Factorisation — Exercise 4.4

Factorise the following:

x² + 5x + 6

Answer

To factorise x² + 5x + 6, we want to find two real numbers whose sum is 5 and product is 6. By trial, we see that 2 + 3 = 5 and 2 × 3 = 6.

∴ x² + 5x + 6 = x² + 2x + 3x + 6

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

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

Hence, x² + 5x + 6 = (x + 3)(x + 2).

Factorise the following:

x² - 8x + 7

Answer

To factorise x² - 8x + 7, we want to find two real numbers whose sum is -8 and product is 7. By trial, we see that (-7) + (-1) = -8 and -7 × -1 = 7.

∴ x² - 8x + 7 = x² - 7x + (-1x) + 7

= x² - 7x - x + 7

= x(x - 7) - 1(x - 7)

= (x - 1)(x - 7).

Hence, x² - 8x + 7 = (x - 1)(x - 7).

Factorise the following:

x² + 6x - 7

Answer

To factorise x² + 6x - 7, we want to find two real numbers whose sum is 6 and product is -7. By trial, we see that 7 - 1 = 6 and 7 × -1 = -7.

∴ x² + 6x - 7 = x² + 7x + (-1x) - 7

= x² + 7x - x - 7

= x(x + 7) - 1(x + 7)

= (x - 1)(x + 7).

Hence, x² + 6x - 7 = (x - 1)(x + 7).

Factorise the following:

y² + 7y - 18

Answer

To factorise y² + 7y - 18, we want to find two real numbers whose sum is 7 and product is -18. By trial, we see that 9 + (-2) = 7 and 9 × -2 = -18.

∴ y² + 7y - 18 = y² + 9y + (-2y) - 18

= y² + 9y - 2y - 18

= y(y + 9) - 2(y + 9)

= (y - 2)(y + 9).

Hence, y² + 7y - 18 = (y - 2)(y + 9).

Factorise the following:

y² - 7y - 18

Answer

To factorise y² - 7y - 18, we want to find two real numbers whose sum is -7 and product is -18. By trial, we see that -9 + 2 = -7 and -9 × 2 = -18.

∴ y² - 7y - 18 = y² - 9y + 2y - 18

= y² - 9y + 2y - 18

= y(y - 9) + 2(y - 9)

= (y - 9)(y + 2).

Hence, y² - 7y - 18 = (y - 9)(y + 2).

Factorise the following:

a² - 3a - 54

Answer

To factorise a² - 3a - 54, we want to find two real numbers whose sum is -3 and product is -54. By trial, we see that -9 + 6 = -3 and -9 × 6 = -54.

∴ a² - 3a - 54 = a² - 9a + 6a - 54

= a(a - 9) + 6(a - 9)

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

Hence, a² - 3a - 54 = (a - 9)(a + 6).

Factorise the following:

2x² - 7x + 6

Answer

To factorise 2x² - 7x + 6, we want to find two real numbers whose sum is -7 and product is 2 × 6 = 12. By trial, we see that (-4) + (-3) = -7 and -4 × -3 = 12.

∴ 2x² - 7x + 6 = 2x² - 4x + (-3x) + 6

= 2x² - 4x - 3x + 6

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

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

Hence, 2x² - 7x + 6 = (x - 2)(2x - 3).

Factorise the following:

6x² + 13x - 5

Answer

To factorise 6x² + 13x - 5, we want to find two real numbers whose sum is 13 and product is 6 × (-5) = -30. By trial, we see that 15 + (-2) = 13 and 15 × -2 = -30.

∴ 6x² + 13x - 5 = 6x² + 15x + (-2x) - 5

= 6x² + 15x - 2x - 5

= 3x(2x + 5) - 1(2x + 5)

= (2x + 5)(3x - 1).

Hence, 6x² + 13x - 5 = (2x + 5)(3x - 1).

Factorise the following:

6x² + 11x - 10

Answer

To factorise 6x² + 11x - 10, we want to find two real numbers whose sum is 11 and product is 6 × (-10) = -60. By trial, we see that 15 + (-4) = 11 and 15 × -4 = -60.

∴ 6x² + 11x - 10 = 6x² + 15x + (-4x) - 10

= 6x² + 15x - 4x - 10

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

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

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

Factorise the following:

6x² - 7x - 3

Answer

To factorise 6x² - 7x - 3, we want to find two real numbers whose sum is -7 and product is 6 × (-3) = -18. By trial, we see that -9 + 2 = -7 and -9 × 2 = -18.

∴ 6x² - 7x - 3 = 6x² - 9x + 2x - 3

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

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

Hence, 6x² - 7x - 3 = (2x - 3)(3x + 1).

Factorise the following:

2x² - x - 6

Answer

To factorise 2x² - x - 6, we want to find two real numbers whose sum is -1 and product is 2 × (-6) = -12. By trial, we see that -4 + 3 = -1 and -4 × 3 = -12.

∴ 2x² - x - 6 = 2x² - 4x + 3x - 6

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

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

Hence, 2x² - x - 6 = (x - 2)(2x + 3).

Factorise the following:

1 - 18y - 63y²

Answer

To factorise 1 - 18y - 63y², we want to find two real numbers whose sum is -18 and product is 1 × (-63) = -63. By trial, we see that -21 + 3 = -18 and -21 × 3 = -63.

∴ 1 - 18y - 63y² = 1 - 21y + 3y - 63y²

= 1(1 - 21y) + 3y(1 - 21y)

= (1 - 21y)(1 + 3y).

Hence, 1 - 18y - 63y² = (1 - 21y)(1 + 3y).

Factorise the following:

2y² + y - 45

Answer

To factorise 2y² + y - 45, we want to find two real numbers whose sum is 1 and product is 2 × (-45) = -90. By trial, we see that 10 - 9 = 1 and 10 × (-9) = -90.

∴ 2y² + y - 45 = 2y² + 10y + (-9y) - 45

= 2y² + 10y - 9y - 45

= 2y(y + 5) - 9(y + 5)

= (y + 5)(2y - 9).

Hence, 2y² + y - 45 = (y + 5)(2y - 9).

Factorise the following:

5 - 4x - 12x²

Answer

To factorise 5 - 4x - 12x², we want to find two real numbers whose sum is -4 and product is 5 × (-12) = -60. By trial, we see that -10 + 6 = -4 and (-10) × 6 = -60.

∴ 5 - 4x - 12x² = 5 - 10x + 6x - 12x²

= 5(1 - 2x) + 6x(1 - 2x)

= (1 - 2x)(5 + 6x).

Hence, 5 - 4x - 12x² = (1 - 2x)(5 + 6x).

Factorise the following:

x(12x + 7) - 10

Answer

x(12x + 7) - 10 = 12x² + 7x - 10.

To factorise 12x² + 7x - 10, we want to find two real numbers whose sum is 7 and product is 12 × (-10) = -120. By trial, we see that 15 + (-8) = 7 and 15 × (-8) = -120.

∴ 12x² + 7x - 10 = 12x² + 15x + (-8x) - 10

= 12x² + 15x - 8x - 10

= 3x(4x + 5) - 2(4x + 5)

= (4x + 5)(3x - 2).

Hence, x(12x + 7) - 10 = (4x + 5)(3x - 2).

Factorise the following:

(4 - x)² - 2x

Answer

We know that,

(a - b)² = a² + b² - 2ab.

∴ (4 - x)² - 2x = 16 + x² - 8x - 2x = 16 + x² - 10x.

Rearranging the above terms we get,

x² - 10x + 16.

To factorise x² - 10x + 16, we want to find two real numbers whose sum is -10 and product is 16. By trial, we see that (-8) + (-2) = -10 and (-8) × (-2) = 16.

∴ x² - 10x + 16 = x² + (-8x) + (-2x) + 16

= x² - 8x - 2x + 16

= x(x - 8) - 2(x - 8)

= (x - 8)(x - 2).

Hence, (4 - x)² - 2x = (x - 8)(x - 2).

Factorise the following:

60x² - 70x - 30

Answer

60x² - 70x - 30 = 10(6x² - 7x - 3).

To factorise 6x² - 7x - 3, we want to find two real numbers whose sum is -7 and product is 6 × (-3) = -18. By trial, we see that (-9) + 2 = -7 and (-9) × (2) = -18.

∴ 6x² - 7x - 3 = 6x² - 9x + 2x - 3

= 6x² - 9x + 2x - 3

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

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

∴ 10(6x² - 7x - 3) = 10(2x - 3)(3x + 1).

Hence, 60x² - 70x - 30 = 10(2x - 3)(3x + 1).

Factorise the following:

x² - 6xy - 7y²

Answer

To factorise x² - 6xy - 7y², we want to find two real numbers whose sum is -6 and product is -7. By trial, we see that (-7) + 1 = -6 and (-7) × 1 = -7.

∴ x² - 6xy - 7y² = x² - 7xy + xy - 7y²

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

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

Hence, x² - 6xy - 7y² = (x - 7y)(x + y).

Factorise the following:

2x² + 13xy - 24y²

Answer

To factorise 2x² + 13xy - 24y², we want to find two real numbers whose sum is 13 and product is 2 × (-24) = -48. By trial, we see that 16 + (-3) = 13 and 16 × (-3) = -48.

∴ 2x² + 13xy - 24y² = 2x² + 16xy + (-3xy) - 24y²

= 2x² + 16xy - 3xy - 24y²

= 2x(x + 8y) - 3y(x + 8y)

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

Hence, 2x² + 13xy - 24y² = (x + 8y)(2x - 3y).

Factorise the following:

6x² - 5xy - 6y²

Answer

To factorise 6x² - 5xy - 6y², we want to find two real numbers whose sum is -5 and product is 6 × (-6) = -36. By trial, we see that -9 + 4 = -5 and (-9) × 4 = -36.

∴ 6x² - 5xy - 6y² = 6x² - 9xy + 4xy - 6y²

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

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

Hence, 6x² - 5xy - 6y² = (2x - 3y)(3x + 2y).

Factorise the following:

5x² + 17xy - 12y²

Answer

To factorise 5x² + 17xy - 12y², we want to find two real numbers whose sum is 17 and product is 5 × (-12) = -60. By trial, we see that 20 + (-3) = 17 and 20 × (-3) = -60.

∴ 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 the following:

x²y² - 8xy - 48

Answer

To factorise x²y² - 8xy - 48, we want to find two real numbers whose sum is -8 and product is -48. By trial, we see that (-12) + 4 = -8 and (-12) × 4 = -48.

∴ x²y² - 8xy - 48 = x²y² - 12xy + 4xy - 48

= xy(xy - 12) + 4(xy - 12)

= (xy - 12)(xy + 4).

Hence, x²y² - 8xy - 48 = (xy - 12)(xy + 4).

Factorise the following:

2a²b² - 7ab - 30

Answer

To factorise 2a²b² - 7ab - 30, we want to find two real numbers whose sum is -7 and product is 2 × (-30) = -60. By trial, we see that (-12) + 5 = -7 and (-12) × 5 = -60.

∴ 2a²b² - 7ab - 30 = 2a²b² - 12ab + 5ab - 30

= 2ab(ab - 6) + 5(ab - 6)

= (ab - 6)(2ab + 5).

Hence, 2a²b² - 7ab - 30 = (ab - 6)(2ab + 5).

Factorise the following:

a(2a - b) - b²

Answer

a(2a - b) - b² = 2a² - ab - b².

To factorise 2a² - ab - b², we want to find two real numbers whose sum is -1 and product is 2 × (-1) = -2. By trial, we see that (-2) + 1 = -1 and (-2) × 1 = -2.

∴ 2a² - ab - b² = 2a² - 2ab + ab - b²

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

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

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

Factorise the following:

(x - y)² - 6(x - y) + 5

Answer

Let (x - y) = a.

(x - y)² - 6(x - y) + 5 = a² - 6a + 5.

Factorising we get,

a² - 6a + 5 = a² - 5a - a + 5.

= a(a - 5) - 1(a - 5)

= (a - 5)(a - 1)

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

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

Factorise the following:

(2x - y)² - 11(2x - y) + 28

Answer

Let (2x - y) = a.

(2x - y)² - 11(2x - y) + 28 = a² - 11a + 28.

Factorising we get,

a² - 11a + 28 = a² - 7a - 4a + 28.

= a(a - 7) - 4(a - 7)

= (a - 7)(a - 4)

= (2x - y - 7)(2x - y - 4).

Hence, (2x - y)² - 11(2x - y) + 28 = (2x - y - 7)(2x - y - 4).

Factorise the following:

4(a - 1)² - 4(a - 1) - 3

Answer

Let (a - 1) = t.

4(a - 1)² - 4(a - 1) - 3 = 4t² - 4t - 3.

Factorising we get,

4t² - 4t - 3 = 4t² - 6t + 2t - 3.

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

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

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

= [2a - 2 - 3][2a - 2 + 1]

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

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

Factorise the following:

1 - 2a - 2b - 3(a + b)².

Answer

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

Let (a + b) = t.

1 - 2a - 2b - 3(a + b)² = 1 - 2t - 3t²

Factorising we get,

1 - 2t - 3t² = 1 - 3t + t - 3t²

= 1(1 - 3t) + t(1 - 3t)

= (1 + t)(1 - 3t)

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

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

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

Factorise the following:

3 - 5a - 5b - 12(a + b)²

Answer

3 - 5a - 5b - 12(a + b)² = 3 - 5(a + b) - 12(a + b)².

Let (a + b) = t.

3 - 5(a + b) - 12(a + b)² = 3 - 5t - 12t²

Factorising we get,

3 - 5t - 12t² = 3 - 9t + 4t - 12t²

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

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

= [1 - 3(a + b)][3 + 4(a + b)]

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

Hence, 3 - 5a - 5b - 12(a + b)² = (1 - 3a - 3b)(3 + 4a + 4b).

Factorise the following:

a⁴ - 11a² + 10

Answer

Factorising we get,

a⁴ - 11a² + 10 = a⁴ - 10a² - a² + 10.

= a²(a² - 10) - 1(a² - 10)

= (a² - 10)(a² - 1)

= (a² - 10)(a - 1)(a + 1)

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

Factorise the following:

(x + 4)² - 5xy - 20y - 6y²

Answer

(x + 4)² - 5xy - 20y - 6y² = (x + 4)² - 5y(x + 4) - 6y².

Let (x + 4) = t.

(x + 4)² - 5y(x + 4) - 6y² = t² - 5yt - 6y².

Factorising we get,

t² - 5yt - 6y² = t² - 6yt + 1yt - 6y².

= t(t - 6y) + y(t - 6y)

= (t - 6y)(t + y)

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

Hence, (x + 4)² - 5xy - 20y - 6y² = (x + y + 4)(x - 6y + 4).

Factorise the following:

(x² - 2x)² - 23(x² - 2x) + 120

Answer

Let (x² - 2x) = t.

(x² - 2x)² - 23(x² - 2x) + 120 = t² - 23t + 120.

t² - 23t + 120 = t² - 15t - 8t + 120.

= t(t - 15) - 8(t - 15)

= (t - 15)(t - 8)

= (x² - 2x - 15)(x² - 2x - 8)

Factorising x² - 2x - 15 we get,

x² - 2x - 15 = x² - 5x + 3x - 15 = x(x - 5) + 3(x - 5) = (x - 5)(x + 3).

Factorising x² - 2x - 8 we get,

x² - 2x - 8 = x² - 4x + 2x - 8 = x(x - 4) + 2(x - 4) = (x - 4)(x + 2).

∴ (x² - 2x - 15)(x² - 2x - 8) = (x - 5)(x + 3)(x - 4)(x + 2).

Hence, (x² - 2x)² - 23(x² - 2x) + 120 = (x - 5)(x + 3)(x - 4)(x + 2).

Factorise the following:

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

Answer

Let (2a - 3) = x and (a - 1) = y,

∴ 4(2a - 3)² - 3(2a - 3)(a - 1) - 7(a - 1)² = 4x² - 3xy - 7y².

Factorising we get,

4x² - 3xy - 7y² = 4x² - 7xy + 4xy - 7y²

= x(4x - 7y) + y(4x - 7y)

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

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

= (3a - 4)(8a - 12 - 7a + 7)

= (3a - 4)(a - 5).

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

Factorise the following:

(2x² + 5x)(2x² + 5x - 19) + 84

Answer

Let 2x² + 5x = y then,

(2x² + 5x)(2x² + 5x - 19) + 84 = y(y - 19) + 84.

= y² - 19y + 84

= y² - 12y - 7y + 84

= y(y - 12) - 7(y - 12)

= (y - 12)(y - 7)

= (2x² + 5x - 12)(2x² + 5x - 7).

Factorising 2x² + 5x - 12 we get,

2x² + 5x - 12 = 2x² + 8x - 3x - 12

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

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

Factorising 2x² + 5x - 7 we get,

2x² + 5x - 7 = 2x² + 7x - 2x - 7

= x(2x + 7) - 1(2x + 7)

= (2x + 7)(x - 1).

Hence, (2x² + 5x)(2x² + 5x - 19) + 84 = (2x - 3)(x + 4)(2x + 7)(x - 1).