Factorisation — Multiple Choice Questions

Factorisation of 12a²b + 15ab² is

1. 3a(4ab + 5b²)

2. 3b(4a² + 5ab)

3. 3ab(4a + 5b)

4. none of these

Answer

H.C.F. of 12a²b and 15ab² is 3ab.

∴ 12a²b + 15ab² = 3ab(4a + 5b).

Hence, Option 3 is the correct option.

Factorisation of 6xy - 4y + 6 - 9x is

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

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

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

4. none of these

Answer

6xy - 4y + 6 - 9x

Rearranging the above terms we get,

6xy - 9x - 4y + 6 = 3x(2y - 3) - 2(y - 3)

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

Hence, Option 2 is the correct option.

Factorisation of 49p³q - 36pq is

1. p(7p + 6q)(7p - 6q)

2. q(7p - 6)(7p + 6)

3. pq(7p + 6)(7p - 6)

4. none of these

Answer

49p³q - 36pq = pq(49p² - 36) = pq[(7p)² - (6)²]

We know that,

a² - b² = (a + b)(a - b).

∴ pq[(7p)² - (6)²] = pq[(7p + 6)(7p - 6)].

Hence, Option 3 is the correct option.

Factorisation of y(y - z) + 9(z - y) is

1. (y - z)(y + 9)

2. (y - z)(y - 9)

3. (z - y)(y + 9)

4. none of these

Answer

y(y - z) + 9(z - y) = y(y - z) + 9(-y + z)

= y(y - z) + 9(-1)(y - z)

= y(y - z) - 9(y - z)

= (y - z)(y - 9).

Hence, Option 2 is the correct option.

Factorisation of (lm + l) + m + 1 is

1. (lm + 1)(m + l)

2. (lm + m)(l + 1)

3. l(m + 1)

4. (l + 1)(m + 1)

Answer

(lm + l) + m + 1

Rearranging the above terms we get,

lm + m + l + 1 = m(l + 1) + 1(l + 1)

= (l + 1)(m + 1).

Hence, Option 4 is the correct option.

Factorisation of 63x² - 112y² is

1. 63(x - 2y)(x + 2y)

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

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

4. none of these

Answer

63x² - 112y² = 7[9x² - 16y²] = 7[(3x)² - (4y)²].

We know that,

a² - b² = (a + b)(a - b).

∴ 7[(3x)² - (4y)²] = 7(3x + 4y)(3x - 4y).

Hence, Option 3 is the correct option.

Factorisation of p⁴ - 81 is

1. (p² - 9)(p² + 9)

2. (p - 3)(p + 3)(p² + 9)

3. (p - 3)²(p + 3)²

4. none of these

Answer

p⁴ - 81 = (p²)² - (9)².

We know that,

a² - b² = (a + b)(a - b).

∴ (p²)² - (9)² = (p² - 9)(p² + 9) = (p - 3)(p + 3)(p² + 9).

Hence, Option 2 is the correct option.

One of the factors of (25x² - 1) + (1 + 5x)² is

1. 5 + x

2. 5 - x

3. 5x - 1

4. 10x

Answer

(25x² - 1) + (1 + 5x)² = 25x² - 1 + 1 + 25x² + 10x

= 50x² + 10x

= 10x(5x + 1).

Hence, Option 4 is the correct option.

Factorisation of x² - 4x - 12 is

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

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

3. (x - 6)(x - 2)

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

Answer

x² - 4x - 12 = x² - 6x + 2x - 12

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

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

Hence, Option 2 is the correct option.

Factorisation of 3x² + 7x - 6

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

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

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

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

Answer

3x² + 7x - 6 = 3x² + 9x - 2x - 6

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

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

Hence, Option 1 is the correct option.

Factorisation of 4x² + 8x + 3 is

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

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

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

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

Answer

4x² + 8x + 3 = 4x² + 6x + 2x + 3

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

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

Hence, Option 2 is the correct option.

Factorisation of 16x² + 40x + 25 is

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

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

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

4. (4x + 5)(4x + 7)

Answer

16x² + 40x + 25 = 16x² + 20x + 20x + 25

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

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

Hence, Option 1 is the correct option.

Factorisation of x² - 4xy + 4y² is

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

2. (x + 2y)(x + 2y)

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

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

Answer

x² - 4xy + 4y² = x² - 2xy - 2xy + 4y²

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

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

Hence, Option 3 is the correct option.

Which of the following is a factor of (x + y)³ - (x³ + y³)?

1. x² + xy + 2xy

2. x² + y² - xy

3. xy²

4. 3xy

Answer

We know that,

(a + b)³ = a³ + b³ + 3ab(a + b).

∴ (x + y)³ - (x³ + y³) = x³ + y³ + 3xy(x + y) - x³ - y³ = 3xy(x + y).

Hence, Option 4 is the correct option.

If $\dfrac{x}{y} + \dfrac{y}{x}$ = -1 (x ≠ 0, y ≠ 0), then the value of x³ - y³ is

1. 1

2. -1

3. 0

4. $\dfrac{1}{2}$

Answer

Given,

$\Rightarrow \dfrac{x}{y} + \dfrac{y}{x} = -1\\[1em] \Rightarrow \dfrac{x^2}{xy} + \dfrac{y^2}{xy} = -1\\[1em] \Rightarrow \dfrac{x^2 + y^2}{xy} = -1\\[1em] \Rightarrow x^2 + y^2 = -xy ..............(1)$

We know that,

x³ - y³ = (x - y)(x² + xy + y²)

Substituting the value of x² + y² in above equation, we get

⇒ x³ - y³ = (x - y)(-xy + xy)

⇒ x³ - y³ = (x - y) × 0 = 0.

Hence, option 3 is the correct option.

If a + b + c = 0, then the value of a³ + b³ + c³ is

1. 0

2. abc

3. 2abc

4. 3abc

Answer

We know that,

a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca)

If the value of (a + b + c) = 0, then

⇒ a³ + b³ + c³ - 3abc = 0.(a² + b² + c² - ab - bc - ca)

⇒ a³ + b³ + c³ - 3abc = 0

⇒ a³ + b³ + c³ = 3abc

Hence, option 4 is the correct option.

If $x^\frac{1}{3} + y^\frac{1}{3} + z^\frac{1}{3}$ = 0, then

1. x³ + y³ + z³ = 0

2. x³ + y³ + z³ = 27xyz

3. (x + y + z)³ = 27xyz

4. x + y + z = 3xyz

Answer

If, $x^\frac{1}{3} + y^\frac{1}{3} + z^\frac{1}{3}$ = 0

If, a + b + c = 0, then a³ + b³ + c³ = 3abc. 

Here, a = $x^\frac{1}{3}$, b = $y^\frac{1}{3}$ and z = $z^\frac{1}{3}$

So,

$\Rightarrow (x^\frac{1}{3})^3 + (y^\frac{1}{3})^3 + (z^\frac{1}{3})^3 = 3(x^\frac{1}{3})(y^\frac{1}{3})(z^\frac{1}{3})\\[1em] \Rightarrow x^\frac{3}{3} + y^\frac{3}{3} + z^\frac{3}{3} = 3(xyz)^\frac{1}{3}\\[1em] \Rightarrow x + y + z = 3(xyz)^\frac{1}{3}$

Cubing both sides we get :

$\Rightarrow (x + y + z)^3 = [3(xyz)^\frac{1}{3}]^3\\[1em] \Rightarrow (x + y + z)^3 = 3^3.(xyz)^\frac{3}{3}\\[1em] \Rightarrow (x + y + z)^3 = 27xyz$

Hence, option 3 is the correct option.

Consider the following two statements.

Statement 1: The factorisation of x² + 2x + 1 is (x - 1)².

Statement 2: (a - b)² = a² + 2ab + b².

Which of the following is valid?

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,

⇒ x² + 2x + 1

⇒ x² + 2.x.1 + 1²

⇒ (x + 1)²

∴ Statement 1 is false.

⇒ (a - b)²

⇒ (a - b)(a - b)

⇒ a(a - b) - b(a - b)

⇒ a² - ab - ab + b²

⇒ a² - 2ab + b²

∴ Statement 2 is false.

∴ Both statements are false.

Hence, option 2 is the correct option.