Class 8 | NCERT Solution Maths Chapter 9 | Algebraic Expressions and Identities | Exercise 9.3

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) a 2 - 9, 4a

(v) pq + qr + rp, 0

Solution:

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

(ii) ab (a - b) = a 2 b - ab 2

(iii)(a + b) (7a 2 b 2 ) = 7a 3 b 2  + 7a 2 b 3

(iv) (a 2  - 9) (4a) = 4a 3  - 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 x 4) ( a 2 x a 22 x 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 x 3 y 3 )

iii) (-10pq3/3) ×(6p3q/5)

= ( -10/3 × 6/5 ) (p × p3× q3 × q)

= (-4p 4 q 4 )

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 + 5l 2

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

= (-4ac + 4bc + 4c 2 ) – (3a 2  + 3ab + 3ac – ( 2ab – 2b 2 + 2bc))

=-4ac + 4bc + 4c 2 – (3a 2  + 3ab + 3ac – 2ab + 2b 2 – 2bc)

= -4ac + 4bc + 4c 2  – 3a 2  – 3ab – 3ac + 2ab – 2b 2 + 2bc

= -7ac + 6bc + 4c2 – 3a2 – ab – 2b2