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

Contents

### Class 8 Maths Chapter 8 Algebraic Expressions and Identities Exercise 9.1

Q1. Identify the terms, and their coefficients for each of the following expressions.

(i) 5xyz2 – 3zy

(ii) 1 + x + x2

(iii) 4x2y2 – 4x2y2z2 + z2

(iv) 3 – pq + qr – p (v) (x/2) + (y/2) – xy (vi) 0.3a – 0.6ab + 0.5b

Solution :

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

Solution:

Let us 1st outline the classifications of those three polynomials:

Monomials Contain only one term.

Binomials, Contain only two terms.

Trinomials, Contain only three terms.

(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

Solution:

i) (ab – bc) + (bc – ca) + (ca-ab)

= 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

= a – a +b – b +c – c + ab + bc + ca

= 0 + 0 + 0 + ab + bc + ca

= ab + bc + ca

iii) 2p2q2 – 3pq + 4, 5 + 7pq – 3p2q2

= (2p2q2 – 3pq + 4) + (5 + 7pq – 3p2q2)

= 2p2q2 – 3p2q2 – 3pq + 7pq + 4 + 5

= – p2q2 + 4pq + 9

iv)(l2 + m2) + (m2 + n2) + (n2 + l2) + (2lm + 2mn + 2nl)

= l2 + l2 + m2 + m2 + n2 + n2 + 2lm + 2mn + 2nl

= 2l2 + 2m2 + 2n2 + 2lm + 2mn + 2nl

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 – 11q + 5pq – 2pq2 + 5p2q

Solution:

(a) (12a – 9ab + 5b – 3) – (4a – 7ab + 3b + 12)

= 12a – 9ab + 5b – 3 – 4a + 7ab – 3b – 12

= 12a – 4a -9ab + 7ab +5b – 3b -3 -12

= 8a – 2ab + 2b – 15

b) (5xy – 2yz – 2zx + 10xyz) – (3xy + 5yz – 7zx)

= 5xy – 2yz – 2zx + 10xyz – 3xy – 5yz + 7zx

=5xy – 3xy – 2yz – 5yz – 2zx + 7zx + 10xyz

= 2xy – 7yz + 5zx + 10xyz

c) (18 – 3p – 11q + 5pq – 2pq2 + 5p2q) – (4p2q – 3pq + 5pq2 – 8p + 7q – 10)

= 18 – 3p – 11q + 5pq – 2pq2 + 5p2q – 4p2q + 3pq – 5pq2 + 8p – 7q + 10

=18+10 -3p+8p -11q – 7q + 5 pq+ 3pq- 2pq^2 – 5pq^2 + 5 p^2 q – 4p^2 q

= 28 + 5p – 18q + 8pq – 7pq2 + p2q

## Class 8 Maths Chapter 8 Algebraic Expressions and Identities 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

Solution:

(i) 4 , 7 p =  4 × 7 × p = 28p

(ii) – 4p × 7p = (-4 × 7 ) × (p × p )= -28p2

(iii) – 4p × 7pq =(-4 × 7 ) (p × pq) =  -28p2q

(iv) 4p3 × – 3p = (4 × -3 ) (p3 × p ) =  -12p4

(v) 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)

Solution:

Area of rectangle = Length x breadth. So, it is multiplication of two monomials.

Results can be written in square units.

(i) p × q = pq

(ii)10m ×  5n = 50mn

(iii) 20x2 ×  5y2 =  100x2y2

(iv) 4x × 3x2 = 12x3

(v) 3mn ×  4np = 12mn2p

3. Complete the following table of products:

Solution:

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

Solution:

Volume of rectangle = length x  breadth x  height. To evaluate volume of rectangular boxes, multiply all the monomials.

(i) 5a x 3a2 x 7a4 = (5 × 3 × 7) (a × a2 × a4 ) = 105a7

(ii) 2p x 4q x 8r = (2 × 4 × 8 ) (p × q × r ) = 64pqr

(iii) y × 2x2y × 2xy2 =(1 × 2 × 2 )( x × x2 × x × y × y × y2 ) =  4x4y4

(iv) a x  2b x 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

Solution:

(i) xy × yz × zx = xyz2

(ii) a × – a2  × a= – a6

(iii) 2 × 4y × 8y2 × 16y= 1024 y6

(iv) a × 2b × 3c × 6abc = 36abc2

(v) m × – mn × mnp = –mnp

### Class 8 Maths Chapter 8 Algebraic Expressions and Identities 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) a– 9, 4a

(v) pq + qr + rp, 0

Solution:

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

(ii)ab(a – b) = ab – a b2

(iii)(a + b) (7a2b2) = 7a3b2 + 7a2b3

(iv) (a2 – 9)(4a) = 4a3 – 36a

(v) (pq + qr + rp) × 0 = 0 ( Anything multiplied by zero is zero )

2. Complete the table.

Solution:

3. Find the product.

i) a2 x (2a22) x (4a26)

ii) (2/3 xy) ×(-9/10 x2y2)

(iii) (-10/3 pq3/) × (6/5 p3q)

(iv) (x) × (x2) × (x3) × (x4)

Solution:

i) a2 x (2a22) x (4a26)

= (2 × 4) ( a2 × a22 × a26 )

= 8 × a2 + 22 + 26

= 8a50

ii) (2xy/3) ×(-9x2y2/10)

=(2/3 × -9/10 ) ( x × x2 × y × y2 )

= (-3/5 x3y3)

iii) (-10pq3/3) ×(6p3q/5)

= ( -10/3 × 6/5 ) (p × p3× q3 × q)

= (-4p4q4)

iv)  ( x) x (x2) x (x3) x (x4)

= x 1 + 2 + 3 + 4

=  x10

4. (a) Simplify 3x (4x – 5) + 3 and find its values for

(i) x = 3 (ii) x =1/2

(b) Simplify a (a2+ a + 1) + 5 and find its value for

(i) a = 0, (ii) a = 1 (iii) a = – 1.

Solution:

a) 3x (4x – 5) + 3

=3x ( 4x) – 3x( 5) +3

=12x2 – 15x + 3

(i) Putting x=3 in the equation we gets 12x2 – 15x + 3 =12(32) – 15 (3) +3

= 108 – 45 + 3

= 66

(ii) Putting x=1/2 in the equation we get

12x2 – 15x + 3 = 12 (1/2)2 – 15 (1/2) + 3

= 12 (1/4) – 15/2 +3

= 3 – 15/2 + 3

= 6- 15/2

= (12- 15 ) /2

= -3/2

b) a(a+a +1)+5

= a x a2 + a x a + a x 1 + 5 =a3+a2+a+ 5

(i) putting a=0 in the equation we get 03+02+0+5=5

(ii) putting a=1 in the equation we get 1+ 1+ 1+5 = 1 + 1 + 1+5 = 8

(iii) Putting a = -1 in the equation we get (-1)3+(-1)+ (-1)+5 = -1 + 1 – 1+5 = 4

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 – 4 m + 5 n) from 4l ( 10 n – 3 m + 2 l )

(d) Subtract: 3a (a + b + c ) – 2 b (a – b + c)  from 4c ( – a + b + c )

Solution:

a) p ( p – q) + q ( q – r) + r ( r – p)

= (p2 – pq) + (q2 – qr) + (r2 – pr)

= p2 + q2 + r2 – pq – qr – pr

b) 2x (z – x – y) + 2y (z – y – x)

= (2xz – 2x2 – 2xy) + (2yz – 2y2 – 2xy)

= 2xz – 4xy + 2yz – 2x2 – 2y2

c) 4l ( 10 n – 3 m + 2 l ) – 3l (l – 4 m + 5 n)

= (40ln – 12lm + 8l2) – (3l2 – 12lm + 15ln)

= 40ln – 12lm + 8l2 – 3l2 +12lm -15 ln

= 25 ln + 5l2

d) 4c ( – a + b + c ) – (3a (a + b + c ) – 2 b (a – b + c))

= (-4ac + 4bc + 4c2) – (3a2 + 3ab + 3ac – ( 2ab – 2b2 + 2bc ))

=-4ac + 4bc + 4c2 – (3a2 + 3ab + 3ac – 2ab + 2b2 – 2bc)

= -4ac + 4bc + 4c2 – 3a2 – 3ab – 3ac +2ab – 2b2 + 2bc

= -7ac + 6bc + 4c2 – 3a2 – ab – 2b2

### Class 8 Maths Chapter 8 Algebraic Expressions and Identities 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) (3/4 a2 + 3b2) and 4( a2 – 2/3 b2)

Solution :

(i) (2x + 5)(4x – 3)

2x x 4x – 2x x 3 + 5 x 4x – 5 x 3

8x² – 6x + 20x -15

8x² + 14x -15

7

ii) ( y – 8)(3y – 4)

= y x 3y – 4y – 8 x 3y + 32

= 3y2 – 4y – 24y + 32

= 3y2 – 28y + 32

5

(iii) (2.5l – 0.5m)(2.5l + 0.5m)

2.5l x 2.5 l + 2.5l x 0.5m – 0.5m x 2.5l – 0.5m x 0.5m

= 6.25l2 + 1.25 lm – 1.25 lm – 0.25 m2

= 6.25l2– 0.25 m2

iv) (a + 3b) (x + 5)

= ax + 5a + 3bx + 15b

v) (2pq + 3q2(3pq – 2q2)

= 2pq x 3pq – 2pq x 2q2 + 3q2 x 3pq – 3q2 x 2q2

= 6p2q2 – 4pq3 + 9pq3 – 6q4

= 6p2q2 + 5pq3 – 6q4

=9

(vi) (3/4 a² + 3b² ) and 4( a² – 2/3 b² )

=(3/4 a² + 3b² ) x 4( a² – 2/3 b² )

=(3/4 a² + 3b² ) x (4a² – 8/3 b² )

=3/4 a² x (4a² – 8/3 b² ) + 3b² x (4a² – 8/3 b² )

=3/4 a² x 4a² -3/4 a² x 8/3 b² + 3b² x 4a² – 3b² x 8/3 b²

=3a4– 2a² b² + 12 a²  b² – 8b4

= 3a4 + 10a²  b² – 8b4

2. Find the product.

(i) (5 – 2x) (3 + x)

(ii) (x + 7y) (7x – y)

(iii) (a2+ b) (a + b2)

(iv) (p– q2) (2p + q)

Solution:

(i) (5 – 2x) (3 + x)

= 5 (3 + x) – 2x (3 + x)

=15 + 5x – 6x – 2x2

= 15 – x -2 x 2

(ii) (x + 7y) (7x – y)

= x(7x-y) + 7y ( 7x-y)

=7x2 – xy + 49xy – 7y2

= 7x2 – 7y2 + 48xy

iii) (a2+ b) (a + b2)

= a2  (a + b2) + b(a + b2)

= a3 + a2b2 + ab + b3

= a3 + b3 + a2b2 + ab

iv) (p2– q2) (2p + q)

= p(2p + q) – q2 (2p + q)

=2p3 + p2q – 2pq2 – q3

= 2p3 – q3 + p2q – 2pq2

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)

Solution :

i) (x2– 5) (x + 5) + 25

= x3 + 5x2 – 5x – 25 + 25

= x3 + 5x2 – 5x

ii) (a2+ 5) (b3+ 3) + 5

= a2b3 + 3a2 + 5b3 + 15 + 5

= a2b3 + 5b3 + 3a2 + 20

iii) (t + s2)(t2 – s)

t (t2 – s) + s2(t2 – s)

= t– st + s2t– s3

= t3 – s3 – st + s2t2

iv) (a + b) (c – d) + (a – b) (c + d) + 2 (ac + bd)

= (a + b) (c – d) + (a – b) (c + d) + 2 (ac + bd)

=(ac – ad + bc – bd) + (ac + ad – bc – bd) + (2ac + 2bd)

= ac – ad + bc – bd + ac + ad – bc – bd + 2ac + 2bd

= 4ac

v) (x + y)(2x + y) + (x + 2y)(x – y)

= 2x2 + xy + 2xy + y2 + x2 – xy + 2xy – 2y2

= 3x2 + 4xy – y2

vi) (x + y)(x2– xy + y2)

= x3 – x2y + xy2 + x2y – xy2 + y3

= x3 + y3

vii) (1.5x – 4y)(1.5x + 4y + 3) – 4.5x + 12y

= 2.25x2 + 6xy + 4.5x – 6xy – 16y2 – 12y – 4.5x + 12y

= 2.25x2 – 16y2

viii) (a + b + c)(a + b – c)

= a2 + ab – ac + ab + b2 – bc + ac + bc – c2

= a2 + b2 – c2 + 2ab

### Class 8 Maths Chapter 8 Algebraic Expressions and Identities 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) (3a – 1/2)(3a – 1/2)

(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) (1/2x + 3/4y) (1/2x + 3/4y)

(x) (7a – 9b) (7a – 9b)

Solution:

(i) (x + 3) (x + 3)

= (x + 3)2

= x2 + 6x + 9

Using (a+b) 2 = a2 + b2 + 2ab

ii) (2y + 5) (2y + 5)

= (2y + 5)2

= 4y2 + 20y + 25

Using (a+b) 2 = a2 + b2 + 2ab

iii) (2a – 7) (2a – 7)

= (2a – 7)2

= 4a2 – 28a + 49

Using (a-b) 2 = a2 + b2 – 2ab

iv) (3a – 1/2)(3a – 1/2)

= (3a – 1/2)2

=  9a2 -3a+(1/4)

Using (a-b) 2  = a2 + b2 – 2ab

v)   (1.1m – 0.4) (1.1m + 0.4)

= 1.21m2 – 0.16

Using (a – b)(a + b) = a2 – b2

vi) (a2+ b2) (– a2+ b2)

= (b2 + a2 ) (b2 – a2)

= -a4 + b4

Using (a – b)(a + b) = a2 – b2

vii) (6x – 7) (6x + 7)

=36x2 – 49

Using (a – b)(a + b) = a2 – b2

viii) (– a + c) (– a + c) = (– a + c)2

= c2 + a2 – 2ac

Using (a-b) 2 = a2 + b2 – 2ab

= (x2/4) + (9y2/16) + (3xy/4)

Using (a+b) 2 = a2 + b2 + 2ab

x) (7a – 9b) (7a – 9b) = (7a – 9b)2

= 49a2 – 126ab + 81b2

Using (a-b) 2 = a2 + b2 – 2ab

2. Use the identity (x + a) (x + b) = x+ (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) (2a+ 9) (2a+ 5)

(vii) (xyz – 4) (xyz – 2)

Solution:

(i)(x + 3) (x + 7)

= x2 + (3+7)x + 21

= x2 + 10x + 21

ii) (4x + 5) (4x + 1)

= 16x2 + 4x + 20x + 5

= 16x2 + 24x + 5

iii) (4x – 5) (4x – 1)

= 16x2 – 4x – 20x + 5

= 16x2 – 24x + 5

iv) (4x + 5) (4x – 1)

= 16x2 + (5-1)4x – 5

= 16x2 +16x – 5

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

= 4x2 + (5y + 3y)2x + 15y2

= 4x2 + 16xy + 15y2

vi) (2a2+ 9) (2a2+ 5)

= 4a4 + (9+5)2a2 + 45

= 4a4 + 28a2 + 45

vii) (xyz – 4) (xyz – 2)

= x2y2z2 + (-4 -2)xyz + 8

= 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) [(2m/3) + (3n/2)]2

(v) (0.4p – 0.5q)2

(vi) (2xy + 5y)2

Solution:

Using identities:

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

(i) (b – 7)= b2 – 14b + 49

(ii) (xy + 3z)= x2y2 + 6xyz + 9z2

(iii) (6x2 – 5y)2 = 36x4 – 60x2y + 25y2

(iv) [(2m/3}) + (3n/2)]= (4m2/9) +(9n2/4) + 2mn

(v) (0.4p – 0.5q)2 = 0.16p2 – 0.4pq + 0.25q2

(vi) (2xy + 5y)2 = 4x2y2 + 20xy2 + 25y2

4. Simplify.

(i) (a– b2)2

(ii) (2x + 5) – (2x – 5)2

(iii) (7m – 8n)+ (7m + 8n)2

(iv) (4m + 5n)+ (5m + 4n)2

(v) (2.5p – 1.5q)– (1.5p – 2.5q)2

(vi) (ab + bc)2– 2ab²c

(vii) (m– n2m)+ 2m3n2

Solution:

i) (a2– b2)2 = a4 + b4 – 2a2b2

ii) (2x + 5) – (2x – 5)2
= 4x2 + 20x + 25 – (4x2 – 20x + 25) = 4x2 + 20x + 25 – 4x2 + 20x – 25 = 40x

iii) (7m – 8n)+ (7m + 8n)2
= 49m2 – 112mn + 64n2 + 49m2 + 112mn + 64n2
= 98m2 + 128n2

iv) (4m + 5n)+ (5m + 4n)2
= 16m2 + 40mn + 25n2 + 25m2 + 40mn + 16n2
= 41m2 + 80mn + 41n2

v) (2.5p – 1.5q)– (1.5p – 2.5q)2
= 6.25p2 – 7.5pq + 2.25q2 – 2.25p2 + 7.5pq – 6.25q2
= 4p2 – 4q2

vi) (ab + bc)2– 2ab²c

= a2b2 + 2ab2c + b2c2 – 2ab2c

= a2b2 + b2c2

vii) (m– n2m)+ 2m3n2
= m4 – 2m3n2 + m2n4 + 2m3n2
= m4 + m2n4

5. Show that.

(i) (3x + 7)– 84x = (3x – 7)2

(ii) (9p – 5q)2+ 180pq = (9p + 5q)2

(iii) (4/3m – 3/4n)2 + 2mn = 16/9 m2 + 9/16 n2

(iv) (4pq + 3q)2– (4pq – 3q)= 48pq2

(v) (a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a) = 0

Solution:

i) LHS = (3x + 7)– 84x

= 9x2 + 42x + 49 – 84x
= 9x2 – 42x + 49
= RHS

LHS = RHS

ii)  LHS = (9p – 5q)2+ 180pq
= 81p2 – 90pq + 25q2 + 180pq
= 81p2 + 90pq + 25q2
RHS = (9p + 5q)2
= 81p2 + 90pq + 25q2
LHS = RHS

LHS = RHS

iv)  LHS = (4pq + 3q)2– (4pq – 3q)2

= 16p2q2 + 24pq2 + 9q2 – 16p2q2 + 24pq2 – 9q2

= 48pq2

= RHS

LHS = RHS

v) LHS = (a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a)

a2 – b2 + b2 – c2 + c2 – a2

= 0

= RHS

6. Using identities, evaluate.

(i) 71²

(ii) 99²

(iii) 1022

(iv) 998²

(v) 5.2²

(vi) 297 x 303

(vii) 78 x 82

(viii) 8.92

(ix) 10.5 x 9.5

Solution:

i) 712

= (70+1)2

= 702 + 140 + 12

= 4900 + 140 +1

= 5041

ii) 99²

= (100 -1)2

= 1002 – 200 + 12

= 10000 – 200 + 1

= 9801

iii) 1022

= (100 + 2)2

= 1002 + 400 + 22

= 10000 + 400 + 4 = 10404

iv) 9982

= (1000 – 2)2

= 10002 – 4000 + 22

= 1000000 – 4000 + 4

= 996004

v) 5.22

= (5 + 0.2)2

= 52 + 2 + 0.22

= 25 + 2 + 0.04 = 27.04

vi) 297 x 303

= (300 – 3 )(300 + 3)

= 3002 – 32

= 90000 – 9

= 89991

vii) 78 x 82

= (80 – 2)(80 + 2)

= 802 – 22

= 6400 – 4

= 6396

viii) 8.92

= (9 – 0.1)2

= 92 – 1.8 + 0.12

= 81 – 1.8 + 0.01

= 79.21

ix) 10.5 x 9.5

= (10 + 0.5)(10 – 0.5)

= 102 – 0.52

= 100 – 0.25

= 99.75

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

(i) 512– 492

(ii) (1.02)2– (0.98)2

(iii) 1532– 1472

(iv) 12.12– 7.92

Solution:

i) 512– 492

= (51 + 49)(51 – 49) = 100 x 2 = 200

ii) (1.02)2– (0.98)2

= (1.02 + 0.98)(1.02 – 0.98) = 2 x 0.04 = 0.08

iii) 153– 1472

= (153 + 147)(153 – 147) = 300 x 6 = 1800

iv) 12.1– 7.92

= (12.1 + 7.9)(12.1 – 7.9) = 20 x 4.2= 84

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

(i) 103 x 104

(ii) 5.1 x 5.2

(iii) 103 x 98

(iv) 9.7 x 9.8

Solution:

i) 103 x 104

= (100 + 3)(100 + 4)

= 1002 + (3 + 4)100 + 12

= 10000 + 700 + 12

= 10712

ii) 5.1 x 5.2

= (5 + 0.1)(5 + 0.2)

= 52 + (0.1 + 0.2)5 + 0.1 x 0.2

= 25 + 1.5 + 0.02

= 26.52

iii) 103 x 98

= (100 + 3)(100 – 2)

= 1002 + (3-2)100 – 6

= 10000 + 100 – 6

= 10094

iv) 9.7 x 9.8

= (9 + 0.7 )(9 + 0.8)

= 92 + (0.7 + 0.8)9 + 0.56

= 81 + 13.5 + 0.56

= 95.06

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

NCERT Solutions for Class 8 CBSE Maths Chapter 9, define basic concepts like terms, coefficients, factors, like and unlike terms, subtraction and addition of algebraic expressions, and multiplication of two or more polynomials. Students also learn in chapter 9 about different algebraic expression identities and solve problems applying these identities. In NCERT Class 8 Math Chapter 9 – Algebraic Expressions and Identities carries a total weightage of 8 to 12 marks in the examination. The Students will utilize the NCERT Solutions for class 8 Maths Chapter 9 while solving the exercise question and getting ready for examination.

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

NCERT Solutions for Class 8 Maths Chapter 9 is mainly based on solving polynomial related problems. The chapter 9 builds a strong foundation for all students to deal with higher grade Maths problems.

### Main Topics of class 8 Math chapter 9 Algebraic Expressions and Identities

Here knowledge glow covered main topics of Algebraic Expressions and Identities chapter:

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## Frequently Asked Questions on NCERT Solutions for Class 8 Maths Chapter 9

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