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

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Exercise 9.1 Page No: 140

Q1. 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 – p (v) (x/2) + (y/2) – xy (vi) 0.3a – 0.6ab + 0.5b

Solution :

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 + 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 first define the classifications of these 3 polynomials:

Monomials, Contain only one term.

Binomials, Contain only two terms.

Trinomials, Contain only three terms.

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

Exercise 9.2 Page No: 143

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.

The 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

Exercise 9.3 Page No: 146

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:

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

Exercise 9.4 Page No: 148

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

ii) ( y – 8)(3y – 4)

= y x 3y – 4y – 8 x 3y + 32

= 3y² – 4y – 24y + 32

= 3y² – 28y + 32

(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

(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

Exercise 9.5 Page No: 151

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 + 49n²
= 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 – (4pq – 3q)² = 48pq²

(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 + 0.4 = 27.4

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, explains basic concepts like terms, factors, coefficients, like and unlike terms, addition and subtraction of algebraic expressions, and multiplication of two or more polynomials. Students will also learn about various algebraic expression identities and solve problems applying these identities. NCERT Class 8 , Chapter 9- Algebraic Expressions and Identities carries a total weightage of 8 to 10 marks in the final examination. NCERT Solutions For Class 8 Maths Chapter 9 Exercises: Get detailed solution for all the questions listed under below exercises:
Exercise 9.1 Solutions : 4 Questions (Short answers)
Exercise 9.2 Solutions : 5 Questions (Short answers)
Exercise 9.3 Solutions : 5 Questions (Short answers)
Exercise 9.4 Solutions : 3 Questions (Short answers)
Exercise 9.5 Solutions : 8 Questions (Short answers)

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

NCERT Solutions for Class 8 Maths Chapter 9 is mainly about the study of solving polynomial related problems. The chapter builds a strong foundation for the students to deal with higher grade Maths problems.
The main topics covered in this chapter include:

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

1. NCERT Solutions provides fully resolved step by step solutions to all textbook questions.
2. Set of solutions contain a list of all important formulas on algebraic identities.
3. These solutions are designed based on the latest syllabus.
4. Solutions are prepared by subject experts.
5. NCERT Solutions are helpful for the preparation of competitive exams.

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

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

NCERT Solutions for Class 8 Maths Chapter 9 provides answers with detailed descriptions as per term limit particularised by the Board for self-evaluation. Solving these solutions will provide excellent practice for the students so they can finish the paper on time. So, it’s clear that the NCERT Solutions for Class 8 Maths Chapter 9 are essential to score high in examinations. Students can get acquainted with writing exams and will be able to face exams more confidently.

Is it necessary to practice all the exercises present in Chapter 9 of NCERT Solutions for Class 8 Maths?

Yes, it is compulsory to practice all the exercises present in Chapter 9 of NCERT Solutions for Class 8 Maths. Because all exercises contain numerous questions to solve which may come in their finals. This makes them more confident towards the syllabus they have.

Mention the topics that are covered in NCERT Solutions for Class 8 Maths Chapter 9?

The topics that are covered in NCERT Solutions for Class 8 Maths Chapter 9 are 9.1 – introduction to 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 – definition and meaning of identity, 9.11 – standard identities and 9.12 – applying identities.