Skip to content
Related Articles

Related Articles

Class 8 RD Sharma Solutions – Chapter 6 Algebraic Expressions And Identities – Exercise 6.7
  • Last Updated : 05 Mar, 2021
GeeksforGeeks - Summer Carnival Banner

Question 1. Find the following products:

(i) (x + 4) (x + 7)

Solution:

By simplifying the given expression, we get

x (x + 7) + 4 (x + 7)

x2 + 7x + 4x + 28

x2+ 11x + 28



(ii) (x – 11) (x + 4)

Solution:

By simplifying the given expression, we get

x (x + 4) – 11 (x + 4)

x2 + 4x – 11x – 44

x2 – 7x – 44

(iii) (x + 7) (x – 5)

Solution:

By simplifying the given expression, we get

x (x – 5) + 7 (x – 5)

x2 – 5x + 7x – 35

x2 + 2x – 35

(iv) (x – 3) (x – 2)

Solution:

By simplifying the given expression, we get

x (x – 2) – 3 (x – 2)

x2 – 2x – 3x + 6

x2 – 5x + 6

(v) (y2 – 4) (y2 – 3)

Solution:

By simplifying the given expression, we get

y2 (y2– 3) – 4 (y2 – 3)



y4 – 3y2 – 4y2 + 12

y4 – 7y2 + 12

(vi) (x + 4/3) (x + 3/4)

Solution:

By simplifying the given expression, we get

x (x + 3/4) + 4/3 (x + 3/4)

x2 + 3x/4 + 4x/3 + 12/12

x2 + 3x/4 + 4x/3 + 1

x2 + 25x/12 + 1

(vii) (3x + 5) (3x + 11)

Solution:

By simplifying the given expression, we get

3x (3x + 11) + 5 (3x + 11)

9x2 + 33x + 15x + 55

9x2 + 48x + 55

(viii) (2x2 – 3) (2x2+ 5)

Solution:

By simplifying the given expression, we get

2x2(2x2 + 5) – 3 (2x2 + 5)

4x4 + 10x2 – 6x2 – 15

4x4 + 4x2 – 15

(ix) (z2 + 2) (z2– 3)

Solution:

By simplifying the given expression, we get

z2 (z2 – 3) + 2 (z2 – 3)

z4 – 3z2 + 2z2 – 6

z4– z2 – 6

(x) (3x – 4y) (2x – 4y)

Solution:

By simplifying the given expression, we get

3x (2x – 4y) – 4y (2x – 4y)

6x2 – 12xy – 8xy + 16y2

6x2 – 20xy + 16y2

(xi) (3x2 – 4xy) (3x2 – 3xy)

Solution:

By simplifying the given expression, we get

3x2 (3x2 – 3xy) – 4xy (3x2 – 3xy)

9x4 – 9x3y – 12x3y + 12x2y2

9x4 – 21x3y + 12x2y2

(xii) (x + 1/5) (x + 5)

Solution:

By simplifying the given expression, we get

x (x + 1/5) + 5 (x + 1/5)

x2 + x/5 + 5x + 1

x2 + 26/5x + 1

(xiii) (z + 3/4) (z + 4/3)

Solution:

By simplifying the given expression, we get

z (z + 4/3) + 3/4 (z + 4/3)

z2 + 4/3z + 3/4z + 12/12

z2+ 4/3z + 3/4z + 1

z2 + 25/12z + 1

(xiv) (x2+ 4) (x2 + 9)

Solution:

By simplifying the given expression, we get

x2 (x2 + 9) + 4 (x2+ 9)

x4 + 9x2 + 4x2 + 36

x4 + 13x2+ 36

(xv) (y2 + 12) (y2+ 6)

Solution:

By simplifying the given expression, we get

y2 (y2+ 6) + 12 (y2 + 6)

y4+ 6y2 + 12y2 + 72

y4 + 18y2 + 72

(xvi) (y2 + 5/7) (y2 – 14/5)

Solution:

By simplifying the given expression, we get

y2 (y2 – 14/5) + 5/7 (y2 – 14/5)

y4 – 14/5y2 + 5/7y2 – 2

y4 – 73/35y2 – 2

(xvii) (p2 + 16) (p2 – 1/4)

Solution:

By simplifying the given expression, we get

p2 (p2 – 1/4) + 16 (p2 – 1/4)

p4 – 1/4p2 + 16p2 – 4

p4 + 63/4p2 – 4

Question 2.  Evaluate the following:

(i) 102 × 106

Solution:

By simplifying the given expression, we get

102 × 106 = (100 + 2) (100 + 6)

= 100 (100 + 6) + 2 (100 + 6)

= 10000 + 600 + 200 + 12

= 10812

(ii) 109 × 107

Solution:

By simplifying the given expression, we get

109 × 107 = (100 + 9) (100 + 7)

= 100 (100 + 7) + 9 (100 + 7)

= 10000 + 700 + 900 + 63

= 11663

(iii) 35 × 37

Solution:

By simplifying the given expression, we get

35 × 37 = (30 + 5) (30 + 7)

= 30 (30 + 7) + 5 (30 + 7)

= 900 + 210 + 150 + 35

= 1295

(iv) 53 × 55

Solution:

By simplifying the given expression, we get

53 × 55 = (50 + 3) (50 + 5)

= 50 (50 + 5) + 3 (50 + 5)

= 2500 + 250 + 150 + 15

= 2915

(v) 103 × 96

Solution:

By simplifying the given expression, we get

103 × 96 = (100 + 3) (100 – 4)

= 100 (100 – 4) + 3 (100 – 4)

= 10000 – 400 + 300 – 12

= 10000 – 112

= 9888

(vi) 34 × 36

Solution:

By simplifying the given expression, we get

34 × 36 = (30 + 4) (30 + 6)

= 30 (30 + 6) + 4 (30 + 6)

= 900 + 180 + 120 + 24

= 1224

(vii) 994 × 1006

Solution:

By simplifying the given expression, we get

994 × 1006 = (1000 – 6) (1000 + 6)

= 1000 (1000 + 6) – 6 (1000 + 6)

= 1000000 + 6000 – 6000 – 36

= 999964

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up
Recommended Articles
Page :