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

  • Last Updated : 20 Nov, 2020

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

(i) 4p, q + r 

Solution:

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(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 expressionProduct
(i)ab + c + dab + ac + ad
(ii)x + y – 55xy5x2y + 5xy2 – 25xy
(iii)p6p2 – 7p + 56p3 – 7p2 + 5p
(iv)4p2q2p2 – q24p4q2 – 4p2q4
(v)a + b + cabca2bc + 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. [ 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

    ⇒       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




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