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NCERT Solutions for Class 8 Maths Chapter 8 Algebraic Expressions

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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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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.


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.



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