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

## RD Sharma Solutions Class 8 Chapter 6 Exercise 6.7

Q1) Find the following products:

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

(ii) (x – 11) (x + 4)

(iii) (x + 7) (x – 5)

(iv) (x – 3) (x – 2)

(v) (y2 – 4) (y2 – 3)

(vi) $(x + \frac{4}{3})\; (x + \frac{3}{4})$

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

(viii) (2x2 – 3) (2x2 + 5)

(ix) (z2 + 2) (z2 – 3)

(x) (3x – 4y) (2x – 4y)

(xi) (3x2 – 4xy) (3x2 – 3xy)

(xii) $(x + \frac{1}{5})\; (x + 5)$

(xiii) $(z + \frac{3}{4})\; (z + \frac{4}{3})$

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

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

(xvi) $(y^{2} + \frac{5}{7})\; (y^{2} – \frac{14}{5})$

(xvii) $(p^{2} + 16)\; (p^{2} – \frac{1}{4})$

Solution:

(i) Here, we will use the identity (x + a) (x + b) = x2 + (a + b)x + ab.

(x + 4) (x + 7)

= $x^{2} + (4+7)x + 4\times 7$

= $x^{2} + 11x + 28$

(ii) Here, we will use the identity (x – a) (x + b) = x2 + (b – a)x – ab.

(x – 11) (x + 4)

= $x^{2} + (4-11)x – 11\times 4$

= $x^{2} – 7x – 44$

(iii) Here, we will use the identity (x + a) (x – b) = x2 + (a – b)x – ab.

(x + 7) (x – 5)

= $x^{2} + (7-5)x – 7\times 5$

= $x^{2} + 2x – 35$

(iv) Here, we will use the identity (x – a) (x – b) = x2 – (a + b)x + ab.

(x – 3) (x – 2)

= $x^{2} – (3+2)x + 3\times 2$

= $x^{2} – 5x + 6$

(v) Here, we will use the identity (x – a) (x – b) = x2 – (a + b)x – ab.

(y2 – 4) (y2 – 3)

= $(y^{2})^{2} – (4 + 3)(y^{2}) + 4\times 3$

= $y^{4} – 7y^{2} + 12$

(vi) Here, we will use the identity (x + a) (x + b) = x2 + (a + b)x + ab.

$(x + \frac{4}{3})\; (x + \frac{3}{4})$

= $x^{2} + (\frac{4}{3} + \frac{3}{4})x + \frac{4}{3}\times \frac{3}{4})$

= $x^{2} + \frac{25}{12}x + 1$

(vii) Here, we will use the identity (x + a) (x + b) = x2 + (a + b)x + ab.

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

= $(3x)^{2} + (5 + 11) (3x) + 5\times 11$

= $9x^{2} + 48x + 55$

(viii) Here, we will use the identity (x – a) (x + b) = x2 + (b – a)x – ab.

(2x2 – 3) (2x2 + 5)

= $(2x^{2})^{2} + (5 – 3)(2x^{2}) – 3\times 5$

= $4x^{4} + 4x^{2} – 15$

(ix) Here, we will use the identity (x + a) (x – b) = x2 + (a – b)x – ab.

(z2 + 2) (z2 – 3)

= $(z^{2})^{2} + (2 – 3)(z^{2}) – 2\times 3$

= $z^{4} – z^{2} – 6$

(x) Here, we will use the identity (x – a) (x – b) = x2 – (a + b)x – ab.

(3x – 4y) (2x – 4y)

= (4y – 3x) (4y – 2x)                           (Taking common -1 from both parentheses)

= $(4y)^{2} – (3x + 2x)\;(4y) + 3x\times 2x$

= $16y^{2} – (12xy + 8xy) + 6x^{2}$

= $16y^{2} – 20xy + 6x^{2}$

(xi) Here, we will use the identity (x – a) (x – b) = x2 – (a + b)x – ab.

(3x2 – 4xy) (3x2 – 3xy)

= $(3x^{2})^{2} – (4xy + 3xy)\; (3x^{2}) + 4xy \times 3xy$

= $9x^{4} – (12x^{3}y + 9x^{3}y) + 12x^{2}y^{2}$

= $9x^{4} – 21x^{3}y + 12x^{2}y^{2}$

(xii) Here, we will use the identity (x + a) (x + b) = x2 + (a + b)x + ab.

$(x + \frac{1}{5})\; (x + 5)$

= $x^{2} + (\frac{1}{5} + 5)x + \frac{1}{5}\times 5$

= $x^{2} + \frac{26}{5}x + 1$

(xiii) Here, we will use the identity (x + a) (x + b) = x2 + (a + b)x + ab.

$(z + \frac{3}{4})\; (z + \frac{4}{3})$

= $z^{2} + (\frac{3}{4} + \frac{4}{3})x + \frac{3}{4}\times \frac{4}{3}$

= $z^{2} + \frac{25}{12}x + 1$

(xiv) Here, we will use the identity (x + a) (x + b) = x2 + (a + b)x + ab.

(x2 + 4) (x2 + 9)

= $(x^{2})^{2} + (4 + 9)(x^{2}) + 4 \times 9$

= $x^{4} + 13x^{2} + 36$

(xv) Here, we will use the identity (x + a) (x + b) = x2 + (a + b)x + ab.

(y2 + 12) (y2 + 6)

= $(y^{2})^{2} + (12 + 6)(y^{2}) + 12 \times 6$

= $y^{4} + 18xy^{2} + 72$

(xvi) Here, we will use the identity (x + a) (x – b) = x2 + (a – b)x –ab.

$(y^{2} + \frac{5}{7})\; (y^{2} – \frac{14}{5})$

= $(y^{2})^{2} + (\frac{5}{7} – \frac{14}{5})(y^{2}) – \frac{5}{7}\times \frac{14}{5}$

= $y^{4} – \frac{73}{35}y^{2} – 2$

(xvii) Here, we will use the identity (x + a) (x – b) = x2 + (a – b)x –ab.

$(p^{2} + 16)\; (p^{2} – \frac{1}{4})$

= $(p^{2})^{2} + (16 – \frac{1}{4})(p^{2}) – 16\times \frac{1}{4}$

= $p^{4} + \frac{63}{4}p^{2} – 4$

Q2. Evaluate the following:

(i) 102 x 106

(ii) 109 x 107

(iii) 35 x 37

(iv) 53 x 55

(v) 103 x 96

(vi) 34 x 36

(vii) 994 x 1006

Solution:

(i) Here, we will use the identity (x + a) (x + b) = x2 + (a + b)x + ab

102 x 106

= (100 + 2) (100 + 6)

= 1002 + (2 + 6)100 + 2 x 6

=10000 + 800 +12 = 10812

(ii) Here, we will use the identity (x + a) (x + b) = x2 + (a + b)x + ab

109 x 107

= (100 + 9) (100 + 7)

= 1002 + (9 + 7)100 + 9 x 7

=10000 + 1600 + 63 = 11663

(iii) Here, we will use the identity (x + a) (x + b) = x2 + (a + b)x + ab

35 x 37

= (30 + 5) (30 + 7)

= 302 + (5 + 7)30 + 5 x 7

= 900 + 360 + 35 = 1295

(iv) Here, we will use the identity (x + a) (x + b) = x2 + (a + b)x + ab

53 x 55

= (50 + 3) (50 + 5)

= 502 + (3 + 5)50 + 3 x 5

= 2500 + 400 + 15 = 2915

(v) Here, we will use the identity (x + a) (x – b) = x2 + (a – b)x – ab

103 x 96

= (100 + 3) (100 – 4)

= 1002 + (3 – 4)100 – 3 x 4

= 10000 – 100 – 12 = 9888

(vi) Here, we will use the identity (x + a) (x + b) = x2 + (a + b)x + ab

34 x 36

= (30 + 4) (30 + 6)

= 302 + (4 + 6)30 + 4 x 6

= 900 + 300 + 24 = 1224

(vii) Here, we will use the identity (x – a) (x + b) = x2 + (b – a)x – ab

994 x 1006

= (1000 – 6) x (1000 + 6)

= 10002 + (6 – 6) x 1000 – 6 x 6

= 1000000 – 36 = 999964

#### Practise This Question

Which of the following is true about multiplication of a vector by a scalar.

1) Scalar multiplication by a positive number other than 1 changes its magnitude but not direction.

2) Scalar multiplication always lead to change in magnitude and direction.

3) Scalar multiplication by -1 will not change the magnitude of the vector but will change its direction.

4) Scalar multiplication by a negative number other than -1 will reverse its direction and change its magnitude as well.