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

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NCERT Solutions CBSE Class 8 Maths Chapter 9 Algebraic Expressions and Identities Part 1
NCERT Solutions CBSE Class 8 Maths Chapter 9 Algebraic Expressions and Identities Part 2
NCERT Solutions CBSE Class 8 Maths Chapter 9 Algebraic Expressions and Identities Part 3
NCERT Solutions CBSE Class 8 Maths Chapter 9 Algebraic Expressions and Identities Part 4
NCERT Solutions CBSE Class 8 Maths Chapter 9 Algebraic Expressions and Identities Part 5
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NCERT Solutions CBSE Class 8 Maths Chapter 9 Algebraic Expressions and Identities Part 7
NCERT Solutions CBSE Class 8 Maths Chapter 9 Algebraic Expressions and Identities Part 8
NCERT Solutions CBSE Class 8 Maths Chapter 9 Algebraic Expressions and Identities Part 9
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Access Answers of Maths NCERT Chapter 9 – Algebraic Expressions and Identities

 

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 :

Sl. No.

Expression

Term

Coefficient

i)

5xyz2 – 3zy

Term: 5xyz2 

Term: -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

3 -pq qr -p

3 -1 1 -1

v)

(x/2) + (y/2) – xy

x/2 Y/2 -xy

½ 1/2 -1

vi)

0.3a – 0.6ab + 0.5b

0.3a -0.6ab 0.5b

0.3 -0.6 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 + 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.

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

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:

  1. 4  7 p =  4 7 × p = 28p
  2. – 4p × 7p = (-4 × 7 ) × (p × p )= -28p2
  3. – 4p × 7pq =(-4 × 7 ) (p × pq) =  -28p2q
  4. 4p3 × – 3p = (4 × -3 ) (p3 × p ) =  -12p4
  5. 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:

ncert solution for class 8 maths chapter 09 fig 1

Solution:

ncert solutions for class 8 maths chapter 09 fig 2

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.

  1. 5a x 3a2 x 7a4 = (5 × 3 × 7) (a × a^2 × a^4 ) = 105a7
  2. 2p x 4q x 8r = (2 × 4 × 8 ) (p × q × r ) = 64pqr
  3. y × 2x2y × 2xy2 =(1 × 2 × 2 )( x × x^2 × x × y × y × y^2 ) =  4x4y4
  4. 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.

ncert solutions for class 8 maths chapter 09 fig 3

Solution:

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 + 5 xy * y – 5 xy * 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

4p^2 q^2 * (p^2 – q^2 )

=4 p^4 q^2 – 4p^2 q^4

(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) ( a^2 × a^22 × a^26 )

= 8 × a^(2+22+26 )

= 8a50

ii) (2xy/3) ×(-9x2y2/10)

=(2/3 × -9/10 ) ( x × x^2 × y × y^2 )

= (-3/5 x3y3)

iii) (-10pq3/3) ×(6p3q/5)

= ( -10/3 × 6/5 ) (p × p^3 × q^3 × q)

= (-4p4q4)

iv)  ( x) x (x2) x (x3) x (x4)

= x^ ( 1+2+3+8)

=  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(3^2) – 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* a^2 + a*a + a*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 – 2b^2 + 2bc ))

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

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

= -7ac + 6bc + 4c2 – 3a2 – ab – 2b^2

 

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 a^2 + 3b^2 ) and 4( a^2 – 2/3 b^2 )

Solution :

(i) (2x + 5)(4x – 3)

2x * 4x – 2x * 3 + 5 * 4x – 5 * 3

8x² – 6x + 20x -15

8x² + 14x -15

ii) ( y – 8)(3y – 4)

= y * 3y – 4y – 8 * 3y + 32

= 3y2 – 4y – 24y + 32

= 3y2 – 28y + 32

(iii) (2.5l – 0.5m)(2.5l + 0.5m)

= 2.5l * 2.5 l + 2.5l * 0.5m – 0.5m * 2.5l – 0.5m * 0.5m

= 6.25l^2 + 1.25 lm – 1.25 lm – 0.25 m^2

= 6.25l^2 – 0.25 m^2

iv) (a + 3b) (x + 5)

= ax + 5a + 3bx + 15b

v) (2pq + 3q2(3pq – 2q2)

= 2pq * 3pq – 2pq * 2q2 + 3q2 * 3pq – 3q2 * 2q2

= 6p2q2 – 4pq3 + 9pq3 – 6q4 

= 6p2q2 + 5pq3 – 6q4

(vi) (3/4 a^2 + 3b^2 ) and 4( a^2 – 2/3 b^2 )

=(3/4 a^2 + 3b^2 ) * 4( a^2 – 2/3 b^2 )

=(3/4 a^2 + 3b^2 ) * (4a^2 – 8/3 b^2 )

=3/4 a^2 * (4a^2 – 8/3 b^2 ) + 3b^2 * (4a^2 – 8/3 b^2 )

=3/4 a^2 * 4a^2 -3/4 a^2 * 8/3 b^2 + 3b^2 * 4a^2 – 3b^2* 8/3 b^2

=3a^4 – 2a^2b^2 + 12 a^2 b^2 – 8b^4

= 3a^4 + 10a^2 b^2 – 8b^4

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 + a2b^2 + ab + b3

= a3 + b3 + a2b^2 + 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

= (3a-1/2)(3a-1/2)=  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

ncert solution for class 8 maths chapter 09 fig 7

= (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) (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 + (5 + 1)4x + 5

= 16x2 + 24x + 5

iii) (4x – 5) (4x – 1)

= 16x2 + (-5-1)4x + 5

= 16x2 – 20x + 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

  1. (b – 7)2 = b2 – 14b + 49
  2. (xy + 3z)2 = x2y2 + 6xyz + 9z2
  3. (6x2 – 5y)2 = 36x4 – 60x2y + 25y2
  4. [(2m/3}) + (3n/2)]2 = (4m2/9) +(9n2/4) + 2mn
  5. (0.4p – 0.5q)2 = 0.16p2 – 0.4pq + 0.25q2
  6. (2xy + 5y)^2 = 4x2y2 + 20xy2 + 25y2

4. Simplify.

(i) (a2 – b2)2 

(ii) (2x + 5) – (2x – 5)2 

(iii) (7m – 8n)2 + (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)2 + (7m + 8n)2
= 49m2 – 112mn + 64n2 + 49m2 + 112mn + 49n2
= 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 – (4pq – 3q)2 = 48pq2

(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

ncert solution for class 8 maths chapter 09 fig 8

LHS = RHS

iv)  LHS = (4pq + 3q)2– (4pq – 3q)2

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

= 48pq2

RHS = 48pq2

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:

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

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 list of all important formulas an algebraic identities.
  3. These solutions are designed based on latest syllabus.
  4. Solutions are prepared by subject experts.
  5. NCERT Solutions are helpful for preparation of competitive exams.

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