NCERT Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities class 8th maths. Here We learn what is in ncert maths class 8 chapter 9 Algebraic Expressions and Identities and how to solve questions with easiest method.  In this chapter we solve the question of NCERT 8th class maths chapter 9. NCERT class 8 maths solutions Algebraic Expressions and Identities are part of NCERT Solutions for class 8 maths chapter 9 solution PDF. Ncert solutions for class 8 chapter 9 Algebraic Expressions and Identities with formula and solution. class 8th maths

NCERT Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities class 8 maths Here we solve class 8th maths ncert solutions chapter 9 Algebraic Expressions and Identities concepts all questions with easy method with expert solutions. It help students in their study, home work and preparing for exam. Soon we provide ncert solutions for class 8 maths chapter 9 Algebraic Expressions and Identities question and answers. Soon we provided ncert class 8 class 8th maths chapter 9 Algebraic Expressions and Identities in free PDF here. ncert solutions for class 8 maths chapter 9 pdf will be provide soon. 8th class maths chapter 9 NCERT Solution and ncert solutions for class 8 maths chapter 9 pdf download book PDF.

NCERT Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities class 8th maths

NCERT Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities
class 8th maths
Ex 9.1

Exercise 9.1

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

(i) 5xyz2 − 3zy	(ii) 1 + x + x2	(iii) 4x2y2 − 4x2y2z2 + z2
(iv) 3 − pq + qr − rp	(v) \(\displaystyle \frac{x}{2}+\frac{y}{2}-xy\)	(vi) 0.3a − 0.6ab + 0.5b

Solutions :

		Terms	Coefficients
(i)	5xyz2 − 3zy	There are two terms in this expressions. 5xyz2 and 3zy	5 and − 3
(ii)	1 + x + x2	There are two terms in this expressions. 1, x and x2	1, 1 and 1
(iii)	4x2y2 − 4x2y2z2 + z2	There are three terms in this expressions. 4x2y2 , − 4x2y2z2 and z2	4, − 4 and 1
(vi)	3 − pq + qr – rp	There are four terms in this expressions. 3, − pq, qr and − rp	3, − 1, 1 and − 1
(v)	\(\displaystyle \frac{x}{2}+\frac{y}{2}-xy\)	There are three terms in this expressions. \(\displaystyle \frac{x}{2}\) , \(\displaystyle \frac{y}{2}\) and − xy	\(\displaystyle \frac{1}{2}\) , \(\displaystyle \frac{1}{2}\) and − 1
(vi)	0.3a − 0.6ab + 0.5b	There are three terms in this expressions. 0.3, − 0.6ab and 0.5b	0.3, − 0.6 and 0.5

2. Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?
x + y, 1000, x + x² + x³, 7 + y + 5x, 2y − 3y², 2y − 3y² + 4y3, 5x − 4y + 3xy, 4z − 15z², ab + bc + cd + da, pqr, p²q, 2p + 2q
Solutions :
Monomials :- 1000, pqr
Binomials :- x + y, 2y − 3y², 4z − 15z², p²q + pq², 2p + 2q
Trinomials :- 7 + y + 5x, 2y − 3y² + 4y³, 5x − 4y + 3xy
These polynomials do not fit in any of these three categories :- x + x² + x³ + x⁴, ab + bc + cd + da

3. Add the following :

(i) ab − bc, bc − ca, ca − ab	(ii) a − b + ab, b − c + bc, c − a + ac
(iii) 2p2q2 − 3pq + 4, 5 + 7pq − 3p2q2	(iv) l2 + m2, m2 + n2, n2 + l2, 2lm + 2mn + 2nl

Solutions :
(i) ab − bc, bc − ca, ca − ab
on writing similar terms together –
= ab − bc + bc − ca + ca − ab
= ab − ab + bc − bc + ca − ca
= 0

(ii) a − b + ab, b − c + bc, c − a + ac
= a − b + ab + b − c + bc + c − a + ac
on writing similar terms together –
= a − a + b − b + c − c + ab + bc + ac
by adding and substracting same terms –
= ab + bc + ac

(iii) 2p²q² − 3pq + 4, 5 + 7pq − 3p²q²
= 2p²q² − 3pq + 4 + 5 + 7pq − 3p²q²
on writing similar terms together –
= 2p²q² − 3p²q² + 7pq − 3pq + 4 + 5
= − p²q² + 4pq + 9

(iv) l² + m², m² + n², n² + l², 2lm + 2mn + 2nl
= l² + m² + m² + n² + n² + l² + 2lm + 2mn + 2nl
on writing similar terms together –
= l² + l² + m² + m² + n² + n² + 2lm + 2mn + 2nl
= 2l² + 2m² + 2n² + 2lm + 2mn + 2nl
= 2(l² + m² + n² + lm + mn + nl)

4. (a) Subtract 4a − 7ab + 3b + 12 from 12a − 9ab + 5b − 3
(b) Subtract 3xy + 5yz − 7zx from 5xy − 2yz − 2zx + 10xyz
(c) Subtract 4p²q − 3pq + 5pq² − 8p + 7q − 10 from 18 − 3p − 11p + 5pq − 2pq² + 5p²q
Solutions :
(a) Subtract 4a − 7ab + 3b + 12 from 12a − 9ab + 5b − 3
= 12a − 9ab + 5b − 3 − (4a − 7ab + 3b + 12)
remove the bracket –
= 12a − 9ab + 5b − 3 − 4a + 7ab − 3b − 12  (If there is a negative sign outside the bracket, the symbols of all the sign inside will change.)
on writing similar terms together –
= 12a − 4a + 5b − 3b + 7ab − 9ab − 4 − 12
= 8a + 2b − 2ab − 16

(b) Subtract 3xy + 5yz − 7zx from 5xy − 2yz − 2zx + 10xyz
= 5xy − 2yz − 2zx + 10xyz − (3xy + 5yz − 7zx)
remove the bracket –
= 5xy − 2yz − 2zx + 10xyz − 3xy − 5yz + 7zx (If there is a negative sign outside the bracket, the symbols of all the sign inside will change.)
on writing similar terms together –
= 5xy − 3xy − 2yz − 5yz + 7zx − 2zx + 10xyz
= 2xy − 7yz + 5zx + 10xyz

(c) Subtract 4p²q − 3pq + 5pq² − 8p + 7q − 10 from 18 − 3p − 11p + 5pq − 2pq² + 5p²q
= 18 − 3p − 11p + 5pq − 2pq² + 5p²q − (4p²q − 3pq + 5pq² − 8p + 7q − 10)
remove the bracket –
= 18 − 3p − 11q + 5pq − 2pq² + 5p²q − 4p²q + 3pq − 5pq² + 8p − 7q + 10)  (If there is a negative sign outside the bracket, the symbols of all the sign inside will change.)
on writing similar terms together –
= − 3p + 8p − 11q − 7q + 5pq + 3pq − 2pq² − 5pq² + 5p²q − 4p²q
= 5p − 18q + 8pq − 7pq² + p²q

NCERT Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities
class 8th maths
Ex 9.2

Exercise 9.2

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

(i) 4, 7p	(ii) − 4p, 7p	(iii) − 4p, 7pq	(iv) 4p3, − 3p
(v) 4p, 0

Solutions : 
(i) 4, 7p
= 4 × 7p
= 28p

(ii)  − 4p, 7p
= − 4p × 7p
= − 28p²

(iii) − 4p, 7pq
= − 4p × 7pq
= − 28p²q

(iv) 4p³, − 3p
= 4p³ × − 3p
= − 12p⁴

(v) 4p, 0
= 4p × 0
= 0

2. Find the areas of rectangles with the following pairs of monomials as their lengths and breadths respectively.  
(p, q); (10m, 5n); (20x², 5y²); (4x, 3x²); (3mn, 4np)
Solutions :
(p, q)
length = p
breadth = q
area of rectangle = length × breadth
= p × q
= pq unit square

(10m, 5n)
length = 10m
breadth = 5n
area of rectangle = length × breadth
= 10m × 5n
= 50mn unit square

(20x², 5y²)
length = 20x²
breadth = 5y²
area of rectangle = length × breadth
= 20x² × 5y²
= 100x²y² unit square

(4x, 3x²)
length = 4x
breadth = 3x²
area of rectangle = length × breadth
= 4x × 3x²
= 12x³ unit square

(3mn, 4np)
length = 3mn
breadth = 4np
area of rectangle = length × breadth
= 3mn × 4np
= 12mn²p unit square

3. Complete the table of products.

First monomial → Second monomial ↓	2x	− 5y	3x2	− 4xy	7x2y	− 9x2y2
2x	4x2
− 5y			− 15x2y
3x2
− 4xy
7x2y
− 9x2y2

Solutions : 

First monomial → Second monomial ↓	2x	− 5y	3x2	− 4xy	7x2y	− 9x2y2
2x	4x2	− 10xy	6×3	− 8x2y	14x3y	− 18x3y2
− 5y	− 10xy	25y2	− 15x2y	20xy2	− 35x2y2	45x2y3
3x2	6x3	− 15x2y	9x4	− 12x3y	21x4y	− 27x4y2
− 4xy	− 8x2y	20xy2	− 12x3y	16x2y2	− 28x3y2	36x3y3
7x2y	14x3y	− 35x2y2	21x4y	− 28x3y2	49x4y2	− 63x4y3
− 9x2y2	− 18x3y2	45x2y2	− 27x4y2	36x3y3	− 63x4y3	81x4y4

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

(i) 5a, 3a2, 7a4

(ii) 2p, 4q, 8r

(iii) xy, 2x2y, 2xy2

(iv) a, 2b, 3c

Solutions : 
(i) 5a, 3a², 7a⁴
volume of rectangular box = length × breadth × height
= 5a × 3a² × 7a⁴
= 5 × 3 × 7 × a × a² × a⁴
= 105a⁷

(ii) 2p, 4q, 8r
volume of rectangular box = length × breadth × height
= 2p × 4q × 8r
= 2 × 4 × 8 × p × q × r
= 64pqr

(iii) xy, 2x²y, 2xy²
volume of rectangular box = length × breadth × height
= xy × 2x²y × 2xy²
= 1 × 2 × 2 × xy × x²y × xy²
= 4x⁴y⁴

(iv) a, 2b, 3c
volume of rectangular box = length × breadth × height
= a × 2b × 3c
= 1 × 2 × 3 × a × b × c
= 6abc

5. Obtain the product of

(i) xy, yz, zx	(ii) a, − a2, a3	(iii) 2, 4y, 8y2, 16y3
(iv) a, 2b, 3c, 6abc	(v) m, − mn, mnp

Solutions :
(i) xy, yz, zx
= xy × yz × zx 
= x²y²z²

(ii) a, − a², a³ 
= a × (− a²) × a³ 
= − a⁶

(iii) 2, 4y, 8y², 16y³
= 2 × 4y × 8y² × 16y³
= 2 × 4 × 8 × 16 × y × y² × y³
= 1024y³

(iv) a, 2b, 3c, 6abc
= 2 × 3 × a × b × c × abc
= 6a²b²c²

(v) m, − mn, mnp
= m × (− mn) × mnp
= − m²n²p

NCERT Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities
class 8th maths
Ex 9.3

Exercise 9.3

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

(i) 4p, q + r	(ii) ab, a − b	(iii) a + b, 7a2b2	(iv) a2 − 9, 4a
(v) pq + qr + rp, 0

Solutions :
(i) 4p, q + r 
= 4p × (q + r)
= 4p × q + 4p × r
= 4pq + 4pr

2. Complete the table.

	First expression	Second expression	Product
(i)	a	b + c + d	____
(ii)	x + y − 5	5xy	____
(iii)	p	6p2 − 7p + 5	____
(iv)	4p2q2	p2 − q2	____
(v)	a + b + c	abc	____

Solutions :

	First expression	Second expression	Product
(i)	a	b + c + d	ab + ac + ad
(ii)	x + y − 5	5xy	5x2y + 5xy2 − 25xy
(iii)	p	6p2 − 7p + 5	6p3 − 7p2 + 5p
(iv)	4p2q2	p2 − q2	4p4q2 − 4p2q4
(v)	a + b + c	abc	a2bc + ab2c + abc2

(i) a × (b + c + d)
= ab + ac + ad

(ii) (x + y − 5) × 5xy
= 5x²y + 5xy² − 25xy

(iii) p × (6p² − 7p + 5)
= 6p³ − 7p² + 5p

(iv) 4p²q² × (p² − q²)
= 4p⁴q² − 4p²q⁴

(v) (a + b + c) × abc
= a²bc + ab²c + abc2

3. Find the product.

(i) (a2) × (2a22) × (4a26)	(ii) \(\displaystyle \left( {\frac{2}{3}xy} \right)\times \left( {\frac{{-9}}{{10}}{{x}^{2}}{{y}^{2}}} \right)\)
(iii) \(\displaystyle \left( {-\frac{{10}}{3}p{{q}^{3}}} \right)\times \left( {\frac{6}{5}{{p}^{3}}q} \right)\)	(iv) x × x2 × x3 × x4

Solutions :
(i) (a²) × (2a²²) × (4a²⁶)
= 2 × 4 × a² × a²² × a²⁶
= 8a⁵⁰

(ii) \(\displaystyle \left( {\frac{2}{3}xy} \right)\times \left( {\frac{{-9}}{{10}}{{x}^{2}}{{y}^{2}}} \right)\)

\(\displaystyle \begin{array}{l}=\frac{2}{3}\times \left( {\frac{{-9}}{{10}}} \right)\left( {xy\times {{x}^{2}}{{y}^{2}}} \right)\\=\frac{{-18}}{{30}}{{x}^{3}}{{y}^{3}}\\=\frac{{-3}}{5}{{x}^{3}}{{y}^{3}}\end{array}\)

(iii) \(\displaystyle \left( {-\frac{{10}}{3}p{{q}^{3}}} \right)\times \left( {\frac{6}{5}{{p}^{3}}q} \right)\)
\(\displaystyle \begin{array}{l}=\left( {-\frac{{10}}{3}} \right)\times \frac{6}{5}\times p{{q}^{3}}\times {{p}^{3}}q\\=-\frac{{60}}{{15}}{{p}^{4}}{{q}^{4}}\\=-4{{p}^{4}}{{q}^{4}}\end{array}\)

(iv) x × x² × x³ × x⁴
= x¹⁰

4. (a) Simplify 3x (4x – 5) + 3 and find its values for (i) x = 3 (ii) \(\displaystyle x=\frac{1}{2}\) 
(b) Simplify a (a² + a + 1) + 5 and find its value for (i) a = 0, (ii) a = 1
(iii) a = – 1.
Solutions :
(a) (i) 3x (4x − 5) + 3
= 3x × 4x − 3x × 5 + 3
= 12x² − 15x + 3 ………………………….. (i)
putting x = 3 equ. in (i)
= 12 × (3)² − 9 × 5 + 3
= 12 × 9 − 45 + 3
= 108 + 3 − 45
= 111 − 45
= 66

(ii) Putting \(\displaystyle x=\frac{1}{2}\) in equ (i) –
\(\displaystyle \begin{array}{l}=12\times {{\left( {\frac{1}{2}} \right)}^{2}}-15\times \frac{1}{2}+3\\=12\times \frac{1}{4}-\frac{{15}}{2}+3\\=3-\frac{{15}}{2}+3\\=\frac{{6-15+6}}{2}\\=\frac{{12-15}}{2}\\=\frac{{-3}}{2}\end{array}\)

5. (a) Add: p (p − q), q (q − r) and r (r − p) 
(b) Add: 2x (z − x − y) and 2y (z − y − x) 
(c) Subtract: 3l (l − 4m + 5n) from 4l (10n − 3m + 2l ) 
(d) Subtract: 3a (a + b + c ) − 2b (a − b + c) from 4c (− a + b + c ) 
Solutions : 
(a) Add: p (p − q), q (q − r) and r (r − p) 
= p (p − q) + q (q − r) + r (r − p)
= p² − pq + q² − qr + r² − rp
= p² + q² + r² − pq − qr − rp

(b) Add: 2x (z − x − y) and 2y (z − y − x) 
= 2x (z − x − y) + 2y (z − y − x)
= 2xz − 2x² − 2xy + 2yz − 2y² − 2xy
= − 2x² − 2y² − 2xy − 2xy + 2yz + 2xz
= − 2x² − 2y² − 4xy + 2yz + 2xz

(c) Subtract: 3l (l − 4m + 5n) from 4l (10n − 3m + 2l ) 
= 4l (10n − 3m + 2l) − 3l (l − 4m + 5n)
= 4l × 10n − 4l × 3m + 4l + 2l − 3l × l − 3l × (− 4m) − 3l × 5n
= 40ln − 12lm + 8l² − 3l² + 12lm − 15ln
= 40ln − 15ln − 12lm + 12lm + 8l² − 3l²
= 25ln − 0lm + 5l²
= 5l² + 25ln

(d) Subtract: 3a (a + b + c ) − 2b (a − b + c) from 4c (− a + b + c ) 
= 4c (− a + b + c) − [3a (a + b + c) − 2b (a − b + c)]
= − 4ac + 4bc + 4c² − [3a² + 3ab + 3ac − 2ab + 2a² – 2bc]
= − 4ac + 4bc + 4c² − 3a² − 3ab − 3ac + 2ab − 2b² + 2bc
= − 4ac − 3ac + 4bc + 2bc − 3ab + 2ab + 4c² − 3a² − 2b²
= − 7ac + 6bc − ab + 4c² − 3a² − 2b²
=  − 3a² − 2b² + 4c² − ab + 6bc − 7ac

NCERT Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities
class 8th maths
Ex 9.4

Exercise 9.4

1. Multiply the binomials.

(i) (2x + 5) and (4x − 3)	(ii) (y − 8) and (3y − 4)
(iii) (2.5l − 0.5m) and (2.5l + 0.5m)	(iv) (a + 3b) and (x + 5)
(v) (2pq + 3q2) and (3pq − 2q2)	(vi) \(\displaystyle \left( {\frac{3}{4}{{a}^{2}}+3{{b}^{2}}} \right)\) and \(\displaystyle \left( {{{a}^{2}}-\frac{2}{3}{{b}^{2}}} \right)\)

Solutions :
(i) (2x + 5) and (4x − 3)
= 2x (4x − 3) + 5 (4x − 3)
= 2x × 4x − 2x × 3 + 5 × 4x − 5 × 3
= 8x² − 6x + 20x − 15
= 8x² + 14x − 15

(ii) (y – 8) and (3y − 4)
= y (3y − 4) − 8 (3y − 4)
= y × 3y − y × 4 − 8 × 3y − 8 × (− 4)
= 3y² − 4y − 24y + 32
= 3y² − 28y + 32

(iii) (2.5l − 0.5m) and (2.5l + 0.5m)
= 2.5l (2.5l + 0.5m) − 0.5m (2.5l + 0.5m)
= 2.5l × 2.5l + 2.5l × 0.5m – 0.5m × 2.5l − 0.5m × 0.5m
= 2.25l² + 1.25lm − 1.25lm − 0.25m²
= 2.25l² − 0.25m²

(iv) (a + 3b) and (x + 5)
= a (x + 5) + 3b (x + 5)
= a × x + a × 5 + 3b × x + 3b × 5
= ax + 5a + 3bx + 15b
= ax + 5a + 3bx + 15b

(v) (2pq + 3q²) and (3pq – 2q²)
= 2pq (3pq − 2q²) + 3q² (3pq − 2q²)
= 2pq × 3pq − 2q² × 2pq + 3q² × 3pq − 3q² × 2q²
= 2p²q² − 4pq³ + 9pq³ − 6q⁴
= 6p²q² + 5pq³ − 6q⁴

2. Find the product.

(i) (5 − 2x) (3 + x)	(ii) (x + 7y) (7x − y)
(iii) (a2 + b) (a + b2)	(iv) (p2 – q2) (2p + q)

Solutions :
(i) (5 − 2x) (3 + x)
= 5 × (3 + x) − 2x × (3 + x)
= 15 + 5x − 6x − 2x²
= 15 − x − 2x²

(ii) (x + 7y) (7x − y)
= x × (7x − y) + 7y × (7x − y)
= 7x² − xy + 49xy − 7y²
= 7x² + 48xy − 7y²

(iii) (a² + b) (a + b²)
= a² × (a + b²) + b × (a + b²)
= a³ + a²b² + ab + b²

(iv) (p² − q²) (2p + q)
= p² × (2p + q) − q² × (2p + q)
= 2p³ + p²q − 2pq² − q³

3. Simplify.

(i) (x2 − 5) (x + 5) + 25	(ii) (a2 + 5) (b3 + 3) + 5
(iii) (t + s2) (t2 − s)
(iv) (a + b) (c − d) + (a − b) (c + d) + 2 (ac + bd)
(v) (x + y) (2x + y) + (x + 2y) (x − y)	(vi) (x + y) (x2 − xy + y2)
(vii) (1.5x − 4y) (1.5x + 4y + 3) − 4.5x + 12y
(viii) (a + b + c) (a + b − c)

Solutions :
(i) (x² – 5) (x + 5) + 25
= x² × (x + 5) − 5 × (x + 5) + 25
= x³ + 5x² − 5x − 25 + 25
= x³ + 5x² − 5x

(ii) (a² + 5) (b³ + 3) + 5
= a² × (b³ + 3) + 5 × (b³ + 3) + 5
= a²b³ + 3a² + 5b³ + 15 + 5
= a²b³ + 3a² + 5b³ + 20

(iii) (t + s²) (t² − s)
= t × (t² − s) + s² × (t² − s)
= t³ − st + s²t² − s³

(iv) (a + b) (c − d) + (a − b) (c + d) + 2 (ac + bd)
= a × (c − d) + b × (c − d) + a × (c + d) − b × (c + d) + 2ac + 2bd
= ac − ad + bc − bd + ac + ad − bc − bd + 2ac + 2bd
= ac + ac + 2ac + bc − bc − bd − bd + 2bd − ad + ad 
= 2ac + 2ac − 2bd + 2bd
= 4ac

(v) (x + y) (2x + y) + (x + 2y) (x – y)
= x × (2x + y) + y × (2x + y) + x × (x − y) + 2y × (x − y)
= 2x² + xy + 2xy + y² + x² − xy + 2xy − 2y²
= 2x² + x² + xy + 2xy − xy + 2xy + y² − 2y²
= 3x² + 5xy − xy − y²
= 3x² + 4xy − y²

(vii) (1.5x − 4y) (1.5x + 4y + 3) − 4.5x + 12y
= 1.5x × (1.5x + 4y + 3) − 4y × (1.5x + 4y + 3) − 4.5x + 12y
= 2.25x² + 6.0xy + 4.5x – 6.0xy − 16y² − 12y − 4.5x − 12y
= 2.25x² + 6.0xy − 6.0xy + 4.5x − 4.5x − 16y² − 12y − 12y
= 2.25x² − 16y²

(viii) (a + b + c) (a + b − c)
= a × (a + b − c) + b × (a + b – c)  + c × (a + b − c)
= a² + ab − ac + ab + b² − bc + ac + bc − c²
= a² + ab + ab − ac + ac + b² − bc + bc − c²
= a² + 2ab + b² − c²
= a² + b² − c² + 2ab

NCERT Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities
class 8th maths
Ex 9.5

Exercise 9.5

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

(i) (x + 3) (x + 3)	(ii) (2y + 5 ) (2y + 5)
(iii) (2a – 7) (2a – 7)	(iv)
(v) (1.1m – 0.4) (1.1m + 0.4)	(vi) (a2 + b2) (− a2 + b2)
(vii) (6x – 7) (6x + 7)	(viii) (– a + c) (– a + c)
(ix)	(x) (7a – 9b) (7a – 9b)

Solutions :
(i) (x + 3) (x + 3)
= (x + 3)²
[Use the Identity (a + b)² = a² + 2ab + b² ]
= x² + 2 × x × 3 + 3²
= x² + 6x + 9

(ii) (2y + 5 ) (2y + 5)
= (2y + 5)²
[Use the Identity (a + b)² = a² + 2ab + b² ]
= (2y)² + 2 × 2y × 5 + 5²
= 4y² + 20y + 25

(iii) (2a – 7) (2a – 7)
= (2a – 7)²
[Use the Identity (a − b)² = a² − 2ab + b² ]
= (2a)² − 2 × 2a × 7 + 7²
= 4a² − 28a + 49

(v) (1.1m – 0.4) (1.1m + 0.4)
= (1.1m)² − (0.4)²
[Use the Identity (a)² − (b)² = (a + b) (a − b) ]
= (1.1m)² − (0.4)²
= 1.21m² − 0.16

(vi) (a² + b²) (− a² + b²)
= (b² + a²) (b² − a²)
= (b²)² − (a²)²
[Use the Identity (a)² − (b)² = (a + b) (a − b) ]
= b⁴ − a⁴

(vii) (6x – 7) (6x + 7)
= (6x)² − (7)²
[Use the Identity (a)² − (b)² = (a + b) (a − b) ]
= 36x² − 49

(viii) (– a + c) (– a + c)
= (c − a) (c − a)
= (c − a)²
[Use the Identity (a − b)² = a² − 2ab + b² ]
= c² − 2 × c × a + a²
= c² − 2ca + a²

(x) (7a − 9b) (7a – 9b)
= (7a – 9b)²
[Use the Identity (a − b)² = a² − 2ab + b² ]
= (7a)² − 2 × 7a × 9b + (9b)²
= 49a² − 126ab + 81b²

2. Use the identity (x + a) (x + b) = x² + (a + b) x + ab to find the following products.

(i) (x + 3) (x + 7)	(ii) (4x + 5) (4x + 1)
(iii) (4x – 5) (4x − 1)	(iv) (4x + 5) (4x – 1)
(v) (2x + 5y) (2x + 3y)	(vi) (2a2 + 9) (2a2 + 5)
(vii) (xyz – 4) (xyz – 2)

Solutions :
(i) (x + 3) (x + 7)
by identity (x + a) (x + b) = x² + (a + b) x + ab 
(x + 3) (x + 7) = x² + (3 + 7) x + 3 × 7
= x² + 10x + 21

(ii) (4x + 5) (4x + 1)
by identity (x + a) (x + b) = x² + (a + b) x + ab 
(4x + 5) (4x + 1) = (4x)² + (5 + 1) 4x + 5 × 1
= 16x² + 6 × 4x + 5
= 16x² + 24x + 5

(iii) (4x – 5) (4x – 1) = 16x − 24x + 5
by identity (x − a) (x − b) = x² − (a + b) x + ab 
(4x – 5) (4x – 1) = (4x)² − (5 + 1) 4x + 5 × 1
= 16x² − 6 × 4x + 5
= 16x² − 24x + 5

(iv) (4x + 5) (4x − 1) = 16x²  − 4x + 20x − 5
by identity (x + a) (x − b) = x² + (a − b) x − ab 
(4x + 5) (4x – 1) = (4x)² + (5 − 1) 4x − 5 × 1
= 16x² + 4 × 4x − 5
= 16x² + 16x − 5 

(v) (2x + 5y) (2x + 3y)
by identity (x + a) (x + b) = x² + (a + b) x + ab 
(2x + 5y) (2x + 3y) = (2x)² + (5y + 3y) 2x + 5y × 3y
= 4x² + 8y × 2x + 15y²
= 4x² + 16xy + 15y²

(vi) (2a² + 9) (2a² + 5)
by identity (x + a) (x + b) = x² + (a + b) x + ab 
(2a² + 9) (2a² + 5) = (2a²)² + (9 + 5) 2a² + 9 × 5
= 4a⁴ + 14 × 2a² + 45
= 4a⁴ + 28a² + 45

(vii) (xyz – 4) (xyz – 2)
by identity (x − a) (x − b) = x² − (a + b) x + ab 
(xyz – 4) (xyz – 2) = (xyz)² − (4 + 2) xyz + 4 × 2
= x²y²z² − 6xyz + 8

3. Find the following squares by using the identities.

(i) (b − 7)2	(ii) (xy + 3z)2	(iii) (6x2 − 5y)2
(iv)	(v) (0.4p − 0.5q)2	(iv) (2xy + 5y)2

Solutions :
(i) (b − 7)²
by identity (a − b)² = a² − 2ab + b² 
(b − 7)² = (b)² − 2 × b × 7 + (7)²
= b² − 14b + 49

(ii) (xy + 3z)²
by identity (a + b)² = a² + 2ab + b² 
(xy + 3z)² = (xy)² + 2 × xy × 3z + (3z)²
= x²y² + 6xyz + 9z²

(iii) (6x² − 5y)²
by identity (a − b)² = a² − 2ab + b²
(6x² − 5y)² = (6x²)² − 2 × 6x² × 5y + (5y)²
= 36x⁴ – 60x²y + 25y²

(v) (0.4p − 0.5q)²
by identity (a − b)² = a² − 2ab + b² 
(0.4p − 0.5q)² = (0.4p)² − 2 × 0.4p × 0.5q + (0.5q)²
= 0.16p² − 0.4pq + 0.25q²

(iv) (2xy + 5y)²
by identity (a + b)² = a² + 2ab + b² 
(2xy + 5y)² = (2xy)² + 2 × 2xy × 5y + (5y)²
= 4x²y² + 20xy + 25y²

4. Simplify.

(i) (a2 − b2)2	(ii) (2x + 5)2 − (2x − 5)2
(iii) (7m – 8n)2 + (7m + 8n)2	(iv) (4m + 5n)2 + (5m + 4n)2
(v) (2.5p − 1.5q)2 − (1.5p − 2.5q)2
(vi) (ab + bc)2 − 2ab2c	(vii) (m2 − n2m)2 + 2m3n2

Solutions :
(i) (a² − b²)²
by identity (a − b)² = a² − 2ab + b² 
(a² − b²)² = (a²)² − 2 × a² × b² + (b²)²
= a⁴ − 2a²b² + b⁴

(ii) (2x + 5)² − (2x − 5)²
[by identity (a)² − (b)² = (a + b) (a − b) ]
(2x + 5)² − (2x − 5)² = (2x + 5 + 2x − 5) {2x + 5 − (2x − 5)}
= 4x (2x + 5 − 2x + 5)
= 4x × 10
= 40x

(iii) (7m – 8n)² + (7m + 8n)²
by identity (a − b)² = a² − 2ab + b² and (a + b)² = a² + 2ab + b² 
= [(7m)² − 2 × 7m × 8n + (8n)²] + [(7m)² + 2 × 7m × 8n + (8n)²]
= 49m² − 112mn + 64n² + 49m² + 112mn + 64n²
= 49m² + 64n² + 49m² + 64n²
= 98m² + 128n²

(iv) (4m + 5n)² + (5m + 4n)²
by identity (a + b)² = a² + 2ab + b² 
= [(4m)² + 2 × 4m × 5n + (5n)²] + [(5m)² + 2 × 5m × 4n + (4n)²]
= 16m² + 40mn + 25n² + 25m² + 20mn + 16n²
= 16m² + 25m² + 80mn + 25n² + 16n²
= 41m² + 80mn + 16n²

(v) (2.5p − 1.5q)² − (1.5p − 2.5q)²
by identity (a − b)² = a² − 2ab + b² 
= [(2.5p)² − 2 × 2.5p × 1.5q + (1.5q)²] − [(1.5p)² – 2 × 1.5p × 2.5q + (2.5q)²]
= 6.25p² − 7.8pq + 2.25q² − [2.25p² − 7.8pq + 6.25q²]
= 6.25p² − 7.8pq + 2.25q² − 2.25p² + 7.8pq − 6.25q²
= 6.25p² − 2.25p² + 2.25q² − 6.25q²
= 4p² − 4q`²

(vi) (ab + bc)² − 2ab²c
by identity (a + b)² = a² + 2ab + b²
= (ab)² + 2 × ab × bc + (bc)² − 2ab²c
= a²b² + 2ab²c + b²c² − 2ab²c
= a²b² + b²c² + 2ab²c − 2ab²c
= a²b² + b²c²

(vii) (m² − n²m)² + 2m³n²
by identity (a − b)² = a² − 2ab + b²  
= (m²)² − 2 × m² × n²m + (n²m)² + 2m³n²
= m⁴ − 2m³n² + n⁴m² + 2m³n²
= m⁴ + n⁴m²

5. Show that.

(i) (3x + 7)2 − 84x = (3x − 7)2	(ii) (9p − 5q)2 + 180pq = (9p + 5q)2
(iii)
(iv) (4pq + 3q)2 − (4pq − 3q)2 = 48pq2
(v) (a − b) (a + b) + (b − c) (b + c) + (c − a) (c + a) = 0

Solutions :
(i) (3x + 7)² − 84x = (3x − 7)²
L.H.S.
(3x + 7)² − 84x
by identity (a + b)² = a² + 2ab + b²  
= (3x)² + 2 × 3x × 7 + 7² − 84x
= 9x² + 42x + 49 − 84x
= 9x² − 42x + 49
(because a² − 2ab + b² = (a − b)² 
therefor 9x² − 42x + 49 = (3x)² − 2 × 3x × 7 + (7)²
= (3x − 7)²
L.H.S. = R.H.S.

(ii) (9p − 5q)² + 180pq = (9p + 5q)²
L.H.S.
(9p − 5q)² + 180pq
by identity (a − b)² = a² − 2ab + b² 
= (9p)² − 2 × 9p × 5q + (5q)² + 180pq
= 81p² − 90pq + 25q² + 180pq
= 81p² + 90pq + 25q²
(because a² + 2ab + b² = (a + b)² 
therefor 81p² + 90pq + 25q² = (9p)² − 2 × 9p × 5q + (5q)²
= (9p + 5q)²
L.H.S. = R.H.S.

(iv) (4pq + 3q)² − (4pq − 3q)² = 48pq²
L.H.S.
(4pq + 3q)² − (4pq − 3q)²
by identity (a + b)² = a² + 2ab + b² and identity (a − b)² = a² − 2ab + b² से – 
= (4pq)² + 2 × 4pq × 3q + (3q)² − [{4pq)² − 2 × 4pq × 3q + (3q)²]
= 16p²q² + 24pq² + 9q² − [16p²q² − 24pq² + 9q²]
= 16p²q² + 24pq² + 9q² − 16p²q² + 24pq² − 9q²
= 16p²q² − 16p²q² + 9q² − 9q² + 24pq² + 24pq²
= 48pq²
L.H.S. = R.H.S.

(v) (a − b) (a + b) + (b − c) (b + c) + (c − a) (c + a) = 0
L.H.S.
(a − b) (a + b) + (b − c) (b + c) + (c − a) (c + a)
(by identity (a)² − (b)² = (a + b) (a − b) 
= (a² − b²) + (b² − c²) + (c² − a²)
= a² − b² + b² − c² + c² − a²
= a² − a² − b² + b² − c² + c²
= 0
L.H.S. = R.H.S.

6. Using identities, evaluate.

(i) 712	(ii) 992	(iii) 1022	(iv) 9982
(v) 5.22	(vi) 297 × 303	(vii) 78 × 82	(viii) 8.92
(ix) 10.5 × 9.5

Solutions :

(i) 71²
= (70 + 1)2
by identity (a + b)² = a² + 2ab + b²  
(70 + 1)² = (70)² + 2 × 70 × 1 + (1)²
= 4900 + 140 + 1
= 5041

(ii) 99²
= (100 − 1)²
by identity (a − b)² = a² − 2ab + b² 
(100 − 1)² = (100)² − 2 × 100 × 1 + (1)²
= 10000 − 200 + 1
= 10001 − 200
= 9801

(iii) 102²
= (100 + 2)²
by identity (a + b)² = a² + 2ab + b² 
(100 + 2)² = (100)² + 2 × 100 × 2 + (2)²
= 10000 + 400 + 4
= 10404

(iv) 998²
= (1000 − 2)²
by identity (a − b)² = a² − 2ab + b² 
(1000 − 2)² = (1000)² − 2 × (1000) × 2 + (2)²
= 1000000 − 4000 + 4
= 1000004 − 4000
= 996004

(vi) 297 × 303
= (300 − 3) (300 + 3)
by identity (a)² − (b)² = (a + b) (a − b) 
= (300)² − (3)²
= 90000 − 9
= 89991

(vii) 78 × 82
= (80 − 2) (80 + 2)
by identity (a)² − (b)² = (a + b) (a − b) 
= (80)² − (2)²
= 6400 − 4
= 6396

(viii) 8.9²
= (10.0 − 1.1)²
by identity (a − b)² = a² − 2ab + b² 
(10.0 − 1.1)² = (10)² − 2 × 10 × 1.1 + (1.1)²
= 100 − 22 + 1.21
= 100.21 − 22
= 79.21

(ix) 10.5 × 9.5
= (10.0 + 0.5) (10.0 – 0.5)
by identity (a)² − (b)² = (a + b) (a − b) 
= (10)² − (0.5)²
= 100 − 0.25
= 99.75

7. Using a² − b² = (a + b) (a − b), find 

(i) 512 − 492	(ii) (1.02)2 − (0.98)2
(iii) 1532 − 1472	(iv) 12.12 − 7.92

Solutions :
(i) 51² − 49²
by identity a² − b² = (a + b) (a − b) 
51² − 49² = (51 + 49) (51 − 49)
= 100 × 2
= 200

(ii) (1.02)² − (0.98)²
by identity a² − b² = (a + b) (a − b) 
(1.02)² − (0.98)² = (1.02 + 0.98) (1.02 − 0.98)
= 2 × 0.04
= 0.08

(iii) 153² − 147²
by identity a² − b² = (a + b) (a − b) 
153² − 147² = (153 + 147) (153 − 147)
= 300 × 6
= 1800

(iv) 12.1² − 7.9²
by identity a² − b² = (a + b) (a − b) 
12.1² − 7.9² = (12.1 + 7.9) (12.1 − 7.9)
= 20 × 4.2
= 84

8. Using (x + a) (x + b) = x² + (a + b) x + ab, find

(i) 103 × 104	(ii) 5.1 × 5.2
(iii) 103 × 98	(iv) 9.7 × 9.8

Solutions :
(i) 103 × 104
= (100 + 3) (100 + 4)
by identityसर्वसमिका (x + a) (x + b) = x² + (a + b) x + ab 
= (100)² + (3 + 4) 100 + 3 × 4
= 10000 + 700 + 12
= 10712

(ii) 5.1 × 5.2
= (5.0 + 0.1) (5.0 + 0.2)
by identity (x + a) (x + b) = x² + (a + b) x + ab 
= (5.0)² + (0.1 + 0.2) 5.0 + 0.1 × 0.2
= 25 + 1.5 + 0.02
= 26.52

(iii) 103 × 98
= (100 + 3) (100 – 2)
by identity (x + a) (x + b) = x² + (a + b) x + ab 
= (100)² + {3 + (− 2} 100 + 3 × (− 2)
= 10000 + 1 × 100 + (− 6)
= 10000 + 100 − 6
= 10100 − 6
= 10094

(iv) 9.7 × 9.8
= (9.0 + 0.7) (9.0 + 0.8)
by identity (x + a) (x + b) = x² + (a + b) x + ab 
= (9)² + (0.7 + 0.8) 9 + 0.7 × 0.8
= 81 + 1.5 × 9 + 0.56
= 81 + 13.5 + 0.56
= 95.06