Factorisation

Factorise the following:

8xy³ + 12x²y²

Answer

H.C.F. of 8xy³ and 12x²y² is 4xy².

∴ 8xy³ + 12x²y² = 4xy²(2y + 3x).

Factorise the following:

15ax³ - 9ax².

Answer

H.C.F. of 15ax³ and 9ax² is 3ax².

∴ 15ax³ - 9ax² = 3ax²(5x - 3).

Factorise the following:

21py² - 56py

Answer

H.C.F. of 21py² and 56py is 7py.

∴ 21py² - 56py = 7py(3y - 8).

Factorise the following:

4x³ - 6x².

Answer

H.C.F. of 4x³ and 6x² is 2x².

∴ 4x³ - 6x² = 2x²(2x - 3).

Factorise the following:

2πr² - 4πr

Answer

H.C.F. of 2πr² and 4πr is 2πr.

∴ 2πr² - 4πr = 2πr(r - 2).

Factorise the following:

18m + 16n

Answer

H.C.F. of 18m and 16n is 2.

∴ 18m + 16n = 2(9m + 8n).

Factorise the following:

25abc² - 15a²b²c

Answer

H.C.F. of 25abc² and 15a²b²c is 5abc.

∴ 25abc² - 15a²b²c = 5abc(5c - 3ab).

Factorise the following:

28p²q²r - 42pq²r²

Answer

H.C.F. of 28p²q²r and 42pq²r² is 14pq²r.

∴ 28p²q²r - 42pq²r² = 14pq²r(2p - 3r).

Factorise the following:

8x³ - 6x² + 10x

Answer

H.C.F. of 8x³, 6x² and 10x is 2x.

∴ 8x³ - 6x² + 10x = 2x(4x² - 3x + 5).

Factorise the following:

14mn + 22m - 62p

Answer

H.C.F. of 14mn, 22m and 62p is 2.

∴ 14mn + 22m - 62p = 2(7mn + 11m - 31p).

Factorise the following:

18p²q² - 24pq² + 30p²q

Answer

H.C.F. of 18p²q², 24pq² and 30p²q is 6pq.

∴ 18p²q² - 24pq² + 30p²q = 6pq(3pq - 4q + 5p).

Factorise the following:

27a³b³ - 18a²b³ + 75a³b²

Answer

H.C.F. of 27a³b³, 18a²b³ and 75a³b² is 3a²b².

∴ 27a³b³ - 18a²b³ + 75a³b² = 3a²b²(9ab - 6b + 25a).

Factorise the following:

15a(2p - 3q) - 10b(2p - 3q)

Answer

H.C.F. of 15a(2p - 3q) and 10b(2p - 3q) is 5(2p - 3q).

∴ 15a(2p - 3q) - 10b(2p - 3q) = 5(2p - 3q)(3a - 2b).

Factorise the following:

3a(x² + y²) + 6b(x² + y²).

Answer

H.C.F. of 3a(x² + y²) and 6b(x² + y²) is 3(x² + y²).

∴ 3a(x² + y²) + 6b(x² + y²) = 3(x² + y²)(a + 2b).

Factorise the following:

6(x + 2y)³ + 8(x + 2y)²

Answer

H.C.F. of 6(x + 2y)³ and 8(x + 2y)² is 2(x + 2y)².

∴ 6(x + 2y)³ + 8(x + 2y)² = 2(x + 2y)²(3(x + 2y) + 4) = 2(x + 2y)²(3x + 6y + 4).

Factorise the following:

14(a - 3b)³ - 21p(a - 3b)

Answer

H.C.F. of 14(a - 3b)³ and 21p(a - 3b) is 7(a - 3b).

∴ 14(a - 3b)³ - 21p(a - 3b) = 7(a - 3b)[2(a - 3b)² - 3p].

Factorise the following:

10a(2p + q)³ - 15b(2p + q)² + 35(2p + q)

Answer

H.C.F. of 10a(2p + q)³, 15b(2p + q)² and 35(2p + q) is 5(2p + q).

∴ 10a(2p + q)³ - 15b(2p + q)² + 35(2p + q) = 5(2p + q)[2a(2p + q)² - 3b(2p + q) + 7].

Factorise the following:

x(x² + y² - z²) + y(-x² - y² + z²) - z(x² + y² - z²).

Answer

The above expression:

x(x² + y² - z²) + y(-x² - y² + z²) - z(x² + y² - z²)

can be written as

x(x² + y² - z²) + y(-1)(x² + y² - z²) - z(x² + y² - z²)
= x(x² + y² - z²) - y(x² + y² - z²) - z(x² + y² - z²).

H.C.F. of x(x² + y² - z²), y(x² + y² - z²) and z(x² + y² - z²) is (x² + y² - z²).

∴ x(x² + y² - z²) + y(-x² - y² + z²) - z(x² + y² - z²) = (x² + y² - z²)(x - y - z).

Factorise the following:

x² + xy - x - y

Answer

x² + xy - x - y

= x(x + y) - 1(x + y)

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

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

Factorise the following:

y² - yz - 5y + 5z

Answer

y² - yz - 5y + 5z

= y(y - z) - 5(y - z)

= (y - z)(y - 5).

Hence, y² - yz - 5y + 5z = (y - z)(y - 5)

Factorise the following:

5xy + 7y - 5y² - 7x

Answer

5xy + 7y - 5y² - 7x

Rearranging the terms:

= 5xy - 7x + 7y - 5y²

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

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

Hence, 5xy + 7y - 5y² - 7x = (x - y)(5y - 7).

Factorise the following:

5p² - 8pq - 10p + 16q

Answer

5p² - 8pq - 10p + 16q

= 5p² - 10p - 8pq + 16q

= 5p(p - 2) - 8q(p - 2)

= (5p - 8q)(p - 2).

Hence, 5p² - 8pq - 10p + 16q = (5p - 8q)(p - 2)

Factorise the following:

a²b - ab² + 3a - 3b

Answer

a²b - ab² + 3a - 3b

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

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

Hence, a²b - ab² + 3a - 3b = (ab + 3)(a - b).

Factorise the following:

x³ - 3x² + x - 3

Answer

x³ - 3x² + x - 3

= x²(x - 3) + 1(x - 3)

= (x² + 1)(x - 3).

Hence, x³ - 3x² + x - 3 = (x² + 1)(x - 3).

Factorise the following:

6xy² - 3xy - 10y + 5

Answer

6xy² - 3xy - 10y + 5

= 3xy(2y - 1) - 5(2y - 1)

= (2y - 1)(3xy - 5).

Hence, 6xy² - 3xy - 10y + 5 = (2y - 1)(3xy - 5).

Factorise the following:

3ax - 6ay - 8by + 4bx

Answer

3ax - 6ay - 8by + 4bx

= 3ax - 6ay + 4bx - 8by

Taking out common terms we get,

3ax - 6ay + 4bx - 8by

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

= (x - 2y)(3a + 4b).

Hence, 3ax - 6ay - 8by + 4bx = (x - 2y)(3a + 4b).

5px - 8qy + 4qx - 10py

Answer

Given,

⇒ 5px - 8qy + 4qx - 10py

⇒ 5px + 4qx - 8qy - 10py

⇒ x(5p + 4q) - 2y(4q + 5p)

⇒ (5p + 4q)(x - 2y).

Hence, 5px - 8qy + 4qx - 10py = (5p + 4q)(x - 2y).

9a²y - 9ay + 6a - 6

Answer

Given,

⇒ 9a²y - 9ay + 6a - 6

⇒ 9ay(a - 1) + 6(a - 1)

⇒ (a - 1)(9ay + 6)

⇒ (a - 1).3.(3ay + 2)

⇒ 3(a - 1)(3ay + 2).

Hence, 9a²y - 9ay + 6a - 6 = 3(a - 1)(3ay + 2).

Factorise the following:

1 - a - b + ab

Answer

1 - a - b + ab

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

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

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

Factorise the following:

a(a - 2b - c) + 2bc

Answer

a(a - 2b - c) + 2bc

= a² - 2ab - ac + 2bc

= a² - ac - 2ab + 2bc

= a(a - c) - 2b(a - c)

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

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

Factorise the following:

x² + xy(1 + y) + y³

Answer

x² + xy(1 + y) + y³

= x² + xy + xy² + y³

= x(x + y) + y²(x + y)

= (x + y)(x + y²).

Hence, x² + xy(1 + y) + y³ = (x + y)(x + y²).

Factorise the following:

y² - xy(1 - x) - x³

Answer

y² - xy(1 - x) - x³

= y² - xy + x²y - x³

= y(y - x) + x²(y - x)

= (y - x)(y + x²).

Hence, y² - xy(1 - x) - x³ = (y - x)(y + x²).

Factorise the following:

ab² + (a - 1)b - 1

Answer

ab² + (a - 1)b - 1

= ab² + ab - b - 1

= ab(b + 1) - 1(b + 1)

= (b + 1)(ab - 1)

Hence, ab² + (a - 1)b - 1 = (b + 1)(ab - 1).

Factorise the following:

2a - 4b - xa + 2bx

Answer

2a - 4b - xa + 2bx

= 2(a - 2b) - x(a - 2b)

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

Hence, 2a - 4b - xa + 2bx = (2 - x)(a - 2b).

Factorise the following:

5ph - 10qk + 2rph - 4qrk

Answer

5ph - 10qk + 2rph - 4qrk

= 5(ph - 2qk) + 2r(ph - 2qk)

= (5 + 2r)(ph - 2qk).

Hence, 5ph - 10qk + 2rph - 4qrk = (5 + 2r)(ph - 2qk).

Factorise the following:

x² - x(a + 2b) + 2ab

Answer

x² - x(a + 2b) + 2ab

= x² - xa - 2bx + 2ab

= x(x - a) - 2b(x - a)

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

Hence, x² - x(a + 2b) + 2ab = (x - a)(x - 2b).

Factorise the following:

ab(x² + y²) - xy(a² + b²)

Answer

ab(x² + y²) - xy(a² + b²)

= abx² + aby² - xya² - xyb²

= abx² - xyb² - xya² + aby²

= bx(ax - by) - ay(ax - by)

= (bx - ay)(ax - by).

Hence, ab(x² + y²) - xy(a² + b²) = (bx - ay)(ax - by).

Factorise the following:

(ax + by)² + (bx - ay)².

Answer

(ax + by)² + (bx - ay)²

= a²x² + b²y² + 2axby + b²x² + a²y² - 2axby

= a²x² + b²x² + b²y² + a²y²

= x²(a² + b²) + y²(b² + a²)

= (x² + y²)(a² + b²).

Hence, (ax + by)² + (bx - ay)² = (x² + y²)(a² + b²).

Factorise the following:

a³ + ab(1 - 2a) - 2b².

Answer

a³ + ab(1 - 2a) - 2b²

= a³ + ab - 2a²b - 2b²

= a³ - 2a²b + ab - 2b²

= a²(a - 2b) + b(a - 2b)

= (a² + b)(a - 2b).

a³ + ab(1 - 2a) - 2b² = (a² + b)(a - 2b).

Factorise the following:

3x²y - 3xy + 12x - 12.

Answer

3x²y - 3xy + 12x - 12.

= 3(x²y - xy + 4x - 4)

= 3[xy(x - 1) + 4(x - 1)]

= 3(x - 1)(xy + 4)

Hence, 3x²y - 3xy + 12x - 12 = 3(x - 1)(xy + 4).

Factorise the following:

a²b + ab² - abc - b²c + axy + bxy.

Answer

a²b + ab² - abc - b²c + axy + bxy

= ab(a + b) - bc(a + b) + xy(a + b)

= (a + b)(ab - bc + xy).

Hence, a²b + ab² - abc - b²c + axy + bxy = (a + b)(ab - bc + xy).

Factorise the following:

ax² - bx² + ay² - by² + az² - bz².

Answer

ax² - bx² + ay² - by² + az² - bz²

= x²(a - b) + y²(a - b) + z²(a - b)

= (a - b)(x² + y² + z²).

Hence, ax² - bx² + ay² - by² + az² - bz² = (a - b)(x² + y² + z²).

Factorise the following:

x - 1 - (x - 1)² + ax - a.

Answer

x - 1 - (x - 1)² + ax - a

= (x - 1) - (x - 1)² + a(x - 1)

= (x - 1)[1 - (x - 1) + a]

= (x - 1)(1 - x + 1 + a)

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

Hence, x - 1 - (x - 1)² + ax - a = (x - 1)(a - x + 2).

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.

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).

Factorise the following:

8x³ + y³

Answer

8x³ + y³ = (2x)³ + (y)³

We know that,

a³ + b³ = (a + b)(a² - ab + b²)

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

= (2x + y)(4x² - 2xy + y²).

Hence, 8x³ + y³ = (2x + y)(4x² - 2xy + y²).

Factorise the following:

64x³ - 125y³

Answer

64x³ - 125y³ = (4x)³ - (5y)³.

We know that,

a³ - b³ = (a - b)(a² + ab + b²)

∴ (4x)³ - (5y)³ = (4x - 5y)[(4x)² + 4x(5y) + (5y)²]

= (4x - 5y)(16x² + 20xy + 25y²).

Hence, 64x³ - 125y³ = (4x - 5y)(16x² + 20xy + 25y²).

Factorise the following:

64x³ + 1

Answer

64x³ + 1 = (4x)³ + (1)³.

We know that,

a³ + b³ = (a + b)(a² - ab + b²)

∴ (4x)³ + (1)³ = (4x + 1)[(4x)² - 4x.1 + 1²]

= (4x + 1)(16x² - 4x + 1).

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

Factorise the following:

7a³ + 56b³

Answer

7a³ + 56b³ = 7(a³ + 8b³)

= 7[(a)³ + (2b)³]

We know that,

a³ + b³ = (a + b)(a² - ab + b²)

∴ 7[(a)³ + (2b)³] = 7(a + 2b)[a² - a.2b + (2b)²]

= 7(a + 2b)(a² - 2ab + 4b²).

Hence, 7a³ + 56b³ = 7(a + 2b)(a² - 2ab + 4b²).

Factorise the following:

$\dfrac{x^6}{343} + \dfrac{343}{x^6}$.

Answer

$\dfrac{x^6}{343} + \dfrac{343}{x^6} = \Big(\dfrac{x^2}{7}\Big)^3 + \Big(\dfrac{7}{x^2}\Big)^3$.

We know that,

a³ + b³ = (a + b)(a² - ab + b²)

$\therefore \Big(\dfrac{x^2}{7}\Big)^3 + \Big(\dfrac{7}{x^2}\Big)^3 = \Big(\dfrac{x^2}{7} + \dfrac{7}{x^2}\Big)\Big[\Big(\dfrac{x^2}{7}\Big)^2 - \dfrac{x^2}{7} \times \dfrac{7}{x^2} + \Big(\dfrac{7}{x^2}\Big)^2\Big] \\[1em] = \Big(\dfrac{x^2}{7} + \dfrac{7}{x^2}\Big)\Big(\dfrac{x^4}{49} - 1 + \dfrac{49}{x^4}\Big).$

Hence, $\dfrac{x^6}{343} + \dfrac{343}{x^6} = \Big(\dfrac{x^2}{7} + \dfrac{7}{x^2}\Big)\Big(\dfrac{x^4}{49} - 1 + \dfrac{49}{x^4}\Big).$

Factorise the following:

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

Answer

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

We know that,

a³ - b³ = (a - b)(a² + ab + b²)

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

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

Factorise the following:

x² + x⁵

Answer

x² + x⁵ = x²(1 + x³) = x²[(1)³ + (x)³].

We know that,

a³ + b³ = (a + b)(a² - ab + b²)

x²[(1)³ + (x)³] = x²(1 + x)(1² - 1.x + x²)

= x²(1 + x)(1 - x + x²).

Hence, x² + x⁵ = x²(1 + x)(1 - x + x²).

Factorise the following:

32x⁴ - 500x

Answer

32x⁴ - 500x = 4x(8x³ - 125) = 4x[(2x)³ - 5³]

We know that,

a³ - b³ = (a - b)(a² + ab + b²)

4x[(2x)³ - 5³] = 4x(2x - 5)[(2x)² + 2x.5 + (5)²]

= 4x(2x - 5)(4x² + 10x + 25).

Hence, 32x⁴ - 500x = 4x(2x - 5)(4x² + 10x + 25).

Factorise the following:

27x³y³ - 8

Answer

27x³y³ - 8 = (3xy)³ - (2)³.

We know that,

a³ - b³ = (a - b)(a² + ab + b²)

(3xy)³ - (2)³ = (3xy - 2)[(3xy)² + 3xy.2 + 2²]

= (3xy - 2)(9x²y² + 6xy + 4).

Hence, 27x³y³ - 8 = (3xy - 2)(9x²y² + 6xy + 4).

Factorise the following:

27(x + y)³ + 8(2x - y)³

Answer

27(x + y)³ + 8(2x - y)³ = [3(x + y)]³ + [2(2x - y)]³

We know that,

a³ + b³ = (a + b)(a² - ab + b²).

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

= [3x + 3y + 4x - 2y][9(x² + y² + 2xy) - 6(x + y)(2x - y) + 4(4x² + y² - 4xy)]

= (7x + y)[9x² + 9y² + 18xy - 6(2x² - xy + 2xy - y²) + 16x² + 4y² - 16xy]

= (7x + y)[9x² + 9y² + 18xy - 12x² + 6xy - 12xy + 6y² + 16x² + 4y² - 16xy]

= (7x + y)(13x² + 19y² - 4xy).

Hence, 27(x + y)³ + 8(2x - y)³ = (7x + y)(13x² + 19y² - 4xy).

Factorise the following:

a³ + b³ + a + b

Answer

We know that,

a³ + b³ = (a + b)(a² - ab + b²).

a³ + b³ + a + b = (a + b)(a² - ab + b²) + (a + b)

= (a + b)(a² - ab + b² + 1).

Hence, a³ + b³ + a + b = (a + b)(a² - ab + b² + 1).

Factorise the following:

a³ - b³ - a + b

Answer

We know that,

a³ - b³ = (a - b)(a² + ab + b²).

a³ - b³ - a + b = (a - b)(a² + ab + b²) - 1(a - b)

= (a - b)(a² + ab + b² - 1).

Hence, a³ - b³ - a + b = (a - b)(a² + ab + b² - 1).

Factorise the following:

x³ + x + 2

Answer

x³ + x + 2 = x³ + x + 1 + 1.

= x³ + 1 + x + 1

= x³ + 1³ + x + 1

We know that,

a³ + b³ = (a + b)(a² - ab + b²).

x³ + 1³ + x + 1 = (x + 1)(x² - x + 1) + (x + 1)

= (x + 1)(x² - x + 1 + 1)

= (x + 1)(x² - x + 2).

Hence, x³ + x + 2 = (x + 1)(x² - x + 2).

Factorise the following:

a³ - a - 120

Answer

a³ - a - 120 = a³ - a - 125 + 5.

Rearranging above terms we get,

a³ - 125 - a + 5

= [a³ - (5)³] - (a - 5)

We know that,

a³ - b³ = (a - b)(a² + ab + b²).

[a³ - (5)³] - (a - 5) = (a - 5)(a² + 5a + 5²) - (a - 5)

= (a - 5)(a² + 5a + 25 - 1)

= (a - 5)(a² + 5a + 24).

Hence, a³ - a - 120 = (a - 5)(a² + 5a + 24).

a³ - 2a - 115

Answer

Given,

⇒ a³ - 2a - 115

⇒ a³ - 2a - 125 + 10

Rearranging above terms we get,

⇒ a³ - 125 - 2a + 10

⇒ (a³ - 5³) - 2(a - 5)

We know that,

a³ - b³ = (a - b)(a² + ab + b²)

⇒ (a - 5)(a² + 5a + 5²) - 2(a - 5)

⇒ (a - 5)[(a² + 5a + 5²) - 2]

⇒ (a - 5)[a² + 5a + 25 - 2]

⇒ (a - 5)(a² + 5a + 23).

Hence, a³ - 2a - 115 = (a - 5)(a² + 5a + 23).

Factorise the following:

x³ + 6x² + 12x + 16

Answer

x³ + 6x² + 12x + 16 can be written as x³ + 6x² + 12x + 8 + 8.

Above terms can be written as,

[(x)³ + (3 × 2 × x²) + (3 × 2² × x) + 2³] + 8

= [(x)³ + 3 × x × 2(x + 2) + 2³] + 2³ ........(i)

We know that,

(a + b)³ = a³ + 3ab(a + b) + b³ .........(ii)

Comparing equation (i) and (ii) we get,

a = x and b = 2.

∴ [(x)³ + 3 × x × 2(x + 2) + 2³] + 2³ = (x + 2)³ + 2³.

We know that,

a³ + b³ = (a + b)(a² - ab + b²).

∴ (x + 2)³ + 2³ = (x + 2 + 2)[(x + 2)² - (x + 2).2 + 2²]

= (x + 4)(x² + 4 + 4x - 2x - 4 + 4)

= (x + 4)(x² + 2x + 4)

Hence, x³ + 6x² + 12x + 16 = (x + 4)(x² + 2x + 4).

Factorise the following:

a³ - 3a²b + 3ab² - 2b³

Answer

a³ - 3a²b + 3ab² - 2b³ = a³ - 3a²b + 3ab² - b³ - b³.

We know that,

(a - b)³ = a³ - b³ - 3a²b + 3ab².

∴ a³ - 3a²b + 3ab² - b³ - b³ = (a - b)³ - b³.

We know that,

a³ - b³ = (a - b)(a² + ab + b²).

∴ (a - b)³ - b³ = (a - b - b)[(a - b)² + (a - b).b + b²]

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

= (a - 2b)(a² + b² - ab).

Hence, a³ - 3a²b + 3ab² - 2b³ = (a - 2b)(a² + b² - ab).

Factorise the following:

2a³ + 16b³ - 5a - 10b

Answer

2a³ + 16b³ - 5a - 10b = 2(a³ + 8b³) - 5(a + 2b)

= 2[a³ + (2b)³] - 5(a + 2b)

We know that,

a³ + b³ = (a + b)(a² - ab + b²).

∴ 2[a³ + (2b)³] - 5(a + 2b) = 2(a + 2b)(a² - 2ab + 4b²) - 5(a + 2b)

= (a + 2b)[2(a² - 2ab + 4b²) - 5]

= (a + 2b)(2a² - 4ab + 8b² - 5).

Hence, 2a³ + 16b³ - 5a - 10b = (a + 2b)(2a² - 4ab + 8b² - 5).

Factorise the following:

$a^3 - \dfrac{1}{a^3} - 2a + \dfrac{2}{a}$.

Answer

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

We know that,

a³ - b³ = (a - b)(a² + ab + b²).

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

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

Factorise the following:

a⁶ - b⁶

Answer

a⁶ - b⁶ = (a²)³ - (b²)³.

We know that,

a³ - b³ = (a - b)(a² + ab + b²).

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

= (a² - b²)(a⁴ + a²b² + b⁴)

= (a - b)(a + b)(a⁴ + a²b² + b⁴)

= (a - b)(a + b)[(a²)² + 2a²b² + (b²)² - a²b²]

=(a - b)(a + b)[(a² + b²)² - a²b²]

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

Hence, a⁶ - b⁶ = (a - b)(a + b)(a² + b² + ab)(a² + b² - ab).

Factorise the following:

x⁶ - 1

Answer

x⁶ - 1 = (x³)² - (1)².

We know that,

a² - b² = (a + b)(a - b).

∴ (x³)² - (1)² = (x³ - 1)(x³ + 1).

We know that,

a³ - b³ = (a - b)(a² + ab + b²).

a³ + b³ = (a + b)(a² - ab + b²).

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

Hence, x⁶ - 1 = (x - 1)(x + 1)(x² + x + 1)(x² - x + 1).

Factorise the following:

64x⁶ - 729y⁶

Answer

64x⁶ - 729y⁶ = [(2x)³]² - [(3y)³]².

We know that,

a² - b² = (a + b)(a - b).

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

We know that,

a³ - b³ = (a - b)(a² + ab + b²).

a³ + b³ = (a + b)(a² - ab + b²).

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

= (2x - 3y)(4x² + 6xy + 9y²)(2x + 3y)(4x² - 6xy + 9y²)

Hence, 64x⁶ - 729y⁶ = (2x - 3y)(2x + 3y)(4x² + 6xy + 9y²)(4x² - 6xy + 9y²).

Factorise the following:

$x^2 - \dfrac{8}{x}$

Answer

Above terms can be written as,

$\Rightarrow x^2 - \dfrac{8}{x} = \dfrac{1}{x}\Big(x^3 - 8\Big) \\[1em] = \dfrac{1}{x}(x^3 - 2^3).$

We know that,

a³ - b³ = (a - b)(a² + ab + b²).

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

We know that,

a³ - b³ = (a - b)(a² + ab + b²).

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

Hence, $x^2 - \dfrac{8}{x} = \dfrac{1}{x}(x - 2)(x^2 + 2x + 4).$

Factorise the following:

250(a - b)³ + 2

Answer

250(a - b)³ + 2 = 2[125(a - b)³ + 1]

= 2[{5(a - b)}³ + 1³]

We know that,

a³ + b³ = (a + b)(a² - ab + b²).

∴ 2[{5(a - b)}³ + 1³] = 2[(5a - 5b + 1){(5a - 5b)² - (5a - 5b)1 + 1²}]

= 2(5a - 5b + 1)(25a² + 25b² - 50ab - 5a + 5b + 1).

Hence, 250(a - b)³ + 2 = 2(5a - 5b + 1)(25a² + 25b² - 50ab - 5a + 5b + 1).

Factorise the following:

32a²x³ - 8b²x³ - 4a²y³ + b²y³.

Answer

32a²x³ - 8b²x³ - 4a²y³ + b²y³ = 8x³(4a² - b²) - y³(4a² - b²)

= (4a² - b²)(8x³ - y³)

= [(2a)² - (b)²][(2x)³ - (y)³]

We know that,

a³ - b³ = (a - b)(a² + ab + b²)

a² - b² = (a - b)(a + b).

∴ [(2a)² - (b)²][(2x)³ - (y)³] = (2a - b)(2a + b)(2x - y)[(2x)² + 2xy + y²]

= (2a - b)(2a + b)(2x - y)(4x² + 2xy + y²).

Hence, 32a²x³ - 8b²x³ - 4a²y³ + b²y³ = (2a - b)(2a + b)(2x - y)(4x² + 2xy + y²).

Factorise the following:

x⁹ + y⁹

Answer

x⁹ + y⁹ = (x³)³ + (y³)³

We know that,

a³ + b³ = (a + b)(a² - ab + b²)

(x³)³ + (y³)³ = (x³ + y³)[(x³)² - x³y³ + (y³)²]

= (x³ + y³)(x⁶ - x³y³ + y⁶)

= (x + y)(x² - xy + y²)(x⁶ - x³y³ + y⁶)

Hence, x⁹ + y⁹ = (x + y)(x² - xy + y²)(x⁶ - x³y³ + y⁶).

Factorise the following:

x⁶ - 7x³ - 8

Answer

Factorising x⁶ - 7x³ - 8 we get,

x⁶ - 7x³ - 8 = x⁶ - 8x³ + x³ - 8

= x³(x³ - 8) + 1(x³ - 8)

= (x³ - 8)(x³ + 1)

= [(x)³ - (2)³][(x)³ + 1³]

We know that,

a³ + b³ = (a + b)(a² - ab + b²)

a³ - b³ = (a - b)(a² + ab + b²)

∴ x³ + 1³ = (x + 1)(x² - x + 1)

and,

(x)³ - (2)³ = (x - 2)(x² + 2x + 4)

∴ [(x)³ - (2)³][(x)³ + 1³] = (x - 2)(x² + 2x + 4)(x + 1)(x² - x + 1).

Hence, x⁶ - 7x³ - 8 = (x - 2)(x² + 2x + 4)(x + 1)(x² - x + 1).

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.

Assertion (A): For the trinomial 2x² + 8x - 9 cannot be factorised.

Reason (R): For trinomial ax² + bx + c to be factorised, b² - 4ac must be a perfect square.

1. Assertion (A) is true, Reason (R) is false.

2. Assertion (A) is false, Reason (R) is true.

3. Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).

4. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason (or explanation) for Assertion (A).

Answer

For trinomial ax² + bx + c to be factorised, the roots are given by the quadratic equation: x = $\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

For the trinomial ax² + bx + c to be factorizable into linear factors with rational coefficients, its roots must be rational numbers. This occurs if and only if the expression the discriminant b² - 4ac, is a perfect square.

∴ Reason (R) is true.

For the trinomial 2x² + 8x - 9, a = 2, b = 8 and c = -9.

D = b² - 4ac

= 8² - 4 × 2 × -9

= 64 + 72

= 136.

Since, 136 is not a perfect square. So, 2x² + 8x - 9 cannot be factorised.

∴ Assertion (A) is true.

∴ Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).

Hence, option 3 is the correct option.

Assertion (A): Factorisation of 4x² + 9y² is (2x - 3y)(2x + 3y).

Reason (R): a² - b² = (a - b)(a + b)

1. Assertion (A) is true, Reason (R) is false.

2. Assertion (A) is false, Reason (R) is true.

3. Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).

4. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason (or explanation) for Assertion (A).

Answer

According to reason: a² - b² = (a - b)(a + b)

Expanding, (a - b)(a + b)

⇒ a(a + b) - b(a + b)

⇒ a² + ab - ab - b²

⇒ a² - b².

Thus, (a - b)(a + b) = a² - b²

∴ Reason (R) is true.

Given,

⇒ (2x - 3y)(2x + 3y)

⇒ 2x(2x + 3y) - 3y(2x + 3y)

⇒ (2x)² + 6xy - 6xy - (3y)²

⇒ 4x² - 9y².

∴ Assertion (A) is false.

∴ Assertion (A) is false, Reason (R) is true.

Hence, option 2 is the correct option.

Assertion (A): Factorisation of x³ - 27 is (x - 3)(x² + 3x + 9).

Reason (R): a³ - b³ = (a - b)(a² + ab + b²).

1. Assertion (A) is true, Reason (R) is false.

2. Assertion (A) is false, Reason (R) is true.

3. Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).

4. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason (or explanation) for Assertion (A).

Answer

We know that,

a³ - b³ = (a - b)(a² + ab + b²)

∴ Reason (R) is true.

Given,

⇒ x³ - 27

⇒ x³ - 3³

⇒ (x - 3)(x² + 3x + 3²)

⇒ (x - 3)(x² + 3x + 9).

∴ Assertion (A) is true.

∴ Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).

Hence, option 3 is the correct option.

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)

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