Factorisation

Factorize :

5x² - 20xy

Answer

Given,

⇒ 5x² - 20xy

⇒ 5x(x - 4y)

Hence, 5x² - 20xy = 5x(x - 4y).

Factorize :

18a²b - 24abc

Answer

Given,

⇒ 18a²b - 24abc

⇒ 6ab(3a - 4c)

Hence, 18a²b - 24abc = 6ab(3a - 4c).

Factorize :

27x³y³ - 45x⁴y²

Answer

Given,

⇒ 27x³y³ - 45x⁴y²

⇒ 9x³y²(3y - 5x)

Hence, 27x³y³ - 45x⁴y² = 9x³y²(3y - 5x).

Factorize :

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

Answer

Given,

⇒ 5a(b + c) - 7b(b + c)

⇒ (b + c)(5a - 7b)

Hence, 5a(b + c) - 7b(b + c) = (b + c)(5a - 7b).

Factorize :

2x(p² + q²) + 4y(p² + q²)

Answer

Given,

⇒ 2x(p² + q²) + 4y(p² + q²)

⇒ (2x + 4y)(p² + q²)

⇒ 2(x + 2y)(p² + q²)

Hence, 2x(p² + q²) + 4y(p² + q²) = 2(x + 2y)(p² + q²).

Factorize :

x(a - 5) + y(5 - a)

Answer

Given,

⇒ x(a - 5) + y(5 - a)

⇒ x(a - 5) - y(a - 5)

⇒ (a - 5)(x - y)

Hence, x(a - 5) + y(5 - a) = (a - 5)(x - y).

Factorize :

4(x + y) - 6(x + y)²

Answer

Given,

⇒ 4(x + y) - 6(x + y)²

⇒ (x + y)[4 - 6(x + y)]

⇒ (x + y).2.[2 - 3(x + y)]

⇒ 2(x + y)(2 - 3x - 3y)

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

Factorize :

6(2a + 3b)² - 8(2a + 3b)

Answer

Given,

⇒ 6(2a + 3b)² - 8(2a + 3b)

⇒ (2a + 3b)[6(2a + 3b) - 8]

⇒ (2a + 3b).2.[3(2a + 3b) - 4]

⇒ 2(2a + 3b)(6a + 9b - 4).

Hence, 6(2a + 3b)² - 8(2a + 3b) = 2(2a + 3b)(6a + 9b - 4).

Factorize :

x(x + y)³ - 3x²y(x + y)

Answer

Given,

⇒ x(x + y)³ - 3x²y(x + y)

⇒ x(x + y)[(x + y)² - 3xy]

⇒ x(x + y)[x² + y² + 2xy - 3xy]

⇒ x(x + y)(x² + y² - xy).

Hence, x(x + y)³ - 3x²y(x + y) = x(x + y)(x² + y² - xy).

Factorize :

a³ + 2a² + 5a + 10

Answer

Given,

⇒ a³ + 2a² + 5a + 10

⇒ a²(a + 2) + 5(a + 2)

⇒ (a + 2)(a² + 5).

Hence, a³ + 2a² + 5a + 10 = (a + 2) (a² + 5).

Factorize :

x² + xy - 2xz - 2yz

Answer

Given,

⇒ x² + xy - 2xz - 2yz

⇒ x(x + y) -2z (x + y)

⇒ (x + y)(x - 2z)

Hence, x² + xy - 2xz - 2yz = (x + y)(x - 2z).

Factorize :

a³b - a²b + 5ab - 5b

Answer

Given,

⇒ a³b - a²b + 5ab - 5b

⇒ a²b(a - 1) + 5b (a - 1)

⇒ (a²b + 5b)(a - 1)

⇒ b(a² + 5)(a - 1).

Hence, a³b - a²b + 5ab - 5b = b(a² + 5)(a - 1).

Factorize :

x² + y - xy - x

Answer

Given,

⇒ x² + y - xy - x

⇒ x² - x - xy + y

⇒ x(x - 1) - y(x - 1)

⇒ (x - 1)(x - y).

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

Factorize :

a(a + b - c) - bc

Answer

Given,

⇒ a(a + b - c) - bc

⇒ a² + ab - ac - bc

⇒ a(a + b) - c(a + b)

⇒ (a - c)(a + b)

Hence, a(a + b - c) - bc = (a - c)(a + b).

Factorize :

(4a - 1)² - 8a + 2

Answer

Given,

⇒ (4a - 1)² - 8a + 2

⇒ (4a - 1)² - 8a + 2

⇒ 16a² - 8a + 1 - 8a + 2

⇒ 16a² -16a + 3

⇒ 16a² -12a - 4a + 3

⇒ 4a(4a - 3) - 1(4a - 3)

⇒ (4a - 1)(4a - 3).

Hence, (4a - 1)² - 8a + 2 = (4a - 1)(4a - 3).

Factorize :

8 - 4a - 2a³ + a⁴

Answer

Given,

⇒ 8 - 4a - 2a³ + a⁴

⇒ 4(2 - a) - a³(2 - a)

⇒ (2 - a)(4 - a³)

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

Factorize :

2a² + bc - 2ab - ac

Answer

Given,

⇒ 2a² + bc - 2ab - ac

⇒ 2a² - 2ab - ac + bc

⇒ 2a(a - b) - c (a - b)

⇒ (2a - c)(a - b).

Hence, 2a² + bc - 2ab - ac = (2a - c)(a - b).

Factorize :

a(a - 2b - c) + 2bc

Answer

Given,

⇒ a(a - 2b - c) + 2bc

⇒ a² - 2ab - ac + 2bc

⇒ a² - ac - 2ab + 2bc

⇒ a(a - c) - 2b(a - c)

⇒ (a - c)(a - 2b)

Hence, a(a - 2b - c) + 2bc = (a - c)(a - 2b).

Factorize :

x² - (a + b)x + ab

Answer

Given,

⇒ x² - (a + b)x + ab

⇒ x² - ax - bx + ab

⇒ x(x - a) - b(x - a)

⇒ (x - a)(x - b).

Hence, x² - (a + b)x + ab = (x - a)(x - b).

Factorize :

3ax - 6ay - 8by + 4bx

Answer

Given,

⇒ 3ax - 6ay - 8by + 4bx

⇒ 3ax + 4bx - 6ay - 8by

⇒ x(3a + 4b) - 2y(3a + 4b)

⇒ (3a + 4b)(x - 2y)

Hence, 3ax - 6ay - 8by + 4bx = (3a + 4b)(x - 2y).

Factorize :

ab(x² + y²) - xy(a² + b²)

Answer

Given,

⇒ ab(x² + y²) - xy(a² + b²)

⇒ abx² + aby² - xya² - xyb²

⇒ abx² - xya² - xyb² + aby²

⇒ ax(bx - ay) - by(bx - ay)

⇒ (bx - ay)(ax - by).

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

Factorize :

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).

Factorize :

a³ + ab(1 - 2a) - 2b²

Answer

Given,

⇒ a³ + ab(1 - 2a) - 2b²

⇒ a³ + ab - 2a²b - 2b²

⇒ a³ - 2a²b + ab - 2b²

⇒ a²(a - 2b) + b(a - 2b)

⇒ (a - 2b)(a² + b).

Hence, a³ + ab(1 - 2a) - 2b² = (a - 2b)(a² + b).

Factorize :

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

Answer

Given,

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

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

Factorize:

x² - 49

Answer

Given,

⇒ x² - 49

⇒ x² - 7²

⇒ (x + 7)(x - 7).

Hence, x² - 49 = (x + 7)(x - 7).

Factorize:

25x² - 64y²

Answer

Given,

⇒ 25x² - 64y²

⇒ (5x)² - (8y)²

⇒ (5x + 8y)(5x - 8y).

Hence, 25x² - 64y² = (5x + 8y)(5x - 8y).

Factorize:

100 - 9p²

Answer

Given,

⇒ 100 - 9p²

⇒ (10)² - (3p)²

⇒ (10 + 3p)(10 - 3p).

Hence, 100 - 9p² = (10 + 3p)(10 - 3p).

Factorize:

80 - 5a²

Answer

Given,

⇒ 80 - 5a²

⇒ 5(16 - a²)

⇒ 5[(4)² - (a)²]

⇒ 5(4 + a)(4 - a).

Hence, 80 - 5a² = 5(4 + a)(4 - a).

Factorize:

32x² - 18y²

Answer

Given,

⇒ 32x² - 18y²

⇒ 2(16x² - 9y²)

⇒ 2[(4x)² - (3y)²]

⇒ 2(4x + 3y)(4x - 3y)

Hence, 32x² - 18y² =2(4x + 3y)(4x - 3y).

Factorize:

3x³ - 48x

Answer

Given,

⇒ 3x³ - 48x

⇒ 3x(x² - 16)

⇒ 3x[(x)² - (4)²]

⇒ 3x(x + 4)(x - 4)

Hence, 3x³ - 48x = 3x(x + 4)(x - 4).

Factorize:

x⁴ - 81

Answer

Given,

⇒ x⁴ - 81

⇒ (x²)² - (9)²

⇒ (x² + 9)(x² - 9)

⇒ (x² + 9)[x² - (3)²]

⇒ (x² + 9)(x + 3)(x - 3).

Hence, x⁴ - 81 = (x² + 9)(x + 3)(x - 3).

Factorize:

2x⁴ - 32

Answer

Given,

⇒ 2x⁴ - 32

⇒ 2(x⁴ - 16)

⇒ 2[(x²)² - (4)²]

⇒ 2(x² + 4)(x² - 4)

⇒ 2(x² + 4)[(x)² - (2)²]

⇒ 2(x² + 4)(x + 2)(x - 2)

Hence, 2x⁴ - 32 = 2(x² + 4)(x + 2)(x - 2).

Factorize:

x³ - 5x² - x + 5

Answer

Given,

⇒ x³ - 5x² - x + 5

⇒ x² (x - 5) - 1(x - 5)

⇒ (x² - 1)(x - 5)

⇒ [(x)² - (1)²] (x - 5)

⇒ (x + 1)(x - 1)(x - 5)

Hence, x³ - 5x² - x + 5 = (x + 1)(x - 1)(x - 5).

Factorize:

9(x + a)² - 4x²

Answer

Given,

⇒ 9(x + a)² - 4x²

⇒ [3(x + a)]² - (2x)²

⇒ [3(x + a) + 2x] [3(x + a) - 2x]

⇒ (3x + 3a + 2x)(3x + 3a - 2x)

⇒ (5x + 3a)(x + 3a).

Hence, 9(x + a)² - 4x² = (5x + 3a)(x + 3a).

Factorize:

9(b + 2a)² - 4a²

Answer

Given,

⇒ 9(b + 2a)² - 4a²

⇒ [3(b + 2a)]² - (2a)²

⇒ [3(b + 2a) + 2a][3(b + 2a) - 2a]

⇒ (3b + 6a + 2a)(3b + 6a - 2a)

⇒ (3b + 8a)(3b + 4a).

Hence, 9(b + 2a)² - 4a² = (3b + 8a)(3b + 4a).

Factorize:

3 - 12(a - b)²

Answer

Given,

⇒ 3 - 12(a - b)²

⇒ 3[1 - 4(a - b)²]

⇒ 3[(1)² - [2(a - b)]²]

⇒ 3[1 + 2(a - b)] [1 - 2(a - b)]

⇒ 3(1 + 2a - 2b)(1 - 2a + 2b)

Hence, 3 - 12(a - b)² = 3(1 + 2a - 2b)(1 - 2a + 2b).

Factorize:

50a² - 2(b - c)²

Answer

Given,

⇒ 50a² - 2(b - c)²

⇒ 2[25a² - (b - c)²]

⇒ 2[(5a)² - (b - c)²]

⇒ 2[5a + (b - c)] [5a - (b - c)]

⇒ 2(5a + b - c)(5a - b + c).

Hence, 50a² - 2(b - c)² = 2(5a + b - c)(5a - b + c).

Factorize:

2(x - 3)² - 32

Answer

Given,

⇒ 2(x - 3)² - 32

⇒ 2[(x - 3)² - 16]

⇒ 2[(x - 3)² - (4)²]

⇒ 2(x - 3 + 4)(x - 3 - 4)

⇒ 2(x - 3 + 4)(x - 3 - 4)

⇒ 2(x + 1)(x - 7).

Hence, 2(x - 3)² - 32 = 2(x + 1)(x - 7).

Factorize:

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).

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

Factorize:

x² - 1 - 2a - a²

Answer

Given,

⇒ x² - 1 - 2a - a²

⇒ x² - (a² + 2a + 1)

⇒ (x)² - (a + 1)²

⇒ (x + a + 1)[x - (a + 1)]

⇒ (x + a + 1)(x - a - 1).

Hence, x² - 1 - 2a - a² = (x + a + 1)(x - a - 1).

Factorize:

x² - y² + 2yz - z²

Answer

Given,

⇒ x² - y² + 2yz - z²

⇒ x² - (y² - 2yz + z²)

⇒ (x)² - (y - z)²

⇒ (x + y - z)[x - (y - z)]

⇒ (x + y - z)(x - y + z).

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

Factorize:

x² - y² - 4xz + 4z²

Answer

Given,

⇒ x² - y² - 4xz + 4z²

⇒ x² - 4xz + 4z² - y²

⇒ (x - 2z)² - (y)²

⇒ (x - 2z + y)(x - 2z - y).

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

Factorize:

x² - 4x + 4y - y²

Answer

Given,

⇒ x² - 4x + 4y - y²

⇒ (x² - 4x) - (y² - 4y)

⇒ (x² - 4x) + 4 - 4 - (y² - 4y)

⇒ (x² - 4x + 4) - (y² - 4y + 4)

⇒ (x - 2)² - (y - 2)²

⇒ (x - 2 + y - 2)[x - 2 - (y - 2)]

⇒ (x - 2 + y - 2)(x - 2 - y + 2)

⇒ (x + y - 4)(x - y).

Hence, x² - 4x + 4y - y² = (x - y)(x + y - 4).

Factorize:

x - y - x² + y²

Answer

Given,

⇒ x - y - x² + y²

⇒ -x² + y² + x - y

⇒ -(x² - y²) + x - y

⇒ -(x + y)(x - y) + x - y

⇒ (x - y)[-(x + y) + 1]

⇒ (x - y)(1 - x - y).

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

Factorize:

x(x + z) - y(y + z)

Answer

Given,

⇒ x(x + z) - y(y + z)

⇒ x² + xz - y² - yz

⇒ x² - y² + xz - yz

⇒ x² - y² + z(x - y)

⇒ (x - y)(x + y) + z(x - y)

⇒ (x - y)(x + y + z).

Hence, x(x + z) - y(y + z) = (x - y)(x + y + z).

Factorize:

x(x - 2) - y(y - 2)

Answer

Given,

⇒ x(x - 2) - y(y - 2)

⇒ x² - 2x - y² + 2y

⇒ x² - y² + 2y - 2x

⇒ (x - y)(x + y) + 2(y - x)

⇒ (x - y)(x + y) - 2(x - y)

⇒ (x - y)(x + y - 2).

Hence, x(x - 2) - y(y - 2) = (x - y)(x + y - 2).

Factorize:

4x²y - 9y³

Answer

Given,

⇒ 4x²y - 9y³

⇒ y(4x² - 9y²)

⇒ y[(2x)² - (3y)²]

⇒ y(2x + 3y)(2x - 3y).

Hence, 4x²y - 9y³ = y(2x + 3y)(2x - 3y).

Factorize:

9x⁴ - x² - 12x - 36

Answer

Given,

⇒ 9x⁴ - x² - 12x - 36

⇒ (9x⁴) - (x² + 12x + 36)

⇒ (3x²)² - (x + 6)²

⇒ (3x² + x + 6)(3x² - x - 6).

Hence, 9x⁴ - x² - 12x - 36 = (3x² + x + 6)(3x² - x - 6).

Factorize:

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

Answer

Given,

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

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

Factorize:

x⁴ + 5x² + 9

Answer

Given,

⇒ x⁴ + 5x² + 9

⇒ (x²)² + 6x² - x² + (3)²

⇒ (x²)² + 6x² + (3)² - x²

⇒ (x²)² + 2 × 3 × x² + (3)² - x²

⇒ (x² + 3)² - (x)²     [As, (a + b)² = a² + b² + 2ab]

⇒ (x² + 3 + x)(x² + 3 - x).

Hence, x⁴ + 5x² + 9 = (x² + 3 + x)(x² + 3 - x).

Factorize:

a² + b² - c² - d² + 2ab - 2cd

Answer

Given,

⇒ a² + b² - c² - d² + 2ab - 2cd

⇒ a² + b² + 2ab - c² - d² - 2cd

⇒ (a² + b² + 2ab) - (c² + d² + 2cd)

⇒ (a + b)² - (c + d)²

⇒ [(a + b) + (c + d)][(a + b) - (c + d)]

⇒ (a + b + c + d)(a + b - c - d).

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

Factorize:

(a² - b²)(c² - d²) - 4abcd

Answer

Given,

⇒ (a² - b²)(c² - d²) - 4abcd

⇒ a²c² - a²d² - b²c² + b²d² - 4abcd

⇒ a²c² + b²d² - 2abcd - b²c² - a²d² - 2abcd

⇒ (a²c² + b²d² - 2 × ac × bd) - (a²d² + b²c² + 2 × ad × bc)

⇒ (ac - bd)² - (ad + bc)²

⇒ [(ac - bd) + (ad + bc)][(ac - bd) - (ad + bc)]

⇒ (ac - bd + ad + bc)(ac - bd - ad - bc).

Hence, (a² - b²)(c² - d²) - 4abcd = (ac - bd + ad + bc)(ac - bd - ad - bc).

Factorize:

4x² - 12ax - y² - z² - 2yz + 9a²

Answer

Given,

⇒ 4x² - 12ax - y² - z² - 2yz + 9a²

⇒ 9a² + 4x² - 12ax - (y² + z² + 2yz)

⇒ (4x² + 9a² - 12ax) - (y² + z² + 2yz)

⇒ [(2x)² + (3a)² - 2 × 3a × 2x] - (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).

Factorize:

9a² + 3a - 8b - 64b²

Answer

Given,

⇒ 9a² + 3a - 8b - 64b²

⇒ 9a² - 64b² + 3a - 8b

⇒ (3a)² - (8b)² + 3a - 8b

⇒ (3a + 8b)(3a - 8b) + 3a - 8b

⇒ (3a - 8b)[(3a + 8b) + 1]

⇒ (3a - 8b)(3a + 8b + 1).

Hence, 9a² + 3a - 8b - 64b² = (3a - 8b)(3a + 8b + 1).

Express (x² + 8x - 15)(x² - 8x - 15) as the difference of two squares.

Answer

Given,

⇒ (x² + 8x - 15)(x² - 8x - 15)

⇒ [(x² - 15) + (8x)] [(x² - 15) - (8x)]

⇒ (x² - 15)² - (8x)².

Hence, (x² + 8x - 15)(x² - 8x - 15) = (x² - 15)² - (8x)².

Evaluate:

(i) (674)² - (326)²

(ii) (18.6)² - (1.4)²

Answer

(i) Given,

⇒ (674)² - (326)²

By using the identity,

(a² - b²) = (a + b)(a - b)

⇒ (674 + 326)(674 - 326)

⇒ (1000)(348)

⇒ 348000.

Hence, (674)² - (326)² = 348000.

(ii) Given,

⇒ (18.6)² - (1.4)²

By using the identity,

(a² - b²) = (a + b)(a - b)

⇒ (18.6 + 1.4)(18.6 - 1.4)

⇒ (20)(17.2)

⇒ 344.

Hence, (18.6)² - (1.4)² = 344.

Factorise:

(x⁴ + x²y² + y⁴)

Answer

Given,

⇒ (x⁴ + x²y² + y⁴)

⇒ x⁴ + 2x²y² - x²y² + y⁴

⇒ x⁴ + 2x²y² + y⁴ - x²y²

⇒ [(x²)² + 2 × x² × y² + (y²)²] - x²y²

By using the identity,

(a + b)² = a² + b² + 2ab

⇒ (x² + y²)² - (xy)²

⇒ (x² + y² + xy)(x² + y² - xy)

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

Factorise:

x³ + 64

Answer

Given,

⇒ x³ + 64

⇒ (x)³ + (4)³

By using the identity,

a³ + b³ = (a + b)(a² - ab + b²)

⇒ (x + 4)(x² - x × 4 + 4²)

⇒ (x + 4)(x² - 4x + 16)

Hence, x³ + 64 = (x + 4)(x² - 4x + 16).

Factorise:

8a³ + 27b³

Answer

Given,

⇒ 8a³ + 27b³

⇒ (2a)³ + (3b)³

By using the identity,

a³ + b³ = (a + b)(a² - ab + b²)

⇒ (2a + 3b)[(2a)² - 2a × 3b + (3b)²]

⇒ (2a + 3b)(4a² - 6ab + 9b²)

Hence, 8a³ + 27b³ = (2a + 3b)(4a² - 6ab + 9b²).

Factorise:

7a³ + 56b³

Answer

Given,

⇒ 7a³ + 56b³

⇒ 7(a³ + 8b³)

⇒ 7[a³ + (2b)³]

By using the identity,

a³ + b³ = (a + b)(a² - ab + b²)

⇒ 7(a + 2b)(a² - a × 2b + (2b)²)

⇒ 7(a + 2b)(a² - 2ab + 4b²).

Hence, 7a³ + 56b³ = 7(a + 2b)(a² - 2ab + 4b²).

Factorise:

x⁵ + x²

Answer

Given,

⇒ x⁵ + x²

⇒ x²(x³ + 1)

⇒ x²[(x)³ + (1)³]

By using the identity,

a³ + b³ = (a + b)(a² - ab + b²)

⇒ x²(x + 1)(x² - x × 1 + 1²)

⇒ x²(x + 1)(x² - x + 1).

Hence, x⁵ + x² = x²(x + 1)(x² - x + 1).

Factorise:

16x⁴ + 54x

Answer

Given,

⇒ 16x⁴ + 54x

⇒ 2x(8x³ + 27)

⇒ 2x[(2x)³ + (3)³]

By using the identity,

a³ + b³ = (a + b)(a² - ab + b²)

⇒ 2x(2x + 3)[(2x)² - 2x × 3 + (3)²]

⇒ 2x(2x + 3)(4x² - 6x + 9).

Hence, 16x⁴ + 54x = 2x(2x + 3)(4x² - 6x + 9).

Factorise:

$216x^3 + \dfrac{1}{27}$

Answer

Given,

⇒ $216x^3 + \dfrac{1}{27}$

By using the identity,

a³ + b³ = (a + b)(a² - ab + b²)

$\Rightarrow (6x)^3 + \Big(\dfrac{1}{3}\Big)^3 \\[1em] \Rightarrow \Big(6x + \dfrac{1}{3}\Big)\Big[(6x)^2 - 6x \times \Big(\dfrac{1}{3}\Big) + \Big(\dfrac{1}{3}\Big)^2\Big] \\[1em] \Rightarrow \Big(6x + \dfrac{1}{3}\Big)\Big(36x^2 - 2x + \dfrac{1}{9}\Big).$

Hence, $216x^3 + \dfrac{1}{27} = \Big(6x + \dfrac{1}{3}\Big)\Big(36x^2 - 2x + \dfrac{1}{9}\Big)$.

Factorise:

a⁶ + b⁶

Answer

Given,

⇒ a⁶ + b⁶

⇒ (a²)³ + (b²)³

⇒ (a² + b²)[(a²)² - a² × b² + (b²)²]

⇒ (a² + b²)(a⁴ - a²b² + b⁴)

Hence, a⁶ + b⁶ = (a² + b²)(a⁴ - a²b² + b⁴).

Factorise:

a⁴ + 343a

Answer

Given,

⇒ a⁴ + 343a

⇒ a(a³ + 343)

⇒ a(a³ + 7³)

By using the identity,

a³ + b³ = (a + b)(a² - ab + b²)

⇒ a(a + 7)[(a)² - a × 7 + (7)²]

⇒ a(a + 7)(a² - 7a + 49)

Hence, a⁴ + 343a = a(a + 7)(a² - 7a + 49).

Factorise:

125x³ + 1

Answer

Given,

⇒ 125x³ + 1

⇒ (5x)³ + (1)³

By using the identity,

a³ + b³ = (a + b)(a² - ab + b²)

⇒ (5x + 1)[(5x)² - 5x × 1 + (1)²]

⇒ (5x + 1)(25x² - 5x + 1)

Hence, 125x³ + 1 = (5x + 1)(25x² - 5x + 1).

Factorise:

2a³ + 16b³ - 3a - 6b

Answer

Given,

⇒ 2a³ + 16b³ - 3a - 6b

⇒ (2a³ + 16b³) - (3a + 6b)

⇒ 2(a³ + 8b³) - 3(a + 2b)

⇒ 2[(a)³ + (2b)³] - 3(a + 2b)

⇒ 2(a + 2b)[(a)² - a × 2b + (2b)²] - 3(a + 2b)

⇒ 2(a + 2b)(a² - 2ab + 4b²) - 3(a + 2b)

⇒ (a + 2b)[2(a² - 2ab + 4b²) - 3]

Hence, 2a³ + 16b³ - 3a - 6b = (a + 2b)[2(a² - 2ab + 4b²) - 3].

Factorise:

a³ - 125 - 2a + 10

Answer

Given,

⇒ a³ - 125 - 2a + 10

⇒ (a³ - 125) - 2a + 10

⇒ [(a)³ - (5)³] - 2(a - 5)

By using the identity,

a³ - b³ = (a - b)(a² + ab + b²)

⇒ (a - 5)(a² + a × 5 + 5²) - 2(a - 5)

⇒ [(a - 5)(a² + 5a + 25)] - 2(a - 5)

⇒ (a - 5)[(a² + 5a + 25) - 2]

⇒ (a - 5)(a² + 5a + 23)

Hence, a³ - 125 - 2a + 10 = (a - 5)(a² + 5a + 23).

Factorise:

x³ - 125

Answer

Given,

⇒ x³ - 125

⇒ (x)³ - (5)³

By using the identity,

a³ - b³ = (a - b)(a² + ab + b²)

⇒ (x - 5)(x² + x × 5 + 5²)

⇒ (x - 5)(x² + 5x + 25).

Hence, x³ - 125 = (x - 5)(x² + 5x + 25).

Factorise:

$8a^3 - \dfrac{1}{27b^3}$

Answer

Given,

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

Hence, $8a^3 - \dfrac{1}{27b^3} = \Big(2a - \dfrac{1}{3b}\Big)\Big(4a^2 + \dfrac{2a}{3b} + \dfrac{1}{9b^2}\Big)$.

Factorise:

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

Answer

$\Rightarrow \dfrac{8a^3}{27} - \dfrac{b^3}{8} \\[1em] \Rightarrow \Big(\dfrac{2a}{3}\Big)^3 - \Big(\dfrac{b}{2}\Big)^3 \\[1em] \Rightarrow \Big(\dfrac{2a}{3} - \dfrac{b}{2}\Big)\Big[\Big(\dfrac{2a}{3}\Big)^2 + \Big(\dfrac{2a}{3}\Big) \times \Big(\dfrac{b}{2}\Big) + \Big(\dfrac{b}{2}\Big)^2\Big] \\[1em] \Rightarrow \Big(\dfrac{2a}{3} - \dfrac{b}{2}\Big)\Big(\dfrac{4a^2}{9} + \dfrac{2ab}{6} + \dfrac{b^2}{4}\Big) \\[1em] \Rightarrow \Big(\dfrac{2a}{3} - \dfrac{b}{2}\Big)\Big(\dfrac{4a^2}{9} + \dfrac{ab}{3} + \dfrac{b^2}{4}\Big).$

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

Factorise:

a - 8ab³

Answer

Given,

⇒ a - 8ab³

⇒ a(1 - 8b³)

⇒ a[(1)³ - (2b)³]

By using the identity,

(a³ - b³) = (a - b)(a² + ab + b²)

⇒ a(1 - 2b)[(1)² + 1 × 2b + (2b)²]

⇒ a(1 - 2b)(1 + 2b + 4b²).

Hence, a - 8ab³ = a(1 - 2b)(1 + 2b + 4b²).

Factorise:

x⁶ - 1

Answer

Given,

⇒ x⁶ - 1

⇒ (x³)² - (1)²

By using the identity,

(a² - b²) = (a + b)(a - b)

⇒ (x³ - 1)(x³ + 1)

⇒ [x³ - (1)³][x³ + (1)³]

⇒ (x - 1)[(x)² + x × 1 + 1²] [(x + 1)(x)² - x × 1 + 1²]

⇒ (x - 1)(x² + x + 1)(x + 1)(x² - x + 1)

⇒ (x - 1)(x + 1)(x² + x + 1)(x² - x + 1).

Hence, x⁶ - 1 = (x - 1)(x + 1)(x² + x + 1)(x² - x + 1).

Factorise:

a³ - 0.064

Answer

Given,

⇒ a³ - 0.064

⇒ (a)³ - (0.4)³

By using the identity,

(a³ - b³) = (a - b)(a² + ab + b²)

⇒ (a - 0.4)[a² + 0.4a + (0.4)²]

⇒ (a - 0.4)(a² + 0.4a + 0.16).

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

Factorise:

24x⁴ - 375x

Answer

Given,

⇒ 24x⁴ - 375x

⇒ 3x(8x³ - 125)

⇒ 3x[(2x)³ - (5)³]

By using the identity,

(a³ - b³) = (a - b)(a² + ab + b²)

⇒ 3x(2x - 5)[(2x)² + 2x × 5 + (5)²]

⇒ 3x(2x - 5)(4x² + 10x + 25).

Hence, 24x⁴ - 375x = 3x(2x - 5)(4x² + 10x + 25).

Factorise:

3a⁷b - 81a⁴b⁴

Answer

Given,

⇒ 3a⁷b - 81a⁴b⁴

⇒ 3a⁴b(a³ - 27b³)

⇒ 3a⁴b[a³ - (3b)³]

⇒ 3a⁴b(a - 3b)[(a)² + a × 3b + (3b)²]

⇒ 3a⁴b(a - 3b)(a² + 3ab + 9b²).

Hence, 3a⁷b - 81a⁴b⁴ = 3a⁴b(a - 3b)(a² + 3ab + 9b²).

Factorise:

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

Answer

Given,

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

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

Factorise:

2x⁷ - 128x

Answer

Given,

⇒ 2x⁷ - 128x

⇒ 2x(x⁶ - 64)

⇒ 2x[(x³)² - (8)²]

⇒ 2x[(x)³ - (8)] [(x)³ + (8)]

⇒ 2x[x³ - 8] [x³ + 8]

⇒ 2x[x³ - 2³] [x³ + 2³]

By using the identity,

(a³ - b³) = (a - b)(a² + ab + b²)

(a³ + b³) = (a + b)(a² - ab + b²)

⇒ 2x[(x - 2)(x² + 2 × x + (2)²)] [(x + 2)(x)² - 2 × x + (2)²]

⇒ 2x(x - 2)(x² + 2x + 4)(x + 2)(x² - 2x + 4)

⇒ 2x(x - 2)(x + 2)(x² + 2x + 4)(x² - 2x + 4).

Hence, 2x⁷ - 128x = 2x(x - 2)(x + 2)(x² + 2x + 4)(x² - 2x + 4).

Factorise:

250(a - b)³ + 2

Answer

Given,

⇒ 250(a - b)³ + 2

⇒ 2[125(a - b)³ + 1]

⇒ 2{[5(a - b)]³ + (1)³}

By using the identity,

(a³ + b³) = (a + b)(a² - ab + b²)

⇒ 2[5(a - b) + 1] {[5(a - b)]² - 5(a - b) × 1 + (1)²}

⇒ 2[(5a - 5b + 1)(25(a - b)² - 5(a - b) + 1)]

⇒ 2[(5a - 5b + 1)(25(a² - 2ab + b²) - 5a + 5b + 1)]

⇒ 2[(5a - 5b + 1)(25a² - 50ab + 25b² - 5a + 5b + 1)]

Hence, 250(a - b)³ + 2 = 2[(5a - 5b + 1)(25a² - 50ab + 25b² - 5a + 5b + 1)].

Factorise:

8a³ - b³ - 4ax + 2bx

Answer

Given,

⇒ 8a³ - b³ - 4ax + 2bx

⇒ 8a³ - b³ - 4ax + 2bx

⇒ (2a)³ - (b)³ - 2x(2a - b)

By using the identity,

(a³ - b³) = (a - b)(a² + ab + b²)

⇒ (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).

Factorise:

a³ - 27b³ + 2a²b - 6ab²

Answer

Given,

⇒ a³ - 27b³ + 2a²b - 6ab²

⇒ (a)³ - (3b)³ + 2ab(a - 3b)

By using the identity,

(a³ - b³) = (a - b)(a² + ab + b²)

⇒ (a - 3b)[a² + 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).

Factorise:

32a²x³ - 8b²x³ - 4a²y³ + b²y³

Answer

Given,

⇒ 32a²x³ - 8b²x³ - 4a²y³ + b²y³

⇒ 8x³(4a² - b²) - y³(4a² - b²)

⇒ (8x³ - y³)(4a² - b²)

⇒ (4a² - b²)[(2x)³ - (y)³]

By using the identity,

(a³ - b³) = (a - b)(a² + ab + b²)

⇒ (4a² - b²){(2x - y)[(2x)² + 2x × y + (y)²]}

⇒ [(2a)² - (b)²][(2x - y)(4x² + 2xy + y²)]

By using the identity,

(a² - b²) = (a + b)(a - b)

⇒ [(2a)² - (b)²] [(2x - y)(4x² + 2xy + y²)]

⇒ (2a + b)(2a - b)(2x - y)(4x² + 2xy + y²).

Hence, 32a²x³ - 8b²x³ - 4a²y³ + b²y³ = (2a + b)(2a - b)(2x - y)(4x² + 2xy + y²).

Factorise:

a² - 4b² + a³ - 8b³ - (a - 2b)²

Answer

Given,

⇒ a² - 4b² + a³ - 8b³ - (a - 2b)²

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

By using the identity,

(a² - b²) = (a + b)(a - b) and (a³ - b³) = (a - b)(a² + ab + b²)

⇒ (a + 2b)(a - 2b) + (a - 2b)(a² + a × 2b + (2b)²) - (a - 2b)(a - 2b)

⇒ (a - 2b)[(a + 2b) + (a² + 2ab + 4b²) - (a - 2b)]

⇒ (a - 2b)[a + 2b - a + 2b + (a² + 2ab + 4b²)]

⇒ (a - 2b)(a² + 2ab + 4b² + 4b).

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

Factorise:

(a + b)³ + (a - b)³

Answer

Given,

⇒ (a + b)³ + (a - b)³

By using the identity,

(a³ + b³) = (a + b)(a² - ab + b²)

⇒ [(a + b) + (a - b)] [(a + b)² - (a + b) × (a - b) + (a - b)²]

⇒ (2a)[a² + 2ab + b² - (a² - b²) + a² - 2ab + b²]

⇒ (2a)[a² + 2ab + b² - a² + b² + a² - 2ab + b²]

⇒ (2a)(a² + 3b²).

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

Factorise:

x³ - 3x² + 3x + 7

Answer

Given,

⇒ x³ - 3x² + 3x + 7

⇒ x³ - 3x² + 3x + 8 - 1

⇒ x³ + 8 - 3x² + 3x - 1

⇒ x³ - 1³ - 3 × 1 x (x + 1) + 8

⇒ (x - 1)³ + 2³

By using the identity,

(a³ + b³) = (a + b)(a² - ab + b²)

⇒ [(x - 1) + 2] [(x - 1)² - (x - 1) × 2 + 2²]

⇒ (x + 1)[(x² - 2x + 1) - (2x - 2) + 4]

⇒ (x + 1)[(x² - 2x + 1) - 2x + 2 + 4]

⇒ (x + 1)(x² - 4x + 7)

Hence, x³ - 3x² + 3x + 7 = (x + 1)(x² - 4x + 7).

Factorise:

x² + 9x + 18

Answer

Given,

⇒ x² + 9x + 18

⇒ x² + 6x + 3x + 18

⇒ x(x + 6) + 3(x + 6)

⇒ (x + 6)(x + 3).

Hence, x² + 9x + 18 = (x + 6)(x + 3).

Factorise:

x² - 5x - 6

Answer

Given,

⇒ x² - 5x - 6

⇒ x² - 6x + x - 6

⇒ x(x - 6) + 1(x - 6)

⇒ (x - 6)(x + 1).

Hence, x² - 5x - 6 = (x - 6)(x + 1).

Factorise:

x² + 6x - 7

Answer

Given,

⇒ x² + 6x - 7

⇒ x² + 7x - x - 7

⇒ x(x + 7) - 1(x + 7)

⇒ (x + 7)(x - 1).

Hence, x² + 6x - 7 = (x + 7)(x - 1).

Factorise:

x² + 7x - 18

Answer

Given,

⇒ x² + 7x - 18

⇒ x² + 9x - 2x - 18

⇒ x(x + 9) - 2(x + 9)

⇒ (x + 9)(x - 2).

Hence, x² + 7x - 18 = (x + 9)(x - 2).

Factorise:

x² - 3x - 54

Answer

Given,

⇒ x² - 3x - 54

⇒ x² - 9x + 6x - 54

⇒ x(x - 9) + 6(x - 9)

⇒ (x - 9)(x + 6).

Hence, x² - 3x - 54 = (x - 9)(x + 6).

Factorise:

x² - 17x - 84

Answer

Given,

⇒ x² - 17x - 84

⇒ x² - 21x + 4x - 84

⇒ x(x - 21) + 4(x - 21)

⇒ (x - 21)(x + 4).

Hence, x² - 17x - 84 = (x - 21)(x + 4).

Factorise:

6x² + 11x - 10

Answer

Given,

⇒ 6x² + 11x - 10

⇒ 6x² + 15x - 4x - 10

⇒ 3x(2x + 5) - 2(2x + 5)

⇒ (2x + 5)(3x - 2).

Hence, 6x² + 11x - 10 = (2x + 5)(3x - 2).

Factorise:

3x² - 4x - 7

Answer

Given,

⇒ 3x² - 4x - 7

⇒ 3x² - 7x + 3x - 7

⇒ x(3x - 7) + 1(3x - 7)

⇒ (3x - 7)(x + 1).

Hence, 3x² - 4x - 7 = (3x - 7)(x + 1).

Factorise:

2x² - 7x - 39

Answer

Given,

⇒ 2x² - 7x - 39

⇒ 2x² + 6x - 13x - 39

⇒ 2x(x + 3) - 13(x + 3)

⇒ (2x - 13)(x + 3).

Hence, 2x² - 7x - 39 = (2x - 13)(x + 3).

Factorise:

2x² + 3x - 90

Answer

Given,

⇒ 2x² + 3x - 90

⇒ 2x² - 12x + 15x - 90

⇒ 2x(x - 6) + 15(x - 6)

⇒ (2x + 15)(x - 6).

Hence, 2x² + 3x - 90 = (2x + 15)(x - 6).

Factorise:

2x² + 5x - 3

Answer

Given,

⇒ 2x² + 5x - 3

⇒ 2x² + 6x - x - 3

⇒ 2x(x + 3) - 1(x + 3)

⇒ (2x - 1)(x + 3).

Hence, 2x² + 5x - 3 = (2x - 1)(x + 3).

Factorise:

10 + 3x - x²

Answer

Given,

⇒ 10 + 3x - x²

⇒ 10 + 3x - x²

⇒ 10 - 2x + 5x - x²

⇒ 2(5 - x) + x(5 - x)

⇒ (2 + x)(5 - x).

Hence, 10 + 3x - x² = (2 + x)(5 - x).

Factorise:

7 - 12x - 4x²

Answer

Given,

⇒ 7 - 12x - 4x²

⇒ 7 - 14x + 2x - 4x²

⇒ 7(1 - 2x) + 2x(1 - 2x)

⇒ (1 - 2x)(7 + 2x).

Hence, 7 - 12x - 4x² = (1 - 2x)(7 + 2x).

Factorise:

5 - 4y - 12y²

Answer

Given,

⇒ 5 - 4y - 12y²

⇒ 5 - 10y + 6y - 12y²

⇒ 5(1 - 2y) + 6y(1 - 2y)

⇒ (1 - 2y)(5 + 6y).

Hence, 5 - 4y - 12y² = (1 - 2y)(5 + 6y).

Factorise:

1 - 18z - 63z²

Answer

Given,

⇒ 1 - 18z - 63z²

⇒ 1 - 21z + 3z - 63z²

⇒ 1(1 - 21z) + 3z(1 - 21z)

⇒ (1 + 3z)(1 - 21z).

Hence, 1 - 18z - 63z² = (1 + 3z)(1 - 21z).

Factorise:

x² - 3ax - 88a²

Answer

Given,

⇒ x² - 3ax - 88a²

⇒ x² + 8ax - 11ax - 88a²

⇒ x(x + 8a) - 11a(x + 8a)

⇒ (x - 11a)(x + 8a).

Hence, x² - 3ax - 88a² = (x - 11a)(x + 8a).

Factorise:

3 - x(4 + 7x)

Answer

Given,

⇒ 3 - x(4 + 7x)

⇒ 3 - 4x - 7x²

⇒ 3 + 3x - 7x - 7x²

⇒ 3(1 + x) - 7x(1 + x)

⇒ (3 - 7x)(1 + x).

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

Factorise:

24x³ + 37x² - 5x

Answer

Given,

⇒ 24x³ + 37x² - 5x

⇒ x(24x² + 37x - 5)

⇒ x(24x² - 3x + 40x - 5)

⇒ x[3x(8x - 1) + 5(8x - 1)]

⇒ x(3x + 5)(8x - 1).

Hence, 24x³ + 37x² - 5x = x(3x + 5)(8x - 1).

Factorise:

x²y² - 12xy - 45

Answer

Given,

⇒ x²y² - 12xy - 45

⇒ (xy)² - 12xy - 45

⇒ (xy)² - 15xy + 3xy - 45

⇒ xy(xy - 15) + 3(xy - 15)

⇒ (xy + 3)(xy - 15).

Hence, x²y² - 12xy - 45 = (xy + 3)(xy - 15).

Factorise:

x(2x - y) - y²

Answer

Given,

⇒ x(2x - y) - y²

⇒ 2x² - xy - y²

⇒ 2x² - 2xy + xy - y²

⇒ 2x(x - y) + y(x - y)

⇒ (2x + y)(x - y).

Hence, x(2x - y) - y² = (2x + y)(x - y).

Factorise:

5x² + 17xy - 12y²

Answer

Given,

⇒ 5x² + 17xy - 12y²

⇒ 5x² + 20xy - 3xy - 12y²

⇒ 5x(x + 4y) - 3y(x + 4y)

⇒ (x + 4y)(5x - 3y).

Hence, 5x² + 17xy - 12y² = (x + 4y)(5x - 3y).

Factorise:

5(3a + b)² + 6(3a + b) - 8

Answer

Given,

⇒ 5(3a + b)² + 6(3a + b) - 8

Let (3a + b) = x,

⇒ 5x² + 6x - 8

⇒ 5x² + 10x - 4x - 8

⇒ 5x(x + 2) - 4(x + 2)

⇒ (5x - 4)(x + 2)

⇒ [5(3a + b) - 4][3a + b + 2]

⇒ (15a + 5b - 4)(3a + b + 2).

Hence, 5(3a + b)² + 6(3a + b) - 8 = (3a + b + 2)(15a + 5b - 4).

Factorise:

3(2a - b)² - 19(2a - b) + 28

Answer

Given,

⇒ 3(2a - b)² - 19(2a - b) + 28

Let 2a - b = x,

⇒ 3x² - 19x + 28

⇒ 3x² - 7x - 12x + 28

⇒ x(3x - 7) - 4(3x - 7)

⇒ (3x - 7)(x - 4)

⇒ [3(2a - b) - 7][2a - b - 4]

⇒ (6a - 3b - 7)(2a - b - 4).

Hence, 3(2a - b)² - 19(2a - b) + 28 = (2a - b - 4)(6a - 3b - 7).

Factorise:

1 - 2(a + 2b) - 3(a + 2b)²

Answer

Given,

⇒ 1 - 2(a + 2b) - 3(a + 2b)²

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

⇒ 1[1 - 3(a + 2b)] + (a + 2b)[1 - 3(a + 2b)]

⇒ (1 + a + 2b)[1 - 3(a + 2b)]

⇒ (1 + a + 2b)(1 - 3a - 6b).

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

Factorise:

(a + 4)² - 5ab - 20b - 6b²

Answer

Given,

⇒ (a + 4)² - 5ab - 20b - 6b²

⇒ (a + 4)² - 5b(a + 4) - 6b²

⇒ (a + 4)² - 6b(a + 4) + b(a + 4) - 6b²

⇒ (a + 4)[(a + 4) - 6b] + b[(a + 4) - 6b]

⇒ (a + 4 + b)(a + 4 - 6b).

Hence, (a + 4)² - 5ab - 20b - 6b² = (a + 4 + b)(a + 4 - 6b).

Factorise:

8(a - 2b)² - 2a + 4b - 1

Answer

Given,

⇒ 8(a - 2b)² - 2a + 4b - 1

⇒ 8(a - 2b)² - 2(a - 2b) - 1

⇒ 8(a - 2b)² - 4(a - 2b) + 2(a - 2b ) - 1

⇒ 4(a - 2b)[2(a - 2b) - 1] + 1[2(a - 2b) - 1]

⇒ [4(a - 2b) + 1][2(a - 2b) - 1]

⇒ (4a - 8b + 1)(2a - 4b - 1).

Hence, 8(a - 2b)² - 2a + 4b - 1 = (4a - 8b + 1)(2a - 4b - 1).

Factorise:

4(2a - 3)² - 3(2a - 3)(a - 1) - 7(a - 1)²

Answer

Given,

⇒ 4(2a - 3)² - 3(2a - 3)(a - 1) - 7(a - 1)²

⇒ 4(2a - 3)² + 4(2a - 3)(a - 1) - 7(2a - 3)(a - 1) - 7(a - 1)²

⇒ 4(2a - 3)[(2a - 3) + (a - 1)] - 7(a - 1)[(2a - 3) + (a - 1)]

⇒ [4(2a - 3) - 7(a - 1)][(2a - 3) + (a - 1)]

⇒ (8a - 12 - 7a + 7)(2a - 3 + a - 1)

⇒ (a - 5)(3a - 4).

Hence, 4(2a - 3)² - 3(2a - 3)(a - 1) - 7(a - 1)² = (a - 5)(3a - 4).

Factorise:

(a² - 3a)(a² - 3a + 7) + 10

Answer

Given,

⇒ (a² - 3a)(a² - 3a + 7) + 10

⇒ (a² - 3a)(a² - 3a) + (a² - 3a)(7) + 10

⇒ (a² - 3a)² + 7(a² - 3a) + 10

Let a² - 3a = x,

⇒ x² + 7x + 10

⇒ x² + 5x + 2x + 10

⇒ x(x + 5) + 2(x + 5)

⇒ (x + 2)(x + 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)(a² - 3a + 5).

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

Factorise:

(a² - a)(4a² - 4a - 5) - 6

Answer

Given,

⇒ (a² - a)(4a² - 4a - 5) - 6

⇒ [4(a² - a)(a² - a) - 5(a² - a)] - 6

⇒ [4(a² - a)² - 5(a² - a) - 6]

⇒ [4(a² - a)² - 8(a² - a) + 3(a² - a) - 6]

⇒ 4(a² - a)[(a² - a) - 2] + 3[(a² - a) - 2]

⇒ [4(a² - a)+ 3] [(a² - a) - 2]

⇒ [4a² - 4a + 3] [a² - 2a + 1a - 2]

⇒ (4a² - 4a + 3) [a(a - 2) + 1(a - 2)]

⇒ (4a² - 4a + 3)(a + 1)(a - 2)

⇒ (a + 1)(a - 2)(4a² - 4a + 3)

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

Factorise:

a⁴ - 11a² + 10

Answer

Given,

⇒ a⁴ - 11a² + 10

⇒ a⁴ - 10a² - 1a² + 10

⇒ a²(a² - 10) - 1(a² - 10)

⇒ (a²- 1)(a² - 10)

⇒ [(a)²- (1)²](a² - 10)

⇒ (a² - 10)(a + 1)(a - 1)

Hence, a⁴ - 11a² + 10 =(a² - 10)(a + 1)(a - 1).

Factorise:

5 - (3x² - 2x)(6 - 3x² + 2x)

Answer

Given,

⇒ 5 - (3x² - 2x)(6 - 3x² + 2x)

⇒ 5 - (3x² - 2x)[6 - (3x² - 2x)]

Let, y = (3x² - 2x)

⇒ 5 - y(6 - y)

⇒ 5 - 6y + y²

⇒ 5 - 5y - y + y²

⇒ 5(1 - y) - y(1 - y)

⇒ (5 - y)(1 - y)

⇒ [5 - (3x² - 2x)] [1 - (3x² - 2x)]

⇒ (5 + 2x - 3x²)(1 + 2x - 3x²)

⇒ (5 + 5x - 3x - 3x²)(1 - x + 3x - 3x²)

⇒ [5(1 + x) - 3x(1 + x)][1(1 - x) + 3x(1 - x)]

⇒ (5 - 3x)(1 + x)(1 + 3x)(1 - x).

Hence, 5 - (3x² - 2x)(6 - 3x² + 2x) = (5 - 3x)(1 + x)(1 + 3x)(1 - x).

b² - ac - bc + ab =

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

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

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

4. (a - b)(b - c)

Answer

Given,

⇒ b² - ac - bc + ab

⇒ b² + ab - ac - bc

⇒ b(b + a) - c(a + b)

⇒ (a + b)(b - c).

Hence, option 1 is correct option.

The value of (1 + x)²(1 + y²) - (1 + x²)(1 + y)² is:

1. 2(x - y)(1 + xy)

2. (x - y)(1 - xy)

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

4. 2(x + y)(1 - xy)

Answer

Given,

⇒ (1 + x)²(1 + y²) - (1 + x²)(1 + y)²

⇒ (1 + 2x + x²)(1 + y²) - (1 + x²)(1 + 2y + y²)

⇒ (1 + y² + 2x + 2xy² + x² + x²y²) - (1 + 2y + y² + x² + 2x²y + x²y²)

⇒ (1 + y² + 2x + 2xy² + x² + x²y² - 1 - 2y - y² - x² - 2x²y - x²y²)

⇒ 2x - 2y - 2x²y + 2xy²

⇒ 2(x - y) - 2xy(x - y)

⇒ (2 - 2xy)(x - y)

⇒ 2(1 - xy)(x - y).

Hence, option 3 is correct option.

x⁴ - y⁴ = ............... (x - y)(x² + y²)

1. 0

2. 1

3. x − y

4. x + y

Answer

Given,

⇒ x⁴ - y⁴

⇒ (x²)² - (y²)²

⇒ (x² - y²)(x² + y²)

⇒ [(x)² - (y)²](x² + y²)

⇒ (x + y)(x - y)(x² + y²)

Hence, option 4 is correct option.

16(2a − b)² − 9(a + b)² =

1. (5a − 7b)(11a − b)

2. (5a + 7b)(11a − b)

3. (5a − 7b)(11a + b)

4. (5a + 7b)(11a + b)

Answer

Given,

⇒ [4(2a − b)]² − [3(a + b)]²

⇒ [4(2a − b) − 3(a + b)][4(2a − b) + 3(a + b)]

⇒ (8a − 4b − 3a - 3b)(8a − 4b + 3a + 3b)

⇒ (5a − 7b)(11a − b).

Hence, option 1 is correct option.

Factorization of 3a(3a + 2c) − 4b(b + c) is:

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

2. (3a − 2b)(3a − 2b + 2c)

3. (3a − 2b)(3a + 2b + 2c)

4. (3a − 2b)(3a + 2b − 2c)

Answer

Given,

⇒ 3a(3a + 2c) − 4b(b + c)

⇒ 9a² + 6ac − 4b² - 4cb

⇒ (3a)² − (2b)² + 6ac - 4cb

⇒ (3a + 2b)(3a - 2b) + 2c(3a - 2b)

⇒ (3a - 2b)(3a + 2b + 2c).

Hence, option 3 is correct option.

12a² - 27b⁴

1. 3(2a + 3b²)(2a - b²)

2. (2a + 3b²)(2a - 3b²)

3. 3(2a + 3b²)(a - 3b²)

4. 3(2a + 3b²)(2a - 3b²)

Answer

Given,

⇒ 3(4a² - 9b⁴)

⇒ 3[(2a)² - (3b²)²]

⇒ 3(2a + 3b²)(2a - 3b²)

Hence, option 4 is correct option.

$\sqrt{3}x^2 + 11x + 6\sqrt{3} =$

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

2. $(\sqrt{3}x + 2)(x + 3\sqrt{3})$

3. $(\sqrt{3}x - 2)(x - 3\sqrt{3})$

4. $(x - 2)(x + 3\sqrt{3})$

Answer

Given,

$\Rightarrow \sqrt{3}x^2 + 11x + 6\sqrt{3} \\[1em] \Rightarrow \sqrt{3}x^2 + 9x + 2x + 6\sqrt{3} \\[1em] \Rightarrow \sqrt{3}x(x + 3\sqrt{3}) + 2(x + 3\sqrt{3}) \\[1em] \Rightarrow (\sqrt{3}x + 2)(x + 3\sqrt{3})$

Hence, option 2 is correct option.

1 − x⁹ = (1 − x)(1 + x + x²) ...............

1. 1 − x³ + x⁶

2. 1 + x³ + x⁶

3. 1 − x³ − x⁶

4. 1 + x³ − x⁶

Answer

Given,

⇒ 1 − x⁹

⇒ (1)³ − (x³)³

By using identity,

a³ - b³ = (a - b)(a² + ab + b²)

⇒ (1 - x³)(1² + x³(1) + (x³)²)

⇒ [(1)³ - (x)³](1 + x³ + x⁶)

⇒ (1 - x)(1² + x(1) + x²)(1 + x³ + x⁶)

⇒ (1 - x)(1 + x + x²)(1 + x³ + x⁶).

Hence, option 2 is correct option.

7 − 12m − 4m² =

1. (1 − 2m)(7 + 2m)

2. (1 + 2m)(7 − 2m)

3. (1 + 2m)(7 + 2m)

4. (2m − 1)(7 + 2m)

Answer

Given,

⇒ 7 − 12m − 4m²

⇒ 7 − 14m + 2m − 4m²

⇒ 7(1 − 2m) + 2m(1 − 2m)

⇒ (1 − 2m)(7 + 2m).

Hence, option 1 is correct option.

Factorization of p² − 2p − (q + 1)(q − 1) is:

1. (p + q − 1)(p − q + 1)

2. (p − q + 1)(p − q − 1)

3. (p + q − 1)(p − q − 1)

4. (p + q + 1)(p − q)

Answer

Given,

⇒ p² − 2p − (q + 1)(q − 1)

⇒ p² − 2p − (q² - 1)

⇒ p² − 2p − q² + 1

⇒ p² − 2p + 1 − q²

⇒ (p - 1)² − q²

⇒ (p - 1 + q)(p - 1 - q)

⇒ (p + q − 1)(p − q - 1).

Hence, option 3 is correct option.

Assertion(A): One of the factors of (5x + 1)² + (25x² - 1) is 2x.

Reason(R): (a + b)² = (a + b)(a + b) and a² - b² = (a + b)(a - b)

1. A is true, R is false

2. A is false, R is true

3. Both A and R are true

4. Both A and R are false

Answer

Given,

⇒ (5x + 1)² + (25x² - 1)

⇒ (5x + 1)² + (5x)² - (1)²

⇒ (5x + 1)² + (5x + 1)(5x - 1)

⇒ (5x + 1)[(5x + 1) + (5x - 1)]

⇒ (5x + 1)(10x)

⇒ 2x(5)(5x + 1).

Assertion (A) is true.

⇒ (a + b)² can be written as (a + b)(a + b)

By identity,

a² - b² = (a + b)(a - b)

Reason (R) is true.

Thus, both A and R are true.

Hence, option 3 is correct option.

Assertion(A): $a^3 - 2\sqrt{2}b^3$ can be factorized as $(a - 2\sqrt{2})(a^2 + \sqrt{2}ab + 2b^2)$

Reason(R): x³ - y³ = (x - y)(x² + xy + y²).

1. A is true, R is false

2. A is false, R is true

3. Both A and R are true

4. Both A and R are false

Answer

Given,

$a^3 - 2\sqrt{2}b^3$

By using identity,

a³ - b³ = (a - b)(a² + ab + b²)

$\Rightarrow (a)^3 - (\sqrt{2}b)^3 \\[1em] \Rightarrow (a - \sqrt{2}b)(a^2 + a(\sqrt{2}b) + (\sqrt{2}b)^2) \\[1em] \Rightarrow (a - \sqrt{2}b)(a^2 + \sqrt{2}ab + 2b^2) \\[1em]$

Assertion (A) is false.

Given,

x³ - y³ = (x - y)(x² + xy + y²)

This is a identity.

Reason (R) is true.

A is false, R is true.

Hence, option 2 is correct option.

One of the factors of (x² - 4x)(x² - 4x - 1) - 20 is :

1. x - 1

2. x - 2

3. x - 4

4. x + 5

Answer

Given,

(x² - 4x)(x² - 4x - 1) - 20

Let us consider y = (x² - 4x)

⇒ (y)(y - 1) - 20

⇒ y² - y - 20

⇒ y² - 5y + 4y - 20

⇒ y(y - 5) + 4(y - 5)

⇒ (y + 4)(y - 5)

⇒ (x² - 4x + 4)(x² - 4x - 5)

⇒ [(x)² - 2(2)(x) + (2)²](x² - 5x + x - 5)

⇒ (x - 2)²[x(x - 5) + 1(x - 5)]

⇒ (x - 2)(x - 2)(x + 1)(x - 5)

Hence, option 2 is correct option.

x⁴ + 7x² + 16 can be factorized as:

1. (x² + x + 4)(x² - x + 4)

2. (x² + x - 4)(x² - x + 4)

3. (x² - x - 4)(x² + x + 4)

4. none of these

Answer

Given,

⇒ x⁴ + 7x² + 16

⇒ x⁴ + 8x² - x² + 16

⇒ x⁴ + 8x² + 16 - x²

⇒ (x²)² + 2(x²)(4) + (4)² - x²

⇒ (x² + 4)² - x²

⇒ (x² + 4 + x)(x² + 4 - x)

⇒ (x² + x + 4)(x² - x + 4).

Hence, option 1 is correct option.

If $x = \dfrac{a + b + c}{3}$, then (x - a)³ + (x - b)³ + (x - c)³ can be factorized as:

1. (x - a)(x - b)(x - c)

2. $\dfrac{(x - a)(x - b)(x - c)}{3}$

3. 3(x - a)(x - b)(x - c)

4. none of these

Answer

Given,

$x = \dfrac{a + b + c}{3}$

Let, p = x - a, q = x - b, r = x - c

Adding,

⇒ p + q + r = (x - a) + (x - b) + (x - c)

⇒ p + q + r = 3x - a - b - c

⇒ p + q + r = 3x - (a + b + c)

⇒ p + q + r = 3 $\Big(\dfrac{a + b + c}{3}\Big)$ - (a + b + c)

⇒ p + q + r = (a + b + c) - (a + b + c)

⇒ p + q + r = 0

If p + q + r = 0, we use identity,

p³ + q³ + r³ = 3pqr

⇒ (x - a)³ + (x - b)³ + (x - c)³ = 3(x - a)(x - b)(x - c).

Hence, option 3 is correct option.

An expression is factorized as (2x³ + 2x² + x)(x² - 2x + 2). Which of the following terms will appear in the simplest form of the above expression?

1. -2x³

2. 3x³

3. 6x³

4. x³

Answer

Given,

⇒ (2x³ + 2x² + x)(x² - 2x + 2)

⇒ 2x³(x² - 2x + 2) + 2x²(x² - 2x + 2) + x(x² - 2x + 2)

⇒ 2x⁵ - 4x⁴ + 4x³ + 2x⁴ - 4x³ + 4x² + x³ - 2x² + 2x

⇒ 2x⁵ - 2x⁴ + 2x² + x³ + 2x

The that term appears in its simplest form is x³

Hence, option 4 is correct option.

If x + y = 7 and x² + y² = 25, then find the value of $\sqrt{xy}$.

Answer

Given,

x + y = 7

x² + y² = 25

By using the identity,

(x + y)² = x² + y² + 2xy

⇒ (7)² = 25 + 2xy

⇒ 49 = 25 + 2xy

⇒ 49 - 25 = 2xy

⇒ 2xy = 24

⇒ xy = $\dfrac{24}{2}$

⇒ xy = 12

⇒ $\sqrt{xy} = \sqrt{12}$

⇒ $\sqrt{xy} = \sqrt{4 × 3}$

⇒ $\sqrt{xy} = 2\sqrt{3}$.

Hence, $\sqrt{xy} = 2\sqrt{3}$.

An expression was factorized as (x - 1)(x - 3)(x - 5) ..... (x - 99). What is the coefficient of x⁴⁹ in the expression ?

Answer

Given,

Polynomial = (x - 1)(x - 3)(x - 5) ........ (x - 99)

Here roots are 1, 3, 5, ....., 99. Let number of roots be n.

The above sequence is in an A.P. with first term (a) = 1, common difference (d) = 2 and last term (a_n) = 99

By formula,

⇒ a_n = a + (n - 1)d

⇒ 99 = 1 + 2(n - 1)

⇒ 99 - 1 = 2(n - 1)

⇒ 98 = 2(n - 1)

⇒ n - 1 = $\dfrac{98}{2}$

⇒ n - 1 = 49

⇒ n = 50.

Sum of roots = $\dfrac{n}{2}(a + l) = \dfrac{50}{2}(1 + 99) = \dfrac{50}{2} \times 100$ = 2500.

For a polynomial of the form,

(a - a₁)(x - a₂)..........

The coefficient of x^{n - 1} is the negative of the sum of all roots.

Thus, the coefficient of x⁴⁹ = -2500.

Hence, coefficient of x⁴⁹ = -2500.

What is the simplified form of the expression

$\dfrac{\dfrac{b^4 - a^4}{a(a + b)} - \dfrac{b^3}{a}}{b^2 - ab + a^2}$?

Answer

Given,

$\dfrac{\dfrac{b^4 - a^4}{a(a + b)} - \dfrac{b^3}{a}}{b^2 - ab + a^2}$

Simplifying the numerator,

$\Rightarrow \dfrac{b^4 - a^4}{a(a + b)} - \dfrac{b^3}{a} \\[1em] \Rightarrow \dfrac{(b^2)^2 - (a^2)^2}{a(a + b)} - \dfrac{b^3}{a} \\[1em] \Rightarrow \dfrac{(b^2 - a^2)(b^2 + a^2)}{a(a + b)} - \dfrac{b^3}{a} \\[1em] \Rightarrow \dfrac{(b - a)(b + a)(b^2 + a^2)}{a(a + b)} - \dfrac{b^3}{a} \\[1em] \Rightarrow \dfrac{(b - a)(b^2 + a^2)}{a} - \dfrac{b^3}{a} \\[1em] \Rightarrow \dfrac{b^3 + ba^2 - ab^2 - a^3}{a} - \dfrac{b^3}{a} \\[1em] \Rightarrow \dfrac{b^3 + ba^2 - ab^2 - a^3 - b^3}{a} \\[1em] \Rightarrow \dfrac{ba^2 - ab^2 - a^3}{a} \\[1em] \Rightarrow \dfrac{a(ba - b^2 - a^2)}{a} \\[1em] \Rightarrow (ab - b^2 - a^2) \\[1em] \Rightarrow -(a^2 + b^2 - ab).$

Substituting value of numerator in given fraction,

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

Hence, $\dfrac{\dfrac{b^4 - a^4}{a(a + b)} - \dfrac{b^3}{a}}{b^2 - ab + a^2} = -1$.