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

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 :

Sl. No.	Expression	Term	Coefficient
i)	5xyz2 – 3zy	Term: 5xyz2Term: -3zy	5 -3
ii)	1 + x + x2	Term: 1 Term: x Term: x2	1 1 1
iii)	4x2y2 – 4x2y2z2 + z2	Term: 4x2y2 Term: -4 x2y2z2 Term :  z2	4 -4 1
iv)	3 – pq + qr – p	Term : 3 -pq qr -p	3 -1 1 -1
v)	(x/2) + (y/2) – xy	Term : x/2 Y/2 -xy	½ 1/2 -1
vi)	0.3a – 0.6ab + 0.5b	Term : 0.3a -0.6ab 0.5b	0.3 -0.6 0.5

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.

x + y	two terms	Binomial
1000	one term	Monomial
x + x2 + x3 + x4	four terms	Polynomial, and it does not fit in listed three categories
2y – 3y2	two terms	Binomial
2y – 3y2 + 4y3	three terms	Trinomial
5x – 4y + 3xy	three terms	Trinomial
4z – 15z2	two terms	Binomial
ab + bc + cd + da	four terms	Polynomial, and it does not fit in listed three categories
pqr	one term	Monomial
p2q + pq2	two terms	Binomial
2p + 2q	two terms	Binomial
7 + y + 5x	three terms	Trinomial

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

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 = x² y² z²

(ii) a × – a²  × a³ = – a⁶

(iii) 2 × 4y × 8y² × 16y³ = 1024 y⁶

(iv) a × 2b × 3c × 6abc = 36a² b² c²

(v) m × – mn × mnp = –m³ n² p

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) a² – 9, 4a

(v) pq + qr + rp, 0

Solution:

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

(ii)ab(a – b) = a² b – 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:

S.No	First expression	Second expression	Product
(i)	a	b + c + d	a(b+c+d)= a×b + a×c + a×d= ab + ac + ad
(ii)	x + y – 5	5xy	5 xy (x + y – 5)= 5 xy x x + 5 xy x y – 5 xy x 5= 5 x2y + 5 xy2 – 25xy
(iii)	p	6p2 – 7p + 5	p (6 p 2-7 p +5)= p× 6 p2 – p× 7 p + p×5= 6 p3 – 7 p2 + 5 p
(iv)	4 p2 q2	P2 – q2	4p2 q2 * (p2 – q2 )=4 p4 q2– 4p2 q4
(v)	a + b + c	abc	abc(a + b + c)= abc × a + abc × b + abc × c= a2bc + ab2c + abc2

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 × a^{2 + 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(a² +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 1³ + 1² + 1+5 = 1 + 1 + 1+5 = 8

(iii) Putting a = -1 in the equation we get (-1)³+(-1)² + (-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) (p² – 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)

= p² (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)

= t³ – st + s²t² – 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) = 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)

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)² = b² – 14b + 49

(ii) (xy + 3z)² = x²y² + 6xyz + 9z²

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

(iv) [(2m/3}) + (3n/2)]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) (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

Solution:

i) (a²– b²)² = a⁴ + b⁴ – 2a²b²

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

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

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

v) (2.5p – 1.5q)² – (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) (m² – n²m)² + 2m³n²
= m⁴ – 2m³n² + m²n⁴ + 2m³n²
= m⁴ + m²n⁴

5. Show that.

(i) (3x + 7)² – 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)² = 48pq²

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

Solution:

i) LHS = (3x + 7)² – 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 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

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) 153² – 147²

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

iv) 12.1² – 7.9²

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

8. Using (x + a) (x + b) = x² + (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:

Exercise	Topic
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
9.8	Multiplying a Monomial by a Polynomial
9.9	Multiplying a Polynomial by a Polynomial
9.10	What is an Identity?
9.11	Standard Identities
9.12	Applying Identities

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

• NCERT Solutions provides entirely resolved step by step solutions to any or all textbook queries.
• Set of solutions contain an inventory of all necessary formulas on algebraical identities.
• These solutions area unit designed supported the most recent program.
• Solutions area unit ready by subject specialists.
• NCERT Solutions area unit helpful for the preparation of competitive exams.

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

How is NCERT Solutions for Class 8 Maths Chapter 9 helpful for board exams?

NCERT Solutions for Class 8 Maths Chapter 9 provides all solutions in detailed as per the term limit particularized by the Board for self-evaluation. By Solving these NCERT solutions will provide excellent practice for the students so they can complete the exam on time. So, it is clear that the NCERT Solutions for Class 8 Maths Chapter 9 are essential to high score in exam. Students can Also get acquainted with writing exams and will be able to face math exams more confidently.

Is it necessary to practice all the exercises of Class 8 Math Chapter 9 Algebraic Expressions and Identities?

Yes, it’s compulsory to practice all the exercises of Class 8 Math Chapter 9 Algebraic Expressions and Identities?. Because all exercises contain good numerous. This Practice makes more confident towards the syllabus they have.