myhelper: RS AGGARWAL CLASS 9 Chapter 3 FACTORISATION OF POLYNOMIAL EXERCISE 3D

EXERCISE 3D

PAGE NO-119

Question 1:

Expand:
(i) (a + 2b + 5c)²
(ii) (2b − b + c)²
(iii) (a − 2b − 3c)²

Answer 1:

$(i) {(a + 2 b + 5 c)}^{2} = {(a)}^{2} + {(2 b)}^{2} + {(5 c)}^{2} + 2 (a) (2 b) + 2 (2 b) (5 c) + 2 (5 c) (a) = a^{2} + 4 b^{2} + 25 c^{2} + 4 a b + 20 b c + 10 a c$

$(ii) {(2 a - b + c)}^{2} = {[(2 a) + (- b) + (c)]}^{2} = {(2 a)}^{2} + {(- b)}^{2} + {(c)}^{2} + 2 (2 a) (- b) + 2 (- b) (c) + 4 (a) (c) = 4 a^{2} + b^{2} + c^{2} - 4 a b - 2 b c + 4 a c$

$(iii) {(a - 2 b - 3 c)}^{2} = {[a + (- 2 b) + (- 3 c)]}^{2} = {(a)}^{2} + {(- 2 b)}^{2} + {(- 3 c)}^{2} + 2 (a) (- 2 b) + 2 (- 2 b) (- 3 c) + 2 (a) (- 3 c) = a^{2} + 4 b^{2} + 9 c^{2} - 4 a b + 12 b c - 6 a c$

Question 2:

Expand:
(i) (2a − 5b − 7c)²
(ii) (−3a + 4b − 5c)²
(iii) ${(\frac{1}{2} a - \frac{1}{4} a + 2)}^{2}$

Answer 2:

$(i) {(2 a - 5 b - 7 c)}^{2} = {[(2 a) + (- 5 b) + (- 7 c)]}^{2} = {(2 a)}^{2} + {(- 5 b)}^{2} + {(- 7 c)}^{2} + 2 (2 a) (- 5 b) + 2 (- 5 b) (- 7 c) + 2 (2 a) (- 7 c) = 4 a^{2} + 25 b^{2} + 49 c^{2} - 20 a b + 70 b c - 28 a c$

$(ii) {(- 3 a + 4 b - 5 c)}^{2} = {[(- 3 a) + (4 b) + (- 5 c)]}^{2} = {(- 3 a)}^{2} + {(4 b)}^{2} + {(- 5 c)}^{2} + 2 (- 3 a) (4 b) + 2 (4 b) (- 5 c) + 2 (- 3 a) (- 5 c) = 9 a^{2} + 16 b^{2} + 25 c^{2} - 24 a b - 40 b c + 30 a c$

$(iii) {(\frac{1}{2} a - \frac{1}{4} b + 2)}^{2} = {[(\frac{a}{2}) + (- \frac{b}{4}) + (2)]}^{2} = {(\frac{a}{2})}^{2} + {(- \frac{b}{4})}^{2} + {(2)}^{2} + 2 (\frac{a}{2}) (\frac{- b}{4}) + 2 (- \frac{b}{4}) (2) + 2 (\frac{a}{2}) (2) = \frac{a^{2}}{4} + \frac{b^{2}}{16} + 4 - \frac{a b}{4} - b + 2 a$

Question 3:

Factorize: 4x² + 9y² + 16z² + 12xy − 24yz − 16xz.

Answer 3:

$We have : 4 x^{2} + 9 y^{2} + 16 z^{2} + 12 x y - 24 y z - 16 x z = {(2 x)}^{2} + {(3 y)}^{2} + {(- 4 z)}^{2} + 2 (2 x) (3 y) + 2 (3 y) (- 4 z) + 2 (- 4 z) (2 x) = {[(2 x) + (3 y) + (- 4 z)]}^{2} = {(2 x + 3 y - 4 z)}^{2}$

Question 4:

Factorize: 9x² + 16y² + 4z² − 24xy + 16yz − 12xz

Answer 4:

$We have : 9 x^{2} + 16 y^{2} + 4 z^{2} - 24 x y + 16 y z - 12 x z = {(- 3 x)}^{2} + {(4 y)}^{2} + {(2 z)}^{2} + 2 (- 3 x) (4 y) + 2 (4 y) (2 z) + 2 (2 z) (- 3 x) = {[(- 3 x) + (4 y) + (2 z)]}^{2} = {(- 3 x + 4 y + 2 z)}^{2}$

Question 5:

Factorize: 25x²+ 4y² + 9z² − 20xy − 12yz + 30xz.

Answer 5:

$We have : 25 x^{2} + 4 y^{2} + 9 z^{2} - 20 x y - 12 y z + 30 x z = {(5 x)}^{2} + {(- 2 y)}^{2} + {(3 z)}^{2} + 2 (5 x) (- 2 y) + 2 (- 2 y) (3 z) + 2 (3 z) (5 x) = {[(5 x) + (- 2 y) + (3 z)]}^{2} = {(5 x - 2 y + 3 z)}^{2}$

Question 6:

16x² + 4y² + 9z² – 16xy – 12yz + 24xz

Answer 6:

$16 x^{2} + 4 y^{2} + 9 z^{2} - 16 x y - 12 y z + 24 x z = {(4 x)}^{2} + {(- 2 y)}^{2} + {(3 z)}^{2} + 2 (4 x) (- 2 y) + 2 (- 2 y) (3 z) + 2 (3 z) (4 x) = {(4 x - 2 y + 3 z)}^{2} [using a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 c a = {(a + b + c)}^{2}]$

Hence, 16x² + 4y² + 9z² – 16xy – 12yz + 24xz = ${(4 x - 2 y + 3 z)}^{2}$ .

Question 7:

Evaluate
(i) (99)²
(ii) (995)²
(iii) (107)²

Answer 7:

$(i) {(99)}^{2} = {(100 - 1)}^{2} = {[(100) + (- 1)]}^{2} = {(100)}^{2} + 2 \times (100) \times (- 1) + {(- 1)}^{2} = 10000 - 200 + 1 = 9801$

$(ii) {(995)}^{2} = {(1000 - 5)}^{2} = {[(1000) + (- 5)]}^{2} = {(1000)}^{2} + 2 \times (1000) \times (- 5) + {(- 5)}^{2} = 1000000 - 10000 + 25 = 990025$

$(iii) {(107)}^{2} = {(100 + 7)}^{2} = {(100)}^{2} + 2 \times (100) \times (7) + {(7)}^{2} = 10000 + 1400 + 49 = 11449$

RS AGGARWAL CLASS 9 Chapter 3 FACTORISATION OF POLYNOMIAL EXERCISE 3D