Factorisation — Exercise 5(A)

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).