# Class 8 RD Sharma Solutions – Chapter 6 Algebraic Expressions And Identities – Exercise 6.7

### (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 â€“ 11) (x + 4)

Solution:

By simplifying the given expression, we get

x (x + 4) â€“ 11 (x + 4)

x2 + 4x â€“ 11x â€“ 44

x2 â€“ 7x â€“ 44

### (iii) (x + 7) (x â€“ 5)

Solution:

By simplifying the given expression, we get

x (x â€“ 5) + 7 (x â€“ 5)

x2 â€“ 5x + 7x â€“ 35

x2 + 2x â€“ 35

### (iv) (x â€“ 3) (x â€“ 2)

Solution:

By simplifying the given expression, we get

x (x â€“ 2) â€“ 3 (x â€“ 2)

x2 â€“ 2x â€“ 3x + 6

x2 â€“ 5x + 6

### (v) (y2 â€“ 4) (y2 â€“ 3)

Solution:

By simplifying the given expression, we get

y2 (y2â€“ 3) â€“ 4 (y2 â€“ 3)

y4 â€“ 3y2 â€“ 4y2 + 12

y4 â€“ 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 â€“ 3) (2x2+ 5)

Solution:

By simplifying the given expression, we get

2x2(2x2 + 5) â€“ 3 (2x2 + 5)

4x4 + 10x2 â€“ 6x2 â€“ 15

4x4 + 4x2 â€“ 15

### (ix) (z2 + 2) (z2â€“ 3)

Solution:

By simplifying the given expression, we get

z2 (z2 â€“ 3) + 2 (z2 â€“ 3)

z4 â€“ 3z2 + 2z2 â€“ 6

z4â€“ z2 â€“ 6

### (x) (3x â€“ 4y) (2x â€“ 4y)

Solution:

By simplifying the given expression, we get

3x (2x â€“ 4y) â€“ 4y (2x â€“ 4y)

6x2 â€“ 12xy â€“ 8xy + 16y2

6x2 â€“ 20xy + 16y2

### (xi) (3x2 â€“ 4xy) (3x2 â€“ 3xy)

Solution:

By simplifying the given expression, we get

3x2 (3x2 â€“ 3xy) â€“ 4xy (3x2 â€“ 3xy)

9x4 â€“ 9x3y â€“ 12x3y + 12x2y2

9x4 â€“ 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 â€“ 14/5)

Solution:

By simplifying the given expression, we get

y2 (y2 â€“ 14/5) + 5/7 (y2 â€“ 14/5)

y4 â€“ 14/5y2 + 5/7y2 â€“ 2

y4 â€“ 73/35y2 â€“ 2

### (xvii) (p2 + 16) (p2 â€“ 1/4)

Solution:

By simplifying the given expression, we get

p2 (p2 â€“ 1/4) + 16 (p2 â€“ 1/4)

p4 â€“ 1/4p2 + 16p2 â€“ 4

p4 + 63/4p2 â€“ 4

### (i) 102 Ã— 106

Solution:

By simplifying the given expression, we get

102 Ã— 106 = (100 + 2) (100 + 6)

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

= 10000 + 600 + 200 + 12

= 10812

### (ii) 109 Ã— 107

Solution:

By simplifying the given expression, we get

109 Ã— 107 = (100 + 9) (100 + 7)

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

= 10000 + 700 + 900 + 63

= 11663

### (iii) 35 Ã— 37

Solution:

By simplifying the given expression, we get

35 Ã— 37 = (30 + 5) (30 + 7)

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

= 900 + 210 + 150 + 35

= 1295

### (iv) 53 Ã— 55

Solution:

By simplifying the given expression, we get

53 Ã— 55 = (50 + 3) (50 + 5)

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

= 2500 + 250 + 150 + 15

= 2915

### (v) 103 Ã— 96

Solution:

By simplifying the given expression, we get

103 Ã— 96 = (100 + 3) (100 â€“ 4)

= 100 (100 â€“ 4) + 3 (100 â€“ 4)

= 10000 â€“ 400 + 300 â€“ 12

= 10000 â€“ 112

= 9888

### (vi) 34 Ã— 36

Solution:

By simplifying the given expression, we get

34 Ã— 36 = (30 + 4) (30 + 6)

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

= 900 + 180 + 120 + 24

= 1224

### (vii) 994 Ã— 1006

Solution:

By simplifying the given expression, we get

994 Ã— 1006 = (1000 â€“ 6) (1000 + 6)

= 1000 (1000 + 6) â€“ 6 (1000 + 6)

= 1000000 + 6000 â€“ 6000 â€“ 36

= 999964

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