# Mathematics – Class 8 – Chapter 9 – Algebraic Expression and Identities – Exercise 9.3 -NCERT Exercise Solution

1. Carry out the multiplication of the expressions in each of the following pairs.

(i) 4p, q + r

Solution:

= 4p(q + r)

= 4pq + 4pr

(ii) ab, a – b

Solution:

= ab(a – b)

= a2 b – a b2

(iii) a + b, 7a²b²

Solution:

= (a + b) (7a2b2)

= 7a3b2 + 7a2b3

(iv) a² – 9, 4a

Solution:

= (a2 – 9)(4a)

= 4a3 – 36a

(v) pq + qr + rp, 0

Solution:

= (pq + qr + rp) × 0

= 0

2. Complete the table.

Solution:

3. Find the product.

i) a2 x (2a22) x (4a26)

Solution:

= a2 x (2a22) x (4a26)

= (2 × 4) (a2 × a22 × a26)

= 8 × a2 + 22 + 26

= 8a50

ii) (2/3 xy) ×(-9/10 x2y2)

Solution:

= (2xy/3) ×(-9x2y2/10)

= (2/3 × -9/10) (x × x2 × y × y2)

= (-3/5 x3y3)

(iii) ((-10/3)pq3) × ((6/5)p3q)

Solution:

= (-10pq3/3) ×(6p3q/5)

= ( -10/3 × 6/5 ) (p × p3× q3 × q)

= (-4p4q4)

(iv) (x) × (x2) × (x3) × (x4)

Solution:

= (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

Solution:

= 3x (4x – 5) + 3

= (3x (4x) – 3x (5)) +3

= 12x2 – 15x + 3

Now,

(i)Putting x=3 in the equation:

= 12x2 – 15x + 3

= 12(32) – 15 (3) +3

= 108 – 45 + 3

= 66

(ii) Putting x=1/2 in the equation:

= 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) Simplify a (a2+ a + 1) + 5 and find its value for (i) a = 0, (ii) a = 1 (iii) a = – 1.

Solution:

= 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:

= 03+02+0+5

=5

(ii) putting a=1 in the equation:

13 + 12 + 1+5

= 1 + 1 + 1+5

= 8

(iii) Putting a = -1 in the equation:

(-1)3+(-1)2 + (-1) +5

= -1 + 1 – 1+5

= 4

5. (a) Add: p ( p – q), q ( q – r) and r ( r – p)

Solution:

= p ( p – q) + q ( q – r) + r ( r – p)

= (p2 – pq) + (q2 – qr) + (r2 – pr)

= p2 + q2 + r2 – pq – qr – pr

(b) Add: 2x (z – x – y) and 2y (z – y – x)

Solution:

= 2x (z – x – y) + 2y (z – y – x)

= (2xz – 2x2 – 2xy) + (2yz – 2y2 – 2xy)

= 2xz – 4xy + 2yz – 2x2 – 2y2

(c) Subtract: 3l (l – 4 m + 5 n) from 4l (10 n – 3 m + 2 l)

Solution:

= 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) Subtract: 3a (a + b + c ) – 2 b (a – b + c)  from 4c ( – a + b + c )

Solution:

= 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

👍👍👍