# NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities

NCERT solutions for class 8 maths chapter 9 Algebraic Expressions and Identities- As you have learned in previous classes that expressions are formed from variables and constants. This chapter will introduce you to the application of algebraic terms and variables to solve various problems. NCERT solutions for class 8 maths chapter 9 algebraic expressions and identities will give you a detailed explanation for every problem of this chapter. Algebra is the most important branch of mathematics which teaches how to form equations and solving them using different kinds of techniques. Important topics like the product of the equation, finding the coefficient of the variable in the equation, subtraction of the equation and creating quadratic equation by the product of its two roots, and division of the equation are covered in this chapter. You will find the questions related to these topics in CBSE NCERT solutions for class 8 maths chapter 9 algebraic expressions and identities which will make your task easy while solving the problems. As it is a new concept, you may find difficulties while dealing with algebraic parts of mathematics but if you practice questions and go through solutions of NCERT for class 8 maths chapter 9 algebraic expressions and identities, you will find it very easy and one of the strongest parts in mathematics. You will get NCERT solutions from class 6 to 12 for science and maths by clicking on the above link.  Here you will solutions to five exercise of this chapter.

Exercise:9.1

Exercise:9.2

Exercise:9.3

Exercise:9.4

Exercise:9.5

## Topics of NCERT for class 8 maths chapter 9 - Algebraic expressions and identities:-

9.1 What are Expressions?

9.2 Terms, Factors and Coefficients

9.3 Monomials, Binomials and Polynomials

9.4 Like and Unlike Terms

9.5 Addition and Subtraction of Algebraic Expressions

9.6 Multiplication of Algebraic Expressions: Introduction

9.7 Multiplying a Monomial by a Monomial

9.7.1 Multiplying two monomials

9.7.2 Multiplying three or more monomials

9.8 Multiplying a Monomial by a Polynomial

9.8.1 Multiplying a monomial by a binomial

9.8.2 Multiplying a monomial by a trinomial

9.9Multiplying a Polynomial by a Polynomial

9.9.1 Multiplying a binomial by a binomial

9.9.2 Multiplying a binomial by a trinomial

9.10 What is an Identity?

9.11 Standard Identities

9.12 Applying Identities

## NCERT solutions for class 8 maths chapter 9 algebraic expressions and identities topic 9.1 what are expressions?

Five examples of expressions containing one variable are:

$x^{^{4}}, y, 3z, p^{^{2}}, -2q^{3}$

Five examples of expressions containing two variables are:

$x + y, 3p-4q,ab,uv^{2},-z^{2}+x^{3}$

Question:

$x$

x on the number line:

Question:

$x-4$

x-4 on the number line:

Question:

2x+1

2x+1 on the number  line:

Question:

$3x-2$

3x - 2 on the number line

## Question:1 Identify the coefficient of each term in the expression.

$x^2y^2-10x^2y+5xy^2-20$

coefficient of each term are given below

$\\The\ coefficient\ of\ x^{2}y^{2}\ is \1\\ \\The\ coefficient\ of\ x^{2}y\ is \ -10\\ \\The\ coefficient\ of\ xy^{2}\ is \5\\$

## Question:1(i) Classify the following polynomials as monomials, binomials, trinomials.

$-z+5$

Binomial since there are two terms with non zero coefficients.

$x+y+z$

Trinomial since there are three terms with non zero coefficients.

$y+z+100$

Trinomial since there are three terms with non zero coefficients.

$ab-ac$

Binomial since there are two terms with non zero coefficients.

$17$

Monomial since there is only one term.

Three binomials with the only x as a variable are:

$\\ \\x+2,\ x +x^{2},\ 3x^{3}-5x^{4}$

Three binomials with x and y as variables are:

$\\ \\x+y,\ x-7y, xy^{2} + 2xy$

Three monomials with x and y as variables are

$\\ xy,\ 3xy^{4},\ -2x^{3}y^{2}$

Question:2(d) Construct

Two polynomials with 4 or more terms are:

$a+b+c+d, x-3xy+2y+4xy^{2}$

## Question:(i) Write two terms which are like

$7xy$

$\\Two\ terms\ like\ 7xy\ are:\\ -3xy\ and\ 5xy$

Question:(ii) Write two terms which are like

$4mn^2$

$\\Two\ terms\ which\ are\ like\ 4mn^{2}\ are:\\ mn^{2}\ and -3mn^{2.}$

we can write more like terms

Question:(iii)

$2l$

$\\Two\ terms\ which\ are\ like\ 2l\ are:\\ l\ and\ -3l$

## NCERT solutions for class 8 maths chapter 9 algebraic expressions and identities-Exercise: 9.1

$5xyz^2-3zy$

following are the terms and coefficient

The terms are$5xyz^{2}\ and\ -3zy$  and the coefficients are 5 and -3.

$1+x+x^2$

the following is the solution

$\\The\ terms\ are\ 1,\ x,\ and\ x^{2}\ and\ the\ coefficients\ are\ 1,\ 1,\ and\ 1\ respectively.$

$4x^2y^2-4x^2y^2z^2+z^2$

$\\The\ terms\ are\ 4x^{2}y^{2},\ -4x^{2}y^{2}z^{2}and\ z^{2}\ and\ the\ coefficients\ are\ 4,\ -4\ and\ 1\ respectively.$

$3-pq+qr-rp$

The terms are 3, -pq, qr,and -rp and the coefficients are 3, -1, 1 and -1 respectively.

$\frac{x}{2}+\frac{y}{2}-xy$

$\\The\ terms\ are\ \frac{x}{2},\ \frac{y}{2}\ and\ -xy\ and\ the\ coefficients\ are\ \frac{1}{2},\ \frac{1}{2}\ and\ -1\ respectively.$

Above are the terms and coefficients

$0.3a-0.6ab+0.5b$

The terms are 0.3a, -0.6ab and 0.5b and the coefficients are 0.3, -0.6 and 0.5.

$x+y$

Binomial.

$1000$

Monomial.

$x+x^2+x^3+x^4$

This polynomial does not fit in any of these three categories.

$7+y-5x$

Trinomial.

$2y-3y^2$

Binomial.

$2y-3y^2+4y^3$

Trinomial.

$5x-4y+3xy$

Trinomial.

$4z-15z^2$

Binomial.

$ab+bc+cd+da$

This polynomial does not fit in any of these three categories.

$pqr$

Monomial.

$p^2q+pq^2$

Binomial.

$2p+2q$

Binomial.

Question:

$ab-bc , bc -ca, ca-ab$

ab-bc+bc-ca+ca-ab=0.

Question:3

$a-b+ab, b-c+bc, c-a+ac$

$\\a-b+ab+b-c+bc+c-a+ac\\ =(a-a)+(b-b)+(c-c)+ab+bc+ac\\ =ab+bc+ca$

$2p^2q^2-3pq+4, 5+7pq-3p^2q^2$

$\\2p^{2}q^{2}-3pq+4+5+7pq-3p^{2}q^{2}\\ =(2-3)p^{2}q^{2} +(-3+7)pq +4+5\\ =-p^{2}q^{2}+4pq+9$

$l^2+m^2+n^2 , n^2+l^2, 2lm+2mn+2nl$

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

12a-9ab+5b-3-(4a-7ab+3b+12)
=(12-4)a +(-9+7)ab+(5-3)b +(-3-12)
=8a-2ab+2b-15

$\\5xy-2yz-2zx+10xyz-(3xy+5yz-7zx)\\ =(5-3)xy+(-2-5)yz+(-2+7)zx+10xyz\\ =2xy-7yz+5zx+10xyz$

$\\18-3p-11q+5pq-2pq^{2}+5p^{2}q-(4p^{2}q-3pq+5pq^{2}-8p+7q-10)\\ =18-(-10)-3p-(-8p)-11q-7q+5pq-(-3pq)-2pq^{2}-5pq^{2}+5p^{2}q-4p^{2}q\\ =28+5p-18q+8pq-7pq^{2}+p^{2}q$

## Solutions for NCERT class 8 maths chapter 9 algebraic expressions and identities topic 9.7.2 multiplying three or more monomials

$\\4x\times 5y\times 7z\\ =(4x\times 5y)\times 7z\\ =20xy\times 7z\\ =140xyz\\ \\4x\times 5y\times 7z\\ =(5y\times 7z)\times 4x\\ =35yz\times 4x\\ =140xyz$

We observe that the result is same in both cases and the result does not depend on the order in which multiplication has been carried out.

## Question:1(i) Find the product of the following pairs of monomials.

$4,7p$

$4\times 7p=28p$

Question:

$-4p,7p$

$\\-4p\times 7p\\=(-4\times 7)p\times p\\=-28p^{2}$

Question:

$-4p,7pq$

$-4p\times 7pq\\=(-4\times 7)p\times pq\\=-28p^{2}q$

Question:

$4p^3,-3p$

$\\4p^{3}\times (-3p)\\ =4\times (-3)p^{3}\times p\\=-12p^{4}$

$4p,0$

$\\4p\times 0=0$

$(p,q)$

The question can be solved as follows

$\\Area=length\times breadth\\ =(p\times q)\\ =pq$

$(10m,5n)$

the area is calculated as follows

$\\Area=length\times breadth\\ =10m\times 5n\\ =50mn$

$(20x^2,5y^2)$

the following is the solution

$\\Area=length\times breadth\\ =20x^{2}\times 5y^{2}\\ =100x^{2}y^{2}$

$(4x,3x^2)$

area of rectangles  is

$\\Area=length\times breadth\\ =4x\times 3x^{2}\\ =12x^{3}$

$(3mn,4np)$

The area is calculated as follows

$\\Area=length\times breadth\\ =3mn\times 4np\\ =12mn^{2}p$

Question:3 Complete the table of products.

 First monomial $\rightarrow$ Second monomial $\downarrow$ $2x$ $-5y$ $3x^2$ $-4xy$ $7x^2y$ $-9x^2y^2$ $2x$ $4x^2$ ... ... ... ... ... $-5y$ ... ... $-15x^2y$ ... ... ... $3x^2$ ... ... ... ... ... ... $-4xy$ ... ... ... ... ... ... $7x^2y$ ... ... ... ... ... ... $-9x^2y^2$ ... ... ... ... ... ...

 First monomial $\rightarrow$ Second monomial $\downarrow$ $2x$ $-5y$ $3x^{2}$ $-4xy$ $7x^{2}y$ $-9x^{2}y^{2}$ $2x$ $4x^{2}$ $-10xy$ $6x^{3}$ $-8x^{2}y$ $14x^{3}y$ $-18x^{3}y^{2}$ $-5y$ $-10xy$ $25y^{2}$ $-15x^{2}y$ $20xy^{2}$ $-35x^{2}y^{2}$ $45x^{2}y^{3}$ $3x^{2}$ $6x^{3}$ $-15x^{2}y^{}$ $9x^{4}$ $-12x^{3}y$ $21x^{4}y$ $-27x^{4}y^{2}$ $-4xy$ $-8x^{2}y$ $20xy^{2}$ $-12x^{3}y$ $16x^{2}y^{2}$ $-28x^{3}y$ $36x^{3}y^{3}$ $7x^{2}y$ $14x^{3}y$ $-35x^{2}y^{2}$ $21x^{4}y$ $-28x^{3}y^{2}$ $49x^{4}y^{2}$ $-63x^{4}y^{3}$ $-9x^{2}y^{2}$ $-18x^{3}y^{2}$ $45x^{2}y^{3}$ $-27x^{4}y^{2}$ $36x^{3}y^{3}$ $-63x^{4}y^{3}$ $81x^{4}y^{4}$

$5a, 3a^2, 7a^4$

$\\Volume=length\times breadth\times height\\ =5a\times 3a^{2}\times 7a^{4}\\ =15a^{3}\times 7a^{4}\\ =105a^{7}$

$2p,4q,8r$

the volume of rectangular boxes with the following length, breadth and height is

$\\Volume=length\times breadth\times height\\ =2p\times 4q\times 8r\\ =8pq\times 8r\\ =64pqr$

$xy, 2x^2y, 2xy^2$

the volume of rectangular boxes with the following length, breadth and height is

$\\Volume=length\times breadth\times height\\ =xy\times 2x^{2}y\times 2xy^{2}\\ =2x^{3}y^{2}\times 2xy^{2}\\ =4x^{4}y^{4}$

$a, 2b, 3c$

the volume of rectangular boxes with the following length, breadth and height is

$\\Volume=length\times breadth\times height\\ =a\times 2b\times 3c\\ =2ab\times 3c\\ =6abc$

Question:5(i) Obtain the product of

$xy,yz,zx$

the product

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

Question:5(ii) Obtain the product of

$a,-a^2,a^3$

the product

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

Question:5(iii) Obtain the product of

$2,\ 4y,\ 8y^{2},\ 16y^{3}$

the product

$\\2\times 4y\times 8y^{2}\times 16y^{3}\\ =8y\times 8y^{2}\times 16y^{3}\\ =64y^{3}\times 16y^{3}\\ =1024y^{6}$

Question:5(iv) Obtain the product of

$a, 2b, 3c, 6abc$

the product

$\\a\times 2b\times 3c\times 6abc\\ =2ab\times 3c\times 6abc\\ =6abc\times 6abc\\ =36a^{2}b^{2}c^{2}$

Question:5(v) Obtain the product of

$m, -mn, mnp$

the product

$\\m\times (-mn)\times mnp\\ =-m^{2}n\times mnp\\ =-m^{3}n^{2}p$

## NCERT solutions for class 8 maths chapter 9 algebraic expressions and identities topic 9.8.1 multiplying a monomial by a binomial

Question:(i) Find the product

$2x(3x+5xy)$

Using distributive law,

$2x(3x + 5xy) = 6x^2 + 10x^2y$

Question:(ii)  Find the product

$a^2(2ab-5c)$

Using distributive law,

We have :       $a^2(2ab-5c) = 2a^3b - 5a^2c$

## NCERT textbook solutions for class 8 maths chapter 9 algebraic expressions and identities topic 9.8.2 multiplying a monomial by a trinomial

Question:1 Find the product:

$(4p^2+5p+7)\times 3p$

By using distributive law,

$(4p^2+5p+7)\times 3p = 12p^3 + 15p^2 + 21p$

## NCERT solutions for class 8 maths chapter 9 algebraic expressions and identities-Exercise: 9.3

$4p, q+r$

Multiplication of the given expression gives :

By distributive law,

$(4p)(q+r) = 4pq + 4pr$

$ab, a-b$

We have ab,  (a-b).

Using distributive law we get,

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

$a+b, 7a^2b^2$

Using distributive law we can obtain multiplication of given expression:

$(a + b)(7a^2b^2) = 7a^3b^2 + 7a^2b^3$

$a^2-9,4a$

We will obtain multiplication of given expression by using distributive law :

$(a^2 - 9 )(4a) = 4a^3 - 36a$

$pq+qr+rp, 0$

Using distributive law :

$(pq + qr + rp)(0) = pq(0) + qr(0) + rp(0) = 0$

Question:2 Complete the table

 First expression Second expression Product (i) $a$ $b+c+d$ ... (ii) $x+y-5$ $5xy$ ... (iii) $p$ $6p^2-7p+5$ ... (iv) $4p^2q^2$ $p^2-q^2$ ... (v) $a+b+c$ $abc$ ...

We will use distributive law to find product in each case.

 First expression Second expression Product (i) $a$ $b+c+d$ $ab + ac+ ad$ (ii) $x+y-5$ $5xy$ $5x^2y + 5xy^2 - 25xy$ (iii) $p$ $6p^2-7p+5$ $6p^3 - 7p^2 + 5p$ (iv) $4p^2q^2$ $p^2-q^2$ $4p^4q^2 - 4p^2q^4$ (v) $a+b+c$ $abc$ $a^2bc + ab^2c + abc^2$

Question:3(i)  Find the product.

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

Opening brackets :

$(a^2)\times (2a^{22})\times (4a^{26}) = (a^2\times2a^{22})\times(4a^{26}) = 2a^{24}\times4a^{26}$

or                                             $=8a^{50}$

Question:3(ii) Find the product.

$(\frac{2}{3}xy)\times (\frac{-9}{10}x^2y^2)$

We have,

$(\frac{2}{3}xy)\times (\frac{-9}{10}x^2y^2) = \frac{-3}{5}x^3y^3$

Question:3(iii) Find the product.

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

We have

$(\frac{-10}{3}pq^3) \times (\frac{6}{5}p^3q) = -4p^4q^4$

Question:3(iv)  Find the product.

$x \times x^2\times x^3\times x^4$

We have      $x \times x^2\times x^3\times x^4$

$x \times x^2\times x^3\times x^4 = (x \times x^2)\times x^3\times x^4$

or          $(x^3)\times x^3\times x^4$

$= x^{10}$

(i) $\small x=3$

(a)  We have

$3x(4x-5)+3 = 12x^2 - 15x + 3$

Put x = 3,

We get :              $12(3)^2 - 15(3) + 3 = 12(9) - 45 + 3 = 108 - 42 = 66$

(ii)  $\small x=\frac{1}{2}$

We have

$\small 3x(4x-5)+3 = 12x^2 -15x + 3$

Put

$x = \frac{1}{2}$

.    So We get,

$12x^2 -15x + 3 = 12(\frac{1}{2})^2 - 15(\frac{1}{2}) + 3 = 6 - \frac{15}{2} = \frac{-3}{2}$

(i) $\small a =0$

We have :                    $\small a(a^2+a+1) +5 = a^3 + a^2 + a +5$

Put a = 0 :                       $= 0^3 + 0^2 + 0 + 5 = 5$

(ii) $\small a=1$

We have   $\small a(a^2+a+1)+5 = a^3 + a^2 + a + 5$

Put a = 1 ,

we get :           $1^3 + 1^2 + 1 + 5 = 1 + 1 + 1+ 5 = 8$

(iii) $\small a=-1$

We have      $\small a(a^2+a+1)+5$.

or                  $\small a(a^2+a+1)+5 = a^3+a^2+a+5$

Put       a = (-1)

$= (-1)^3+(-1)^2+(-1)+5 = -1 + 1 -1 +5 = 4$

(a)First we will solve each brackets individually.

$p(p-q) = p^2 - pq$;                  $q(q-r) = q^2 - qr$;                     $r(r-p) = r^2 - rp$

Addind all we get :             $p^2 - pq + q^2 - qr + r^2 - rp$

$= p^2 + q^2 + r^2 -pq-qr-rp$

Question:5(b) Add:  $\small 2x(z-x-y)$   and  $\small 2y(z-y-x)$

Firstly, open the brackets:

$\small 2x(z-x-y) = 2xz -2x^2-2xy$

and               $\small 2y(z-y-x) = 2yz-2y^2-2xy$

$\small 2xz -2x^2-2xy +2yz-2y^2-2xy$

or                  $\small = -2x^2-2y^2-4xy + 2xz+2yz$

Question:5(c) Subtract: $\small 3l(l-4m+5n)$  from  $\small 4l(10n-3m+2l)$

At first we will solve each bracket individually,

$\small 3l(l-4m+5n) = 3l^2 - 12lm + 15ln$

and            $\small 4l(10n-3m+2l) = 40ln - 12ml + 8l^2$

Subtracting:

$\small 40ln - 12ml + 8l^2 - (3l^2 - 12lm+15ln)$

or                 $\small = 40ln - 12ml + 8l^2 - 3l^2 + 12lm-15ln$

or                 $\small = 25ln + 5l^2$

Question:5(d) Subtract:  $\small 3a(a+b+c)-2b(a-b+c)$  from  $\small 4c(-a+b+c)$

Solving brackets :

$3a(a+b+c)-2b(a-b+c) = 3a^2+3ab+3ac - 2ab+2b^2-2bc$

$= 3a^2+ab+3ac+ 2b^2-2bc$

and                 $\small 4c(-a+b+c) = -4ac +4bc + 4c^2$

Subtracting :                 $\small -4ac +4bc + 4c^2 -(3a^2 + ab + 3ac+2b^2-2bc)$

$\small = -4ac + 4bc+4c^2-3a^2-ab-3ac-2b^2+2bc$

$\small =-3a^2 -2b^2+4c^2-ab+ 6bc-7ac$

## NCERT solutions for class 8 maths chapter 9 algebraic expressions and identities-Exercise: 9.4

Question:1(i)  Multiply the binomials.

$\small (2x+5)$  and  $\small (4x-3)$

We have    (2x + 5) and (4x - 3)
(2x + 5)  X   (4x - 3)   =  (2x)(4x) + (2x)(-3) + (5)(4x) + (5)(-3)
= 8$x^{2}$ - 6x + 20x - 15
= 8$x^{2}$ + 14x -15

Question:1(ii) Multiply the binomials.

$\small (y-8)$  and $\small (3y-4)$

We need to multiply (y - 8) and (3y - 4)
(y - 8) X  (3y - 4)  =   (y)(3y) +  (y)(-4) + (-8)(3y) + (-8)(-4)
=   3$y^{2}$ - 4y - 24y  + 32
= 3$y^{2}$ - 28y + 32

Question:1(iii)  Multiply the binomials

$\small (2.5l-0.5m)$  and  $\small (2.5l+0.5m)$

We need to multiply (2.5l - 0.5m) and (2.5l + 0..5m)
(2.5l - 0.5m) X (2.5l + 0..5m)    = $(2.5l)^{2} - (0.5m)^{2}$                                    using    $(a-b)(a+b) = (a)^{2} - (b)^{2}$
= 6.25$l^{2}$  - 0.25$m^{2}$

Question:1(iv)  Multiply the binomials.

$\small (a+3b)$   and    $\small (x+5)$

(a + 3b) X (x + 5) = (a)(x) + (a)(5) + (3b)(x) + (3b)(5)
= ax + 5a + 3bx + 15b

Question:1(v) Multiply the binomials.

$\small (2pq+3q^2)$  and  $\small (3pq-2q^2)$

(2pq + 3q2) X (3pq - 2q2)  =  (2pq)(3pq) + (2pq)(-2q2) + ( 3q2)(3pq) + (3q2)(-2q2)
= 6p2q2 - 4pq3  + 9pq3 - 6q4
= 6p2q2 +5pq3 - 6q4

Question:1(vi) Multiply the binomials.

$\small (\frac{3}{4}a^2+3b^2)$   and   $\small 4(a^2-\frac{2}{3}b^2)$

Multiplication can be done as follows

$\small (\frac{3}{4}a^2+3b^2)$   X     $\small (4a^2-\frac{8}{3}b^2)$     =    $\frac{3a^{2}}{4} \times 4a^{2} + \frac{3a^{2}}{4} \times (-\frac{8b^{2}}{3}) + 3b^{2} \times 4a^{2} + 3b^{2} \times (-\frac{8b^{2}}{3})$

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

=  $3a^{4} + 10a^{2}b^{2} - 8b^{4}$

Question:2(i) Find the product.

$\small (5-2x)$ $\small (3+x)$

(5 - 2x) X (3 + x) =  (5)(3) + (5)(x) +(-2x)(3) + (-2x)(x)
= 15 + 5x - 6x - 2$x^{2}$
= 15  - x - 2$x^{2}$

Question:2(ii) Find the product.

$\small (x+7y)(7x-y)$

(x + 7y) X (7x - y) =  (x)(7x) + (x)(-y) + (7y)(7x) + (7y)(-y)
= 7$x^{2}$ - xy + 49xy - 7$y^{2}$
=  7$x^{2}$ + 48xy - 7$y^{2}$

Question:2(iii) Find the product.

$\small (a^2+b)(a+b^2)$

($a^{2}$ + b) X (a + $b^{2}$)  = ($a^{2}$)(a) + ($a^{2}$)($b^{2}$) + (b)(a) + (b)($b^{2}$)
= $a^{3 } + a^{2}b^{2} + ab + b^{3}$

Question:2(iv) Find the product.

$\small (p^2-q^2)(2p+q)$

following is the solution

($p^{2}- q^{2}$) X (2p + q)  = $(p^{2})(2p) + (p^{2})(q) + (-q^{2})(2p) + (-q^{2})(q)$
$2p^{3} + p^{2}q - 2q^{2}p - q^{3}$

Question:3(i) Simplify.

$\small (x^2-5)(x+5)+25$

this can be simplified as follows

($x^{2}$ -5) X (x + 5) + 25  = ($x^{2}$)(x) + ($x^{2}$)(5) + (-5)(x) + (-5)(5)  + 25
= $x^{3} + 5x^{2} - 5x -25 + 25$
= $x^{3} + 5x^{2} - 5x$

Question:3(ii) Simplify.

$(a^2+5)(b^3+3)+5$

This can be simplified as

($a^{2}$ + 5) X ($b^{3}$ + 3) + 5 = ($a^{2}$)($b^{3}$) + ($a^{2}$)(3) + (5)($b^{3}$) + (5)(3) + 5
=  $a^{2}b^{3} + 3a^{2} + 5b^{3} + 15+5$
= $a^{2}b^{3} + 3a^{2} + 5b^{3} + 20$

Question:3(iii) Simplify.

$(t+s^2)(t^2-s)$

simplifications can be

(t + $s^{2}$)($t^{2}$ - s) =  (t)($t^{2}$) + (t)(-s)  + ($s^{2}$)($t^{2}$)  + ($s^{2}$)(-s)
=  $t^{3} - ts + s^{2}t^{2} - s^{3}$

Question:3(iv) Simplify.

$(a+b)(c-d)+(a-b)(c+d)+2 (ac+bd)$

(a + b) X ( c -d) + (a - b) X (c + d) + 2(ac + bd )
=   (a)(c) + (a)(-d) + (b)(c) + (b)(-d) +   (a)(c) + (a)(d) + (-b)(c) + (-b)(d)  + 2(ac + bd )
= ac - ad + bc - bd + ac +ad -bc - bd   + 2(ac + bd )
=  2(ac - bd )  + 2(ac +bd )
=  2ac - 2bd + 2ac  + 2bd
= 4ac

Question:3(v) Simplify.

$(x+y)(2x+y)+(x+2y)(x-y)$

(x + y) X ( 2x + y) + (x + 2y) X (x - y)
=(x)(2x) + (x)(y) + (y)(2x) + (y)(y)  + (x)(x) + (x)(-y) + (2y)(x) + (2y)(-y)
= 2$x^{2}$ + xy + 2xy + $y^{2}$ + $x^{2}$ - xy + 2xy - 2$y^{2}$
=3$x^{2}$ + 4xy -  $y^{2}$

Question:3(vi) Simplify.

$(x+y)(x^2-xy+y^2)$

simplification is done as follows

(x + y) X ($x^{2} -xy + y^{2}$)  = x X ($x^{2} -xy + y^{2}$) + y ($x^{2} -xy + y^{2}$)
= $x^{3} -x^{2}y + xy^{2} + yx^{2} - xy^{2} + y^{3}$
=$x^{3}+ y^{3}$

Question:3(vii) Simplify.

$(1.5x-4y)(1.5x+4y+3)-4.5x+12y$

(1.5x - 4y) X (1.5x + 4y + 3) - 4.5x + 12y =   (1.5x)  X (1.5x + 4y + 3)  -4y X (1.5x + 4y + 3)   - 4.5x + 12y
=  2.25$x^{2}$ + 6xy + 4.5x  - 6xy - 16$y^{2}$ - 12y   -4.5x + 12 y
= 2.25$x^{2}$    - 16$y^{2}$

Question:3(viii) Simplify.

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

(a + b + c) X (a + b - c)  = a  X (a + b - c) + b  X (a + b - c) + c  X (a + b - c)
= $a^{2}$ + ab - ac + ab + $b^{2}$ -bc + ac + bc - $c^{2}$
= $a^{2}$ + $b^{2}$  - $c^{2}$ + 2ab

## NCERT solutions for class 8 maths chapter 9 algebraic expressions and identities topic 9.11 standerd identities

Identity 1 $\Rightarrow (a+b)^{2} = a^{2} + 2ab + b^{2}$
If we replace b with -b in identity 1
We get,
$a^{2} + 2a(-b) + (-b)^{2} = a^{2} - 2ab + b^{2}$
which is equal to
$(a-b)^{2}$    which is identity 2
So, we get identity 2 by replacing b with -b in identity 1

## NCERT free solutions for class 8 maths chapter 9 algebraic expressions and identities topic 9.11 standard identities

Identity IV
(a + x)(b + x) = $x^{2} + (a+b)x + ab$
So, it is given that a = 2, b = 3 and x = 5
Lets put these value in identity  IV
(2 + 5)(3 + 5) = $5^{2}$ + (2 + 3)5 +2 X 3
7 X 8 = 25 + 5 X 5 + 6
56 = 25 + 25 + 6
= 56
L.H.S. = R.H.S.
So,  by this we can say that identity IV satisfy with given value of a,b and x

Identity IV is  $\Rightarrow (a +x)(b+x) = x^{2} + (a+b)x + ab$
If a =b than

(a + x)(a + x) = $x^{2} + (a+a)x + a\times a$
$(a+x)^{2} = x^{2} + 2ax + a^{2}$
Which is identity I

Identity IV is $\Rightarrow (a+x )(b +x) = x^{2} + (a+b)x + ab$
If a = b = -c  than,
(x - c)(x - c) = $x^{2} + (-c + (-c))x + (-c) \times (-c)$
$(x-c)^{2} = x^{2} + -2cx + c^{2}$
Which is identity II

Identity IV is $\Rightarrow (a+x )(b +x) = x^{2} + (a+b)x + ab$
If b = -a than,

(x + a)(x - a) = $x^{2} + (a +(-a))x + (-a) \times a$
=$x^{2} - a^{2}$
Which is identity III

## NCERT solutions for class 8 maths chapter 9 algebraic expressions and identities-Exercise: 9.5

$(x+3)(x+3)$

(x + 3) X (x +3) = $(x +3)^{2}$
So, we use identity I for this which is
$(a+b)^{2} = a^{2} + 2ab + b^{2}$
In  this  a=x and b = x
$(x+3)^{2} = x^{2} + 2(x)(3)+ 3^{2}$
=$x^{2} + 6x+ 9$

$(2y+5)(2y+5)$

(2y + 5) X ( 2y + 5) =  $(2y +5)^{2}$
We use identity I for this which is
$(a+b)^{2} = a^{2} + 2ab + b^{2}$
IN this  a = 2y and b = 5
$(2y+5)^{2} = (2y)^{2} + 2(2y)(5) + 5^{2}$
= $(2y+5)^{2} = 4y^{2} + 20y + 25$

$(2a-7)(2a-7)$

(2a -7) X (2a - 7)  = $(2a - 7)^{2}$
We use identity II for this which is
$(a-b)^{2} = a^{2} - 2ab + b^{2}$
in this a = 2a and b = 7
$(2a-7)^{2} = (2a)^{2} - 2(2a)(7) + 7^{2}$
=  $4a^{2} - 28a + 49$

$(3a - \frac{1}{2}) (3a -\frac{1}{2} )$

$(3a - \frac{1}{2}) \times (3a -\frac{1}{2} ) = ((3a - \frac{1}{2}))^{2}$
We use identity II for this which is
$(a-b)^{2} = a^{2} -2ab + b^{2}$
in this a = 3a and b = -1/2
$(3a-\frac{1}{2})^{2} = (3a)^{2} -2(3a)(\frac{1}{2}) + (\frac{1}{2})^{2}$
= $9a^{2} -3a + \frac{1}{4}$

$(1.1m - 4)(1.1m+4)$

$(1.1m - 4)(1.1m+4)$
We use identity III for this which is
(a - b)(a + b) =  $a^{2} - b^{2}$
In this a = 1.1m  and b = 4
$(1.1m - 4)(1.1m+4)$ =  $(1.1m)^{2} - (4)^{2}$
=  1.21$m^{2}$ - 16

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

take the (-)ve sign common so our question becomes
-$-(a^{2}+b^{2})(a^{2}-b^{2})$
We use identity III for this which is
(a - b)(a + b) =  $a^{2} - b^{2}$
In this a = $a^{2}$ and b = $b^{2}$

$-(a^{2}+b^{2})(a^{2}-b^{2})$ = $-((a^{2})^{2} -(b^{2})^{2}) = -a^{4} + b^{4}$

Question:1(vii)  Use a suitable identity to get each of the following.

$(6x-7) (6x+7)$

(6x -7) X (6x - 7) = $(6x-7)^{2}$
We use identity III for this which is
(a - b)(a + b) =  $a^{2} - b^{2}$
In this a = 6x and b = 7
(6x -7) X (6x - 7) = $(6x)^{2} - (7)^{2} = 36x^{2} - 49$

Question:1(viii) Use a suitable identity to get each of the following product.

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

take (-)ve sign common from both the brackets So, our question become
(a -c) X (a -c) =$(a -c)^{2}$
We use identity II for this which is
$(a-b)^{2} =a^{2} -2ab + b^{2}$
In this a = a and b = c
$(a-c)^{2} =a^{2} -2ac + c^{2}$

$(\frac{x}{2}+ \frac{3y}{4})(\frac{x}{2}+ \frac{3y}{4})$

$(\frac{x}{2}+ \frac{3y}{4}) \times (\frac{x}{2}+ \frac{3y}{4}) = (\frac{x}{2}+ \frac{3y}{4})^{2}$

We use identity I for this which is
$(a+b)^{2} =a^{2}+2ab + b^{2}$
In this a = $\frac{x}{2}$ and b = $\frac{3y}{4}$

$(\frac{x}{2}+ \frac{3y}{4})^{2} = (\frac{x}{2})^{2} + 2 (\frac{x}{2})(\frac{3y}{4}) + (\frac{3y}{4})^{2}$
=  $\frac{x^{2}}{4} + \frac{3xy}{4} + \frac{9y^{2}}{16}$

$(7a-9b)(7a-9b)$

$(7a-9b) \times (7a-9b) = (7a-9b)^{2}$

We use identity II for this which is
$(a-b)^{2} =a^{2}-2ab + b^{2}$
In this a = 7a and b = 9b
$(7a-9b)^{2} =(7a)^{2}-2(7a)(9b) + (9b)^{2}$
=  $49a^{2}-126ab + 81b^{2}$

$(x+3)(x+7)$

We use identity $(x+a) (x+b) = x^2+(a+b)x+ab$
in this a = 3 and b = 7
$(x+3)(x+7)$ = $x^2+(3+7)x+3 \times 7$
= $x^2+10x+ 21$

$(4x+5)(4x+1)$

We use identity  $(x+a)(x+b)=x^2+(a+b)x+ab$
In this a= 5 , b = 1 and x = 4x
$(4x+5)(4x+1)$   =  $(4x)^2+(5+1)4x+(5)(1)$
= $16x^2+24x+5$

$(4x-5)(4x-1)$

We use  identity   $(x+a)(x+b)= x^2+(a+b)x+ab$
in this x = 4x , a = -5 and b = -1
$(4x-5)(4x-1)$   = $(4x)^2+(-5-1)4x+(-5)(-1)$
=  $16x^2 - 24x+ 5$

$(4x+5)(4x-1)$

We use identity  $(x+a)(x+b)=x^2+(a+b)x+ab$
In this a = 5 , b = -1 and x = 4x
$(4x+5)(4x-1)$ =  $(4x)^2+(5+(-1))4x+(5)(-1)$
= $16x^2+16x- 5$

$(2x+5y)(2x+3y)$

We use identity  $(x+a)(x+b)=x^2+(a+b)x+ab$
In this a = 5y , b = 3y and x = 2x
$(2x+5y)(2x+3y)$ =  $(2x)^2+(5y+3y)(2x)+(5y)(3y)$
= $4x^2+16xy + 15y^{2}$

$(2a^2+9)(2a^2+5)$

We use identity  $(x+a)(x+b)=x^2+(a+b)x+ab$
In this a = 9 , b = 5 and x = $2a^{2}$
$(2a^{2}+9)(2a^{2}+5)$ =  $(2a^{2})^2+(9+5)2a^{2}+(9)(5)$
= $4a^{4} + 28a^{2} + 45$

$(xyz-4) (xyz-2)$

We use identity  $(x+a)(x+b)=x^2+(a+b)x+ab$
In this a = -4 , b = -2 and x = xyz
$(xyz-4)(xyz-2)$ =  $(xyz)^2+((-4)+(-2))xyz+(-4)(-2)$
= $x^{2}y^{2}z^{2} -6xyz + 8$

Question:3(i) Find the following squares by using the identities.

$(b-7)^2$

We use identity
$(a-b)^{2} = a^{2} - 2ab + b^{2}$
In this a =b and b = 7
$(b-7)^{2} = b^{2} - 2(b)(7) + 7^{2}$
= $b^{2} - 14b + 49$

Question:3(ii) Find the following squares by using the identities.

$(xy+3z)^2$

We use
$(a+b)^{2} = a^{2} + 2ab + b^{2}$

In this a = xy and b = 3z
$(xy+3z)^{2} = (xy)^{2} + 2(xy)(3z) + (3z)^{2}$
=  $x^{2}y^{2} + 6xyz+ 9z^{2}$

Question:3(iii)Find the following squares by using the identities.

$(6x^2-5y)^2$

We use
$(a-b)^{2} = a^{2} - 2ab + b^{2}$
In this a = $6x^{2}$ and b = $5y^{2}$
$(6x-5y)^{2} = (6x)^{2} - 2(6x)(5y) + (5y)^{2}$
= $36x^{2} - 60xy + 25y^{2}$

Question:3(iv)  Find the following squares by using the identities.

$(\frac{2}{3}m+\frac{3}{2}n)^2$

we use the identity
$(a+b)^{2} = a^{2} + 2ab + b^{2}$
In this a = $\frac{2m}{3}$  and b =$\frac{3n}{2}$
$(\frac{2m}{3} + \frac{3n}{2})^{2} = (\frac{2m}{3})^{2} + 2(\frac{2m}{3})( \frac{3n}{2}) + ( \frac{3n}{2})^{2}$

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

Question:3(v) Find the following squares by using the identities.

$(0.4p-0.5q)^2$

we use
$(a-b)^{2} = a^{2} - 2ab + b^{2}$
In this a = 0.4p  and b =0.5q
$(0.4p-0.5q)^{2} = (0.4p)^{2} - 2(0.4p)(0.5q) + (0.5q)^{2}$
=  $0.16p^{2} - 0.4pq + 0.25q^{2}$

Question:3(vi) Find the following squares by using the identities.

$(2xy+5y)^2$

we use the identity
$(a+b)^{2} = a^{2} + 2ab + b^{2}$
In this a = 2xy  and b =5y
$(2xy+5y)^{2} = (2xy)^{2} + 2(2xy)(5y) + (5y)^{2}$
=  $4x^{2}y^{2} + 20xy^{2} + 25y^{2}$

Question:4(i) Simplify:

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

we use
$(a-b)^{2} = a^{2} - 2ab + b^{2}$
In this a = $a^{2}$ and b = $b^{2}$
$(a^{2}-b^{2})^{2} = (a^{2})^{2} - 2(a^{2})(b^{2}) + (b^{2})^{2}$
=$a^{4} - 2a^{2}b^{2} + b^{4}$

Question:4(ii) Simplify.

$(2x+5)^2-(2x-5)^2$

we use
$a^{2} - b^{2} = (a-b)(a+b)$
In this a = (2x + 5) and b = (2x - 5)
$(2x + 5)^{2} - (2x - 5)^{2} = ( (2x + 5)- (2x - 5))( (2x + 5)+ (2x - 5))$
=  $( 2x + 5- 2x + 5)( 2x + 5+ 2x - 5)$
= (4x)(10)
=40x

or

remember that

$(a+b)^2-(a-b)^2=4ab$

here a= 2x, b= 5

$4ab=4\times 2x \times 5=40x$

Question:4(iii)  Simplify.

$(7m-8n)^2+(7m+8n)^2$

we use
$(a-b)^{2} = a^{2} -2ab + b^{2}$       and     $(a+b)^{2} = a^{2} +2ab + b^{2}$
In this a = 7m and b = 8n
$(7m-8n)^{2} = (7m)^{2} -2(7m)(8n) + (8n)^{2}$
=$49m^{2} -112mn + 64n^{2}$
and
$(7m+8n)^{2} = (7m)^{2} +2(7m)(8n) + (8n)^{2}$
=$49m^{2} +112mn + 64n^{2}$

So,  $(7m - 8n)^{2} + (7m + 8n)^{2}$   =   $49m^{2} -112mn + 64n^{2}$  +  $49m^{2} +112mn + 64n^{2}$
=$2(49m^{2} + 64n^{2})$

or

remember that

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

Question: 4(iv) Simplify.

$(4m+5n)^2+(5m+4n)^2$

we use
$(a+b)^{2} = a^{2} +2ab + b^{2}$
1 )   In this a = 4m and b = 5n

$(4m+5n)^{2} = (4m)^{2} +2(4m)(5n) + (5n)^{2}$
= $16m^{2} +40mn + 25n^{2}$
2 )      in this  a = 5m and b = 4n
$(5m+4n)^{2} = (5m)^{2} +2(5m)(4n) + (4n)^{2}$
=    $25m^{2} +40mn + 16n^{2}$

So,      $(4m + 5n)^{2} + (5m + 4n)^{2}$  =   $16m^{2} +40mn + 25n^{2}$  +  $25m^{2} +40mn + 16n^{2}$
=   $41m^{2} +80mn + 41n^{2}$

Question: 4(v) Simplify.

$(2.5p-1.5q)^2-(1.5p-2.5q)^2$

we use
$a^{2}- b^{2} = (a-b)(a+b)$
1 )   In this a = (2.5p- 1.5q) and b = (1.5p - 2.5q)
$(2.5p- 1.5q)^{2}- (1.5p- 2.5q)^{2} = ( (2.5p- 1.5q)- (1.5p- 2.5q))( (2.5p- 1.5q)+ (1.5p- 2.5q))$
= $( 2.5p- 1.5q- 1.5p + 2.5q)(2.5p- 1.5q+ 1.5p- 2.5q)$
= 4(p + q ) (p - q)
= 4$(p^{2} - q^{2})$

Question:4(vi) Simplify.

$(ab+bc)^2-2ab^2c$

We use identity
$(a+b)^{2} = a^{2} + 2ab + b^{2}$
In this a =  ab and b = bc
$(ab+bc)^{2} = (ab)^{2} + 2(ab)(bc) + (bc)^{2}$
= $a^{2}b^{2} + 2ab^{2}c + b^{2}c^{2}$
Now,    $a^{2}b^{2} + 2ab^{2}c + b^{2}c^{2}$ - $2ab^{2}c$
=  $a^{2}b^{2} + b^{2}c^{2}$

Question:4(vii)  Simplify.

$(m^2 -n^2m)^2+2m^3n^2$

We use identity
$(a-b)^{2} = a^{2} - 2ab + b^{2}$
In this a =  $m^{2}$ and b = $n^{2}m$
$(m^{2}-n^{2}m)^{2} = (m^{2})^{2} - 2(m^{2})(n^{2}m) + (n^{2}m)^{2}$
=  $m^{4} - 2m^{3}n^{2} + n^{4}m^{2}$
Now,    $m^{4} - 2m^{3}n^{2} + n^{4}m^{2}$  + $2m^{3}n^{2}$
=  $m^{4} + n^{4}m^{2}$

Question:5(i) Show that

$(3x+7)^2-84x=(3x-7)^2$

L.H.S. =     $(3x+7)^2 - 84x = 9x^2 + 42x + 49 - 84x$

$= 9x^2 - 42 x +49$

$= (3x - 7)^2$

=  R.H.S.

Hence it is prooved

Question:5(ii) Show that

$(9p-5q)^2+180pq=(9p+5q)^2$

L.H.S. =  $(9p-5q)^2+180pq = 81p^2 - 90pq + 25q^2 + 180pq$                (Using $(a-b)^2 = a^2 - 2ab + b^2$)

$= 81p^2 +90pq + 25q^2$

$= (9p + 5q)^2$                                                      $\left ( (a+b)^2 = a^2 + 2ab + b^2 \right )$

= R.H.S.

Question:5(iii) Show that.

$(\frac{4}{3}m-\frac{3}{4}n)^2 +2mn=\frac{16}{9}m^2+\frac{9}{16}n^2$

First we will solve the LHS :

$= (\frac{4}{3}m-\frac{3}{4}n)^2 +2mn = \frac{16}{9}m^2 - 2mn + \frac{9}{16}n^2 + 2mn$

or                $= \frac{16}{9}m^2 + \frac{9}{16}n^2$

=  RHS

Question:5(iv)  Show that.

$(4pq+3q)^2-(4pq-3q)^2=48pq^2$

Opening both brackets we get,

$(4pq+3q)^2-(4pq-3q)^2 = 16p^2q^2 + 24pq^2 + 9q^2 - (16p^2q^2 - 24pq^2 + 9q^2)$

$= 16p^2q^2 + 24pq^2 + 9q^2 - 16p^2q^2 + 24pq^2 - 9q^2)$

$= 48pq^2$

= R.H.S.

Question:5(v) Show that

$(a-b)(a+b) + (b-c) ( b +c)+(c-a)(c+a)=0$

Opening all brackets from the LHS, we get :

$(a-b)(a+b) + (b-c) ( b +c)+(c-a)(c+a)\\\\ =\ a^2 +ab - ab- b^2 + b^2+bc - bc -c^2 + c^2 +ca - ac -a^2$

$= 0$              =  RHS

Question:6(i) Using identities, evaluate.

$71^2$

We will use the identity:

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

So,             $71^2 = (70 +1)^2 = 70^2 + 2(70)(1) + 1^2$

$= 4900 + 140 + 1$

$= 5041$

Question:6(ii)  Using identities, evaluate.

$99^2$

Here we will use the identity :

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

So :                         $99^2 = (100 - 1) ^2 = 100^2 - 2(100)(1) + 1^2$

or                                                                 $= 10000 - 200 + 1$

$= 9801$

Question:6(iii) Using identities, evaluate.

$102^2$

Here we will use the identity :

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

So :

$102^2 = (100 + 2)^2 = 100^2 + 2(100)(2) + 2^2$

or                                                                                               $= 10000 + 400 + 4$

$= 10404$

Question:6(iv) Using identities, evaluate.

$998^2$

Here we will the identity :

$998^2=(1000 - 2)^2 = 1000^2 - 2(1000)(2) + 2^2$

or                                                                                              $= 1000000 - 4000+ 4$

or                                                                                              $= 996004$

Question:6(v) Using identities, evaluate.

$5.2^2$

Here we will use :

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

Thus

$(5.2)^2 = (5.0 + 0.2)^2 = 5^2 + 2(5)(0.2) + (0.2)^2$

or                                                                                      $= 25 + 2 + 0.04$

$= 27.04$

Question:6(vi) Using identities, evaluate.

$297 \times 303$

$297\times303 = (300-3)\times(300+3)$
using                                     $(a-b)(a+b)=a^2-b^2$