# Class 8 NCERT Solutions – Chapter 9 Algebraic Expressions and Identities – Exercise 9.3

### Question 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 – ab2

(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

Explanation: Anything multiplied to 0 will give zero.

Solution:

### Question 3. Find the product.

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

Solution:

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

= (8) (a50)

= 8a50

Explanation: when two number numbers of same base are multiplied  their power gets added up. [ ax x ay = ax+y ]

(ii) (2/3 xy) x (-9/10 x2y2)

Solution:

(2/3 x -9/10) (xy x x2y2)

= (-3/5) (x3y3)

= -3/5 x3y3

Explanation: when two number numbers of same base are multiplied  their power gets added up. [ ax x ay = ax+y ]

(iii) (-10/3pq3) * (6/5p3q)

Solution:

(-10/3 x 6/5) (pq3 x p3q)

= (-4) (p4q4)

= -4p4q4

Explanation: when two number numbers of same base are multiplied  their power gets added up. [ ax x ay = ax+y]

(iv) x * x2 * x3 * x4

Solution:

(x) (x2) (x3) (x4)

= x10

Explanation: when two number numbers of same base are multiplied  their power gets added up. [ ax x ay = ax+y]

### Question 4.

(a) Simplify 3x (4x – 5) + 3 and find its values for

(i) x = 3

Solution:

First we will simplify the given equation and the put the value of x as required.

3x (4x – 5) + 3

â‡’       12x2 – 15x + 3

â‡’        12 (3)2 – 15 (3) + 3    [ putting the value of x = 3 ]

â‡’        12 (9) – 15 (3) + 3

â‡’        108 – 45 + 3

â‡’        66

(ii) x = 1/2

Solution:

3x (4x – 5) + 3

â‡’    12x2 – 15x + 3

â‡’    12 (1/2)2 – 15 (1/2) + 3

â‡’    12 (1/4) – 15 (1/2) +3

â‡’    3 – 15/2 + 3

â‡’    -3/2

(b) Simplify a (a2 + a + 1) + 5 and find its value for

(i) a = 0

Solution:

a (a2 + a + 1) + 5

â‡’  a3 + a2 + a + 5

â‡’  (0)3 + (0)2 + (0) + 5   [ anything to the power 0 gives 0 only ]

â‡’  5

(ii) a = 1

Solution:

a (a2 + a + 1) + 5

â‡’   a3 + a2 + a + 5

â‡’  (1)3 + (1)2 + (1) + 5   [ 1 to the power any number gives 1

â‡’  8

(iii) a = -1

Solution:

a (a2 + a + 1) + 5

â‡’      a3 + a2 + a + 5

â‡’      (-1)3 + (-1)2 + (-1) + 5    [ if -1 has even power then it is 1 or else if it has odd power it is -1]

â‡’      -1 + 1 -1 + 5

â‡’       4

### Question 5.

(a) Add: p (p â€“ q), q (q â€“ r) and r (r â€“ p)

Solution:

p (p â€“ q) + q (q â€“ r) + r (r â€“ p)

â‡’   p2 – pq + q2 – qr + r2 – rp

â‡’   p2 + q2+ r2 – pq – rp – qr

(b) Add: 2x (z â€“ x â€“ y) and 2y (z â€“ y â€“ x)

Solution:

2x (z â€“ x â€“ y) + 2y (z â€“ y â€“ x)

â‡’  2xz – 2x2 – 2xy + 2yz – 2y2 – 2yx

â‡’  2xz â€“ 4xy + 2yz â€“ 2x2 â€“ 2y2

(c) Subtract: 3l (l â€“ 4m + 5n) from 4l (10n â€“ 3m + 2l)

Solution:

4l (10n â€“ 3m + 2l) – 3l (l  â€“  4m + 5n)

â‡’    40ln – 12lm + 8l2 – 3l2 + 12lm – 15ln

â‡’   25ln + 5l2

(d) Subtract: 3a (a + b + c) â€“ 2b (a â€“ b + c) from 4c (â€“a + b + c)

Solution:

4c (â€“ a + b + c) – [3a (a + b + c) â€“ 2b (a â€“ b + c)]

â‡’   -4ac + 4bc + 4c2 – [3a2 + 3ab + 3ac – 2ab + 2b2 – 2bc]

â‡’   -4ac + 4bc + 4c2 – 3a2 – ab – 3ac – 2b2 + 2bc

â‡’   -3a2 – 2b2 + 4c2 – 7ac + 6bc – ab

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