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NCERT Solutions for Class 8 Maths Chapter 9 - Algebraic Expressions And Identities

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NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities - Free PDF

Vedantu’s NCERT Solutions Class 8 Maths is like an all-time available solution to the problems of students whether it is a Class test or the final exam. These NCERT solutions are the best solutions for the questions given in the NCERT books and will help the students to get the best marks in exams. To get the best and better solution, you may always rely upon our NCERT Solution Class 8 Chapter 9 as they are prepared in a very easy to understand language by our subject experts. All the Mathematical steps of solutions to different maths problems are explained in a language that is easily comprehensible by the students. Science Students who are looking for NCERT Solutions for Class 8 Science will also find the Solutions curated by our Master Teachers really Helpful. You can also download NCERT Solutions Class 8 Maths to help you to revise complete syllabus ans score more marks in your examinations.  Learning has never been so simple!


Class:

NCERT Solutions For Class 8

Subject:

Class 8 Maths

Chapter Name:

Chapter 9 - Algebraic Expressions and Identities

Content Type:

Text, Videos, Images and PDF Format

Academic Year:

2023-24

Medium:

English and Hindi

Available Materials:

  • Chapter Wise

  • Exercise Wise

Other Materials

  • Important Questions

  • Revision Notes


Topics Covered Under NCERT Class 8 Math Chapter 9 Algebraic Expressions and Identities 

Chapter 9 Algebraic Expressions and Identities is an important topic for Class 8 as well as for higher mathematics. Algebra is an interesting concept of mathematics, and the NCERT Class 8 Math Chapter 9 Algebraic Expressions and Identities covers an introduction to this unit in detail. 

 

With several identities, expressions, examples, and exercise questions, Chapter 9 of NCERT Class 8 Math is a lengthy one. So, students can maintain their motivation by knowing beforehand what topics would be covered in each exercise of the chapter.

 

The important concepts covered under Math Class 8 Algebraic Expressions and Identities can be seen in the table below.

Sl. No.

Topics

1

What are Expressions?

2

Terms, Factors, and Coefficients

3

Monomials, Binomials, and Polynomials

4

Like and Unlike Terms

5

Addition and Subtraction of Algebraic Expressions

6

Multiplication of Algebraic Expressions: Introduction

7

Multiplying a Monomial by a Monomial

8

Multiplying a Monomial by a Polynomial

9

Multiplying a Polynomial by a Polynomial

10

What is an Identity?

11

Standard Identities

12

Applying Identities

Access NCERT Solutions for Class 8 Mathematics Chapter 9 – Algebraic Expressions and Identities

Exercise – 9.1

1. Identify the terms, their coefficient for each of the following expressions.

(i) $5{\text{xy}}{{\text{z}}^2} - 3{\text{zy}}$

Ans:  

Terms 

Coefficients

$5{\text{xy}}{{\text{z}}^2} $

$   - 3{\text{zy}} $

5

 - 3 


(ii) $1 + {\text{x}} + {{\text{x}}^2}$

Ans:  

Terms 

Coefficients

1

x

${{\text{x}}^2}$

1

1


(iii) ${\text{4}}{{\text{x}}^{\text{2}}}{{\text{y}}^{\text{2}}}{\text{ - 4}}{{\text{x}}^{\text{2}}}{{\text{y}}^{\text{2}}}{{\text{z}}^{\text{2}}}{\text{ + }}{{\text{z}}^{\text{2}}}$

Ans:  

Terms 

Coefficients

${\text{4}}{{\text{x}}^{\text{2}}}{{\text{y}}^{\text{2}}}$

$ {\text{ - 4}}{{\text{x}}^{\text{2}}}{{\text{y}}^{\text{2}}}{{\text{z}}^{\text{2}}}$

${{\text{z}}^{\text{2}}}$

- 4 

 4 


(iv) ${\text{3 - pq + qr - rp}}$

Ans:  

Terms 

Coefficients

3

$  {\text{ - pq}} $

$ {\text{qr}} $

$  {\text{ - rp}} $

3

- 1 

- 1 


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

Ans:  

Terms 

Coefficients

$\dfrac{{\text{x}}}{{\text{2}}} $

$\dfrac{{\text{y}}}{{\text{2}}} $

${\text{ - xy}} $

$\dfrac{1}{2} $

$ \dfrac{1}{2} $

 - 1  


(vi) ${\text{0}}{\text{.3a - 0}}{\text{.6ab + 0}}{\text{.5b}}$

Ans:  

Terms 

Coefficients

${\text{0}}{\text{.3a}} $

$  {\text{ - 0}}{\text{.6ab}} $

${\text{0}}{\text{.5b}} $

${\text{0}}{\text{.3}} $

$ {\text{ - 0}}{\text{.6}}$

${\text{0}}{\text{.5}} $


2. Classify the following polynomials as monomials, binomials, trinomials. In which polynomials do not fit in any of these three categories?

${x + y}$, 1000, ${x + {{x}^{2}} + {{x}^{3}} + {{x}^{4}}}$, ${7 + y + 5x}$, ${2y - 3{y}^{2}}$, ${2y - 3{y}^{2} + 4{y}^{3}}$, ${5x - 4y + 3xy}$, ${4z -15{z}^{2}}$, ${ab + bc + cd + da}$, pqr, ${{p}^{2}q + p{q}^{2}}$, ${2p + 2q}$

Ans:  The given expressions are classified as 

Monomials: ${\text{1000,}}$${\text{pqr}}$

Binomials: ${\text{x + y,2y - 3}}{{\text{y}}^{\text{2}}}{\text{,4z - 15}}{{\text{z}}^{\text{2}}}{\text{,}}{{\text{p}}^{\text{2}}}{\text{q + p}}{{\text{q}}^{\text{2}}}{\text{,2p + 2q}}$

Trinomials: ${\text{7 + y + 5x,2y - 3}}{{\text{y}}^{\text{2}}}{\text{ + 4}}{{\text{y}}^{\text{3}}}{\text{,5x - 4y + 3xy}}$

Polynomials that do not fit in any categories are 

${\text{x + }}{{\text{x}}^{\text{2}}}{\text{ + }}{{\text{x}}^{\text{3}}}{\text{ + }}{{\text{x}}^{\text{4}}}{\text{,ab + bc + cd + da}}$


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


4. 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 – 9.2

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 – 9.3

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


  1. 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

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

(a) ${\text{ x  =  3}}$

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 }} $


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

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}}}  $


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

(a) ${\text{a  =  0}}$

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


(b) ${\text{a  =  1}}$

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 }}  $


(c) ${\text{a  =   - 1}}$

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 }}  $


  1. 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 – 9.4

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}}  $


Exercise – 9.5

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

(i) ${\text{(x  +  3) (x  +  3) }}$

Ans:  The products will be as follows.

$  {\text{ =  (x}}{{\text{)}}^2}{\text{  +  2(x) (3)  +  (3}}{{\text{)}}^2}{\text{ [(a  +  b}}{{\text{)}}^2}{\text{  =  }}{{\text{a}}^2}{\text{  +  2ab  +  }}{{\text{b}}^2}{\text{]}} $

$  {\text{ =  }}{{\text{x}}^2}{\text{  +  6x  +  9}} $


(ii) ${\text{(2y  +  5) (2y  +  5) }}$

Ans:  The products will be as follows.

${\text{(2y  +  5) (2y  +  5)  =  (2y  +  5}}{{\text{)}}^2}$

$  {\text{ =  (2y}}{{\text{)}}^2}{\text{  +  2(2y) (5)  +  (5}}{{\text{)}}^2}{\text{ [(a  +  b}}{{\text{)}}^2}{\text{  =  }}{{\text{a}}^2}{\text{  +  2ab  +  }}{{\text{b}}^2}{\text{]}} $

$  {\text{ =  4}}{{\text{y}}^2}{\text{  +  20y  +  25}}  $


(iii) ${\text{(2a  -  7) (2a  -  7) }}$

Ans:  The products will be as follows.

${\text{(2a  -  7) (2a  -  7)  =  (2a  -  7}}{{\text{)}}^2}$

${\text{[(a  -  b}}{{\text{)}}^2}{\text{  =  }}{{\text{a}}^2}{\text{  -  2ab  +  }}{{\text{b}}^2}{\text{]}}$


(iv) $\left( {3a - \dfrac{1}{2}} \right)\left( {3a - \dfrac{1}{2}} \right)$

Ans:  $\left( {\dfrac{x}{2} + \dfrac{{3y}}{4}} \right)\left( {\dfrac{x}{2} + \dfrac{{3y}}{4}} \right) = {\left( {\dfrac{x}{2} + \dfrac{{3y}}{4}} \right)^2}$

${\left( {{\text{3a}}} \right)^{\text{2}}}{\text{ - 2(3a)}}\left( {\dfrac{{\text{1}}}{{\text{2}}}} \right){\text{ + }}{\left( {\dfrac{{\text{1}}}{{\text{2}}}} \right)^{\text{2}}}$        

${\text{[(a  -  b}}{{\text{)}}^2}{\text{  =  }}{{\text{a}}^2}{\text{  -  2ab  +  }}{{\text{b}}^2}{\text{]}}$

${\text{ = 9}}{{\text{a}}^{\text{2}}}{\text{ - 3a + }}\dfrac{{\text{1}}}{{\text{4}}}$

(v) ${\text{(1}}{\text{.1m  -  0}}{\text{.4) (1}}{\text{.1 m  +  0}}{\text{.4)}}$

Ans:  ${\text{(1}}{\text{.1m  -  0}}{\text{.4) (1}}{\text{.1 m  +  0}}{\text{.4)}}$

$  {\text{ =  (1}}{\text{.1m}}{{\text{)}}^2}{\text{  -  (0}}{\text{.4}}{{\text{)}}^2}{\text{ [(a  +  b) (a  -  b)  =  }}{{\text{a}}^2}{\text{  -  }}{{\text{b}}^2}{\text{]}} $

$  {\text{ =  1}}{\text{.21}}{{\text{m}}^2}{\text{  -  0}}{\text{.16}}  $


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

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

$  {\text{ =  (}}{{\text{b}}^2}{{\text{)}}^2}{\text{  -  (}}{{\text{a}}^2}{{\text{)}}^2}{\text{ [(a  +  b) (a  -  b)  =  }}{{\text{a}}^2}{\text{  -  }}{{\text{b}}^2}{\text{]}} $

$  {\text{ =  }}{{\text{b}}^4}{\text{  -  }}{{\text{a}}^4}  $


(vii) ${\text{(6x  -  7) (6x  +  7)}}$

Ans:  ${\text{(6x  -  7) (6x  +  7)  =  (6x}}{{\text{)}}^2}{\text{  -  (7}}{{\text{)}}^2}{\text{ [(a  +  b) (a  -  b)  =  }}{{\text{a}}^2}{\text{  -  }}{{\text{b}}^2}{\text{]}}$

${\text{ =  36}}{{\text{x}}^2}{\text{  -  49}}$


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

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

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

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


(ix) $\left( {\dfrac{x}{2} + \dfrac{{3y}}{4}} \right)\left( {\dfrac{x}{2} + \dfrac{{3y}}{4}} \right)$

Ans:  $\left( {\dfrac{{\text{x}}}{{\text{2}}}{\text{ + }}\dfrac{{{\text{3y}}}}{{\text{4}}}} \right)\left( {\dfrac{{\text{x}}}{{\text{2}}}{\text{ + }}\dfrac{{{\text{3y}}}}{{\text{4}}}} \right){\text{ = }}{\left( {\dfrac{{\text{x}}}{{\text{2}}}{\text{ + }}\dfrac{{{\text{3y}}}}{{\text{4}}}} \right)^{\text{2}}}$  

${\text{ =  16}}{{\text{x}}^2}{\text{  +  24x  +  5}}$   ${\text{[(a  +  b}}{{\text{)}}^2}{\text{  =  }}{{\text{a}}^2}{\text{  +  2ab  +  }}{{\text{b}}^2}{\text{]}}$

${\dfrac{{\text{x}}}{{\text{4}}}^{\text{2}}}{\text{ + }}\dfrac{{{\text{3xy}}}}{{\text{4}}}{\text{ + }}\dfrac{{{\text{9}}{{\text{y}}^{\text{2}}}}}{{{\text{16}}}}$


(x) ${\text{(7a  -  9b) (7a  -  9b)}}$

Ans:  ${\text{(7a  -  9b) (7a  -  9b)  =  (7a  -  9b}}{{\text{)}}^2}$

$  {\text{ =  (7a}}{{\text{)}}^2}{\text{  -  2(7a)(9b)  +  (9b}}{{\text{)}}^2}{\text{ [(a  -  b}}{{\text{)}}^2}{\text{  =  }}{{\text{a}}^2}{\text{  -  2ab  +  }}{{\text{b}}^2}{\text{]}} $

$  {\text{ =  49}}{{\text{a}}^2}{\text{  -  126ab  +  81}}{{\text{b}}^2}  $


2. Use the identity ${\text{(x  +  a) (x  +  b)  =  }}{{\text{x}}^2}{\text{  +  (a  +  b)x  +  ab}}$ to find the following products.

(i) ${\text{(x  +  3) (x  +  7)}}$

Ans:  ${\text{(x  +  3) (x  +  7) =  }}{{\text{x}}^2}{\text{  +  (3  +  7) x  +  (3) (7)}}$

${\text{ =  }}{{\text{x}}^2}{\text{  +  10x  +  21}}$


(ii) ${\text{(4x  +  5) (4x  +  1) }}$

Ans:  ${\text{(4x  +  5) (4x  +  1)  =  (4x}}{{\text{)}}^2}{\text{  +  (5  +  1) (4x)  +  (5) (1)}}$

${\text{ =  16}}{{\text{x}}^2}{\text{  +  24x  +  5}}$


(iii) ${\text{(4x  +  5) (4x  -  1) }}$

Ans:  ${\text{(4x  -  5) (4x  -  1)  =  (4x}}{{\text{)}}^2}{\text{  +  }}\left[ {{\text{( - 5)  +  ( - 1)}}} \right]{\text{ (4x)  +  ( - 5) ( - 1)}}$

$  {\text{ =  (}}{{\text{m}}^2}{{\text{)}}^2}{\text{  -  2(}}{{\text{m}}^2}{\text{) (}}{{\text{n}}^2}{\text{m)  +  (}}{{\text{n}}^2}{\text{m}}{{\text{)}}^2}{\text{  +  2}}{{\text{m}}^3}{{\text{n}}^2}{\text{ [(a  -  b}}{{\text{)}}^2}{\text{  =  }}{{\text{a}}^2}{\text{  -  2ab  +  }}{{\text{b}}^2}{\text{ ]}} $

$  {\text{ =  }}{{\text{m}}^4}{\text{  -  2}}{{\text{m}}^3}{{\text{n}}^2}{\text{  +  }}{{\text{n}}^4}{{\text{m}}^2}{\text{  +  2}}{{\text{m}}^3}{{\text{n}}^2} $

$  {\text{ =  }}{{\text{m}}^4}{\text{  +  }}{{\text{n}}^4}{{\text{m}}^2} $


(iv) ${\text{(4x  +  5) (4x  -  1) }}$

Ans:  ${\text{(4x  +  5) (4x  -  1)  =  (4x}}{{\text{)}}^2}{\text{  +  }}\left[ {{\text{(5)  +  ( - 1)}}} \right]{\text{ (4x)  +  (5) ( - 1)}}$

${\text{ =  16}}{{\text{x}}^2}{\text{  +  16x  -  5}}$


(v) ${\text{(2x  + 5y) (2x  +  3y)}}$

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

${\text{ =  4}}{{\text{x}}^2}{\text{  +  16xy  +  15}}{{\text{y}}^2}$


(vi) ${\text{(2}}{{\text{a}}^2}{\text{  + 9) (2}}{{\text{a}}^2}{\text{  +  5)}}$

Ans:  ${\text{(2}}{{\text{a}}^2}{\text{  + 9) (2}}{{\text{a}}^2}{\text{  +  5) =  (2}}{{\text{a}}^2}{{\text{)}}^2}{\text{  +  (9  +  5) (2}}{{\text{a}}^2}{\text{)  +  (9) (5)}}$

${\text{ =  4}}{{\text{a}}^4}{\text{  +  28}}{{\text{a}}^2}{\text{  +  45}}$


(vii) ${\text{(xyz  -  4) (xyz  -  2)}}$

Ans:  ${{\text{(xyz)}}^2}{\text{ + }}\left[ {( - 4) + ( - 2)} \right]{\text{ (xyz)  +  ( - 4) ( - 2)}}$

${\text{ =  }}{{\text{x}}^2}{{\text{y}}^2}{{\text{z}}^2}{\text{  -  6xyz  +  8}}$


  1. Find the following squares by suing the identities

(i) ${{\text{(b  -  7)}}^2}$

Ans:  ${{\text{(b  -  7)}}^2}{\text{  =  (b}}{{\text{)}}^2}{\text{  -  2(b) (7)  +  (7}}{{\text{)}}^2}{\text{ [(a  -  b}}{{\text{)}}^2}{\text{  =  }}{{\text{a}}^2}{\text{  -  2ab  +  }}{{\text{b}}^2}{\text{]}}$

${\text{ =  }}{{\text{b}}^2}{\text{  -  14b  +  49}}$


(ii) ${{\text{(xy  +  3z)}}^2}$

Ans:  ${{\text{(xy  +  3z)}}^2}{\text{ =  (xy}}{{\text{)}}^2}{\text{  +  2(xy) (3z)  +  (3z}}{{\text{)}}^2}{\text{ [(a  +  b}}{{\text{)}}^2}{\text{  =  }}{{\text{a}}^2}{\text{  +  2ab  +  }}{{\text{b}}^2}{\text{]}}$

${\text{ =  }}{{\text{x}}^2}{{\text{y}}^2}{\text{  +  6xyz  +  9}}{{\text{z}}^2}$


(iii) ${{\text{(6}}{{\text{x}}^2}{\text{  -  5y)}}^2}$

Ans:  ${\text{ =  (6}}{{\text{x}}^2}{{\text{)}}^2}{\text{  -  2(6}}{{\text{x}}^2}{\text{) (5y)  +  (5y}}{{\text{)}}^2}{\text{ [(a  -  b}}{{\text{)}}^2}{\text{  =  }}{{\text{a}}^2}{\text{  -  2ab  +  }}{{\text{b}}^2}{\text{]}}$

${\text{ =  36}}{{\text{x}}^4}{\text{  -  60}}{{\text{x}}^2}{\text{y  +  25}}{{\text{y}}^2}$


(iv) ${\left( {\dfrac{{\text{2}}}{{\text{3}}}{\text{m + }}\dfrac{{\text{3}}}{{\text{2}}}{\text{n}}} \right)^{\text{2}}}$

Ans:  ${\left( {\dfrac{{\text{2}}}{{\text{3}}}{\text{m + }}\dfrac{{\text{3}}}{{\text{2}}}{\text{n}}} \right)^{\text{2}}}{\text{ = }}{\left( {\dfrac{{\text{2}}}{{\text{3}}}{\text{m}}} \right)^{\text{2}}}{\text{ + 2}}\left( {\dfrac{{\text{2}}}{{\text{3}}}{\text{m}}} \right)\left( {\dfrac{{\text{3}}}{{\text{2}}}{\text{n}}} \right){\text{ + }}{\left( {\dfrac{{\text{3}}}{{\text{2}}}{\text{n}}} \right)^{\text{2}}}\left[ {{{{\text{(a + b)}}}^{\text{2}}}{\text{ = }}{{\text{a}}^{\text{2}}}{\text{ + 2ab + }}{{\text{b}}^{\text{2}}}} \right]$

${\text{ = }}\dfrac{{\text{4}}}{{\text{9}}}{{\text{m}}^{\text{2}}}{\text{ + 2mn + }}\dfrac{{\text{9}}}{{\text{4}}}{{\text{n}}^{\text{2}}}$


(v) ${{\text{(0}}{\text{.4p  -  0}}{\text{.5q)}}^2}$

Ans:  ${{\text{(0}}{\text{.4p  -  0}}{\text{.5q)}}^2}{\text{ =  (0}}{\text{.4p}}{{\text{)}}^2}{\text{  -  2 (0}}{\text{.4p) (0}}{\text{.5q)  +  (0}}{\text{.5q}}{{\text{)}}^2}$

$  {\text{[(a  -  b}}{{\text{)}}^2}{\text{  =  }}{{\text{a}}^2}{\text{  -  2ab  +  }}{{\text{b}}^2}{\text{]}} $

$  {\text{ =  0}}{\text{.16}}{{\text{p}}^2}{\text{  -  0}}{\text{.4pq  +  0}}{\text{.25}}{{\text{q}}^2} $


(vi) ${{\text{(2xy  +  5y)}}^2}$

Ans:  ${{\text{(2xy  +  5y)}}^2}{\text{ =  (2xy}}{{\text{)}}^2}{\text{  +  2(2xy) (5y)  +  (5y}}{{\text{)}}^2}$

$  {\text{[(a  +  b}}{{\text{)}}^2}{\text{  =  }}{{\text{a}}^2}{\text{  +  2ab  +  }}{{\text{b}}^2}{\text{]}} $

$  {\text{ =  4}}{{\text{x}}^2}{{\text{y}}^2}{\text{  +  20x}}{{\text{y}}^2}{\text{  +  25}}{{\text{y}}^2}  $


  1. Simplify.

(i) ${{\text{(}}{{\text{a}}^2}{\text{  -  }}{{\text{b}}^2}{\text{)}}^2}$

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

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


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

Ans:  ${{\text{(2x  + 5)}}^2}{\text{ -  (2x  -  5}}{{\text{)}}^2}{\text{  =  (2x}}{{\text{)}}^2}{\text{  +  2(2x) (5)  +  (5}}{{\text{)}}^2}{\text{  -  [(2x}}{{\text{)}}^2}{\text{  -  2(2x) (5)  +  (5}}{{\text{)}}^2}{\text{]}}$

$  {\text{[(a  -  b}}{{\text{)}}^2}{\text{  =  }}{{\text{a}}^2}{\text{  -  2ab  +  }}{{\text{b}}^2}{\text{]}} $

$  {\text{[(a  +  b}}{{\text{)}}^2}{\text{  =  }}{{\text{a}}^2}{\text{  +  2ab  +  }}{{\text{b}}^2}{\text{]}} $

$  {\text{ =  4}}{{\text{x}}^2}{\text{  +  20x  +  25  -  [4}}{{\text{x}}^2}{\text{  -  20x  +  25]}} $

$  {\text{ =  4}}{{\text{x}}^2}{\text{  +  20x  +  25  -  4}}{{\text{x}}^2}{\text{  +  20x  -  25  =  40x}}  $


(iii) ${{\text{(7m  -  8n)}}^2}{\text{  +  (7m  +  8n}}{{\text{)}}^2}$

Ans:  ${{\text{(7m  -  8n)}}^2}{\text{  +  (7m  +  8n}}{{\text{)}}^2}$

$  {\text{ =  (7m}}{{\text{)}}^2}{\text{  -  2(7m) (8n)  +  (8n}}{{\text{)}}^2}{\text{  +  (7m}}{{\text{)}}^2}{\text{  +  2(7m) (8n)  +  (8n}}{{\text{)}}^2} $

$  {\text{[(a  -  b}}{{\text{)}}^2}{\text{  =  }}{{\text{a}}^2}{\text{  -  2ab  +  }}{{\text{b}}^2}{\text{ and (a  +  b}}{{\text{)}}^2}{\text{  =  }}{{\text{a}}^2}{\text{  +  2ab  +  }}{{\text{b}}^2}{\text{]}} $

$  {\text{ =  49}}{{\text{m}}^2}{\text{  -  112mn  +  64}}{{\text{n}}^2}{\text{  +  49}}{{\text{m}}^2}{\text{  +  112mn  +  64}}{{\text{n}}^2} $

$  {\text{ =  98}}{{\text{m}}^2}{\text{  +  128}}{{\text{n}}^2}  $


(iv) ${{\text{(4m  +  5n)}}^2}{\text{  +  (5m  +  4n}}{{\text{)}}^2}$

Ans:  ${{\text{(4m  +  5n)}}^2}{\text{  +  (5m  +  4n}}{{\text{)}}^2}$

$  {\text{ =  (4m}}{{\text{)}}^2}{\text{  +  2(4m) (5n)  +  (5n}}{{\text{)}}^2}{\text{  +  (5m}}{{\text{)}}^2}{\text{  +  2(5m) (4n)  +  (4n}}{{\text{)}}^2} $

$  {\text{[ (a  +  b}}{{\text{)}}^2}{\text{  =  }}{{\text{a}}^2}{\text{  +  2ab  +  }}{{\text{b}}^2}{\text{]}}  $

$  {\text{ =  16}}{{\text{m}}^2}{\text{  +  40mn  +  25}}{{\text{n}}^1}{\text{  +  25}}{{\text{m}}^2}{\text{  +  40mn  +  16}}{{\text{n}}^2} $

$  {\text{ =  41}}{{\text{m}}^2}{\text{  +  80mn  +  41}}{{\text{n}}^2} $


(v) ${{\text{(2}}{\text{.5p  -  1}}{\text{.5q)}}^2}{\text{  -  (1}}{\text{.5p  -  2}}{\text{.5q}}{{\text{)}}^2}$

Ans:  ${{\text{(2}}{\text{.5p  -  1}}{\text{.5q)}}^2}{\text{  -  (1}}{\text{.5p  -  2}}{\text{.5q}}{{\text{)}}^2}$

$  {\text{ =  (2}}{\text{.5p}}{{\text{)}}^2}{\text{  -  2(2}}{\text{.5p) (1}}{\text{.5q)  +  (1}}{\text{.5q}}{{\text{)}}^2}{\text{  -  [(1}}{\text{.5p}}{{\text{)}}^2}{\text{  -  2(1}}{\text{.5p)(2}}{\text{.5q)  +  (2}}{\text{.5q}}{{\text{)}}^2}{\text{]}} $

$  {\text{[(a  -  b}}{{\text{)}}^2}{\text{  =  }}{{\text{a}}^2}{\text{  -  2ab  +  }}{{\text{b}}^2}{\text{ ]}} $

$  {\text{ =  6}}{\text{.25}}{{\text{p}}^2}{\text{  -  7}}{\text{.5pq  +  2}}{\text{.25}}{{\text{q}}^2}{\text{  -  [2}}{\text{.25}}{{\text{p}}^2}{\text{  -  7}}{\text{.5pq  +  6}}{\text{.25}}{{\text{q}}^2}{\text{]}} $

$  {\text{ =  6}}{\text{.25}}{{\text{p}}^2}{\text{  -  7}}{\text{.5pq  +  2}}{\text{.25}}{{\text{q}}^2}{\text{  -  2}}{\text{.25}}{{\text{p}}^2}{\text{  +  7}}{\text{.5pq  -  6}}{\text{.25}}{{\text{q}}^2}{\text{]}} $

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


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

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

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

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

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


(vii) ${{\text{(}}{{\text{m}}^2}{\text{  -  }}{{\text{n}}^2}{\text{m)}}^2}{\text{  +  2}}{{\text{m}}^3}{{\text{n}}^2}$

Ans:  ${{\text{(}}{{\text{m}}^2}{\text{  -  }}{{\text{n}}^2}{\text{m)}}^2}{\text{  +  2}}{{\text{m}}^3}{{\text{n}}^2}$

$  {\text{ =  (}}{{\text{m}}^2}{{\text{)}}^2}{\text{  -  2(}}{{\text{m}}^2}{\text{) (}}{{\text{n}}^2}{\text{m)  +  (}}{{\text{n}}^2}{\text{m}}{{\text{)}}^2}{\text{  +  2}}{{\text{m}}^3}{{\text{n}}^2}{\text{ [(a  -  b}}{{\text{)}}^2}{\text{  =  }}{{\text{a}}^2}{\text{  -  2ab  +  }}{{\text{b}}^2}{\text{ ]}} $

$  {\text{ =  }}{{\text{m}}^4}{\text{  -  2}}{{\text{m}}^3}{{\text{n}}^2}{\text{  +  }}{{\text{n}}^4}{{\text{m}}^2}{\text{  +  2}}{{\text{m}}^3}{{\text{n}}^2} $

$  {\text{ =  }}{{\text{m}}^4}{\text{  +  }}{{\text{n}}^4}{{\text{m}}^2}  $


  1. Show that

(i) ${{\text{(3x  +  7)}}^2}{\text{  -  84x  =  (3x  -  7}}{{\text{)}}^2}$

Ans:  Taking L.H.S.

${{\text{(3x  +  7)}}^2}{\text{  -  84x}}$

$  {\text{ =  (3x}}{{\text{)}}^2}{\text{  +  2(3x)(7)  +  (7}}{{\text{)}}^2}{\text{  -  84x}} $

$  {\text{ =  9}}{{\text{x}}^2}{\text{  +  42x  +  49  -  84x}} $

$  {\text{ =  9}}{{\text{x}}^2}{\text{  -  42x  +  49}}  $

Now considering R.H.S.

$  {\text{ =  (3x  -  7}}{{\text{)}}^2}{\text{  =  (3x}}{{\text{)}}^2}{\text{  -  2(3x)(7)  + (7}}{{\text{)}}^2} $

$  {\text{ =  9}}{{\text{x}}^2}{\text{  -  42x  +  49}}  $

Therefore, L.H.S. = R.H.S.


(ii) ${{\text{(9p  -  5q)}}^2}{\text{  +  180pq  =  (9p  +  5q}}{{\text{)}}^2}$

Ans:  Taking L.H.S.

${{\text{(9p  -  5q)}}^2}{\text{  +  180pq}}$

$  {\text{ =  (9p}}{{\text{)}}^2}{\text{  -  2(9p)(5q)  +  (5q}}{{\text{)}}^2}{\text{  -  180pq}} $

$  {\text{ =  81}}{{\text{p}}^2}{\text{  -  90pq  +  25}}{{\text{q}}^2}{\text{  +  180pq}} $

$  {\text{ =  81}}{{\text{p}}^2}{\text{  +  90pq  +  25}}{{\text{q}}^2}  $

Now, considering R.H.S.

${{\text{(9p  +  5q)}}^2}$

$  {\text{ =  (9p}}{{\text{)}}^2}{\text{  +  2(9p)(5q)  +  (5q}}{{\text{)}}^2} $

$  {\text{ =  81}}{{\text{p}}^2}{\text{  +  90pq  +  25}}{{\text{q}}^2}  $

Therefore, L.H.S. = R.H.S.


(iii) ${\left( {\dfrac{{\text{4}}}{{\text{3}}}{\text{m - }}\dfrac{{\text{3}}}{{\text{4}}}{\text{n}}} \right)^{\text{2}}}{\text{ + 2mn = }}\dfrac{{{\text{16}}}}{{\text{9}}}{{\text{m}}^{\text{2}}}{\text{ + }}\dfrac{{\text{9}}}{{{\text{16}}}}{{\text{n}}^{\text{2}}}$

Ans:  Taking L.H.S.

${\left( {\dfrac{{\text{4}}}{{\text{3}}}{\text{m - }}\dfrac{{\text{3}}}{{\text{4}}}{\text{n}}} \right)^{\text{2}}}$

$  {\text{ = }}{\left( {\dfrac{{\text{4}}}{{\text{3}}}{\text{m}}} \right)^{\text{2}}}{\text{ - 2}}\left( {\dfrac{{\text{4}}}{{\text{3}}}{\text{m}}} \right)\left( {\dfrac{{\text{3}}}{{\text{4}}}{\text{n}}} \right){\text{ + }}{\left( {\dfrac{{\text{3}}}{{\text{4}}}{\text{n}}} \right)^{\text{2}}}{\text{ + 2mn}} $

$  {\text{ = }}\dfrac{{{\text{16}}}}{{\text{9}}}{\text{m}}{{\text{r}}^{\text{2}}}{\text{ - 2mn + }}\dfrac{{\text{9}}}{{{\text{16}}}}{{\text{n}}^{\text{2}}}{\text{ + 2mn}} $

$  {\text{ = }}\dfrac{{{\text{16}}}}{{\text{9}}}{{\text{m}}^{\text{2}}}{\text{ + }}\dfrac{{\text{9}}}{{{\text{16}}}}{{\text{n}}^{\text{2}}}{\text{ =  R}}{\text{.H}}{\text{.S}} $ 

Therefore, L.H.S.=R.H.S.


(iv) ${{\text{(4pq  +  3q)}}^2}{\text{  -  (4pq  -  3q}}{{\text{)}}^2}{\text{  =  48p}}{{\text{q}}^2}$

Ans:  Taking L.H.S.

$  {\text{ =  (4pq}}{{\text{)}}^2}{\text{  +  2(4pq)(3q)  +  (3q}}{{\text{)}}^2}{\text{  -  [(4pq}}{{\text{)}}^2}{\text{  -  2(4pq) (3q)  +  (3q}}{{\text{)}}^2}{\text{]}} $

$  {\text{ =  16}}{{\text{p}}^2}{{\text{q}}^2}{\text{  +  24p}}{{\text{q}}^2}{\text{  +  9}}{{\text{q}}^2}{\text{  -  [16}}{{\text{p}}^2}{{\text{q}}^2}{\text{  -  24p}}{{\text{q}}^2}{\text{  +  9}}{{\text{q}}^2}{\text{]}} $

$  {\text{ =  16}}{{\text{p}}^2}{{\text{q}}^2}{\text{  +  24p}}{{\text{q}}^2}{\text{  +  9}}{{\text{q}}^2}{\text{  - 16}}{{\text{p}}^2}{{\text{q}}^2}{\text{  +  24p}}{{\text{q}}^2}{\text{  -  9}}{{\text{q}}^2} $

$ {\text{ =  48p}}{{\text{q}}^2}{\text{  =  R}}{\text{.H}}{\text{.S}}  $


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

Ans:  Taking L.H.S

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

${\text{ =  (}}{{\text{a}}^2}{\text{  -  }}{{\text{b}}^2}{\text{)  +  (}}{{\text{b}}^2}{\text{  -  }}{{\text{c}}^2}{\text{)  +  (}}{{\text{c}}^2}{\text{  -  }}{{\text{a}}^2}{\text{)  =  0  =  R}}{\text{.H}}{\text{.S}}{\text{.  =  0}}$


6. Using identities, evaluate.

(i) ${\text{7}}{{\text{1}}^2}$

Ans:  ${\text{7}}{{\text{1}}^2}{\text{  =  (70  +  1}}{{\text{)}}^2}$    ${\text{[(a  +  b}}{{\text{)}}^2}{\text{  =  }}{{\text{a}}^2}{\text{  +  2ab  +  }}{{\text{b}}^2}{\text{ ]}}$

$  {\text{ =  (70}}{{\text{)}}^2}{\text{  +  2(70) (1)  +  (1}}{{\text{)}}^2}{\text{ }} $

$  {\text{ =  4900  +  140  +  1  =  5041}}  $


(ii) ${\text{9}}{{\text{9}}^2}$

Ans:   ${\text{9}}{{\text{9}}^2}{\text{  =  (100  -  1}}{{\text{)}}^2}$   ${\text{[(a  -  b}}{{\text{)}}^2}{\text{  =  }}{{\text{a}}^2}{\text{  -  2ab  +  }}{{\text{b}}^2}{\text{ ]}}$

${\text{ =  (100}}{{\text{)}}^2}{\text{  -  2(100) (1)  +  (1}}{{\text{)}}^2}{\text{ }}$

$ {\text{ =  10000  -  200  +  1  =  9801}}  $


(iii) ${\text{10}}{{\text{2}}^2}{\text{ }}$

Ans:  ${\text{10}}{{\text{2}}^2}{\text{  =  (100  +  2}}{{\text{)}}^2}$    ${\text{[(a  +  b}}{{\text{)}}^2}{\text{  =  }}{{\text{a}}^2}{\text{  +  2ab  +  }}{{\text{b}}^2}{\text{ ]}}$

$  {\text{ =  (100}}{{\text{)}}^2}{\text{  +  2(100)(2)  +  (2}}{{\text{)}}^2} $

$  {\text{ =  10000  +  400  +  4  =  10404}}  $


(iv) ${\text{99}}{{\text{8}}^2}$

Ans:  ${\text{99}}{{\text{8}}^2}{\text{  =  (1000  -  2}}{{\text{)}}^2}$ ${\text{[(a  +  b}}{{\text{)}}^2}{\text{  =  }}{{\text{a}}^2}{\text{  +  2ab  +  }}{{\text{b}}^2}{\text{ ]}}$

$  {\text{ =  (1000}}{{\text{)}}^2}{\text{  -  2(1000)(2)  +  (2}}{{\text{)}}^2} $

$  {\text{ =  1000000  -  4000  +  4  =  996004}}  $


(v) ${{\text{(5}}{\text{.2)}}^2}{\text{ }}$

Ans:  ${{\text{(5}}{\text{.2)}}^2}{\text{  =  (5}}{\text{.0  +  0}}{\text{.2}}{{\text{)}}^2}$ ${\text{[(a  +  b}}{{\text{)}}^2}{\text{  =  }}{{\text{a}}^2}{\text{  +  2ab  +  }}{{\text{b}}^2}{\text{ ]}}$

$  {\text{ =  (5}}{\text{.0}}{{\text{)}}^2}{\text{  +  2(5}}{\text{.0) (0}}{\text{.2)  +  (0}}{\text{.2}}{{\text{)}}^2}{\text{ }} $

$  {\text{ =  25  +  2  +  0}}{\text{.04  =  27}}{\text{.04}}  $


(vi) ${\text{297 }} \times {\text{ 303}}$

Ans:  ${\text{297 }} \times {\text{ 303  =  (300  -  3) }} \times {\text{ (300  +  3) }}$[${{\text{a}}^2}{\text{  -  }}{{\text{b}}^2}{\text{  =  (a  +  b) (a  -  b),}}$]

${\text{ =  (300}}{{\text{)}}^2}{\text{  -  (3}}{{\text{)}}^2}{\text{ }}$

${\text{ =  90000  -  9  =  89991}}$


(vii) ${\text{78 }} \times {\text{ 82}}$

Ans:  ${\text{78 }} \times {\text{ 82  =   (80  -  2) (80  +  2)}}$   [${{\text{a}}^2}{\text{  -  }}{{\text{b}}^2}{\text{  =  (a  +  b) (a  -  b),}}$]

$  {\text{ =  (80}}{{\text{)}}^2}{\text{  -  (2}}{{\text{)}}^2}{\text{ }} $

$  {\text{ =  6400  -  4  =  6396}}  $


(viii) ${\text{8}}{\text{.}}{{\text{9}}^2}{\text{ }}$

Ans:  ${\text{8}}{\text{.}}{{\text{9}}^2}{\text{  =  (9}}{\text{.0  -  0}}{\text{.1}}{{\text{)}}^2}$   ${\text{[(a  -  b}}{{\text{)}}^2}{\text{  =  }}{{\text{a}}^2}{\text{  -  2ab  +  }}{{\text{b}}^2}{\text{ ]}}$

$  {\text{ =  (9}}{\text{.0}}{{\text{)}}^2}{\text{  -  2(9}}{\text{.0) (0}}{\text{.1)  +  (0}}{\text{.1}}{{\text{)}}^2} $

$  {\text{ =  81  -  1}}{\text{.8  +  0}}{\text{.01  =  79}}{\text{.21}}  $


(ix) ${\text{1}}{\text{.05 }} \times \,{\text{9}}{\text{.5}}$

Ans:  ${\text{1}}{\text{.05 }} \times \,{\text{9}}{\text{.5 =  1}}{\text{.05 }} \times 0.95 \times 10$

${\text{ = (1  +  0}}{\text{.05) (1 -  0}}{\text{.05) }}$[${{\text{a}}^2}{\text{  -  }}{{\text{b}}^2}{\text{  =  (a  +  b) (a  -  b),}}$

${\text{ =  [(1}}{{\text{)}}^2}{\text{  -  (0}}{\text{.05}}{{\text{)}}^2}{\text{] }}$

${\text{ =  [1  -  0}}{\text{.0025] }} \times {\text{ 10}}$

${\text{ =  0}}{\text{.9975 }} \times \,{\text{10  = 9}}{\text{.975}}$


  1. Using ${{\text{a}}^2}{\text{  -  }}{{\text{b}}^2}{\text{  =  (a  +  b) (a  -  b),}}$ find

(i) ${\text{5}}{{\text{1}}^2}{\text{  -  4}}{{\text{9}}^2}$

Ans:  ${\text{5}}{{\text{1}}^2}{\text{  -  4}}{{\text{9}}^2}$

${\text{5}}{{\text{1}}^2}{\text{  -  4}}{{\text{9}}^2}{\text{ =  (51  +  49) (51  -  49)}}$

${\text{ =  (100) (2)  =  200}}$


(ii) ${{\text{(1}}{\text{.02)}}^2}{\text{  -  (0}}{\text{.98}}{{\text{)}}^2}{\text{ }}$

Ans: ${{\text{(1}}{\text{.02)}}^2}{\text{  -  (0}}{\text{.98}}{{\text{)}}^2}{\text{ }}$

${{\text{(1}}{\text{.02)}}^2}{\text{  -  (0}}{\text{.98}}{{\text{)}}^2}{\text{  =  (1}}{\text{.02  +  0}}{\text{.98) (1}}{\text{.02  -  0}}{\text{.98)}}$


(iii) ${\text{15}}{{\text{3}}^2}{\text{  -  14}}{{\text{7}}^2}$

Ans:  ${\text{15}}{{\text{3}}^2}{\text{  -  14}}{{\text{7}}^2}$

${\text{15}}{{\text{3}}^2}{\text{  -  14}}{{\text{7}}^2}{\text{  =  (153  +  147) (153  -  147)}}$

${\text{ =  (300) (6)  =  1800}}$


(iv) ${\text{12}}{\text{.}}{{\text{1}}^2}{\text{  -  7}}{\text{.}}{{\text{9}}^2}$

Ans:  ${\text{12}}{\text{.}}{{\text{1}}^2}{\text{  -  7}}{\text{.}}{{\text{9}}^2}$

${\text{12}}{\text{.}}{{\text{1}}^2}{\text{  -  7}}{\text{.}}{{\text{9}}^2}{\text{  =  (12}}{\text{.1  +  7}}{\text{.9) (12}}{\text{.1  -  7}}{\text{.9)}}$


8. Using ${\text{(x  +  a) (x  +  b)  =  }}{{\text{x}}^2}{\text{  +  (a  +  b) x  +  ab,}}$ find

(i) ${\text{103 }} \times \,{\text{104}}$

Ans:  ${\text{103 }} \times \,{\text{104  =  (100  +  3) (100  +  4)}}$

$  {\text{ =  (100}}{{\text{)}}^2}{\text{  +  (3  +  4) (100)  +  (3) (4)}} $

$  {\text{ =  10000  +  700  +  12  =  10712}}  $


(ii) ${\text{5}}{\text{.1 }} \times {\text{ 5}}{\text{.2}}$

Ans:  ${\text{5}}{\text{.1 }} \times {\text{ 5}}{\text{.2  =  (5  +  0}}{\text{.1) (5  +  0}}{\text{.2)}}$

$  {\text{ =  (5}}{{\text{)}}^2}{\text{  +  (0}}{\text{.1  +  0}}{\text{.2) (5)  +  (0}}{\text{.1) (0}}{\text{.2)}} $

$  {\text{ =  25  +  1}}{\text{.5  +  0}}{\text{.02  =  26}}{\text{.52}} $

 

(iii) ${\text{103 }} \times \,{\text{98}}$

Ans:  ${\text{103 }} \times \,{\text{98  =  (100  +  3) (100  -  2)}}$

$  {\text{ =  (100}}{{\text{)}}^2}{\text{  +  [3  +  ( -  2)] (100)  +  (3) ( -  2)}} $

$  {\text{ =  10000  +  100  -  6}} $

$  {\text{ =  10094}}  $


(iv) ${\text{9}}{\text{.7 }} \times \,\,9.8$

Ans:  ${\text{9}}{\text{.7 }} \times \,\,9.8{\text{  =  (10  -  0}}{\text{.3) (10  -  0}}{\text{.2)}}$

$  {\text{ =  (10}}{{\text{)}}^2}{\text{  +  [( -  0}}{\text{.3)  +  ( -  0}}{\text{.2)] (10)  +  ( -  0}}{\text{.3) ( -  0}}{\text{.2)}} $

$  {\text{ =  100  +  ( -  0}}{\text{.5)10  +  0}}{\text{.06  =  100}}{\text{.06  -  5  =  95}}{\text{.06}}


NCERT Solutions for Class 8 Maths Chapter 9 - Free PDF Download

If you are worried about internet connectivity then don’t worry as  NCERT Solutions Class 8 Maths Chapter 9 are available in pdf format. They are easy to download and after downloading, these solutions can be accessed as per students' wish. These NCERT Solutions Class 8 are available on our website and our app. NCERT Solutions Class 8 is entirely free of cost. So if you are going to have a test or exam near, our NCERT Solutions Class 8 is there for you. These solutions help in the last minute revision of the chapters and ensure that one does not miss the most important questions of NCERT.


NCERT Solutions for Class 8 Maths Chapter 9

Chapter - 9 Algebraic Expressions and Identities

In the curriculum of Class 8 Maths Chapter 9 is Algebraic Expressions and Identities.

Algebra is introduced in Class 8th to make the students know about the algebraic terms, algebraic identities, variables, constants, algebraic expressions, monomial, binomial and trinomial expressions. Also, the mathematical operations like addition, subtraction, multiplication, and division of algebraic expressions are explained in this chapter.

Algebra and its real-life applications are an important part of the maths syllabus. Various Exercises are given in Chapter 9 Maths for students to solve and become familiar with algebra.

With our Grade 8 NCERT solutions, it becomes easy for students to understand the algebraic concepts and solve the questions. These solutions are prepared by our subject experts.


Class 8 Maths Chapter 9 Marks Weightage

Chapter 9 Algebraic Expressions and Identities is a very important chapter from any exam point of view whether it's school exams or competitive exams thus going through these NCERT Solutions will help the student to get a good score on their exams.

In this chapter, a total of 5 Exercises are given with different types of questions and our solutions will help the students to solve these questions.


We Cover all Exercises in the Chapter Given Below:

Chapter 9 All Exercises in PDF Format

Exercise 9.1

4 Question & Solutions

Exercise 9.2

5 Questions & Solutions

Exercise 9.3

5 Questions & Solutions

Exercise 9.4

3 Questions with Solutions

Exercise 9.5

8 Questions with Solutions


Why are NCERT Solutions Class 8 Chapter 9 Important?

  • Preparing from our NCERT Solutions Class 8  helps students to gain confidence over the chapter through detailed explanations given in our NCERT solutions.

  • NCERT solutions ensure that students have gone through all the questions given in NCERT which are important from an exam point of view.

  • Last-minute revision and preparation can be easily done by these solutions.

  • NCERT Solutions are made in simple language which makes it understandable for the students.

  • NCERT solutions also give the students an idea of how a mathematical solution must be written.


NCERT Class 8 Maths Chapter wise Solutions in Hindi


NCERT Solutions for Class 8 Maths - Chapterwise Solutions


Conclusion 

The NCERT Solutions for Class 8 Maths Chapter 9 on Algebraic Expressions and Identities by Vedantu are a helpful resource for students. This chapter focuses on understanding and working with algebraic expressions, which are mathematical phrases involving numbers, variables, and operations. It's an important foundation for higher-level math. One crucial section is likely the exploration of identities, where students learn about equations that hold true for any value of the variables. Mastering this concept is key for solving complex mathematical problems. Vedantu's solutions provide clear explanations, making the learning process more accessible and supporting students in grasping the fundamentals of algebra.

FAQs on NCERT Solutions for Class 8 Maths Chapter 9 - 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.