Factorisation — Chapter Test

Factorise the following:

15(2x - 3)³ - 10(2x - 3)

Answer

H.C.F. of 15(2x - 3)³ and 10(2x - 3) = 5(2x - 3).

∴ 15(2x - 3)³ - 10(2x - 3) = 5(2x - 3)[3(2x - 3)² - 2].

Factorise the following:

a(b - c)(b + c) - d(c - b)

Answer

a(b - c)(b + c) - d(c - b) = a(b - c)(b + c) - d(-1)(b - c)

= a(b - c)(b + c) + d(b - c)

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

Hence, a(b - c)(b + c) - d(c - b) = (b - c)[a(b + c) + d].

Factorise the following:

2a²x - bx + 2a² - b

Answer

Rearranging the above terms we get,

2a²x + 2a² - bx - b

= 2a²(x + 1) - b(x + 1)

= (x + 1)(2a² - b).

Hence, 2a²x - bx + 2a² - b = (x + 1)(2a² - b).

Factorise the following:

p² - (a + 2b)p + 2ab

Answer

p² - (a + 2b)p + 2ab = p² - ap - 2bp + 2ab

= p² - 2bp - ap + 2ab

= p(p - 2b) - a(p - 2b)

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

Hence, p² - (a + 2b)p + 2ab = (p - 2b)(p - a).

Factorise the following:

(x² - y²)z + (y² - z²)x

Answer

(x² - y²)z + (y² - z²)x = x²z - y²z + y²x - z²x

= x²z - xz² + y²x - y²z

= xz(x - z) + y²(x - z)

= (x - z)(xz + y²).

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

Factorise the following:

5a⁴ - 5a³ + 30a² - 30a

Answer

5a⁴ - 5a³ + 30a² - 30a = 5a(a³ - a² + 6a - 6)

= 5a[a²(a - 1) + 6(a - 1)]

= 5a(a - 1)(a² + 6).

Hence, 5a⁴ - 5a³ + 30a² - 30a = 5a(a - 1)(a² + 6).

Factorise the following:

b(c - d)² + a(d - c) + 3c - 3d

Answer

b(c - d)² + a(d - c) + 3c - 3d = b(c - d)² + a(-1)(c - d) + 3(c - d)

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

= (c - d)(bc - bd - a + 3).

Hence, b(c - d)² + a(d - c) + 3c - 3d = (c - d)(bc - bd - a + 3).

Factorise the following:

x³ - x² - xy + x + y - 1

Answer

Rearrange the above terms we get,

x³ - x² - xy + y + x - 1

= x²(x - 1) - y(x - 1) + 1(x - 1)

= (x - 1)(x² - y + 1).

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

Factorise the following:

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

Answer

x(x + z) - y(y + z) = x² + xz - y² - yz.

Rearranging the above terms we get,

x² - y² + xz - yz

We know that,

a² - b² = (a + b)(a - b).

∴ x² - y² + xz - yz = (x + y)(x - y) + z(x - y)

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

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

Factorise the following:

a¹²x⁴ - a⁴x¹²

Answer

a¹²x⁴ - a⁴x¹² = a⁴x⁴(a⁸ - x⁸)

= a⁴x⁴[(a⁴)² - (x⁴)²]

We know that,

a² - b² = (a + b)(a - b).

= a⁴x⁴(a⁴ + x⁴)(a⁴ - x⁴)

= a⁴x⁴(a⁴ + x⁴)[(a²)² - (x²)²]

= a⁴x⁴(a⁴ + x⁴)(a² + x²)(a² - x²)

= a⁴x⁴(a⁴ + x⁴)(a² + x²)(a + x)(a - x).

Hence, a¹²x⁴ - a⁴x¹² = a⁴x⁴(a⁴ + x⁴)(a² + x²)(a + x)(a - x).

Factorise the following:

9x² + 12x + 4 - 16y²

Answer

9x² + 12x + 4 - 16y² = (3x)² + (2 × 3x × 2) + 2² - (4y)²

We know that,

(a + b)² = a² + b² + 2ab.

∴ (3x)² + (2 × 3x × 2) + 2² - (4y)² = (3x + 2)² - (4y)²

We know that,

a² - b² = (a + b)(a - b).

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

Hence, 9x² + 12x + 4 - 16y² = (3x + 2 + 4y)(3x + 2 - 4y).

Factorise the following:

x⁴ + 3x² + 4

Answer

x⁴ + 3x² + 4

Above terms can be written as,

(x²)² + 3(x²) + 4 = (x²)² + 4x² - x² + 2²

= (x² + 2²)² - (x²)

We know that,

(a² - b²) = (a + b)(a - b)

∴ (x² + 2²)² - (x²) = (x² + 2 + x)(x² + 2 - x)

= (x² + x + 2)(x² - x + 2)

Hence, x⁴ + 3x² + 4 = (x² + x + 2)(x² - x + 2).

Factorise the following:

21x² - 59xy + 40y²

Answer

21x² - 59xy + 40y² = 21x² - 35xy - 24xy + 40y²

= 7x(3x - 5y) - 8y(3x - 5y)

= (3x - 5y)(7x - 8y).

Hence, 21x² - 59xy + 40y² = (3x - 5y)(7x - 8y).

Factorise the following:

4x³y - 44x²y + 112xy

Answer

4x³y - 44x²y + 112xy = 4xy(x² - 11x + 28)

= 4xy(x² - 7x - 4x + 28)

= 4xy[x(x - 7) - 4(x - 7)]

= 4xy(x - 7)(x - 4).

Hence, 4x³y - 44x²y + 112xy = 4xy(x - 7)(x - 4).

Factorise the following:

x²y² - xy - 72

Answer

x²y² - xy - 72 = x²y² - 9xy + 8xy - 72

= xy(xy - 9) + 8(xy - 9)

= (xy - 9)(xy + 8).

Hence, x²y² - xy - 72 = (xy - 9)(xy + 8).

Factorise the following:

9x³y + 41x²y² + 20xy³

Answer

9x³y + 41x²y² + 20xy³ = xy(9x² + 41xy + 20y²)

= xy(9x² + 36xy + 5xy + 20y²)

= xy[9x(x + 4y) + 5y(x + 4y)]

= xy(x + 4y)(9x + 5y).

Hence, 9x³y + 41x²y² + 20xy³ = xy(x + 4y)(9x + 5y).

Factorise the following:

(3a - 2b)² + 3(3a - 2b) - 10

Answer

Let (3a - 2b) = t.

∴ (3a - 2b)² + 3(3a - 2b) - 10 = t² + 3t - 10

= t² + 5t - 2t - 10

= t(t + 5) - 2(t + 5)

= (t + 5)(t - 2)

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

Hence, (3a - 2b)² + 3(3a - 2b) - 10 = (3a - 2b + 5)(3a - 2b - 2).

Factorise the following:

(x² - 3x)(x² - 3x + 7) + 10

Answer

Let us assume, x² - 3x = t.

∴ (x² - 3x)(x² - 3x + 7) + 10 = t(t + 7) + 10

= t² + 7t + 10

= t² + 5t + 2t + 10

= t(t + 5) + 2(t + 5)

= (t + 5)(t + 2)

= (x² - 3x + 5)(x² - 3x + 2)

= (x² - 3x + 5)(x² - 2x - x + 2)

= (x² - 3x + 5)[x(x - 2) - 1(x - 2)]

= (x² - 3x + 5)(x - 2)(x - 1).

Hence, (x² - 3x)(x² - 3x + 7) + 10 = (x² - 3x + 5)(x - 2)(x - 1).

Factorise the following:

(x² - x)(4x² - 4x - 5) - 6

Answer

(x² - x)(4x² - 4x - 5) - 6 = (x² - x)[4(x² - x) - 5] - 6

Let x² - x = p.

(x² - x)[4(x² - x) - 5] - 6 = p(4p - 5) - 6

= 4p² - 5p - 6

= 4p² - 8p + 3p - 6

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

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

= (x² - x - 2)[4(x² - x) + 3]

= (x² - x - 2)(4x² - 4x + 3)

= [x² - 2x + x - 2](4x² - 4x + 3)

= [x(x - 2) + 1(x - 2)](4x² - 4x + 3)

= (x - 2)(x + 1)(4x² - 4x + 3).

Hence, (x² - x)(4x² - 4x - 5) - 6 = (x - 2)(x + 1)(4x² - 4x + 3).

Factorise the following:

x⁴ + 9x²y² + 81y⁴

Answer

x⁴ + 9x²y² + 81y⁴ = x⁴ + 18x²y² - 9x²y² + 81y⁴

= (x²)² + (2 × x² × 9y²) + (9y²)² - 9x²y²

We know that,

(a + b)² = a² + b² + 2ab

∴ (x²)² + (2 × x² × 9y²) + (9y²)² - 9x²y² = (x² + 9y²)² - 9x²y²

= (x² + 9y²)² - (3xy)²

We know that,

a² - b² = (a + b)(a - b)

∴ (x² + 9y²)² - (3xy)² = (x² + 9y² + 3xy)(x² + 9y² - 3xy).

Hence, x⁴ + 9x²y² + 81y⁴ = (x² + 9y² + 3xy)(x² + 9y² - 3xy).

Factorise the following:

$\dfrac{8}{27}x^3 - \dfrac{1}{8}y^3$

Answer

$\dfrac{8}{27}x^3 - \dfrac{1}{8}y^3 = \Big(\dfrac{2}{3}x\Big)^3 - \Big(\dfrac{1}{2}y\Big)^3$

We know that,

a³ - b³ = (a - b)(a² + ab + b²)

$\therefore \Big(\dfrac{2}{3}x\Big)^3 - \Big(\dfrac{1}{2}y\Big)^3 = \Big(\dfrac{2}{3}x - \dfrac{1}{2}y\Big)\Big[\Big(\dfrac{2}{3}x\Big)^2 + \dfrac{2}{3}x \times \dfrac{1}{2}y + \Big(\dfrac{1}{2}y\Big)^2\Big] \\[1em] = \Big(\dfrac{2}{3}x - \dfrac{1}{2}y\Big)\Big(\dfrac{4}{9}x^2 + \dfrac{1}{3}xy + \dfrac{1}{4}y^2\Big).$

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

Factorise the following:

x⁶ + 63x³ - 64

Answer

x⁶ + 63x³ - 64 = x⁶ + 64x³ - x³ - 64

= x³(x³ + 64) - 1(x³ + 64)

= (x³ + 64)(x³ - 1)

= (x³ + 4³)(x³ - 1³)

We know that,

a³ + b³ = (a + b)(a² + b² - ab)

a³ - b³ = (a - b)(a² + b² + ab)

∴ (x³ + 4³)(x³ - 1³) = (x + 4)(x² - 4.x + 4²)(x - 1)(x² + x.1 + 1²)

= (x + 4)(x² - 4x + 16)(x - 1)(x² + x + 1).

Hence, x⁶ + 63x³ - 64 = (x + 4)(x² - 4x + 16)(x - 1)(x² + x + 1).

Factorise the following:

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

Answer

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

We know that,

a² - b² = (a + b)(a - b)

a³ + b³ = (a + b)(a² - ab + b²)

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

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

Factorise the following:

(x + 1)⁶ - (x - 1)⁶

Answer

(x + 1)⁶ - (x - 1)⁶ = [(x + 1)³]² - [(x - 1)³]²

We know that,

a² - b² = (a + b)(a - b)

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

We know that,

a³ + b³ = (a + b)(a² + b² - ab)

a³ - b³ = (a - b)(a² + b² + ab)

∴ [(x + 1)³ + (x - 1)³][(x + 1)³ - (x - 1)³] = [(x + 1 + x - 1){(x + 1)² - (x + 1)(x - 1) + (x - 1)²}][(x + 1) - (x - 1)][(x + 1)² + (x + 1)(x - 1) + (x - 1)²]

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

= 2x[x² + 1 + 2x - x² + 1 + x² + 1 - 2x]2[x² + 1 + 2x + x² - 1 + x² + 1 - 2x]

= 4x(x² + 3)(3x² + 1).

Hence, (x + 1)⁶ - (x - 1)⁶ = 4x(x² + 3)(3x² + 1).

Factorise (x + 1)(x - 3) + (x + 1)(x + 4)

Answer

Given,

⇒ (x + 1)(x - 3) + (x + 1)(x + 4)

⇒ (x + 1)[(x - 3) + (x + 4)]

⇒ (x + 1)(x - 3 + x + 4)

⇒ (x + 1)(2x + 1).

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

Show that 97³ + 14³ is divisible by 111.

Answer

We know that,

a³ + b³ = (a + b)(a² + b² - ab)

∴ 97³ + 14³ = (97 + 14)[97² + 14² - 97 × 14]

= 111[9409 + 196 - 1358]

= 111 × 8247.

∴ 111 × 8247 is divisible by 111

Hence, proved that 97³ + 14³ is divisible by 111.

If a + b = 8 and ab = 15, find the value of a⁴ + a²b² + b⁴

Answer

Given,

a + b = 8 and ab = 15

∴ (a + b)² = 8²

⇒ a² + b² + 2ab = 64

⇒ a² + b² + 2(15) = 64

⇒ a² + b² = 64 - 30 = 34.

a⁴ + a²b² + b⁴ = a⁴ + 2a²b² - a²b² + b⁴

= (a² + b²)² - a²b²

We know that,

a² - b² = (a + b)(a - b)

∴ (a² + b²)² - a²b² = (a² + b² + ab)(a² + b² - ab)

= (34 + 15)(34 - 15)

= 49 × 19

= 931.

Hence, the value of a⁴ + a²b² + b⁴ = 931.