Factorisation

4x² + 16 in the form of factors is :

1. 4(x² + 4)

2. 4(x + 2)(x - 2)

3. (2x + 4)(2x - 4)

4. 16(4x² + 1)

Answer

   4x² + 16

= 4(x² + 4).

Hence, Option 1 is the correct option.

7(x + 3y)² - 2x - 6y in the form of factors is :

1. (9x² + 21xy)(x + 6y)

2. 9(x + 3y)(x + 3y - 1)

3. (x + 3y)(7x + 21y - 2)

4. 7(x + 3y)(x + 3y - 3)

Answer

   7(x + 3y)² - 2x - 6y

= 7(x + 3y)² - 2(x + 3y)

= (x + 3y)[7(x + 3y) - 2]

= (x + 3y)(7x + 21y - 2)

Hence, Option 3 is the correct option.

8(x - y) + 5(y - x) in the form of factors is :

1. 3(x + y)

2. 13(x - y)

3. 3(x - y)

4. 13(x + y)

Answer

   8(x - y) + 5(y - x)

= 8(x - y) - 5(x - y)

= (x - y)[8 - 5]

= 3(x - y).

Hence, Option 3 is the correct option.

$4x^2 + \dfrac{1}{4x^2} - 2 - 6x + \dfrac{3}{2x}$ in the form of factors is :

1. $\Big(2x + \dfrac{1}{2x}\Big)\Big(2x - \dfrac{1}{2x} - 3\Big)$

2. $\Big(2x - \dfrac{1}{2x}\Big)\Big(2x - \dfrac{1}{2x} + 3\Big)$

3. $3\Big(2x + \dfrac{1}{2x}\Big)\Big(2x + \dfrac{1}{2x} - 3\Big)$

4. $\Big(2x - \dfrac{1}{2x}\Big)\Big(2x - \dfrac{1}{2x} - 3\Big)$

Answer

$\phantom{\Rightarrow} 4x^2 + \dfrac{1}{4x^2} - 2 - 6x + \dfrac{3}{2x} \\[1em] = \Big(4x^2 + \dfrac{1}{4x^2} - 2\Big) - 6x + \dfrac{3}{2x} \\[1em] = \Big(2x - \dfrac{1}{2x}\Big)^2 - 3\Big(2x - \dfrac{1}{2x}\Big) \\[1em] = \Big(2x - \dfrac{1}{2x}\Big)\Big(2x - \dfrac{1}{2x} - 3\Big)$

Hence, Option 4 is the correct option.

3ab - 6b + 4a² - 8a in the form of factors is :

1. (a + 2)(4a + 3b)

2. (a - 2)(4a + 3b)

3. (a - 2)(4a - 3b)

4. (a + 2)(4a - 3b)

Answer

  3ab - 6b + 4a² - 8a

= 3b(a - 2) + 4a(a - 2)

= (a - 2)(3b + 4a).

Hence, Option 2 is the correct option.

Factorise by taking out the common factors :

xy(3x² - 2y²) - yz(2y² - 3x²) + zx(15x² - 10y²)

Answer

Given,

xy(3x² - 2y²) - yz(2y² - 3x²) + zx(15x² - 10y²)

= xy(3x² - 2y²) - yz[-(3x² - 2y²)] + 5zx(3x² - 2y²)

= xy(3x² - 2y²) + yz(3x² - 2y²) + 5zx(3x² - 2y²)

= (3x² - 2y²)(xy + yz + 5zx).

Hence, xy(3x² - 2y²) - yz(2y² - 3x²) + zx(15x² - 10y²) = (3x² - 2y²)(xy + yz + 5zx).

Factorise by taking out the common factors :

2x(a - b) + 3y(5a - 5b) + 4z(2b - 2a)

Answer

Given,

   2x(a - b) + 3y(5a - 5b) + 4z(2b - 2a)

= 2x(a - b) + 3y × 5(a - b) + 4z × -2(a - b)

= 2x(a - b) + 15y(a - b) - 8z(a - b)

= (a - b)(2x + 15y - 8z).

Hence, 2x(a - b) + 3y(5a - 5b) + 4z(2b - 2a) = (a - b)(2x + 15y - 8z).

Factorise by grouping method :

16(a + b)² - 4a - 4b

Answer

Given,

   16(a + b)² - 4a - 4b

= 16(a + b)² - 4(a + b)

= 4(a + b)[4(a + b) - 1]

= 4(a + b)[4a + 4b - 1]

= 4(a + b)(4a + 4b - 1).

Hence, 16(a + b)² - 4a - 4b = 4(a + b)(4a + 4b - 1).

Factorise by grouping method :

a⁴ - 2a³ - 4a + 8.

Answer

Given,

   a⁴ - 2a³ - 4a + 8

= a⁴ - 4a - 2a³ + 8

= a(a³ - 4) - 2(a³ - 4)

= (a - 2)(a³ - 4).

Hence, a⁴ - 2a³ - 4a + 8 = (a - 2)(a³ - 4).

Factorise by grouping method :

ab(x² + 1) + x(a² + b²).

Answer

Given,

   ab(x² + 1) + x(a² + b²)

= abx² + ab + xa² + xb²

= abx² + xa² + xb² + ab

= ax(bx + a) + b(bx + a)

= (ax + b)(bx + a).

Hence, ab(x² + 1) + x(a² + b²) = (ax + b)(bx + a).

Factorise by grouping method :

(ax + by)² + (bx - ay)²

Answer

Given,

   (ax + by)² + (bx - ay)²

= (ax)² + (by)² + 2 × ax × by + (bx)² + (ay)² - 2 × bx × ay

= a²x² + b²y² + 2abxy + b²x² + a²y² - 2abxy

= a²x² + b²x² + a²y² + b²y²

= x²(a² + b²) + y²(a² + b²)

= (x² + y²)(a² + b²).

Hence, (ax + by)² + (bx - ay)² = (x² + y²)(a² + b²).

Factorise by grouping method :

a²x² + (ax² + 1)x + a

Answer

Given,

   a²x² + (ax² + 1)x + a

= a²x² + ax³ + x + a

= a²x² + ax³ + a + x

= ax²(a + x) + 1(a + x)

= (a + x)(ax² + 1).

Hence, a²x² + (ax² + 1)x + a = (a + x)(ax² + 1).

Factorise by grouping method :

y² - (a + b)y + ab

Answer

Given,

   y² - (a + b)y + ab

= y² - ay - by + ab

= y(y - a) - b(y - a)

= (y - a)(y - b).

Hence, y² - (a + b)y + ab = (y - a)(y - b).

x² + 2(5x + 12) in the form of factors is :

1. (x - 6)(x + 4)

2. (x - 6)(x - 4)

3. (x + 6)(x + 4)

4. (x + 6)(x - 4)

Answer

Given,

   x² + 2(5x + 12)

= x² + 10x + 24

= x² + 6x + 4x + 24

= x(x + 6) + 4(x + 6)

= (x + 4)(x + 6).

Hence, Option 3 is the correct option.

x(2x - 5) + 3 in the form of factors is :

1. (x + 1)(2x - 3)

2. (x + 1)(2x + 3)

3. (x - 1)(2x + 3)

4. (x - 1)(2x - 3)

Answer

Given,

   x(2x - 5) + 3

= 2x² - 5x + 3

= 2x² - 3x - 2x + 3

= x(2x - 3) - 1(2x - 3)

= (x - 1)(2x - 3).

Hence, Option 4 is the correct option.

6x² - 2 - 4x in the form of factors is :

1. 2(x - 1)(3x - 1)

2. 2(x - 1)(3x + 1)

3. (2x - 1)(3x - 1)

4. (6x - 2)(x + 1)

Answer

Given,

   6x² - 2 - 4x

= 6x² - 4x - 2

= 6x² - 6x + 2x - 2

= 6x(x - 1) + 2(x - 1)

= (x - 1)(6x + 2)

= 2(x - 1)(3x + 1).

Hence, Option 2 is the correct option.

x(x - 1) - 20 in the form of factors is :

1. (x - 5)(x + 4)

2. (x - 5)(x - 4)

3. (x + 5)(x - 4)

4. (x + 5)(x + 4)

Answer

Given,

   x(x - 1) - 20

= x² - x - 20

= x² - 5x + 4x - 20

= x(x - 5) + 4(x - 5)

= (x - 5)(x + 4).

Hence, Option 1 is the correct option.

5 - x(6 - x) in the form of factors is :

1. (1 - x)(5 + x)

2. (1 + x)(5 + x)

3. (1 - x)(5 - x)

4. (1 + x)(5 - x)

Answer

Given,

   5 - x(6 - x)

= 5 - 6x + x²

= x² - 6x + 5

= x² - 5x - x + 5

= x(x - 5) - 1(x - 5)

= (x - 1)(x - 5)

= [-(1 - x).-(5 - x)]

= (1 - x)(5 - x).

Hence, Option 3 is the correct option.

Factorise :

1 - 2a - 3a²

Answer

Given,

   1 - 2a - 3a²

= -(3a² + 2a - 1)

= -[3a² + 3a - a - 1]

= -[3a(a + 1) - 1(a + 1)]

= -(3a - 1)(a + 1)

= (1 - 3a)(a + 1).

Hence, 1 - 2a - 3a² = (1 - 3a)(a + 1).

Factorise :

24a³ + 37a² - 5a

Answer

Given,

   24a³ + 37a² - 5a

= 24a³ + 40a² - 3a² - 5a

= 8a²(3a + 5) - a(3a + 5)

= (8a² - a)(3a + 5)

= a(8a - 1)(3a + 5).

Hence, 24a³ + 37a² - 5a = a(8a - 1)(3a + 5).

Factorise :

3 - a(4 + 7a)

Answer

Given,

   3 - a(4 + 7a)

= 3 - 4a - 7a²

= -7a² - 4a + 3

= -(7a² + 4a - 3)

= -[7a² + 7a - 3a - 3]

= -[7a(a + 1) - 3(a + 1)]

= -[(a + 1)(7a - 3)]

= -[-(a + 1)(3 - 7a)]

= (a + 1)(3 - 7a).

Hence, 3 - a(4 + 7a) = (a + 1)(3 - 7a).

Factorise :

(2a + b)² - 6a - 3b - 4

Answer

Given,

   (2a + b)² - 6a - 3b - 4

= (2a + b)² - 3(2a + b) - 4

Substituting (2a + b) = x, we get :

= x² - 3x - 4

= x² - 4x + x - 4

= x(x - 4) + 1(x - 4)

= (x - 4)(x + 1)

= (2a + b - 4)(2a + b + 1).

Hence, (2a + b)² - 6a - 3b - 4 = (2a + b - 4)(2a + b + 1).

Factorise :

1 - 2a - 2b - 3(a + b)²

Answer

Given,

   1 - 2a - 2b - 3(a + b)²

= 1 - 2(a + b) - 3(a + b)²

Substituting (a + b) = x, we get :

= 1 - 2x - 3x²

= -3x² - 2x + 1

= -[3x² + 2x - 1]

= -[3x² + 3x - x - 1]

= -[3x(x + 1) - 1(x + 1)]

= -[(x + 1)(3x - 1)]

= -[-(x + 1)(1 - 3x)]

= (x + 1)(1 - 3x)

= (a + b + 1)[1 - 3(a + b)]

= (a + b + 1)(1 - 3a - 3b).

Hence, 1 - 2a - 2b - 3(a + b)² = (a + b + 1)(1 - 3a - 3b).

Factorise :

3a² - 1 - 2a

Answer

Given,

   3a² - 1 - 2a

= 3a² - 2a - 1

= 3a² - 3a + a - 1

= 3a(a - 1) + 1(a - 1)

= (a - 1)(3a + 1).

Hence, 3a² - 1 - 2a = (a - 1)(3a + 1).

Factorise :

x² + 3x + 2 + ax + 2a

Answer

Given,

   x² + 3x + 2 + ax + 2a

= x² + 3x + 2 + a(x + 2)

= x² + 2x + x + 2 + a(x + 2)

= x(x + 2) + 1(x + 2) + a(x + 2)

= (x + 2)(x + 1 + a).

Hence, x² + 3x + 2 + ax + 2a = (x + 2)(x + 1 + a).

Factorise :

(3x - 2y)² + 3(3x - 2y) - 10

Answer

Given,

   (3x - 2y)² + 3(3x - 2y) - 10

Substituting (3x - 2y) = a, we get :

⇒ a² + 3a - 10

= a² + 5a - 2a - 10

= a(a + 5) - 2(a + 5)

= (a + 5)(a - 2)

= (3x - 2y + 5)(3x - 2y - 2).

Hence, (3x - 2y)² + 3(3x - 2y) - 10 = (3x - 2y + 5)(3x - 2y - 2).

Factorise :

5 - (3a² - 2a)(6 - 3a² + 2a)

Answer

Given,

   5 - (3a² - 2a)(6 - 3a² + 2a)

= 5 - [-(3a² - 2a)(3a² - 2a - 6)]

= 5 + (3a² - 2a)(3a² - 2a - 6)

Substituting 3a² - 2a = x, we get :

= 5 + x(x - 6)

= 5 + x² - 6x

= x² - 6x + 5

= x² - 5x - x + 5

= x(x - 5) - 1(x - 5)

= (x - 5)(x - 1)

= (3a² - 2a - 5)(3a² - 2a - 1)

= [3a² - 5a + 3a - 5] [3a² - 3a + a - 1]

= [a(3a - 5) + 1(3a - 5)] [3a(a - 1) + 1(a - 1)]

= (3a - 5)(a + 1)(a - 1)(3a + 1).

Hence, 5 - (3a² - 2a)(6 - 3a² + 2a) = (3a - 5)(a + 1)(a - 1)(3a + 1).

x⁴ - 1 in the form of factors is :

1. (x + 1)(x - 1)(x² - 1)

2. (x - 1)(x + 1)(x² + 1)

3. (x - 1)(x - 2)(x² + 2)

4. (x + 1)(x + 2)(x² - 1)

Answer

Given,

   x⁴ - 1

= (x²)² - 1²

= (x² - 1)(x² + 1)

= (x - 1)(x + 1)(x² + 1).

Hence, Option 2 is the correct option.

2a² - 18 in the form of factors is :

1. 2(a + 3)(a - 3)

2. 2(a + 1)(a - 9)

3. 2(a - 1)(a - 9)

4. (2a - 1)(a - 9)

Answer

Given,

   2a² - 18

= 2(a² - 9)

= 2[(a)² - (3)²]

= 2(a + 3)(a - 3).

Hence, Option 1 is the correct option.

27x³ - 48x in the form of factors is :

1. 3(3x - 4)(3x - 4)

2. 3x(3x + 4)(3x + 4)

3. 3(3x² - 4x)(3x + 4)

4. 3x(3x - 4)(3x + 4)

Answer

Given,

   27x³ - 48x

= 3x(9x² - 16)

= 3x[(3x)² - (4)²]

= 3x(3x - 4)(3x + 4).

Hence, Option 4 is the correct option.

x² - (x - 4y)² in the form of factors is :

1. 8y(x - 2y)

2. 8y(x + 2y)

3. 4y(x + 2y)

4. 4y(x - 2y)

Answer

Given,

   x² - (x - 4y)²

= [x + (x - 4y)][x - (x - 4y)]

= (2x - 4y)[x - x + 4y]

= 4y(2x - 4y)

= 8y(x - 2y).

Hence, Option 1 is the correct option.

$9x^2 + \dfrac{1}{9x^2} + 1$ in the form of factors is :

1. $\Big(3x + \dfrac{1}{3x} + 1\Big)\Big(3x + \dfrac{1}{3x} - 1\Big)$

2. $\Big(3x - \dfrac{1}{3x} + 1\Big)\Big(3x - \dfrac{1}{3x} - 1\Big)$

3. $\Big(3x + \dfrac{1}{3x}\Big)\Big(3x - \dfrac{1}{3x} + 1\Big)$

4. $\Big(3x - \dfrac{1}{3x}\Big)\Big(3x + \dfrac{1}{3x} - 1\Big)$

Answer

Given,

$\phantom{=} 9x^2 + \dfrac{1}{9x^2} + 1 \\[1em] = 9x^2 + \dfrac{1}{9x^2} + 2 - 1 \\[1em] = (3x)^2 + \Big(\dfrac{1}{3x}\Big)^2 + 2 - 1 \\[1em] = \Big(3x + \dfrac{1}{3x}\Big)^2 - 1^2 \\[1em] = \Big(3x + \dfrac{1}{3x} + 1\Big)\Big(3x + \dfrac{1}{3x} - 1\Big).$

Hence, Option 1 is the correct option.

Factorise :

a² - (2a + 3b)²

Answer

Given,

   a² - (2a + 3b)²

= (a + 2a + 3b)[a - (2a + 3b)]

= (3a + 3b)(a - 2a - 3b)

= (3a + 3b)(-a - 3b)

= -3(a + b)(a + 3b)

Hence, a² - (2a + 3b)² = -3(a + b)(a + 3b).

Factorise :

25(2a - b)² - 81b²

Answer

Given,

   25(2a - b)² - 81b²

= [5(2a - b)]² - (9b)²

= [5(2a - b) + 9b][5(2a - b) - 9b]

= (10a - 5b + 9b)(10a - 5b - 9b)

= (10a + 4b)(10a - 14b)

= 2(5a + 2b) × 2(5a - 7b)

= 4(5a + 2b)(5a - 7b).

Hence, 25(2a - b)² - 81b² = 4(5a + 2b)(5a - 7b).

Factorise :

50a³ - 2a

Answer

Given,

   50a³ - 2a

= 2a[25a² - 1]

= 2a[(5a)² - 1²]

= 2a(5a + 1)(5a - 1).

Hence, 50a³ - 2a = 2a(5a + 1)(5a - 1).

Factorise :

4a²b - 9b³

Answer

Given,

   4a²b - 9b³

= b[4a² - 9b²]

= b[(2a)² - (3b)²]

= b(2a + 3b)(2a - 3b).

Hence, 4a²b - 9b³ = b(2a + 3b)(2a - 3b).

Factorise :

9(a - 2)² - 16(a + 2)²

Answer

Given,

   9(a - 2)² - 16(a + 2)²

= {[3(a - 2)]² - [4(a + 2)]²}

= {[3(a - 2) + 4(a + 2)][3(a - 2) - 4(a + 2)]}

= [(3a - 6 + 4a + 8)(3a - 6 - 4a - 8)]

= (7a + 2)(-a - 14)

= -(7a + 2)(a + 14).

Hence, 9(a - 2)² - 16(a + 2)² = -(7a + 2)(a + 14).

(a + b)³ - a - b

Answer

Given,

   (a + b)³ - a - b

= (a + b)³ - (a + b)

= (a + b)[(a + b)² - 1]

= (a + b)[(a + b)² - 1²]

= (a + b)(a + b + 1)(a + b - 1).

Hence, (a + b)³ - a - b = (a + b)(a + b + 1)(a + b - 1).

a(a - 1) - b(b - 1)

Answer

Given,

   a(a - 1) - b(b - 1)

= a² - a - b² + b

= a² - b² - a + b

= (a + b)(a - b) - 1(a - b)

= (a - b)(a + b - 1).

Hence, a(a - 1) - b(b - 1) = (a - b)(a + b - 1).

4a² - (4b² + 4bc + c²)

Answer

Given,

   4a² - (4b² + 4bc + c²)

= 4a² - [(2b)² + 2 × 2b × c + c²]

= (2a)² - (2b + c)²

= (2a + 2b + c)[2a - (2b + c)]

= (2a + 2b + c)(2a - 2b - c).

Hence, 4a² - (4b² + 4bc + c²) = (2a + 2b + c)(2a - 2b - c).

4a² - 49b² + 2a - 7b

Answer

Given,

   4a² - 49b² + 2a - 7b

= (2a)² - (7b)² + 2a - 7b

= (2a + 7b)(2a - 7b) + (2a - 7b)

= (2a - 7b)(2a + 7b + 1).

Hence, 4a² - 49b² + 2a - 7b = (2a - 7b)(2a + 7b + 1).

4a² - 12a + 9 - 49b²

Answer

Given,

   4a² - 12a + 9 - 49b²

= (2a)² - 2 × 2a × 3 + (3)² - (7b)²

= (2a - 3)² - (7b)²

= (2a - 3 + 7b)(2a - 3 - 7b).

Hence, 4a² - 12a + 9 - 49b² = (2a - 3 + 7b)(2a - 3 - 7b).

4xy - x² - 4y² + z²

Answer

Given,

   4xy - x² - 4y² + z²

= -[x² + 4y² - 4xy - z²]

= -[(x)² + (2y)² - 2 × x × 2y - (z)²]

= -[(x - 2y)² - (z)²]

= (z)² - (x - 2y)²

= (z + x - 2y)(z - x + 2y).

Hence, 4xy - x² - 4y² + z² = (z + x - 2y)(z - x + 2y).

a² + b² - c² - d² + 2ab - 2cd

Answer

Given,

   a² + b² - c² - d² + 2ab - 2cd

= a² + b² + 2ab - c² - d² - 2cd

= (a + b)² - (c² + d² + 2cd)

= (a + b)² - (c + d)²

= (a + b + c + d)(a + b - c - d).

Hence, a² + b² - c² - d² + 2ab - 2cd = (a + b + c + d)(a + b - c - d).

4x² - 12ax - y² - z² - 2yz + 9a²

Answer

Given,

   4x² - 12ax - y² - z² - 2yz + 9a²

= 4x² - 12ax + 9a² - y² - z² - 2yz

= (2x)² - 2 × 2x × 3a + (3a)² - (y² + z² + 2yz)

= (2x - 3a)² - (y + z)²

= (2x - 3a + y + z)[2x - 3a - (y + z)]

= (2x - 3a + y + z)(2x - 3a - y - z).

Hence, 4x² - 12ax - y² - z² - 2yz + 9a² = (2x - 3a + y + z)(2x - 3a - y - z).

(a² - 1)(b² - 1) + 4ab

Answer

Given,

   (a² - 1)(b² - 1) + 4ab

= a²b² - a² - b² + 1 + 4ab

= a²b² - a² - b² + 1 + 2ab + 2ab

= a²b² + 2ab + 1 - a² - b² + 2ab

= (ab + 1)² - (a² + b² - 2ab)

= (ab + 1)² - (a - b)²

= (ab + 1 + a - b)(ab + 1 - a + b).

Hence, (a² - 1)(b² - 1) + 4ab = (ab + 1 + a - b)(ab + 1 - a + b).

x⁴ + x² + 1

Answer

Given,

   x⁴ + x² + 1

Adding and subtracting x² in the polynomial,

⇒ x⁴ + x² + 1 + x² - x²

= x⁴ + 2x² + 1 - x²

= (x²)² + 2 × x² × 1 + (1)² - (x)²

= (x² + 1)² - (x)²

= (x² + 1 + x)(x² + 1 - x).

Hence, x⁴ + x² + 1 = (x² + 1 + x)(x² + 1 - x).

(a² + b² - 4c²)² - 4a²b²

Answer

Given,

   (a² + b² - 4c²)² - 4a²b²

= (a² + b² - 4c²)² - (2ab)²

= (a² + b² - 4c² + 2ab)(a² + b² - 4c² - 2ab)

= (a² + b² + 2ab - 4c²)(a² + b²- 2ab - 4c²)

= [(a + b)² - (2c)²][(a - b)² - (2c)²]

= (a + b + 2c)(a + b - 2c)(a - b + 2c)(a - b - 2c).

Hence, (a² + b² - 4c²)² - 4a²b² = (a + b + 2c)(a + b - 2c)(a - b + 2c)(a - b - 2c).

(x² + 4y² - 9z²)² - 16x²y²

Answer

Given,

   (x² + 4y² - 9z²)² - 16x²y²

= (x² + 4y² - 9z²)² - (4xy)²

= (x² + 4y² - 9z² + 4xy)(x² + 4y² - 9z² - 4xy)

= [x² + (2y)² + 4xy - 9z²][x² + (2y)² - 4xy - 9z²]

= [(x + 2y)² - (3z)²][(x - 2y)² - (3z)²]

= (x + 2y + 3z)(x + 2y - 3z)(x - 2y + 3z)(x - 2y - 3z).

Hence, (x² + 4y² - 9z²)² - 16x²y² = (x + 2y + 3z)(x + 2y - 3z)(x - 2y + 3z)(x - 2y - 3z).

(a + b)² - a² + b²

Answer

Given,

   (a + b)² - a² + b²

= (a + b)² - (a² - b²)

= (a + b)² - (a + b)(a - b)

= (a + b)[(a + b) - (a - b)]

= (a + b)[a + b - a + b]

= 2b(a + b).

Hence, (a + b)² - a² + b² = 2b(a + b).

a² - b² - (a + b)²

Answer

Given,

   a² - b² - (a + b)²

= (a + b)(a - b) - (a + b)²

= (a + b)[(a - b) - (a + b)]

= (a + b)[a - b - a - b]

= -2b(a + b).

Hence, a² - b² - (a + b)² = -2b(a + b).

a³ - 8 in the form of factors is :

1. (a - 2)(a² - 2a + 4)

2. (a - 2)(a² + 2a + 4)

3. (a - 2)(a² + 2a - 4)

4. (a + 2)(a² - 2a + 4)

Answer

Given,

   a³ - 8

= (a)³ - (2)³

= (a - 2)[(a)² + 2 × a + (2)²]

= (a - 2)(a² + 2a + 4)

Hence, Option 2 is the correct option.

27 + 8x³ in the form of factors is :

1. (3 + 2x)(9 + 6x + 4x²)

2. (3 - 2x)(9 + 6x + 4x²)

3. (3 + 2x)(9 - 6x + 4x²)

4. (3 - 2x)(9 - 6x + 4x²)

Answer

Given,

   27 + 8x³

= (3)³ + (2x)³

= (3 + 2x)[(3)² - 3 × 2x + (2x)²]

= (3 + 2x)(9 - 6x + 4x²)

Hence, Option 3 is the correct option.

8a³ + 1 in the form of factors is :

1. (2a + 1)(4a² - 2a + 1)

2. (2a - 1)(4a² - 2a + 1)

3. (2a + 1)(4a² + 2a + 1)

4. (2a - 1)(4a² + 2a + 1)

Answer

Given,

   8a³ + 1

= (2a)³ + (1)³

= (2a + 1)[(2a)² - 2a × 1 + (1)²]

= (2a + 1)(4a² - 2a + 1)

Hence, Option 1 is the correct option.

$\Big(\dfrac{x^3}{8} - \dfrac{y^3}{27}\Big)$ in the form of factors is :

1. $\Big(\dfrac{x}{2} - \dfrac{y}{3}\Big)\Big(\dfrac{x^2}{4} - \dfrac{xy}{6} + \dfrac{y^2}{9}\Big)$

2. $\Big(\dfrac{x}{2} + \dfrac{y}{3}\Big)\Big(\dfrac{x^2}{4} - \dfrac{xy}{6} + \dfrac{y^2}{9}\Big)$

3. $\Big(\dfrac{x}{2} - \dfrac{y}{3}\Big)\Big(\dfrac{x^2}{4} + \dfrac{xy}{6} + \dfrac{y^2}{9}\Big)$

4. $\Big(\dfrac{x}{2} + \dfrac{y}{3}\Big)\Big(\dfrac{x^2}{4} + \dfrac{xy}{6} + \dfrac{y^2}{9}\Big)$

Answer

Given,

$\phantom{=} \Big(\dfrac{x^3}{8} - \dfrac{y^3}{27}\Big) \\[1em] = \Big(\dfrac{x}{2}\Big)^3 - \Big(\dfrac{y}{3}\Big)^3 \\[1em] = \Big(\dfrac{x}{2} - \dfrac{y}{3}\Big)\Big[\Big(\dfrac{x}{2}\Big)^2 + \dfrac{x}{2} \times \dfrac{y}{3} + \Big(\dfrac{y}{3}\Big)^2\Big] \\[1em] = \Big(\dfrac{x}{2} - \dfrac{y}{3}\Big)\Big(\dfrac{x^2}{4} + \dfrac{xy}{6} + \dfrac{y^2}{9}\Big).$

Hence, Option 3 is the correct option.

Factorise :

64 - a³b³

Answer

Given,

   64 - a³b³

= (4)³ - (ab)³

= (4 - ab)[(4)² + 4 × ab + (ab)²]

= (4 - ab)(16 + 4ab + a²b²).

Hence, 64 - a³b³ = (4 - ab)(16 + 4ab + a²b²).

a⁶ + 27b³

Answer

Given,

   a⁶ + 27b³

= (a²)³ + (3b)³

= (a² + 3b)[(a²)² - a² × 3b + (3b)²]

= (a² + 3b)(a⁴ - 3a²b + 9b²).

Hence, a⁶ + 27b³ = (a² + 3b)(a⁴ - 3a²b + 9b²).

3x⁷y - 81x⁴y⁴

Answer

Given,

   3x⁷y - 81x⁴y⁴

= 3x⁴y[x³ - 27y³]

= 3x⁴y[(x)³ - (3y)³]

= 3x⁴y(x - 3y)[(x)² + x × 3y + (3y)²]

= 3x⁴y(x - 3y)(x² + 3xy + 9y²).

Hence, 3x⁷y - 81x⁴y⁴ = 3x⁴y(x - 3y)(x² + 3xy + 9y²).

a³ - $\dfrac{27}{a^3}$

Answer

Given,

$\phantom{=} a^3 - \dfrac{27}{a^3} \\[1em] = (a)^3 - \Big(\dfrac{3}{a}\Big)^3 \\[1em] = \Big(a - \dfrac{3}{a}\Big)\Big[(a)^2 + a \times \dfrac{3}{a} + \Big(\dfrac{3}{a}\Big)^2\Big] \\[1em] = \Big(a - \dfrac{3}{a}\Big)\Big(a^2 + 3 + \dfrac{9}{a^2}\Big).$

Hence, $a^3 - \dfrac{27}{a^3} = \Big(a - \dfrac{3}{a}\Big)\Big(a^2 + 3 + \dfrac{9}{a^2}\Big).$

a³ + 0.064

Answer

Given,

    a³ + 0.064

= (a)³ + (0.4)³

= (a + 0.4)[a² - 0.4 × a + (0.4)²]

= (a + 0.4)(a² - 0.4a + 0.16)

Hence, a³ + 0.064 = (a + 0.4)(a² - 0.4a + 0.16)

(x - y)³ - 8x³

Answer

Given,

    (x - y)³ - 8x³

= (x - y)³ - (2x)³

= (x - y - 2x)[(x - y)² + 2x(x - y) + (2x)²]

= (-x - y)[x² + y² - 2xy + 2x² - 2xy + 4x²]

= -(x + y)(7x² - 4xy + y²).

Hence, (x - y)³ - 8x³ = -(x + y)(7x² - 4xy + y²).

$\Big(\dfrac{8a^3}{27} - \dfrac{b^3}{8}\Big)$

Answer

Given,

$\phantom{=}\Big(\dfrac{8a^3}{27} - \dfrac{b^3}{8}\Big) \\[1em] = \Big(\dfrac{2a}{3}\Big)^3 - \Big(\dfrac{b}{2}\Big)^3 \\[1em] = \Big(\dfrac{2a}{3} - \dfrac{b}{2}\Big)\Big[\Big(\dfrac{2a}{3}\Big)^2 + \dfrac{2a}{3} \times \dfrac{b}{2} + \Big(\dfrac{b}{2}\Big)^2\Big] \\[1em] = \Big(\dfrac{2a}{3} - \dfrac{b}{2}\Big)\Big(\dfrac{4a^2}{9} + \dfrac{ab}{3} + \dfrac{b^2}{4}\Big) \\[1em]$

Hence, $\Big(\dfrac{8a^3}{27} - \dfrac{b^3}{8}\Big) = \Big(\dfrac{2a}{3} - \dfrac{b}{2}\Big)\Big(\dfrac{4a^2}{9} + \dfrac{ab}{3} + \dfrac{b^2}{4}\Big).$

a⁶ - b⁶

Answer

Given,

    a⁶ - b⁶

= (a³)² - (b³)²

= (a³ + b³)(a³ - b³)

= (a + b)(a² - ab + b²)(a - b)(a² + ab + b²)

= (a + b)(a - b)(a² - ab + b²)(a² + ab + b²).

Hence, a⁶ - b⁶ = (a + b)(a - b)(a² - ab + b²)(a² + ab + b²).

a⁶ - 7a³ - 8

Answer

Given,

    a⁶ - 7a³ - 8

= (a³)² - 7a³ - 8

Substituting a³ = x, we get :

= x² - 7x - 8

= x² - 8x + x - 8

= x(x - 8) + 1(x - 8)

= (x - 8)(x + 1)

= (a³ - 8)(a³ + 1)

= (a³ - 2³)(a³ + 1³)

= (a - 2)(a² + a × 2 + 2²)(a + 1)(a² - a × 1 + 1²)

= (a - 2)(a² + 2a + 4)(a + 1)(a² - a + 1)

= (a - 2)(a + 1)(a² + 2a + 4)(a² - a + 1).

Hence, a⁶ - 7a³ - 8 = (a - 2)(a + 1)(a² + 2a + 4)(a² - a + 1).

a³ - 27b³ + 2a²b - 6ab²

Answer

Given,

    a³ - 27b³ + 2a²b - 6ab²

= (a)³ - (3b)³ + 2ab(a - 3b)

= (a - 3b)[(a)² + a × 3b + (3b)²] + 2ab(a - 3b)

= (a - 3b)[a² + 3ab + 9b²] + 2ab(a - 3b)

= (a - 3b)(a² + 3ab + 9b² + 2ab)

= (a - 3b)(a² + 5ab + 9b²).

Hence, a³ - 27b³ + 2a²b - 6ab² = (a - 3b)(a² + 5ab + 9b²).

8a³ - b³ - 4ax + 2bx

Answer

Given,

    8a³ - b³ - 4ax + 2bx

= (2a)³ - (b)³ - 2x(2a - b)

= (2a - b)[(2a)² + 2a × b + (b)²] - 2x(2a - b)

= (2a - b)(4a² + 2ab + b²) - 2x(2a - b)

= (2a - b)(4a² + 2ab + b² - 2x).

Hence, 8a³ - b³ - 4ax + 2bx = (2a - b)(4a² + 2ab + b² - 2x).

a - b - a³ + b³

Answer

Given,

    a - b - a³ + b³

= (a - b) - (a³ - b³)

= (a - b) - (a - b)(a² + ab + b²)

= (a - b)(1 - a² - ab - b²).

Hence, a - b - a³ + b³ = (a - b)(1 - a² - ab - b²).

$x^4 + \dfrac{1}{x^4} - 2$ in the form of factors is :

1. $\Big(x - \dfrac{1}{x}\Big)^2\Big(x + \dfrac{1}{x} + 1\Big)^2$

2. $\Big(x - \dfrac{1}{x}\Big)^2\Big(x + \dfrac{1}{x}\Big)^2$

3. $\Big(x + \dfrac{1}{x}\Big)^2\Big(x - \dfrac{1}{x} + 1\Big)^2$

4. $\Big(x + \dfrac{1}{x}\Big)^2\Big(x + \dfrac{1}{x} - 1\Big)^2$

Answer

Given,

$\phantom{=}x^4 + \dfrac{1}{x^4} - 2 \\[1em] = (x^2)^2 + \Big(\dfrac{1}{x^2}\Big)^2 - 2 \times x^2 \times \dfrac{1}{x^2} \\[1em] = \Big(x^2 - \dfrac{1}{x^2}\Big)^2 \\[1em] = \Big[\Big(x - \dfrac{1}{x}\Big)\Big(x + \dfrac{1}{x}\Big)\Big]^2 \\[1em] = \Big(x - \dfrac{1}{x}\Big)^2\Big(x + \dfrac{1}{x}\Big)^2.$

Hence, Option 2 is the correct option.

$x^2 + \dfrac{1}{x^2} - 3$ in the form of factors is :

1. $\Big(x - \dfrac{1}{x}\Big)\Big(x + \dfrac{1}{x}\Big)$ - 1

2. $\Big(x + \dfrac{1}{x} + 1\Big)\Big(x - \dfrac{1}{x} - 1\Big)$

3. $\Big(x - \dfrac{1}{x}\Big)\Big(x - \dfrac{1}{x} - 1\Big)$

4. $\Big(x - \dfrac{1}{x} + 1\Big)\Big(x - \dfrac{1}{x} - 1\Big)$

Answer

Given,

$\phantom{=}x^2 + \dfrac{1}{x^2} - 3 \\[1em] = x^2 + \dfrac{1}{x^2} - 2 - 1 \\[1em] = x^2 + \dfrac{1}{x^2} - \Big(2 \times x \times \dfrac{1}{x}\Big) - 1 \\[1em] = \Big[x^2 + \dfrac{1}{x^2} - \Big(2 \times x \times \dfrac{1}{x}\Big)\Big] - 1 \\[1em] = \Big(x - \dfrac{1}{x}\Big)^2 - 1 \\[1em] = \Big(x - \dfrac{1}{x}\Big)^2 - 1^2 \\[1em] = \Big(x - \dfrac{1}{x} + 1\Big)\Big(x - \dfrac{1}{x} - 1\Big).$

Hence, Option 4 is the correct option.

Factorise :

$x^2 + \dfrac{1}{4x^2} + 1 - 7x - \dfrac{7}{2x}$

Answer

Given,

$\phantom{=} x^2 + \dfrac{1}{4x^2} + 1 - 7x - \dfrac{7}{2x} \\[1em] = x^2 + \Big(\dfrac{1}{2x}\Big)^2 + 1 - 7\Big(x + \dfrac{1}{2x}\Big) \\[1em] = x^2 + \Big(\dfrac{1}{2x}\Big)^2 + 2 \times x \times \dfrac{1}{2x} - 7\Big(x + \dfrac{1}{2x}\Big) \\[1em] = \Big(x + \dfrac{1}{2x}\Big)^2 - 7\Big(x + \dfrac{1}{2x}\Big) \\[1em] = \Big(x + \dfrac{1}{2x}\Big)\Big(x + \dfrac{1}{2x} - 7\Big).$

Hence, $x^2 + \dfrac{1}{4x^2} + 1 - 7x - \dfrac{7}{2x} = \Big(x + \dfrac{1}{2x}\Big)\Big(x + \dfrac{1}{2x} - 7\Big).$

$9a^2 + \dfrac{1}{9a^2} - 2 - 12a + \dfrac{4}{3a}$

Answer

Given,

$\phantom{=} 9a^2 + \dfrac{1}{9a^2} - 2 - 12a + \dfrac{4}{3a} \\[1em] = (3a)^2 + \Big(\dfrac{1}{3a}\Big)^2 - 2 - 4\Big(3a - \dfrac{1}{3a}\Big) \\[1em] = \Big(3a - \dfrac{1}{3a}\Big)^2 - 4\Big(3a - \dfrac{1}{3a}\Big) \\[1em] = \Big(3a - \dfrac{1}{3a}\Big)\Big(3a - \dfrac{1}{3a} - 4\Big).$

Hence, $9a^2 + \dfrac{1}{9a^2} - 2 - 12a + \dfrac{4}{3a} = \Big(3a - \dfrac{1}{3a}\Big)\Big(3a - \dfrac{1}{3a} - 4\Big).$

$x^2 + \dfrac{a^2 + 1}{a}x + 1$

Answer

Given,

$\phantom{=} x^2 + \dfrac{a^2 + 1}{a}x + 1 \\[1em] = \dfrac{ax^2 + (a^2 + 1)x + a}{a} \\[1em] = \dfrac{ax^2 + a^2x + x + a}{a} \\[1em] = \dfrac{ax(x + a) + (x + a)}{a} \\[1em] = \dfrac{(x + a)(ax + 1)}{a} \\[1em] = (x + a)\dfrac{ax + 1}{a} \\[1em] = (x + a)\Big(x + \dfrac{1}{a}\Big).$

Hence, $x^2 + \dfrac{a^2 + 1}{a}x + 1 = (x + a)\Big(x + \dfrac{1}{a}\Big).$

x⁴ + y⁴ - 27x²y²

Answer

Given,

    x⁴ + y⁴ - 27x²y²

= x⁴ + y⁴ - 2x²y² - 25x²y²

= (x² - y²)² - (5xy)²

= (x² - y² + 5xy)(x² - y² - 5xy).

Hence, x⁴ + y⁴ - 27x²y² = (x² - y² + 5xy)(x² - y² - 5xy).

4x⁴ + 9y⁴ + 11x²y²

Answer

Given,

    4x⁴ + 9y⁴ + 11x²y²

= (2x²)² + (3y²)² + 12x²y² - x²y²

= (2x²)² + (3y²)² + 2 × 2x² × 3y² - x²y²

= (2x² + 3y²)² - (xy)²

= (2x² + 3y² + xy)(2x² + 3y² - xy).

Hence, 4x⁴ + 9y⁴ + 11x²y² = (2x² + 3y² + xy)(2x² + 3y² - xy).

$x^2 + \dfrac{1}{x^2} - 3$

Answer

Given,

$\phantom{=}x^2 + \dfrac{1}{x^2} - 3 \\[1em] = x^2 + \dfrac{1}{x^2} - 2 - 1 \\[1em] = \Big(x - \dfrac{1}{x}\Big)^2 - 1^2 \\[1em] = \Big(x - \dfrac{1}{x} + 1\Big)\Big(x - \dfrac{1}{x} - 1\Big).$

Hence, $x^2 + \dfrac{1}{x^2} - 3 = \Big(x - \dfrac{1}{x} + 1\Big)\Big(x - \dfrac{1}{x} - 1\Big).$

a - b - 4a² + 4b²

Answer

Given,

    a - b - 4a² + 4b²

= a - b - 4(a² - b²)

= (a - b) - 4(a + b)(a - b)

= (a - b)[1 - 4(a + b)]

= (a - b)(1 - 4a - 4b).

Hence, a - b - 4a² + 4b² = (a - b)(1 - 4a - 4b).

(2a - 3)² - 2(2a - 3)(a - 1) + (a - 1)²

Answer

Given,

    (2a - 3)² - 2(2a - 3)(a - 1) + (a - 1)²

= [(2a - 3) - (a - 1)]²

= (2a - 3 - a + 1)²

= (a - 2)².

Hence, (2a - 3)² - 2(2a - 3)(a - 1) + (a - 1)² = (a - 2)².

(a² - 3a)(a² - 3a + 7) + 10

Answer

Given,

    (a² - 3a)(a² - 3a + 7) + 10

Substituting a² - 3a = x, we get :

⇒ x(x + 7) + 10

= x² + 7x + 10

= x² + 2x + 5x + 10

= x(x + 2) + 5(x + 2)

= (x + 5)(x + 2)

= (a² - 3a + 5)(a² - 3a + 2)

= (a² - 3a + 5)(a² - 2a - a + 2)

= (a² - 3a + 5)[a(a - 2) -1(a - 2)]

= (a² - 3a + 5)(a - 1)(a - 2)

Hence, (a² - 3a)(a² - 3a + 7) + 10 = (a² - 3a + 5)(a - 1)(a - 2).

(a² - a)(4a² - 4a - 5) - 6

Answer

Given,

    (a² - a)(4a² - 4a - 5) - 6

= (a² - a)[4(a² - a) - 5] - 6

Substituting a² - a = x, we get :

= x(4x - 5) - 6

= 4x² - 5x - 6

= 4x² - 8x + 3x - 6

= 4x(x - 2) + 3(x - 2)

= (x - 2)(4x + 3)

= (a² - a - 2)[4(a² - a) + 3]

= (a² - a - 2)(4a² - 4a + 3)

= (a² - 2a + a - 2)(4a² - 4a + 3)

= [a(a - 2) + 1(a - 2)](4a² - 4a + 3)

= (a - 2)(a + 1)(4a² - 4a + 3).

Hence, (a² - a)(4a² - 4a - 5) - 6 = (a - 2)(a + 1)(4a² - 4a + 3).

x⁴ + y⁴ - 3x²y²

Answer

Given,

    x⁴ + y⁴ - 3x²y²

= (x²)² + (y²)² - 2x²y² - x²y²

= (x² - y²)² - (xy)²

= (x² - y² + xy)(x² - y² - xy).

Hence, x⁴ + y⁴ - 3x²y² = (x² - y² + xy)(x² - y² - xy).

5a² - b² - 4ab + 7a - 7b

Answer

Given,

    5a² - b² - 4ab + 7a - 7b

= 5a² - 4ab - b² + 7a - 7b

= 5a² - 5ab + ab - b² + 7a - 7b

= 5a(a - b) + b(a - b) + 7(a - b)

= (a - b)(5a + b + 7).

Hence, 5a² - b² - 4ab + 7a - 7b = (a - b)(5a + b + 7).

12(3x - 2y)² - 3x + 2y - 1

Answer

Given,

    12(3x - 2y)² - 3x + 2y - 1

= 12(3x - 2y)² - (3x - 2y) - 1

Substituting (3x - 2y) = a, we get :

= 12a² - a - 1

= 12a² - 4a + 3a - 1

= 4a(3a - 1) + 1(3a - 1)

= (4a + 1)(3a - 1)

= [4(3x - 2y) + 1][3(3x - 2y) - 1]

= (12x - 8y + 1)(9x - 6y - 1).

Hence, 12(3x - 2y)² - 3x + 2y - 1 = (12x - 8y + 1)(9x - 6y - 1).

4(2x - 3y)² - 8x + 12y - 3

Answer

Given,

    4(2x - 3y)² - 8x + 12y - 3

= 4(2x - 3y)² - 4(2x - 3y) - 3

Substituting (2x - 3y) = a, we get :

= 4a² - 4a - 3

= 4a² - 6a + 2a - 3

= 2a(2a - 3) + 1(2a - 3)

= (2a + 1)(2a - 3)

= [2(2x - 3y) + 1][2(2x - 3y) - 3]

= (4x - 6y + 1)(4x - 6y - 3).

Hence, 4(2x - 3y)² - 8x + 12y - 3 = (4x - 6y + 1)(4x - 6y - 3).

3 - 5x + 5y - 12(x - y)²

Answer

Given,

    3 - 5x + 5y - 12(x - y)²

= 3 - 5(x - y) - 12(x - y)²

Substituting x - y = a, we get :

= 3 - 5a - 12a²

= 3 - 9a + 4a - 12a²

= 3(1 - 3a) + 4a(1 - 3a)

= (1 - 3a)(3 + 4a)

= [1 - 3(x - y)][3 + 4(x - y)]

= (1 - 3x + 3y)(3 + 4x - 4y).

Hence, 3 - 5x + 5y - 12(x - y)² = (1 - 3x + 3y)(3 + 4x - 4y).

$\dfrac{(a + b)(a^2 - ab + b^2)}{(a^2b + ab^2)}$ in simplest form is equal to:

1. $\dfrac{a}{b} - 1 + \dfrac{b}{a}$

2. $\dfrac{a^3 - b^3}{a^2b + ab^2}$

3. $\dfrac{a^2 + b^2}{ab}$

4. none of these

Answer

Given,

$\Rightarrow \dfrac{(a + b)(a^2 - ab + b^2)}{(a^2b + ab^2)}\\[1em] \Rightarrow \dfrac{(a + b)(a^2 - ab + b^2)}{ab(a + b)}\\[1em] \Rightarrow \dfrac{(a^2 - ab + b^2)}{ab}\\[1em] \Rightarrow \dfrac{a^2}{ab} - \dfrac{ab}{ab} + \dfrac{b^2}{ab} \\[1em] \Rightarrow \dfrac{a}{b} - 1 + \dfrac{b}{a}$

Hence, option 1 is the correct option.

L.C.M. of x² + 3x + 2 and x² - 2x - 3 in simplest form is:

1. (x + 1)²(x + 2)(x - 3)

2. (x² + 3x + 2)(x² - 2x - 3)

3. (x + 1)(x + 2)(x - 3)

4. none of these

Answer

Given, x² + 3x + 2 and x² - 2x - 3

The factors of x² + 3x + 2

⇒ x² + 2x + x + 2

⇒ x(x + 2) + 1(x + 2)

⇒ (x + 2)(x + 1).

The factors of x² - 2x - 3

⇒ x² - 3x + x - 3

⇒ x(x - 3) + 1(x - 3)

⇒ (x - 3)(x + 1)

L.C.M. = (x + 1)(x + 2)(x - 3)

Hence, option 3 is the correct option.

H.C.F of x² + 3x + 2 and x² - 2x - 3 is :

1. (x + 1)

2. (x + 1)(x + 2)(x - 3)

3. 1

4. none of these

Answer

Given, x² + 3x + 2 and x² - 2x - 3

The factors of x² + 3x + 2

⇒ x² + 2x + x + 2

⇒ x(x + 2) + 1(x + 2)

⇒ (x + 2)(x + 1)

The factors of x² - 2x - 3

⇒ x² - 3x + x - 3

⇒ x(x - 3) + 1(x - 3)

⇒ (x - 3)(x + 1)

H.C.F. = (x + 1)

Hence, option 1 is the correct option.

(3a - 1)² - 6a + 2 is equal to:

1. (3a - 1)(a - 1)

2. 3(3a - 1)(a - 1)

3. (3a - 1)(a + 1)

4. 3(3a - 1)(a + 1)

Answer

Solving,

⇒ (3a - 1)² - 6a + 2

⇒ (3a)² + 1² - 2 x 3a x 1 - 6a + 2

⇒ 9a² + 1 - 6a - 6a + 2

⇒ 9a² - 12a + 3

⇒ 3(3a² - 4a + 1)

⇒ 3(3a² - 3a - a + 1)

⇒ 3[3a(a - 1) - 1(a - 1)]

⇒ 3(3a - 1)(a - 1).

Hence, option 2 is the correct option.

$\dfrac{1}{3}$ x² - 2x - 9 is equal to:

1. $\dfrac{1}{3}$ (x - 9)(x + 3)

2. $\dfrac{1}{3}$ (x - 9)(x - 3)

3. $\dfrac{1}{3}$ (x + 9)(x - 3)

4. $\dfrac{1}{3}$ (x + 9)(x + 3)

Answer

Given,

$\dfrac{1}{3} x^2 - 2x - 9\\[1em] \Rightarrow \dfrac{1}{3} x^2 - \dfrac{6}{3}x - \dfrac{27}{3}\\[1em] \Rightarrow \dfrac{1}{3} (x^2 - 6x - 27)\\[1em] \Rightarrow \dfrac{1}{3} (x^2 - 9x + 3x - 27)\\[1em] \Rightarrow \dfrac{1}{3} [x(x - 9) + 3(x - 9)]\\[1em] \Rightarrow \dfrac{1}{3} (x - 9)(x + 3)$

Hence, option 1 is the correct option.

(x² + 3x) men can do a piece of work in (x² - 2x) days, then one day work of 1 man is :

1. $\dfrac{x + 3}{x - 2}$

2. (x² + 3x)(x² - 2x)

3. $\dfrac{x - 2}{x + 3}$

4. none of these

Answer

Given, total number of men = (x² + 3x)

Total number of days = (x² - 2x)

Total work = (x² + 3x)(x² - 2x)

One day work of 1 man = $\dfrac{1}{(x^2 + 3x)(x^2 - 2x)}$

Hence, option 4 is the correct option.

Statement 1: ₹ (x³ - x) is spent in buying some identical articles at ₹ (x - 1) each. Number of articles bought = $\dfrac{x^3 - x}{x - 1}$.

Statement 2: The number of articles bought

= $\dfrac{x^3 - x}{x - 1} = \dfrac{x(x - 1)(x + 1)}{x - 1} = x^2 + x$

1. Both the statements are true.

2. Both the statements are false.

3. Statement 1 is true, and statement 2 is false.

4. Statement 1 is false, and statement 2 is true.

Answer

Given, cost of each article = ₹ (x - 1)

Total cost = ₹ (x³ - x)

$\Rightarrow \text{No. of articles bought} = \dfrac{\text{Total cost}}{\text{Cost of each article}} \\[1em] = \dfrac{x^3 - x}{x - 1}\\[1em] = \dfrac{x(x^2 - 1)}{x - 1}\\[1em] = \dfrac{x(x^2 - 1^2)}{x - 1}\\[1em] = \dfrac{x(x - 1)(x + 1)}{x - 1}\\[1em] = x(x + 1)\\[1em] = x^2 + x.$

∴ Both the statements are true.

Hence, option 1 is the correct option.

Statement 1: The area of rectangle is x² - 5x + 6 and the longer side of the rectangle is (x - 2).

Statement 2: x² - 5x + 6

= (x - 2) (x - 3)

⇒ for every positive value of x(x > 3), (x - 2) is greater.

1. Both the statements are true.

2. Both the statements are false.

3. Statement 1 is true, and statement 2 is false.

4. Statement 1 is false, and statement 2 is true.

Answer

Give, Area of rectangle = x² - 5x + 6

Longer side = (x - 2)

Factorise the area,

⇒ x² - 5x + 6 = 0

⇒ x² - 3x - 2x + 6 = 0

⇒ x(x - 3) - 2(x - 3) = 0

⇒ (x - 2) (x - 3) = 0

So, the given two sides of rectangle are:

(x - 2) (x - 3)

Since x > 3,

∴ (x - 2) > (x - 3)

So, (x - 2) is the longer side.

∴ Both the statements are true.

Hence, option 1 is the correct option.

Assertion (A): Distance of (x² - 7x + 12) km is covered in (x² - 16) hrs.

Speed = $\dfrac{x^2 - 7x + 12}{x^2 - 16}$ km/hr

Reason (R): Speed = Distance x Time

= (x² - 7x + 12)(x² - 16) km/hr

1. A is true, but R is false.

2. A is false, but R is true.

3. Both A and R are true, and R is the correct reason for A.

4. Both A and R are true, and R is the incorrect reason for A.

Answer

Given,

Distance = (x² - 7x + 12) km

Time = (x² - 16) hrs

By formula,

Speed = $\dfrac{\text{Distance}}{\text{Time}}$

= $\dfrac{x^2 - 7x + 12}{x^2 - 16}$ km/hr

∴ A is true, but R is false.

Hence, option 1 is the correct option.

Factorise :

$a^2 + \dfrac{1}{a^2} - 2 - 3a + \dfrac{3}{a}$

Answer

Given,

$\phantom{=}a^2 + \dfrac{1}{a^2} - 2 - 3a + \dfrac{3}{a} \\[1em] = \Big(a - \dfrac{1}{a}\Big)^2 - 3\Big(a - \dfrac{1}{a}\Big) \\[1em] = \Big(a - \dfrac{1}{a}\Big)\Big(a - \dfrac{1}{a} - 3\Big).$

Hence, $a^2 + \dfrac{1}{a^2} - 2 - 3a + \dfrac{3}{a} = \Big(a - \dfrac{1}{a}\Big)\Big(a - \dfrac{1}{a} - 3\Big).$

x² + y² + x + y + 2xy

Answer

Given,

    x² + y² + x + y + 2xy

= x² + y² + 2xy + x + y

= (x + y)² + (x + y)

= (x + y)(x + y + 1).

Hence, x² + y² + x + y + 2xy = (x + y)(x + y + 1).

a² + 4b² - 3a + 6b - 4ab

Answer

Given,

    a² + 4b² - 3a + 6b - 4ab

= a² + 4b² - 4ab - 3a + 6b

= a² + (2b)² - 2 × a × 2b - 3a + 6b

= (a - 2b)² - 3(a - 2b)

= (a - 2b)(a - 2b - 3).

Hence, a² + 4b² - 3a + 6b - 4ab =(a - 2b)(a - 2b - 3).

m(x - 3y)² + n(3y - x) + 5x - 15y

Answer

Given,

    m(x - 3y)² + n(3y - x) + 5x - 15y

= m(x - 3y)² - n(x - 3y) + 5(x - 3y)

= (x - 3y)[m(x - 3y) - n + 5]

= (x - 3y)(mx - 3my - n + 5).

Hence, m(x - 3y)² + n(3y - x) + 5x - 15y = (x - 3y)(mx - 3my - n + 5).

x(6x - 5y) - 4(6x - 5y)²

Answer

Given,

    x(6x - 5y) - 4(6x - 5y)²

= (6x - 5y)[x - 4(6x - 5y)]

= (6x - 5y)(x - 24x + 20y)

= (6x - 5y)(20y - 23x).

Hence, x(6x - 5y) - 4(6x - 5y)² = (6x - 5y)(20y - 23x).

$\dfrac{1}{35} + \dfrac{12}{35}a + a^2$

Answer

Given,

$\phantom{=}\dfrac{1}{35} + \dfrac{12}{35}a + a^2 \\[1em] = a^2 + \dfrac{12}{35}a + \dfrac{1}{35} \\[1em] = \dfrac{35a^2 + 12a + 1}{35} \\[1em] = \dfrac{35a^2 + 5a + 7a + 1}{35} \\[1em] = \dfrac{5a(7a + 1) + 1(7a + 1)}{35} \\[1em] = \dfrac{(7a + 1)(5a + 1)}{35}.$

Hence, $\dfrac{1}{35} + \dfrac{12}{35}a + a^2 = \dfrac{(7a + 1)(5a + 1)}{35}$.

(x² - 3x)(x² - 3x - 1) - 20

Answer

Given,

    (x² - 3x)(x² - 3x - 1) - 20

Substituting x² - 3x = a, we get :

⇒ a(a - 1) - 20

= a² - a - 20

= a² - 5a + 4a - 20

= a(a - 5) + 4(a - 5)

= (a - 5)(a + 4)

= (x² - 3x - 5)(x² - 3x + 4).

Hence, (x² - 3x)(x² - 3x - 1) - 20 = (x² - 3x - 5)(x² - 3x + 4).

For each trinomial (quadratic expression), given below, find whether it is factorisable or not. Factorise, if possible.

(i) x² - 3x - 54

(ii) 2x² - 7x - 15

(iii) 2x² + 2x - 75

(iv) 3x² + 4x - 10

(v) x(2x - 1) - 1

Answer

(i) Given,

    x² - 3x - 54

= x² - 9x + 6x - 54

= x(x - 9) + 6(x - 9)

= (x - 9)(x + 6).

Hence, the above equation is factorisable and x² - 3x - 54 = (x - 9)(x + 6).

(ii) Given,

    2x² - 7x - 15

= 2x² - 10x + 3x - 15

= 2x(x - 5) + 3(x - 5)

= (2x + 3)(x - 5).

Hence, the above equation is factorisable and 2x² - 7x - 15 = (2x + 3)(x - 5).

(iii) Given,

    2x² + 2x - 75

Hence, the above equation is not factorisable.

(iv) Given,

    3x² + 4x - 10

Hence, the above equation is not factorisable.

(v) Given,

    x(2x - 1) - 1

= 2x² - x - 1

= 2x² - 2x + x - 1

= 2x(x - 1) + 1(x - 1)

= (x - 1)(2x + 1).

Hence, the above equation is factorisable and x(2x - 1) - 1 = (x - 1)(2x + 1).

Factorise :

(i) $4\sqrt{3}x^2 + 5x - 2\sqrt{3}$

(ii) $7\sqrt{2}x^2 - 10x - 4\sqrt{2}$

Answer

(i) Given,

$\phantom{=}4\sqrt{3}x^2 + 5x - 2\sqrt{3} \\[1em] = 4\sqrt{3}x^2 + 8x - 3x - 2\sqrt{3}\\[1em] = 4x(\sqrt{3}x + 2) - \sqrt{3}(\sqrt{3}x + 2) \\[1em] = (\sqrt{3}x + 2)(4x - \sqrt{3}).$

Hence, $4\sqrt{3}x^2 + 5x - 2\sqrt{3} = (\sqrt{3}x + 2)(4x - \sqrt{3}).$

(ii) Given,

$\phantom{=}7\sqrt{2}x^2 - 10x - 4\sqrt{2} \\[1em] = 7\sqrt{2}x^2 - 14x + 4x - 4\sqrt{2}\\[1em] = 7\sqrt{2}x(x - \sqrt{2}) + 4(x - \sqrt{2}) \\[1em] = (x - \sqrt{2})(7\sqrt{2}x + 4).$

Hence, $7\sqrt{2}x^2 - 10x - 4\sqrt{2} = (x - \sqrt{2})(7\sqrt{2}x + 4).$

Give possible expressions for the length and the breadth of the rectangle whose area is

12x² - 35x + 25.

Answer

Given,

Area = 12x² - 35x + 25

⇒ lb = 12x² - 35x + 25

⇒ lb = 12x² - 15x - 20x + 25

⇒ lb = 3x(4x - 5) - 5(4x - 5)

⇒ lb = (4x - 5)(3x - 5).

Hence, if length = (4x - 5) then breadth = (3x - 5) and if length = (3x - 5) then breadth = (4x - 5).

Factorise :

9a² - (a² - 4)²

Answer

Given,

    9a² - (a² - 4)²

= (3a)² - (a² - 4)²

= (3a + a² - 4)[3a - (a² - 4)]

= (a² + 3a - 4)(4 - a² + 3a)

= (a² + 4a - a - 4).-(a² - 3a - 4)

= [a(a + 4) - 1(a + 4)].-[a² - 4a + a - 4]

= (a + 4)(a - 1).-[a(a - 4) + 1(a - 4)]

= (a + 4)(a - 1).-(a - 4)(a + 1)

= (a + 4)(a - 1)(4 - a)(a + 1).

Hence, 9a² - (a² - 4)² = (a + 4)(a - 1)(4 - a)(a + 1).

$x^2 + \dfrac{1}{x^2} - 11$

Answer

Given,

$\phantom{=} x^2 + \dfrac{1}{x^2} - 11 \\[1em] = x^2 + \dfrac{1}{x^2} - 2 - 9 \\[1em] = \Big(x - \dfrac{1}{x}\Big)^2 - 9 \\[1em] = \Big(x - \dfrac{1}{x}\Big)^2 - 3^2 \\[1em] = \Big(x - \dfrac{1}{x} + 3\Big)\Big(x - \dfrac{1}{x} - 3\Big).$

Hence, $x^2 + \dfrac{1}{x^2} - 11 = \Big(x - \dfrac{1}{x} + 3\Big)\Big(x - \dfrac{1}{x} - 3\Big).$

$4x^2 + \dfrac{1}{4x^2} + 1$

Answer

Given,

$\phantom{=} 4x^2 + \dfrac{1}{4x^2} + 1 \\[1em] = 4x^2 + \dfrac{1}{4x^2} + 2 - 1 \\[1em] = (2x)^2 + \Big(\dfrac{1}{2x}\Big)^2 + 2 \times 2x \times \dfrac{1}{2x} - 1 \\[1em] = \Big(2x + \dfrac{1}{2x}\Big)^2 - 1^2 \\[1em] = \Big(2x + \dfrac{1}{2x} + 1\Big)\Big(2x + \dfrac{1}{2x} - 1\Big).$

Hence, $4x^2 + \dfrac{1}{4x^2} + 1 = \Big(2x + \dfrac{1}{2x} + 1\Big)\Big(2x + \dfrac{1}{2x} - 1\Big).$

4x⁴ - x² - 12x - 36

Answer

Given,

    4x⁴ - x² - 12x - 36

= 4x⁴ - [x² + 12x + 36]

= 4x⁴ - [x² + 6x + 6x + 36]

= 4x⁴ - [x(x + 6) + 6(x + 6)]

= 4x⁴ - (x + 6)(x + 6)

= (2x²)² - (x + 6)²

= (2x² + x + 6)(2x² - x - 6).

= (2x² + x + 6)(2x² - 4x + 3x - 6)

= (2x² + x + 6)[2x(x - 2) + 3(x - 2)]

= (2x² + x + 6)(x - 2)(2x + 3).

Hence, 4x⁴ - x² - 12x - 36 = (2x² + x + 6)(x - 2)(2x + 3).

a²(b + c) - (b + c)³

Answer

Given,

    a²(b + c) - (b + c)³

= (b + c)[a² - (b + c)²]

= (b + c)(a + b + c)[a - (b + c)]

= (b + c)(a + b + c)(a - b - c).

Hence, a²(b + c) - (b + c)³ = (b + c)(a + b + c)(a - b - c).

2x³ + 54y³ - 4x - 12y

Answer

Given,

    2x³ + 54y³ - 4x - 12y

= 2(x³ + 27y³) - 4(x + 3y)

= 2[(x)³ + (3y)³] - 4(x + 3y)

= 2(x + 3y)(x² - 3xy + 9y²) - 4(x + 3y) [∵ a³ + b³ = (a + b)(a² - ab + b²)]

= 2(x + 3y)(x² - 3xy + 9y² - 2).

Hence, 2x³ + 54y³ - 4x - 12y = 2(x + 3y)(x² - 3xy + 9y² - 2).

1029 - 3x³

Answer

Given,

    1029 - 3x³

= 3(343 - x³)

= 3[(7)³ - (x)³]

= 3(7 - x)[(7)² + 7x + x²] [∵ a³ - b³ = (a - b)(a² + ab + b²)]

= 3(7 - x)(x² + 7x + 49).

Hence, 1029 - 3x³ = 3(7 - x)(x² + 7x + 49).

Show that :

(i) 13³ - 5³ is divisible by 8.

(ii) 35³ + 27³ is divisible by 62.

Answer

(i) We know that

a³ - b³ = (a - b)(a² + ab + b²)

Factorising 13³ - 5³, we get :

⇒ 13³ - 5³ = (13 - 5)[13² + 13 × 5 + 5²]

= 8(169 + 65 + 25)

= 8 × 259, which is divisible by 8.

Hence, proved that 13³ - 5³ is divisible by 8.

(ii) We know that

a³ + b³ = (a + b)(a² - ab + b²)

Factorising 35³ + 27³, we get :

⇒ 35³ + 27³ = (35 + 27)[(35)² - 35 × 27 + (27)²]

= 62[1225 - 945 + 729]

= 62 × 1009, which is divisible by 62.

Hence, proved that 35³ + 27³ is divisible by 62.

Evaluate :

$\dfrac{5.67 \times 5.67 \times 5.67 + 4.33 \times 4.33 \times 4.33}{5.67 \times 5.67 - 5.67 \times 4.33 + 4.33 \times 4.33}$

Answer

Substituting a = 5.67 and b = 4.33, we get :

$\Rightarrow \dfrac{a \times a \times a + b \times b \times b}{a \times a - a \times b + b \times b} \\[1em] \Rightarrow \dfrac{a^3 + b^3}{a^2 - ab + b^2} \\[1em] \Rightarrow \dfrac{(a + b)(a^2 - ab + b^2)}{a^2 - ab + b^2} \\[1em] \Rightarrow (a + b) \\[1em] \Rightarrow (5.67 + 4.33) \\[1em] \Rightarrow 10.$

Hence, $\dfrac{5.67 \times 5.67 \times 5.67 + 4.33 \times 4.33 \times 4.33}{5.67 \times 5.67 - 5.67 \times 4.33 + 4.33 \times 4.33} = 10$.

9x² + 3x - 8y - 64y²

Answer

Given,

    9x² + 3x - 8y - 64y²

= 9x² - 64y² + 3x - 8y

= (3x)² - (8y)² + 3x - 8y

= (3x + 8y)(3x - 8y) + (3x - 8y)

= (3x - 8y)(3x + 8y + 1).

Hence, 9x² + 3x - 8y - 64y² = (3x - 8y)(3x + 8y + 1).

$2\sqrt{3}x^2 + x - 5\sqrt{3}$

Answer

Given,

$\phantom{=} 2\sqrt{3}x^2 + x - 5\sqrt{3} \\[1em] = 2\sqrt{3}x^2 + 6x - 5x - 5\sqrt{3} \\[1em] = 2\sqrt{3}x(x + \sqrt{3}) - 5(x + \sqrt{3}) \\[1em] = (x + \sqrt{3})(2\sqrt{3}x - 5).$

Hence, $2\sqrt{3}x^2 + x - 5\sqrt{3} = (x + \sqrt{3})(2\sqrt{3}x - 5).$

$\dfrac{1}{4}(a + b)^2 - \dfrac{9}{16}(2a - b)^2$

Answer

Given,

$\phantom{=}\dfrac{1}{4}(a + b)^2 - \dfrac{9}{16}(2a - b)^2 \\[1em] = \Big[\dfrac{1}{2}(a + b)\Big]^2 - \Big[\dfrac{3}{4}(2a - b)\Big]^2 \\[1em] = \Big[\dfrac{1}{2}(a + b) + \dfrac{3}{4}(2a - b)\Big]\Big[\dfrac{1}{2}(a + b) - \dfrac{3}{4}(2a - b)\Big] \\[1em] = \Big[\dfrac{2(a + b) + 3(2a - b)}{4}\Big]\Big[\dfrac{2(a + b) - 3(2a - b)}{4}\Big] \\[1em] = \Big(\dfrac{2a + 2b + 6a - 3b}{4}\Big)\Big(\dfrac{2a + 2b - 6a + 3b}{4}\Big)\\[1em] = \dfrac{8a - b}{4} \times \dfrac{5b - 4a}{4} \\[1em] = \dfrac{1}{16}(8a - b)(5b - 4a).$