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.

Question 2. Complete the table.

 Solution:      

	First expression	Second expression	Product
(i)	a	b + c + d	ab + ac + ad
(ii)	x + y – 5	5xy	5x2y + 5xy2 – 25xy
(iii)	p	6p2 – 7p + 5	6p3 – 7p2 + 5p
(iv)	4p2q2	p2 – q2	4p4q2 – 4p2q4
(v)	a + b + c	abc	a2bc + ab2c + abc2

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. [ a^x x a^y = a^x+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. [ a^x x a^y = a^x+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. [ a^x x a^y = a^x+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. [ a^x x a^y = a^x+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

    ⇒       12x² – 15x + 3

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

    ⇒        12 (9) – 15 (3) + 3

    ⇒        108 – 45 + 3

    ⇒        66

(ii) x = 1/2

Solution:

3x (4x – 5) + 3

    ⇒    12x² – 15x + 3

    ⇒    12 (1/2)² – 15 (1/2) + 3

    ⇒    12 (1/4) – 15 (1/2) +3

    ⇒    3 – 15/2 + 3

    ⇒    -3/2

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

(i) a = 0

Solution:

a (a² + a + 1) + 5

    ⇒  a³ + a² + a + 5

    ⇒  (0)³ + (0)² + (0) + 5   [ anything to the power 0 gives 0 only ]

    ⇒  5

(ii) a = 1 

Solution:

          a (a² + a + 1) + 5

    ⇒   a³ + a² + a + 5

    ⇒  (1)³ + (1)² + (1) + 5   [ 1 to the power any number gives 1 ] 

    ⇒  8

(iii) a = -1

Solution:

a (a² + a + 1) + 5

    ⇒      a³ + a² + a + 5

    ⇒      (-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]

    ⇒      -1 + 1 -1 + 5

    ⇒       4

Question 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 – rp

  ⇒   p² + q²+ r² – pq – rp – qr

(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 – 2yx

  ⇒  2xz – 4xy + 2yz – 2x² – 2y²

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

Solution:

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

 ⇒    40ln – 12lm + 8l² – 3l² + 12lm – 15ln

 ⇒   25ln + 5l²

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

Solution:

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

 ⇒   -4ac + 4bc + 4c² – [3a² + 3ab + 3ac – 2ab + 2b² – 2bc]

 ⇒   -4ac + 4bc + 4c² – 3a² – ab – 3ac – 2b² + 2bc

 ⇒   -3a² – 2b² + 4c² – 7ac + 6bc – ab