Factorisation — Exercise 4.3

Factorise the following:

4x² - 25y²

Answer

4x² - 25y² = (2x)² - (5y)².

Using identity,

a² - b² = (a + b)(a - b).

(2x)² - (5y)² = (2x + 5y)(2x - 5y).

Hence, 4x² - 25y² = (2x + 5y)(2x - 5y).

Factorise the following:

9x² - 1

Answer

9x² - 1 = (3x)² - (1)².

Using identity,

a² - b² = (a + b)(a - b).

(3x)² - (1)² = (3x + 1)(3x - 1).

Hence, 9x² - 1 = (3x + 1)(3x - 1).

Factorise the following:

150 - 6a²

Answer

150 - 6a² = 6(25 - a²) = 6(5² - a²).

Using identity,

a² - b² = (a + b)(a - b).

6(5² - a²) = 6(5 + a)(5 - a).

Hence, 150 - 6a² = 6(5 + a)(5 - a).

Factorise the following:

32x² - 18y²

Answer

32x² - 18y² = 2(16x² - 9y²) = 2[(4x)² - (3y)²].

Using identity,

a² - b² = (a + b)(a - b).

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

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

Factorise the following:

(x - y)² - 9

Answer

(x - y)² - 9 = (x - y)² - (3)².

Using identity,

a² - b² = (a + b)(a - b).

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

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

Factorise the following:

9(x + y)² - x²

Answer

9(x + y)² - x² = [3(x + y)]² - x².

Using identity,

a² - b² = (a + b)(a - b).

[3(x + y)]² - x² = [3(x + y) - x][3(x + y) + x] = (3x + 3y + x)(3x + 3y - x) = (4x + 3y)(2x + 3y).

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

Factorise the following:

20x² - 45y²

Answer

20x² - 45y² = 5(4x² - 9y²) = 5[(2x)² - (3y)²].

Using identity,

a² - b² = (a + b)(a - b).

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

Hence, 20x² - 45y² = 5(2x + 3y)(2x - 3y).

Factorise the following:

9x² - 4(y + 2x)²

Answer

9x² - 4(y + 2x)² = (3x)² - [2(y + 2x)]².

Using identity,

a² - b² = (a + b)(a - b).

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

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

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

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

Hence, 9x² - 4(y + 2x)² = -(7x + 2y)(x + 2y).

Factorise the following:

2(x - 2y)² - 50y²

Answer

2(x - 2y)² - 50y²

= 2[(x - 2y)² - 25y²]

= 2[(x - 2y)² - (5y)²].

Using identity,

a² - b² = (a + b)(a - b).

2[(x - 2y)² - (5y)²]

= 2(x - 2y + 5y)(x - 2y - 5y) = 2(x + 3y)(x - 7y).

Hence, 2(x - 2y)² - 50y² = 2(x + 3y)(x - 7y).

Factorise the following:

32 - 2(x - 4)²

Answer

32 - 2(x - 4)²

= 2[16 - (x - 4)²]

= 2[(4)² - (x - 4)²].

Using identity,

a² - b² = (a + b)(a - b).

2[(4)² - (x - 4)²]

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

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

= 2(x)(8 - x)

= 2x(8 - x).

Hence, 32 - 2(x - 4)² = 2x(8 - x).

Factorise the following:

108a² - 3(b - c)²

Answer

108a² - 3(b - c)²

= 3[36a² - (b - c)²]

= 3[(6a)² - (b - c)²].

Using identity,

a² - b² = (a - b)(a + b).

3[(6a)² - (b - c)²]

= 3[6a - (b - c)][6a + (b - c)]

= 3(6a - b + c)(6a + b - c).

Hence, 108a² - 3(b - c)² = 3(6a - b + c)(6a + b - c).

Factorise the following:

πa⁵ - π³ab²

Answer

πa⁵ - π³ab²

= πa(a⁴ - π²b²)

= πa[(a²)² - (πb)²].

Using identity,

a² - b² = (a - b)(a + b).

πa[(a²)² - (πb)²] = πa(a² - πb)(a² + πb).

Hence, πa⁵ - π³ab² = πa(a² - πb)(a² + πb).

Factorise the following:

50x² - 2(x - 2)²

Answer

50x² - 2(x - 2)²

= 2[25x² - (x - 2)²]

= 2[(5x)² - (x - 2)²].

Using identity,

a² - b² = (a - b)(a + b).

2[(5x)² - (x - 2)²] = 2[5x - (x - 2)][5x + (x - 2)]

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

= 2(4x + 2)(6x - 2)

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

= 8(2x + 1)(3x - 1).

Hence, 50x² - 2(x - 2)² = 8(2x + 1)(3x - 1).

Factorise the following:

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

Answer

Using identity,

(a - b)(a + b) = (a² - b²).

(x - 2)(x + 2) + 3 = (x² - 4) + 3 = (x² - 1).

Using identity,

a² - b² = (a - b)(a + b).

(x² - 1) = (x - 1)(x + 1).

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

Factorise the following:

x - 2y - x² + 4y²

Answer

x - 2y - x² + 4y²

= x - 2y - (x² - 4y²)

= x - 2y - [x² - (2y)²].

Using identity,

a² - b² = (a - b)(a + b).

(x - 2y) - [x² - (2y)²] = (x - 2y) - (x - 2y)(x + 2y)

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

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

Hence, x - 2y - x² + 4y² = (x - 2y)(1 - x - 2y).

Factorise the following:

4a² - b² + 2a + b

Answer

4a² - b² + 2a + b = (2a)² - b² + 2a + b.

Using identity,

a² - b² = (a - b)(a + b).

(2a)² - b² + 2a + b = (2a - b)(2a + b) + (2a + b)

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

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

Factorise the following:

a(a - 2) - b(b - 2)

Answer

a(a - 2) - b(b - 2) = a² - 2a - b² + 2b

= a² - b² - 2a + 2b

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

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

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

Factorise the following:

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

Answer

a(a - 1) - b(b - 1) = a² - a - b² + b

= a² - b² - a + b

= a² - b² - (a - b).

Using identity,

a² - b² = (a - b)(a + b).

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

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

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

Factorise the following:

9 - x² + 2xy - y².

Answer

9 - x² + 2xy - y²

Above terms can be written as,

9 - x² + xy + xy - y².

or,

9 - x² + xy + 3x - 3x + 3y - 3y + xy - y²

Rearranging above terms, we get,

9 - 3x + 3y + 3x - x² + xy + xy - 3y - y².

Take out common in all terms we get,

3(3 - x + y) + x(3 - x + y) + y(-3 - y + x)

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

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

Hence, 9 - x² + 2xy - y² = (3 + x - y)(3 - x + y).

Factorise the following:

9x⁴ - (x² + 2x + 1)

Answer

9x⁴ - (x² + 2x + 1) = (3x²)² - (x + 1)².

Using identity,

a² - b² = (a - b)(a + b).

(3x²)² - (x + 1)² = (3x² - x - 1)(3x² + x + 1).

Hence, 9x⁴ - (x² + 2x + 1) = (3x² - x - 1)(3x² + x + 1).

Factorise the following:

9x⁴ - x² - 12x - 36

Answer

9x⁴ - x² - 12x - 36 = 9x⁴ - (x² + 12x + 36).

The above equation can be written as,

9x⁴ - [x² + (2 × 6 × x) + (6)²]

As, (a + b)² = a² + 2ab + b².

∴ 9x⁴ - [x² + (2 × 6 × x) + (6)²] = (3x²)² - (x + 6)².

Using identity,

a² - b² = (a - b)(a + b)

(3x²)² - (x + 6)² = (3x² + x + 6)(3x² - x - 6).

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

Factorise the following:

x³ - 5x² - x + 5

Answer

x³ - 5x² - x + 5 = x²(x - 5) - 1(x - 5)

= (x² - 1)(x - 5).

Using identity,

a² - b² = (a - b)(a + b).

(x² - 1)(x - 5) = (x - 1)(x + 1)(x - 5).

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

Factorise the following:

a⁴ - b⁴ + 2b² - 1

Answer

a⁴ - b⁴ + 2b² - 1

Above terms can be written as,

a⁴ - (b⁴ - 2b² + 1)

= a⁴ - [(b²)² - (2 × b² × 1) + 1²]

We know that,

(a - b)² = a² - 2ab + b².

a⁴ - [(b²)² - (2 × b² × 1) + 1²] = (a²)² - (b² - 1)².

Using identity,

a² - b² = (a - b)(a + b).

(a²)² - (b² - 1)² = (a² + b² - 1)(a² - b² + 1).

Hence, a⁴ - b⁴ + 2b² - 1 = (a² + b² - 1)(a² - b² + 1).

Factorise the following:

x³ - 25x

Answer

x³ - 25x

Taking out common in all terms,

x(x² - 25) = x(x² - 5²).

Using identity,

a² - b² = (a - b)(a + b).

x(x² - 5²) = x(x - 5)(x + 5).

Hence, x³ - 25x = x(x - 5)(x + 5).

Factorise the following:

2x⁴ - 32

Answer

2x⁴ - 32

Taking out common in all terms,

2(x⁴ - 16) = 2[(x²)² - 4²].

Using identity,

a² - b² = (a - b)(a + b).

2[(x²)² - 4²] = 2(x² - 4)(x² + 4)

= 2(x - 2)(x + 2)(x² + 4).

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

Factorise the following:

a²(b + c) - (b + c)³

Answer

a²(b + c) - (b + c)³

Taking out common in all terms,

(b + c)[a² - (b + c)²]

Using identity,

a² - b² = (a - b)(a + b).

(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).

Factorise the following:

(a + b)³ - a - b

Answer

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

Taking out common in all terms,

(a + b)[(a + b)² - 1].

Using identity,

a² - b² = (a - b)(a + b).

(a + b)[(a + b)² - 1] = (a + b)(a + b - 1)(a + b + 1).

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

Factorise the following:

x² - 2xy + y² - a² - 2ab - b².

Answer

x² - 2xy + y² - a² - 2ab - b² = (x² - 2xy + y²) - (a² + 2ab + b²)

We know that,

(a + b)² = a² + 2ab + b²

and

(a - b)² = a² - 2ab + b²

∴ (x² - 2xy + y²) - (a² + 2ab + b²) = (x - y)² - (a + b)².

Using identity,

a² - b² = (a - b)(a + b).

(x - y)² - (a + b)² = (x - y - a - b)(x - y + a + b).

Hence, x² - 2xy + y² - a² - 2ab - b² = (x - y - a - b)(x - y + a + b).

Factorise the following:

(a² - b²)(c² - d²) - 4abcd

Answer

(a² - b²)(c² - d²) - 4abcd

= a²(c² - d²) - b²(c² - d²) - 4abcd

= a²c² - a²d² - b²c² + b²d² - 4abcd

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

= a²c² + b²d² - 2abcd - a²d² - b²c² - 2abcd.

= a²c² + b²d² - 2abcd - (a²d² + b²c² + 2abcd).

We know that,

(a + b)² = a² + 2ab + b²

and

(a - b)² = a² - 2ab + b²

∴ a²c² + b²d² - 2abcd - (a²d² + b²c² + 2abcd) = (ac - bd)² - (ad + bc)².

Using identity,

a² - b² = (a - b)(a + b).

(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).

Factorise the following:

4x² - y² - 3xy + 2x - 2y

Answer

4x² - y² - 3xy + 2x - 2y

Above terms can be written as,

x² + 3x² - y² - 3xy + 2x - 2y

Rearranging the above terms, we get,

(x² - y²) + (3x² - 3xy) + (2x - 2y)

We know that, a² - b² = (a - b)(a + b) and taking out common terms we get,

(x² - y²) + (3x² - 3xy) + (2x - 2y) = (x - y)(x + y) + 3x(x - y) + 2(x - y)

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

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

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

Factorise the following:

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

Answer

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

We know that,

(a - b)² = a² - 2ab + b²

and

a² - b² = (a - b)(a + b)

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

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

Factorise the following:

x⁴ + 5x² + 9

Answer

x⁴ + 5x² + 9 = x⁴ + 6x² - x² + 9

= (x⁴ + 6x² + 9) - x²

= [(x²)² + 2(3x²) + 3²] - x²

We know that,

(a + b)² = a² + 2ab + b²

and

a² - b² = (a - b)(a + b)

∴ [(x²)² + 2(3x²) + 3²] - x² = (x² + 3)² - x²

= (x² + 3 - x)(x² + 3 + x)

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

(i) x² + $\dfrac{4}{x^2}$ - 5

(ii) x⁴ + 12x² + 11

Answer

(i) Given,

x² + $\dfrac{4}{x^2}$ - 5

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

Hence, x² + $\dfrac{4}{x^2} - 5 = \Big(x - \dfrac{2}{x} - 1\Big)\Big(x - \dfrac{2}{x} + 1\Big)$

(ii) Given,

⇒ x⁴ + 12x² + 11

⇒ x⁴ + 11x² + x² + 11

⇒ x²(x² + 11) + 1(x² + 11)

⇒ (x² + 11)(x² + 1).

Hence, x⁴ + 12x² + 11 = (x² + 11)(x² + 1).

Factorise the following:

a⁴ + b⁴ - 7a²b².

Answer

Above terms can be written as,

a⁴ + b⁴ + 2a²b² - 9a²b²

= [(a²)² + (b²)² + 2(a²b²)] - (3ab)²

We know that,

(a + b)² = a² + 2ab + b²

and

a² - b² = (a - b)(a + b)

∴ [(a²)² + (b²)² + 2(a²b²)] - (3ab)² = (a² + b²)² - (3ab)²

= (a² + b² + 3ab)(a² + b² - 3ab)

Hence, a⁴ + b⁴ - 7a²b² = (a² + b² + 3ab)(a² + b² - 3ab).

Factorise the following:

x⁴ - 14x² + 1

Answer

Above terms can be written as,

x⁴ + 2x² - 16x² + 1

= x⁴ + 2x² + 1 - 16x²

= [(x²)² + 2x² + 1] - (4x)²

We know that,

(a + b)² = a² + 2ab + b²

and

a² - b² = (a - b)(a + b)

∴ [(x²)² + 2x² + 1] - (4x)² = (x² + 1)² - (4x)²

= (x² + 1 - 4x)(x² + 1 + 4x).

Hence, x⁴ - 14x² + 1 = (x² + 4x + 1)(x² - 4x + 1).

Express each of the following as the difference of two squares:

(x² - 5x + 7)(x² + 5x + 7)

Answer

(x² - 5x + 7)(x² + 5x + 7)

Rearranging the above terms, we get,

[(x² + 7) - 5x][(x² + 7) + 5x]

We know that

a² - b² = (a - b)(a + b).

∴ [(x² + 7) - 5x][(x² + 7) + 5x] = (x² + 7)² - (5x)²

Hence, (x² - 5x + 7)(x² + 5x + 7) = (x² + 7)² - (5x)².

Express each of the following as the difference of two squares:

(x² - 5x + 7)(x² - 5x - 7)

Answer

(x² - 5x + 7)(x² - 5x - 7)

= [(x² - 5x) + 7][(x² - 5x) - 7].

As, we know that,

a² - b² = (a - b)(a + b).

∴ [(x² - 5x) + 7][(x² - 5x) - 7] = (x² - 5x)² - 7²

Hence, (x² - 5x + 7)(x² - 5x - 7) = (x² - 5x)² - 7².

Express each of the following as the difference of two squares:

(x² + 5x - 7)(x² - 5x + 7)

Answer

(x² + 5x - 7)(x² - 5x + 7) = [{x² + (5x - 7)}{x² - (5x - 7)}]

As, we know that,

a² - b² = (a - b)(a + b).

∴ [{x² + (5x - 7)}{x² - (5x - 7)}] = (x²)² - (5x - 7)².

Hence, (x² + 5x - 7)(x² - 5x + 7) = (x²)² - (5x - 7)².

Evaluate the following by using factors:

(979)² - (21)²

Answer

We know that,

a² - b² = (a - b)(a + b).

∴ (979)² - (21)² = (979 - 21)(979 + 21)

= 958 x 1000

= 958000.

Hence, (979)² - (21)² = 958000.

Evaluate the following by using factors:

(99.9)² - (0.1)²

Answer

We know that,

a² - b² = (a - b)(a + b).

∴ (99.9)² - (0.1)² = (99.9 - 0.1)(99.9 + 0.1)

= 99.8 x 100

= 9980.

Hence, (99.9)² - (0.1)² = 9980.