text{y}}^{\text{2}}}{\text{ }}} \right){\text{; }}\left( {{\text{4x, 3}}{{\text{x}}^{\text{2}}}{\text{ }}} \right){\text{; }}\left( {{\text{3mn, 4np}}} \right){\text{ }}$

Ans:  We know that,

Area of rectangle = length x breadth

Area of 1st rectangle = p x q = pq

Area of 2nd rectangle = ${{10m  \times  5n  =  10  \times  5  \times  m  \times  n   =  50mn}}$

Area of 3rd rectangle = ${\text{20}}{{\text{x}}^{\text{2}}}{{  \times  5}}{{\text{y}}^{\text{2}}}{{ =  20 \times 5 \times }}{{\text{x}}^{\text{2}}}{{ \times }}{{\text{y}}^{\text{2}}}{\text{ = 100}}{{\text{x}}^{\text{2}}}{{\text{y}}^{\text{2}}}$

Area of 4th rectangle = ${{4x }} \times {\text{ 3}}{{\text{x}}^{\text{2}}}{{  =  4 \times 3}} \times {{x}} \times {{\text{x}}^2}{\text{ = 12}}{{\text{x}}^3}$

Area of 5th rectangle ${{ =  3mn  \times  4np  =  3  \times  4  \times  m  \times  n  \times  n  \times  p  =  12m}}{{\text{n}}^{\text{2}}}{\text{p}}$


3. Complete the table of products.

$\dfrac{{{\text{first monomial}} \to }}{{{\text{second monomial}} \downarrow }}$

2x

-5y

3x2

-4xy

7x2y

-9x2y

2x

4x2

..

-5y

15x2

3x2

-4xy

7x2y

-9x2y2


Ans:

The table can be completed as follows.

$\dfrac{{{\text{first monomial}} \to }}{{{\text{second monomial}} \downarrow }}$

2x

-5y

3x2

4xy

7x2y

-9x2y

2x

4x2

-10xy

6x2

-8x2y

14x3y

-18x3y2

-5y

-10xy

25y2

-15x2

20xy2

-35x2y2

45x2y3

3x2

6x3

-15x2y

9x4

-12x3

21x4y

-27x4y2

-4xy

-8x2y

20xy2

-12x3y

16x2y2

-28x3y2

36x3y3

7x2y

14x3y

-35x2y2

21x4y

-28x3y2

49x4y2

-63x4y3

-9x2y2

-18x3y2

45x2y3

-27x4y2

36x3y3

-63x4y3

81x4y4


4. Obtain the volume of rectangular boxes with the following length, breadth and height respectively.

(i) ${\text{5a,3}}{{\text{a}}^{\text{2}}}{\text{,7}}{{\text{a}}^{\text{4}}}$

Ans:  We know that 

Volume= length x breadth x height

Volume =${{5a \times 3}}{{\text{a}}^{\text{2}}}{{ \times 7}}{{\text{a}}^{\text{4}}}{\text{ = 105}}{{\text{a}}^{\text{7}}}$


(ii) ${\text{2p,4q,8r}}$

Ans:  We know that 

Volume = length x breadth x height

Volume = ${{2p \times 4q \times 8r = 64pqr}}$


(iii) ${\text{xy,2}}{{\text{x}}^{\text{2}}}{\text{y,2x}}{{\text{y}}^{\text{2}}}$

Ans:  We know that 

Volume = length x breadth x height

Volume = ${{xy \times 2}}{{\text{x}}^{\text{2}}}{{y \times 2x}}{{\text{y}}^{\text{2}}}{\text{ = 4}}{{\text{x}}^{\text{4}}}{{\text{y}}^{\text{4}}}$


(iv) ${\text{a,2b,3c}}$

Ans:  We know that 

Volume = length x breadth x height

Volume = ${{a}} \times {\text{2b}} \times {\text{3c = 6abc}}$


5. Obtain the product of

(i) ${\text{xy, yz, zx }}$

Ans: ${{xy  \times  yz  \times  zx  =  }}{{\text{x}}^{\text{2}}}{\text{ }}{{\text{y}}^{\text{2}}}{\text{ }}{{\text{z}}^{\text{2}}}$


(ii) ${\text{a,  -  }}{{\text{a}}^{\text{2}}}{\text{ , }}{{\text{a}}^{\text{3}}}{\text{ }}$

Ans:  ${{a}} \times ({\text{ -  }}{{\text{a}}^{{2}}}) \times {\text{ }}{{\text{a}}^{\text{3}}}{\text{ =  - }}{{\text{a}}^6}{\text{ }}$


(iii) ${\text{2, 4y, 8}}{{\text{y}}^2}{\text{ , 16}}{{\text{y}}^3}$

Ans:  ${{2}} \times {{ 4y}} \times {\text{8}}{{\text{y}}^2} \times {\text{ 16}}{{\text{y}}^3} = 1024{y^6}$


(iv) ${\text{a, 2b, 3c, 6abc}}$

Ans:  ${{a  \times  2b  \times  3c  \times  6abc  = }}$${\text{36}}{{\text{a}}^{\text{2}}}{{\text{b}}^{\text{2}}}{{\text{c}}^{\text{2}}}$


(v) ${\text{m,  -  mn, mnp}}$

Ans:  ${{m  \times  }}\left( {{\text{ -  mn}}} \right){{  \times  mnp  =   -  }}{{\text{m}}^{\text{3}}}{{\text{n}}^{\text{2}}}$


Exercise - 8.3

1. Carry out the multiplication of the expressions in each of the following pairs.

(i) ${\text{4p, q  +  r }}$

Ans:  $\left( {{\text{4p}}} \right){{  \times  }}\left( {{\text{q  +  r}}} \right){\text{  =  }}\left( {{{4p  \times  q}}} \right){\text{  +  }}\left( {{{4p  \times  r}}} \right){\text{  =  4pq  +  4pr}}$


(ii) ${\text{ab, a  -  b }}$

Ans:  $\left( {{\text{ab}}} \right){{  \times  }}\left( {{\text{a  -  b}}} \right){\text{  =  }}\left( {{{ab  \times  a}}} \right){\text{  +  }}\left[ {{{ab  \times  }}\left( {{\text{ -  b}}} \right)} \right]{\text{  =  }}{{\text{a}}^{\text{2}}}{\text{b  -  a}}{{\text{b}}^{\text{2}}}$


(iii) ${\text{a  +  b, 7}}{{\text{a}}^{\text{2}}}{{\text{b}}^{\text{2}}}$

Ans:  $\left( {{\text{a  +  b}}} \right){{  \times  }}\left( {{\text{7a 2 b 2 }}} \right){\text{  =  }}\left( {{{a  \times  7}}{{\text{a}}^{\text{2}}}{{\text{b}}^{\text{2}}}{\text{ }}} \right){\text{  +  }}\left( {{{b  \times  7}}{{\text{a}}^{\text{2}}}{{\text{b}}^{\text{2}}}{\text{ }}} \right){\text{  =  7}}{{\text{a}}^{\text{3}}}{{\text{b}}^{\text{2}}}{\text{  +  7}}{{\text{a}}^{\text{2}}}{{\text{b}}^{\text{3}}}$


(iv) ${{\text{a}}^{\text{2}}}{\text{  -  9, 4a}}$

Ans:  $\left( {{{\text{a}}^2}{\text{  -  9}}} \right){{  \times  }}\left( {{\text{4a}}} \right){\text{  =  }}\left( {{{\text{a}}^{\text{2}}}{{  \times  4a}}} \right){\text{  +  }}\left( {{\text{ -  9}}} \right){{  \times  }}\left( {{\text{4a}}} \right){\text{  =  4}}{{\text{a}}^{\text{3}}}{\text{  -  36a}}$


(v) ${\text{pq  +  qr  +  rp, 0}}$

Ans:  $\left( {{\text{pq  +  qr  +  rp}}} \right){{  \times  0  =  }}\left( {{{pq  \times  0}}} \right){\text{  +  }}\left( {{{qr  \times  0}}} \right){\text{  +  }}\left( {{{rp  \times  0}}} \right){\text{  =  0 }}$


2. Complete the table

--

First expression

Second expression

Product


a

b+c+d

-


x+y-5

5xy

-


p

${\text{6}}{{\text{p}}^{\text{2}}}{\text{  -  7p  +  5 }}$

-


${\text{4}}{{\text{p}}^{\text{2}}}{{\text{q}}^{\text{2}}}$

${{\text{p}}^{\text{2}}}{\text{ - }}{{\text{q}}^{\text{2}}}$

-


a+b+c

abc

-


Ans:  The table can be completed as follows:

---

First expression

Second expression

Product


a

b+c+d

ab+ac+ad


x+y-5

5xy

5x2y+5xy2-25xy


p

${\text{6}}{{\text{p}}^{\text{2}}}{\text{  -  7p  +  5 }}$

6p3-7p2+5p


${\text{4}}{{\text{p}}^{\text{2}}}{{\text{q}}^{\text{2}}}$

${{\text{p}}^{\text{2}}}{\text{ - }}{{\text{q}}^{\text{2}}}$

4p4q2-4p2q4


a+b+c

abc

a2bc+ab2c+abc2


3. Find the product:

(i) $\left( {{{\text{a}}^{\text{2}}}} \right){{  \times  }}\left( {{\text{2}}{{\text{a}}^{{\text{22}}}}} \right){{  \times  }}\left( {{\text{4}}{{\text{a}}^{{\text{26}}}}} \right)$

Ans:  $\left( {{{\text{a}}^{\text{2}}}} \right){{  \times  }}\left( {{\text{2}}{{\text{a}}^{{\text{22}}}}} \right){{  \times  }}\left( {{\text{4}}{{\text{a}}^{{\text{26}}}}} \right){{  =  2  \times  4  \times }}{{\text{a}}^{\text{2}}}{{  \times  }}{{\text{a}}^{{\text{22}}}}{{  \times  }}{{\text{a}}^{{\text{26}}}}{\text{  =  8}}{{\text{a}}^{{\text{50}}}}$


(ii) $\left( {\dfrac{{\text{2}}}{{\text{3}}}{\text{xy}}} \right){{  \times  }}\left( {\dfrac{{{\text{ - 9}}}}{{{\text{10}}}}{{\text{x}}^{\text{2}}}{{\text{y}}^{\text{2}}}} \right)$

Ans:  $\left( {\dfrac{{\text{2}}}{{\text{3}}}{\text{xy}}} \right){{  \times  }}\left( {\dfrac{{{\text{ - 9}}}}{{{\text{10}}}}{{\text{x}}^{\text{2}}}{{\text{y}}^{\text{2}}}} \right){\text{  =  }}\left( {\dfrac{{\text{2}}}{{\text{3}}}} \right){{ \times }}\left( {\dfrac{{{\text{ - 9}}}}{{{\text{10}}}}} \right){{ \times x \times y \times }}{{\text{x}}^{\text{2}}}{{ \times }}{{\text{y}}^{\text{2}}}{\text{  =  }}\dfrac{{{\text{ - 3}}}}{{\text{5}}}{{\text{x}}^{\text{3}}}{{\text{y}}^{\text{3}}}$


(iii) $\left( {{\text{ - }}\dfrac{{{\text{10}}}}{{\text{3}}}{\text{p}}{{\text{q}}^{\text{3}}}} \right){{  \times  }}\left( {\dfrac{{\text{6}}}{{\text{5}}}{{\text{p}}^{\text{3}}}{\text{q}}} \right)$

Ans:  $\left( {{\text{ - }}\dfrac{{{\text{10}}}}{{\text{3}}}{\text{p}}{{\text{q}}^{\text{3}}}} \right){{  \times  }}\left( {\dfrac{{\text{6}}}{{\text{5}}}{{\text{p}}^{\text{3}}}{\text{q}}} \right){\text{  =  }}\left( {\dfrac{{{\text{ - 10}}}}{{\text{3}}}} \right){{ \times }}\left( {\dfrac{{\text{6}}}{{\text{5}}}} \right){{ \times p}}{{\text{q}}^{\text{3}}}{{ \times }}{{\text{p}}^{\text{3}}}{\text{q  =   - 4}}{{\text{p}}^{\text{4}}}{{\text{q}}^{\text{4}}}$


(iv) ${{x  \times  }}{{\text{x}}^{\text{2}}}{{  \times  }}{{\text{x}}^{\text{3}}}{{  \times  }}{{\text{x}}^{\text{4}}}$

Ans:  ${{x  \times  }}{{\text{x}}^{\text{2}}}{{  \times  }}{{\text{x}}^{\text{3}}}{{  \times  }}{{\text{x}}^{\text{4}}}{\text{  =  }}{{\text{x}}^{10}}$


4. Solve the following

(a) Simplify ${\text{3x }}\left( {{\text{4x  - 5}}} \right){\text{  +  3}}$and find its values for 

(i) ${\text{ x  =  3}}$

Ans:  ${\text{3x }}\left( {{\text{4x  -  5}}} \right){\text{  +  3  =  12}}{{\text{x}}^{\text{2}}}{\text{  -  15x  +  3 }}$

$  {\text{ For x  =  3, 12}}{{\text{x}}^{\text{2}}}{\text{  -  15x  +  3  =  12 }}{\left( {\text{3}} \right)^{\text{2}}}{\text{  -  15}}\left( {\text{3}} \right){\text{  +  3 }} $

$  {\text{ =  108  -  45  +  3 }} $

$  {\text{ =  66 }} $


(ii) ${\text{x = }}\dfrac{{\text{1}}}{{\text{2}}}$

Ans:  

$  {\text{ For x  =  }}\dfrac{{\text{1}}}{{\text{2}}}{\text{, 12}}{{\text{x}}^{\text{2}}}{\text{  -  15x  +  3  =  12 }}{\left( {\dfrac{{\text{1}}}{{\text{2}}}} \right)^{\text{2}}}{\text{  -  15}}\left( {\dfrac{{\text{1}}}{{\text{2}}}} \right){\text{  +  3 }} $

$  {\text{ =  3  -  }}\dfrac{{{\text{15}}}}{{\text{2}}}{\text{  +  3 }} $

$  {\text{ =  6 - }}\dfrac{{{\text{15}}}}{{\text{2}}}{\text{ = }}\dfrac{{{\text{12 - 15}}}}{{\text{2}}}{\text{ = }}\dfrac{{{\text{ - 3}}}}{{\text{2}}}  $


(b) ${\text{a }}\left( {{{\text{a}}^{\text{2}}}{\text{  +  a  +  1}}} \right){\text{  +  5}}$ and find its value for 

(i) ${\text{a  =  0}}$

Ans:  ${\text{For a  =  0, }}{{\text{a}}^{\text{3}}}{\text{  +  }}{{\text{a}}^{\text{2}}}{\text{  +  a  +  5  =  0  +  0  +  0  +  5  =  5}}$


(ii) ${\text{a  =  1}}$

Ans:  $ {\text{For a  =  1, }}{{\text{a}}^{\text{3}}}{\text{  +  }}{{\text{a}}^{\text{2}}}{\text{  +  a  +  5  =  }}{\left( {\text{1}} \right)^{\text{3}}}{\text{  +  }}{\left( {\text{1}} \right)^{\text{2}}}{\text{  +  1  +  5}} $

$  {\text{  =  1  +  1  +  1  +  5  =  8 }}  $


(iii) ${\text{a  =   - 1}}$

Ans:  $  {\text{For a  =   - 1, }}{{\text{a}}^{\text{3}}}{\text{  +  }}{{\text{a}}^{\text{2}}}{\text{  +  a  +  5  =  }}{\left( {{\text{ - 1}}} \right)^{\text{3}}}{\text{  +  }}{\left( {{\text{ - 1}}} \right)^{\text{2}}}{\text{  +  }}\left( {{\text{ - 1}}} \right){\text{  +  5 }} $

$  {\text{ =   -  1  +  1  -  1  +  5  =  4 }}  $


5. Solve the following 

(i) Add: ${\text{p (p  -  q), q (q  -  r)}}$ and ${\text{r (r  -  p)}}$

Ans:  

$  {\text{First expression  =  p }}\left( {{\text{p  -  q}}} \right){\text{  =  }}{{\text{p}}^2}{\text{  -  pq }} $

$  {\text{Second expression  =  q }}\left( {{\text{q  -  r}}} \right){\text{  =  }}{{\text{q}}^2}{\text{  -  qr}} $

$  {\text{Third expression  =  r }}\left( {{\text{r  -  p}}} \right){\text{  =  }}{{\text{r}}^2}{\text{  -  pr}} $

Adding the three expressions, we obtain

$  {\text{       }}{{\text{p}}^{\text{2}}}{\text{  -  pq }} $

$  {\text{ +                   }}{{\text{q}}^{\text{2}}}{\text{  -  qr}} $

$  {\text{ +                                }}{{\text{r}}^{\text{2}}}{\text{  -  pr}} $

$  \overline {{\text{      }}{{\text{p}}^{\text{2}}}{\text{ - pq     + }}{{\text{q}}^{\text{2}}}{\text{ - qr   + }}{{\text{r}}^{\text{2}}}{\text{ - pr}}}  $

Therefore, the sum is ${{\text{p}}^{\text{2}}}{\text{ - pq + }}{{\text{q}}^{\text{2}}}{\text{ - qr  + }}{{\text{r}}^{\text{2}}}{\text{ - pr}}$


(ii) Add: ${\text{2x }}\left( {{\text{z  -  x  -  y}}} \right){\text{ and 2y }}\left( {{\text{z  -  y  -  x}}} \right){\text{ }}$

Ans:  

$  {\text{First expression  =  2x }}\left( {{\text{z  -  x  -  y}}} \right){\text{  =  2xz  -  2}}{{\text{x}}^{\text{2}}}{\text{  -  2xy }} $

$  {\text{Second expression  =  2y }}\left( {{\text{z  -  y  -  x}}} \right){\text{  =  2yz  -  2}}{{\text{y}}^{\text{2}}}{\text{  -  2yx }} $

Adding the two expressions, we obtain

$  {\text{    2xz  -  2}}{{\text{x}}^{\text{2}}}{\text{  -  2xy }} $

$  {\text{ +                    -  2yx  + 2yz -  2}}{{\text{y}}^{\text{2}}}{\text{  }} $

$  \overline {\,\,{\text{   2xz  -  2}}{{\text{x}}^{\text{2}}}{\text{ - 4xy    + 2yz  - 2}}{{\text{y}}^{\text{2}}}}  $

Therefore, the sum is ${\text{2xz  -  2}}{{\text{x}}^{\text{2}}}{\text{ - 4xy  + 2yz  - 2}}{{\text{y}}^{\text{2}}}$


(iii) Subtract ${\text{3l }}\left( {{\text{l  -  4m  +  5n}}} \right){\text{ from 4l }}\left( {{\text{10n  -  3m  +  2l}}} \right){\text{ }}$

Ans: 

$  {\text{3l }}\left( {{\text{l  -  4m  +  5n}}} \right){\text{  =  3}}{{\text{l}}^{\text{2}}}{\text{  -  12lm  +  15ln }} $

$  {\text{4l }}\left( {{\text{10n  -  3m  +  2l}}} \right){\text{  =  40ln  -  12lm  +  8}}{{\text{l}}^{\text{2}}}{\text{ }} $

Subtracting these expressions, we obtain

$  {\text{   8}}{{\text{l}}^{\text{2}}}{\text{  -  12lm  +  40ln}} $

$  {\text{   3}}{{\text{l}}^{\text{2}}}{\text{  -  12lm  +  15ln}} $

$  ( - )\,{\text{   }}( + ){\text{     }}( - ) $

$  \overline {{\text{   5}}{{\text{l}}^2}{\text{              + 25ln     }}} {\text{ }}  $

Therefore, the result is ${\text{5}}{{\text{l}}^2}{\text{ + 25ln}}$


(iv) Subtract ${\text{3a }}\left( {{\text{a  +  b  +  c}}} \right){\text{  -  2b }}\left( {{\text{a  -  b  +  c}}} \right){\text{ from 4c }}\left( {{\text{ -  a  +  b  +  c}}} \right)$

Ans:  

$  {\text{                 +  4}}{{\text{c}}^{\text{2}}}{\text{          -  4ac  +  4bc }} $

$  {\text{ 3}}{{\text{a}}^{\text{2}}}{\text{  +  2}}{{\text{b}}^{{\text{2 }}}}{\text{         +  ab  +  3ac  -  2bc}} $

$  {\text{( - )   ( - )              ( - )     ( - )    ( + )}} $

$  \overline {{\text{ - 3}}{{\text{a}}^{\text{2}}}{\text{ - 2}}{{\text{b}}^{\text{2}}}\,{\text{ + 4}}{{\text{c}}^{\text{2}}}{\text{ + ab  -  7ac  + 6bc}}}  $

Therefore, the result is ${\text{ - 3}}{{\text{a}}^{\text{2}}}{\text{ - 2}}{{\text{b}}^{\text{2}}}\,{\text{ + 4}}{{\text{c}}^{\text{2}}}{\text{ + ab  -  7ac  + 6bc}}$


Exercise - 8.4

1. Multiply the binomials.

(i) ${\text{(2x  +  5)}}$and ${\text{(4x  -  3)}}$

Ans: ${\text{(2x  +  5) }} \times {\text{ (4x  -  3)  =  2x }} \times {\text{(4x  -  3)  +  5}} \times {\text{(4x  -  3)}}$

${\text{ =  8}}{{\text{x}}^2}{\text{  -  6x  +  20x  -  15}}$

${\text{ =  8x2  +  14x  - 15 (By adding like terms)}}$


(ii) ${\text{(y  -  8)}}$and ${\text{(3y  -  4)}}$

Ans: ${{ (y  -  8)  \times  (3y  -  4) =  y \times (3y  -  8)  -  8 \times (3y  -  4)}}$

${\text{ =  3}}{{\text{y}}^2}{\text{ -  4y  -  24y  +  32}}$

${\text{ =  3}}{{\text{y}}^{\text{2}}}{\text{  -  28y  +  32 (By adding like terms)}}$


(iii) ${\text{(2}}{\text{.5l  -  0}}{\text{.5m)}}$and ${\text{(2}}{\text{.5l  +  0}}{\text{.5m)}}$

Ans: ${\text{(2}}{\text{.5l  -  0}}{\text{.5m)(2}}{\text{.5l  +  0}}{\text{.5m) = 2}}{{.5l \times (2}}{\text{.5l  +  0}}{\text{.5m) - 0}}{\text{.5m(2}}{\text{.5l  +  0}}{\text{.5m)}}$

${\text{ =  6}}{\text{.25}}{{\text{l}}^2}{\text{  +  1}}{\text{.25lm  -  1}}{\text{.25lm  -  0}}{\text{.25}}{{\text{m}}^2}$

${\text{ =  6}}{\text{.25}}{{\text{l}}^2}{\text{  -  0}}{\text{.25}}{{\text{m}}^2}$


(iv) $\left( {{\text{a  +  3b}}} \right)$and ${\text{(x  +  5)}}$

Ans:  ${\text{(a  +  3b) }} \times {\text{ (x  +  5)  =  a}} \times {{(x  +  5)  +  3b }} \times {\text{(x  +  5)}}$

${\text{ =  ax  +  5a  +  3bx  +  15b}}$


(v) ${\text{(2pq  +  3}}{{\text{q}}^2}{\text{)}}$and ${\text{(3pq  -  2}}{{\text{q}}^2}{\text{)}}$

Ans:  ${\text{(2pq  +  3}}{{\text{q}}^2}{\text{)}} \times {\text{(3pq  -  2}}{{\text{q}}^2}{\text{) =  2pq }} \times {\text{(3pq  -  2}}{{\text{q}}^2}{\text{) +  3}}{{\text{q}}^2} \times {\text{(3pq  -  2}}{{\text{q}}^2}{\text{)}}$

${\text{ =  6p2}}{{\text{q}}^2}{\text{  -  4p}}{{\text{q}}^3}{\text{  +  9p}}{{\text{q}}^3}{\text{  -  6}}{{\text{q}}^4}$

${\text{ =  6p2}}{{\text{q}}^2}{\text{  +  5p}}{{\text{q}}^3}{\text{  -  6}}{{\text{q}}^4}$


(vi) $\left( {\dfrac{3}{4}{a^2} + 3{b^2}} \right)$and $4\left( {{a^2} - \dfrac{2}{3}{b^2}} \right)$

Ans:  $\left( {\dfrac{{\text{3}}}{{\text{4}}}{{\text{a}}^{\text{2}}}{\text{ + 3}}{{\text{b}}^{\text{2}}}} \right){{ \times }}\left[ {{\text{4}}\left( {{{\text{a}}^{\text{2}}}{\text{ - }}\dfrac{{\text{2}}}{{\text{3}}}{{\text{b}}^{\text{2}}}} \right)} \right]{\text{ = }}\left( {\dfrac{{\text{3}}}{{\text{4}}}{{\text{a}}^{\text{2}}}{\text{ + 3}}{{\text{b}}^{\text{2}}}} \right){{ \times }}\left( {{\text{4}}{{\text{a}}^{\text{2}}}{\text{ - }}\dfrac{{\text{8}}}{{\text{3}}}{{\text{b}}^{\text{2}}}} \right)$

${\text{ = }}\dfrac{{\text{3}}}{{\text{4}}}{{\text{a}}^{\text{2}}}{{ \times }}\left( {{\text{4}}{{\text{\alpha }}^{\text{2}}}{\text{ - }}\dfrac{{\text{8}}}{{\text{3}}}{{\text{b}}^{\text{2}}}} \right){\text{ + 3}}{{\text{b}}^{\text{2}}}{{ \times }}\left( {{\text{4}}{{\text{a}}^{\text{2}}}{\text{ - }}\dfrac{{\text{8}}}{{\text{3}}}{{\text{b}}^{\text{2}}}} \right) $

$  {\text{ = 3}}{{\text{a}}^{\text{4}}}{\text{ - 2}}{{\text{a}}^{\text{2}}}{{\text{b}}^{\text{2}}}{\text{ + 12}}{{\text{b}}^{\text{2}}}{{\text{a}}^{\text{2}}}{\text{ - 8}}{{\text{b}}^{\text{4}}} $

$  {\text{ = 3}}{{\text{a}}^{\text{4}}}{\text{ + 10}}{{\text{a}}^{\text{2}}}{{\text{b}}^{\text{2}}}{\text{ - 8}}{{\text{b}}^{\text{4}}}  $


2. Find the product.

(i) ${\text{(5  -  2x) (3  +  x)}}$

Ans:  ${\text{(5  -  2x) (3  +  x) =  5 (3  +  x)  -  2x (3  +  x)}}$

$  {\text{ =  15  +  5x  -  6x  -  2}}{{\text{x}}^2} $

$ {\text{ =  15  -  x  -  2}}{{\text{x}}^2} $


(ii) ${\text{(x  +  7y) (7x  -  y)}}$

Ans:  ${\text{(x  +  7y) (7x  -  y) =  x (7x  -  y)  +  7y (7x  -  y)}}$

$  {\text{ =  7}}{{\text{x}}^2}{\text{  -  xy  +  49xy  -  7}}{{\text{y}}^2} $

$  {\text{ =  7}}{{\text{x}}^2}{\text{  +  48xy  -  7}}{{\text{y}}^2} $


(iii) ${\text{(}}{{\text{a}}^2}{\text{  +  b) (a  +  }}{{\text{b}}^2}{\text{)}}$

Ans:  ${\text{(}}{{\text{a}}^2}{\text{  +  b) (a  +  }}{{\text{b}}^2}{\text{) =  }}{{\text{a}}^2}{\text{ (a  +  }}{{\text{b}}^2}{\text{)  +  b (a  +  }}{{\text{b}}^2}{\text{)}}$

${\text{ =  }}{{\text{a}}^3}{\text{  +  }}{{\text{a}}^2}{{\text{b}}^2}{\text{  +  ab  +  }}{{\text{b}}^3}$


(iv) ${\text{(}}{{\text{p}}^2}{\text{  -  }}{{\text{q}}^2}{\text{) (2p  +  q)}}$

Ans:  ${\text{(a  -  b) (a  +  b)  +  (b  -  c) (b  +  c)  +  (c  -  a) (c  +  a)  =  0}}$

${\text{ =  2}}{{\text{p}}^3}{\text{  +  }}{{\text{p}}^2}{\text{q  -  2p}}{{\text{q}}^2}{\text{  -  }}{{\text{q}}^3}$


3. Simplify.

(i) ${\text{(}}{{\text{x}}^2}{\text{  -  5) (x  +  5)  +  25}}$

Ans:  ${\text{(}}{{\text{x}}^2}{\text{  -  5) (x  +  5)  +  25}}$

$  {{\text{x}}^2}{\text{ (x  +  5)  -  5 (x  +  5)  +  25}} $

$  {\text{ =  }}{{\text{x}}^3}{\text{  +  5}}{{\text{x}}^2}{\text{  -  5x  -  25  +  25}} $

$  {\text{ =  }}{{\text{x}}^3}{\text{  +  5}}{{\text{x}}^2}{\text{  -  5x}}  $


(ii) ${\text{(}}{{\text{a}}^2}{\text{  +  5) (}}{{\text{b}}^3}{\text{  +  3)  +  5}}$

Ans:   ${\text{(}}{{\text{a}}^2}{\text{  +  5) (}}{{\text{b}}^3}{\text{  +  3)  +  5}}$

$  {\text{ =  }}{{\text{a}}^2}{\text{ (}}{{\text{b}}^3}{\text{  +  3)  +  5 (}}{{\text{b}}^3}{\text{  +  3)  +  5}} $

$  {\text{ =  }}{{\text{a}}^2}{{\text{b}}^3}{\text{  +  3}}{{\text{a}}^2}{\text{  +  5}}{{\text{b}}^3}{\text{  +  15  +  5}} $

$  {\text{ =  }}{{\text{a}}^2}{{\text{b}}^3}{\text{  +  3}}{{\text{a}}^2}{\text{  +  5}}{{\text{b}}^3}{\text{  +  20}}  $


(iii) ${\text{(t  +  }}{{\text{s}}^2}{\text{) (}}{{\text{t}}^2}{\text{  -  s)}}$

Ans:  ${\text{(t  +  }}{{\text{s}}^2}{\text{) (}}{{\text{t}}^2}{\text{  -  s)}}$

$  {\text{ =  t (}}{{\text{t}}^2}{\text{  -  s)  +  }}{{\text{s}}^2}{\text{ (}}{{\text{t}}^2}{\text{  -  s)}} $

$  {\text{ =  }}{{\text{t}}^3}{\text{  -  st  +  }}{{\text{s}}^2}{{\text{t}}^2}{\text{  -  }}{{\text{s}}^3}  $


(iv) ${\text{(a  +  b) (c  -  d)  +  (a  -  b) (c  +  d)  +  2 (ac  +  bd)}}$

Ans:  ${\text{(a  +  b) (c  -  d)  +  (a  -  b) (c  +  d)  +  2 (ac  +  bd)}}$

$  {\text{ =  a (c  -  d)  +  b (c  -  d)  +  a (c  +  d)  -  b (c  +  d)  +  2 (ac  +  bd)}} $

$  {\text{ =  ac  -  ad  +  bc  -  bd  +  ac  +  ad  -  bc  -  bd  +  2ac  +  2bd}} $

$  {\text{ =  (ac  +  ac  +  2ac)  +  (ad  -  ad)  +  (bc  -  bc)  +  (2bd  -  bd  -  bd)}} $

$  {\text{ =  4ac}}  $


(v) ${\text{(x  +  y) (2x  +  y)  +  (x  +  2y) (x  -  y)}}$

Ans:  ${\text{(x  +  y) (2x  +  y)  +  (x  +  2y) (x  -  y)}}$

$  {\text{ =  x (2x  +  y)  +  y (2x  +  y)  +  x (x  -  y)  +  2y (x  -  y)}} $

$  {\text{ =  2}}{{\text{x}}^2}{\text{  +  xy  +  2xy  +  }}{{\text{y}}^2}{\text{  +  }}{{\text{x}}^2}{\text{  -  xy  +  2xy  -  2}}{{\text{y}}^2} $

$  {\text{ =  (2}}{{\text{x}}^2}{\text{  +  }}{{\text{x}}^2}{\text{)  +  (}}{{\text{y}}^2}{\text{  -  2}}{{\text{y}}^2}{\text{)  +  (xy  +  2xy  -  xy  +  2xy)}} $

$  {\text{ =  3}}{{\text{x}}^2}{\text{  -  }}{{\text{y}}^2}{\text{  +  4xy}} $


(vi) ${\text{(x  +  y) (}}{{\text{x}}^2}{\text{  -  xy  +  }}{{\text{y}}^2}{\text{)}}$

Ans:  ${\text{(x  +  y) (}}{{\text{x}}^2}{\text{  -  xy  +  }}{{\text{y}}^2}{\text{)}}$

$  {\text{ =  x (}}{{\text{x}}^2}{\text{  -  xy  +  }}{{\text{y}}^2}{\text{)  +  y (}}{{\text{x}}^2}{\text{  -  xy  +  }}{{\text{y}}^2}{\text{)}} $

$  {\text{ =  }}{{\text{x}}^3}{\text{  -  }}{{\text{x}}^2}{\text{y  +  x}}{{\text{y}}^2}{\text{  +  }}{{\text{x}}^2}{\text{y  -  x}}{{\text{y}}^2}{\text{  +  }}{{\text{y}}^3} $

$  {\text{ =  }}{{\text{x}}^3}{\text{  +  }}{{\text{y}}^3}{\text{  +  (x}}{{\text{y}}^2}{\text{  -  x}}{{\text{y}}^2}{\text{)  +  (}}{{\text{x}}^2}{\text{y  -  }}{{\text{x}}^2}{\text{y)}} $

$  {\text{ =  }}{{\text{x}}^3}{\text{  +  }}{{\text{y}}^3}  $


(vii) ${\text{(1}}{\text{.5x  -  4y) (1}}{\text{.5x  +  4y  +  3)  -  4}}{\text{.5x  +  12y}}$

Ans:  ${\text{(1}}{\text{.5x  -  4y) (1}}{\text{.5x  +  4y  +  3)  -  4}}{\text{.5x  +  12y}}$

$  {\text{ =  1}}{\text{.5x (1}}{\text{.5x  +  4y  +  3)  -  4y (1}}{\text{.5x  +  4y  +  3)  -  4}}{\text{.5x  +  12y}} $

$  {\text{ =  2}}{\text{.25 }}{{\text{x}}^2}{\text{  +  6xy  +  4}}{\text{.5x  -  6xy  -  16}}{{\text{y}}^2}{\text{  -  12y  -  4}}{\text{.5x  +  12y}} $

$  {\text{ =  2}}{\text{.25 }}{{\text{x}}^2}{\text{  +  (6xy  -  6xy)  +  (4}}{\text{.5x  -  4}}{\text{.5x)  -  16}}{{\text{y}}^2}{\text{  +  (12y  -  12y)}} $

$  {\text{ =  2}}{\text{.25}}{{\text{x}}^2}{\text{  -  16}}{{\text{y}}^2}  $


(viii) ${\text{(a  +  b  +  c) (a  +  b  -  c)}}$

Ans:  ${\text{(a  +  b  +  c) (a  +  b  -  c)}}$

$  {\text{ =  a (a  +  b  -  c)  +  b (a  +  b  -  c)  +  c (a  +  b  -  c)}} $

$  {\text{ =  }}{{\text{a}}^2}{\text{  +  ab  -  ac  +  ab  + }}{{\text{b}}^2}{\text{  -  bc  +  ca  +  bc  -  }}{{\text{c}}^2}  $

$  {\text{ =  }}{{\text{a}}^2}{\text{  +  }}{{\text{b}}^2}{\text{  -  }}{{\text{c}}^2}{\text{  +  (ab  +  ab)  +  (bc  -  bc)  +  (ca  -  ca)}} $

$  {\text{ =  }}{{\text{a}}^2}{\text{  +  }}{{\text{b}}^2}{\text{  -  }}{{\text{c}}^2}{\text{  +  2ab}}  $


NCERT Solutions for Class 8 Maths chapter 8 PDF

Overview of Deleted Syllabus for CBSE Class 8 Maths Algebraic Expressions and Identities

Chapter

Dropped Topics

Algebraic Expressions and Identities

8.1 Introduction 

8.2 Terms, Factors and Coefficients

8.3 Monomials, Binomials and Polynomials

8.4 Like and Unlike Terms

8.10 What is an Identity?

8.11 Standard Identities

8.12 Applying Identities


Class 8 Maths chapter 8: Exercises Breakdown

Exercise

Number of Questions

Exercise 8.1

2 Questions and Solutions

Exercise 8.2

5 Questions and Solutions

Exercise 8.3

5 Questions and Solutions

Exercise 8.4

3 Questions with Solutions


Conclusion 

NCERT Solutions for Maths Algebraic Expressions Class 8 Chapter 8  by Vedantu are essential for building a strong foundation in algebra. This chapter introduces you to the basics of forming and simplifying algebraic expressions, and understanding and applying various algebraic identities.


In previous year exams, around 3–4 questions have been asked from this chapter, highlighting its significance in the overall curriculum. By thoroughly practising the problems and understanding the step-by-step solutions provided by Vedantu, you can confidently tackle algebraic expressions and identities.


Other Study Material for CBSE Class 8 Maths chapter 8


Chapter-Specific NCERT Solutions for Class 8 Maths

Given below are the chapter-wise NCERT Solutions for Class 8 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.



Important Related Links for CBSE Class 8 Maths

FAQs on NCERT Solutions Class 8 Maths chapter 8 Algebraic Expressions and Identities

Q1.What are the Topics Covered in the Class 8 Chapter 9 Maths?

Ans: Chapter 9 of Grade 8 Maths is Algebraic Expressions and Identities. 


The topics covered in this chapter in the NCERT book are as follows:

  • Meaning of algebraic expressions

  • What are the terms, factors, and coefficients

  • What are monomials, binomials, and polynomials

  • Like and Unlike terms

  • Addition and subtraction of algebraic terms

  • Multiplication of algebraic terms

  • Identities, standard identities, and application of identities.

All these topics are the core topics covered. Subtopics like multiplication of monomial to monomial, polynomial to polynomial, monomial to polynomial are also covered in the chapter.

Q2. Are these NCERT Solutions Helpful in Scoring Good Marks in School Exams?

Ans: Our 8th standard Maths algebra NCERT solutions are prepared by experts who are highly qualified and experienced. These solutions are prepared in such a manner that the methods and mathematical steps can be easily understood by the students. When students can gain the concept they will be able to solve the questions related to algebra that may come in their school exams and this will increase their score in their exams. All the Exercises given in the NCERT book in Chapter 9 are included in our algebraic expressions NCERT Class 8 solutions.

Q3. What are algebraic expressions in Class 8?

Ans: In Class 8, algebraic expressions are equations or expressions formed by the combination of variables, numbers and algebraic operations (addition, subtraction, multiplication etc.). An Algebraic expression is a term consisting of variables, coefficients and a constant. For example: 4x + 7y - 2. Here, 4 and 7 are coefficients, x and y are variables, and ‘2’ is a constant. Algebraic expressions are of three types - monomial, binomial, and trinomial. 

Q4. What is the formula for algebraic expression?

Ans: Class 8 Maths introduces the concept of algebraic expressions and algebraic identities to the students. Algebraic identities are algebraic equations that are valid for all variable values in them.  Following are the formulas for algebraic expressions:

(a+b)2 = a2+2ab +b2 

(a-b)2= a2-2ab +b2

(a+b)(a-b)= a2-b2

(x+a)(x+b)= x2+(a+b) x+ab

(x+a)(x-b)= x2+(a-b) x-ab

(x-a)(x+b)= x2+(b-a) x-ab

(x-a)(x-b)= x2-(a+b) x+ab

(a+b)3= a3+3ab(a+b) +b3

(a-b)3= a3-3ab(a-b) -b3 

You can learn more about algebraic formulas and their applications from Vedantu.

Q5. How can I prepare myself to score good marks in Maths Class 8?

Ans: To score good grades in Maths Class 8, students should refer to the study material available on Vedantu. Vedantu provides you with notes and NCERT Solutions. In these solutions, all the exercises from the Class 8 Maths NCERT textbook have been thoroughly addressed. Also, NCERT Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities are available on the Vedantu app for free of cost. Furthermore, these solutions might help you prepare for a variety of competitive exams.

Q6. Where can I get NCERT Solutions for Class 8 Maths Chapter 9?

Ans: Students can find the NCERT Solutions for Class 8 Maths Chapter 9 “Algebraic Expressions and Identities” on Vedantu. The NCERT Solutions provided by Vedantu are the best because of the accuracy and precision with which they are prepared. Vedantu offers free chapter-by-chapter NCERT Solutions for Class 8 Maths in PDF format. These solutions are created by subject matter experts with years of experience. 

Q7. What concepts can I learn using the NCERT Solutions for Class 8 Maths Chapter 9?

Ans: NCERT Solutions for Class 8 Maths Chapter 9 provided in the website and the app of Vedantu ensure to clear all the concepts given in the NCERT Maths Textbook Class 8. It provides an in-depth explanation of what an Algebraic expression is and what algebraic identities are. It provides students with not only the theoretical part but also the sample questions for practice.