NCERT Solutions for class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.3

NCERT Solutions has been carefully compiled and developed keeping into consideration the latest CBSE syllabus. CBSE Class 8 Maths, Chapter 9 Exercise 9.3, exercise questions and answers helps students to understand multiplication of the two polynomials as well as simplification of expression at particular values. NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities exercise 9.3 are prepared using step by step approach and aim to improve problem-solving skills.

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NCERT Solutions CBSE Class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.3 Part 1
NCERT Solutions CBSE Class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.3 Part 2
NCERT Solutions CBSE Class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.3 Part 3
NCERT Solutions CBSE Class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.3 Part 4

 

Access Answers of Maths NCERT class 8 Chapter 9 Algebraic Expressions and Identities Exercise 9.3 Page number 146

 

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 )

  1. 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:

  1. 3x (4x – 5) + 3

=3x ( 4x) – 3x( 5) +3

=12x2 – 15x + 3

  1. 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

  1. 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

Access other exercise solutions of class 8 Maths Chapter 9 Algebraic Expressions and Identities

Exercise 9.1 Solutions : 4 Questions (Short answers)

Exercise 9.2 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 Exercise 9.3

Algebraic Expressions and Identities Exercise 9.3 is about multiply a monomial by a binomial, multiply a monomial by a trinomial and simplification of the expressions and their values for given points. Practice exercise problems and get going with your homework.

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