Factorisation

ICSE Class 9 Mathematics · ML Aggarwal Solutions Understanding ICSE Mathematics Class 9 · 192 solved questions

Exercise

Marks

1(i)Exercise 4.1

Factorise the following:

8xy³ + 12x²y²

Show SolutionHide Solution↓

H.C.F. of 8xy³ and 12x²y² is 4xy².

∴ 8xy³ + 12x²y² = 4xy²(2y + 3x).

1(ii)Exercise 4.1

Factorise the following:

15ax³ - 9ax².

Show SolutionHide Solution↓

H.C.F. of 15ax³ and 9ax² is 3ax².

∴ 15ax³ - 9ax² = 3ax²(5x - 3).

2(i)Exercise 4.1

Factorise the following:

21py² - 56py

Show SolutionHide Solution↓

H.C.F. of 21py² and 56py is 7py.

∴ 21py² - 56py = 7py(3y - 8).

2(ii)Exercise 4.1

Factorise the following:

4x³ - 6x².

Show SolutionHide Solution↓

H.C.F. of 4x³ and 6x² is 2x².

∴ 4x³ - 6x² = 2x²(2x - 3).

3(i)Exercise 4.1

Factorise the following:

2πr² - 4πr

Show SolutionHide Solution↓

H.C.F. of 2πr² and 4πr is 2πr.

∴ 2πr² - 4πr = 2πr(r - 2).

3(ii)Exercise 4.1

Factorise the following:

18m + 16n

Show SolutionHide Solution↓

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

∴ 18m + 16n = 2(9m + 8n).

4(i)Exercise 4.1

Factorise the following:

25abc² - 15a²b²c

Show SolutionHide Solution↓

H.C.F. of 25abc² and 15a²b²c is 5abc.

∴ 25abc² - 15a²b²c = 5abc(5c - 3ab).

4(ii)Exercise 4.1

Factorise the following:

28p²q²r - 42pq²r²

Show SolutionHide Solution↓

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

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

5(i)Exercise 4.1

Factorise the following:

8x³ - 6x² + 10x

Show SolutionHide Solution↓

H.C.F. of 8x³, 6x² and 10x is 2x.

∴ 8x³ - 6x² + 10x = 2x(4x² - 3x + 5).

5(ii)Exercise 4.1

Factorise the following:

14mn + 22m - 62p

Show SolutionHide Solution↓

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

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

6(i)Exercise 4.1

Factorise the following:

18p²q² - 24pq² + 30p²q

Show SolutionHide Solution↓

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

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

6(ii)Exercise 4.1

Factorise the following:

27a³b³ - 18a²b³ + 75a³b²

Show SolutionHide Solution↓

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

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

7(i)Exercise 4.1

Factorise the following:

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

Show SolutionHide Solution↓

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).

7(ii)Exercise 4.1

Factorise the following:

3a(x² + y²) + 6b(x² + y²).

Show SolutionHide Solution↓

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).

8(i)Exercise 4.1

Factorise the following:

6(x + 2y)³ + 8(x + 2y)²

Show SolutionHide Solution↓

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).

8(ii)Exercise 4.1

Factorise the following:

14(a - 3b)³ - 21p(a - 3b)

Show SolutionHide Solution↓

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].

9(i)Exercise 4.1

Factorise the following:

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

Show SolutionHide Solution↓

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].

9(ii)Exercise 4.1

Factorise the following:

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

Show SolutionHide Solution↓

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).

1(i)Exercise 4.2

Factorise the following:

x² + xy - x - y

Show SolutionHide Solution↓

x² + xy - x - y

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

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

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

1(ii)Exercise 4.2

Factorise the following:

y² - yz - 5y + 5z

Show SolutionHide Solution↓

y² - yz - 5y + 5z

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

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

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

2(i)Exercise 4.2

Factorise the following:

5xy + 7y - 5y² - 7x

Show SolutionHide Solution↓

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).

2(ii)Exercise 4.2

Factorise the following:

5p² - 8pq - 10p + 16q

Show SolutionHide Solution↓

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)

3(i)Exercise 4.2

Factorise the following:

a²b - ab² + 3a - 3b

Show SolutionHide Solution↓

a²b - ab² + 3a - 3b

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

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

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

3(ii)Exercise 4.2

Factorise the following:

x³ - 3x² + x - 3

Show SolutionHide Solution↓

x³ - 3x² + x - 3

= x²(x - 3) + 1(x - 3)

= (x² + 1)(x - 3).

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

4(i)Exercise 4.2

Factorise the following:

6xy² - 3xy - 10y + 5

Show SolutionHide Solution↓

6xy² - 3xy - 10y + 5

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

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

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

4(ii)Exercise 4.2

Factorise the following:

3ax - 6ay - 8by + 4bx

Show SolutionHide Solution↓

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).

5(i)Exercise 4.2

5px - 8qy + 4qx - 10py

Show SolutionHide Solution↓

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).

5(ii)Exercise 4.2

9a²y - 9ay + 6a - 6

Show SolutionHide Solution↓

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).

6(i)Exercise 4.2

Factorise the following:

1 - a - b + ab

Show SolutionHide Solution↓

1 - a - b + ab

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

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

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

6(ii)Exercise 4.2

Factorise the following:

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

Show SolutionHide Solution↓

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).

7(i)Exercise 4.2

Factorise the following:

x² + xy(1 + y) + y³

Show SolutionHide Solution↓

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).

7(ii)Exercise 4.2

Factorise the following:

y² - xy(1 - x) - x³

Show SolutionHide Solution↓

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).

8(i)Exercise 4.2

Factorise the following:

ab² + (a - 1)b - 1

Show SolutionHide Solution↓

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).

8(ii)Exercise 4.2

Factorise the following:

2a - 4b - xa + 2bx

Show SolutionHide Solution↓

2a - 4b - xa + 2bx

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

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

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

9(i)Exercise 4.2

Factorise the following:

5ph - 10qk + 2rph - 4qrk

Show SolutionHide Solution↓

5ph - 10qk + 2rph - 4qrk

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

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

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

9(ii)Exercise 4.2

Factorise the following:

x² - x(a + 2b) + 2ab

Show SolutionHide Solution↓

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).

10(i)Exercise 4.2

Factorise the following:

ab(x² + y²) - xy(a² + b²)

Show SolutionHide Solution↓

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).

10(ii)Exercise 4.2

Factorise the following:

(ax + by)² + (bx - ay)².

Show SolutionHide Solution↓

(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).

11(i)Exercise 4.2

Factorise the following:

a³ + ab(1 - 2a) - 2b².

Show SolutionHide Solution↓

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).

11(ii)Exercise 4.2

Factorise the following:

3x²y - 3xy + 12x - 12.

Show SolutionHide Solution↓

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).

12Exercise 4.2

Factorise the following:

a²b + ab² - abc - b²c + axy + bxy.

Show SolutionHide Solution↓

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).

13Exercise 4.2

Factorise the following:

ax² - bx² + ay² - by² + az² - bz².

Show SolutionHide Solution↓

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).

14Exercise 4.2

Factorise the following:

x - 1 - (x - 1)² + ax - a.

Show SolutionHide Solution↓

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).

1(i)Exercise 4.3

Factorise the following:

4x² - 25y²

Show SolutionHide Solution↓

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).

1(ii)Exercise 4.3

Factorise the following:

9x² - 1

Show SolutionHide Solution↓

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).

2(i)Exercise 4.3

Factorise the following:

150 - 6a²

Show SolutionHide Solution↓

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).

2(ii)Exercise 4.3

Factorise the following:

32x² - 18y²

Show SolutionHide Solution↓

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).

3(i)Exercise 4.3

Factorise the following:

(x - y)² - 9

Show SolutionHide Solution↓

(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).

3(ii)Exercise 4.3

Factorise the following:

9(x + y)² - x²

Show SolutionHide Solution↓

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).

4(i)Exercise 4.3

Factorise the following:

20x² - 45y²

Show SolutionHide Solution↓

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).

4(ii)Exercise 4.3

Factorise the following:

9x² - 4(y + 2x)²

Show SolutionHide Solution↓

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).

5(i)Exercise 4.3

Factorise the following:

2(x - 2y)² - 50y²

Show SolutionHide Solution↓

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).

5(ii)Exercise 4.3

Factorise the following:

32 - 2(x - 4)²

Show SolutionHide Solution↓

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).

6(i)Exercise 4.3

Factorise the following:

108a² - 3(b - c)²

Show SolutionHide Solution↓

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).

6(ii)Exercise 4.3

Factorise the following:

πa⁵ - π³ab²

Show SolutionHide Solution↓

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).

7(i)Exercise 4.3

Factorise the following:

50x² - 2(x - 2)²

Show SolutionHide Solution↓

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).

7(ii)Exercise 4.3

Factorise the following:

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

Show SolutionHide Solution↓

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).

8(i)Exercise 4.3

Factorise the following:

x - 2y - x² + 4y²

Show SolutionHide Solution↓

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).

8(ii)Exercise 4.3

Factorise the following:

4a² - b² + 2a + b

Show SolutionHide Solution↓

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).

9(i)Exercise 4.3

Factorise the following:

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

Show SolutionHide Solution↓

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).

9(ii)Exercise 4.3

Factorise the following:

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

Show SolutionHide Solution↓

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).

10(i)Exercise 4.3

Factorise the following:

9 - x² + 2xy - y².

Show SolutionHide Solution↓

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).

10(ii)Exercise 4.3

Factorise the following:

9x⁴ - (x² + 2x + 1)

Show SolutionHide Solution↓

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).

11(i)Exercise 4.3

Factorise the following:

9x⁴ - x² - 12x - 36

Show SolutionHide Solution↓

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).

11(ii)Exercise 4.3

Factorise the following:

x³ - 5x² - x + 5

Show SolutionHide Solution↓

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).

12(i)Exercise 4.3

Factorise the following:

a⁴ - b⁴ + 2b² - 1

Show SolutionHide Solution↓

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).

12(ii)Exercise 4.3

Factorise the following:

x³ - 25x

Show SolutionHide Solution↓

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).

13(i)Exercise 4.3

Factorise the following:

2x⁴ - 32

Show SolutionHide Solution↓

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).

13(ii)Exercise 4.3

Factorise the following:

a²(b + c) - (b + c)³

Show SolutionHide Solution↓

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).

14(i)Exercise 4.3

Factorise the following:

(a + b)³ - a - b

Show SolutionHide Solution↓

(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).

14(ii)Exercise 4.3

Factorise the following:

x² - 2xy + y² - a² - 2ab - b².

Show SolutionHide Solution↓

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).

15(i)Exercise 4.3

Factorise the following:

(a² - b²)(c² - d²) - 4abcd

Show SolutionHide Solution↓

(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).

15(ii)Exercise 4.3

Factorise the following:

4x² - y² - 3xy + 2x - 2y

Show SolutionHide Solution↓

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).

16(i)Exercise 4.3

Factorise the following:

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

Show SolutionHide Solution↓

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

We know that,

(a - b)² = a² - 2ab + b²

and

a² - b² = (a - b)(a + b)

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

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

16(ii)Exercise 4.3

Factorise the following:

x⁴ + 5x² + 9

Show SolutionHide Solution↓

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)

17Exercise 4.3

(i) x² + $\frac{4}{x^2}$ - 5

(ii) x⁴ + 12x² + 11

Show SolutionHide Solution↓

(i) Given,

x² + $\frac{4}{x^2}$ - 5

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

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

(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).

18(i)Exercise 4.3

Factorise the following:

a⁴ + b⁴ - 7a²b².

Show SolutionHide Solution↓

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).

18(ii)Exercise 4.3

Factorise the following:

x⁴ - 14x² + 1

Show SolutionHide Solution↓

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).

19(i)Exercise 4.3

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

(x² - 5x + 7)(x² + 5x + 7)

Show SolutionHide Solution↓

(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.

19(ii)Exercise 4.3

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

(x² - 5x + 7)(x² - 5x - 7)

Show SolutionHide Solution↓

(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.

19(iii)Exercise 4.3

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

(x² + 5x - 7)(x² - 5x + 7)

Show SolutionHide Solution↓

(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.

20(i)Exercise 4.3

Evaluate the following by using factors:

(979)² - (21)²

Show SolutionHide Solution↓

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.

20(ii)Exercise 4.3

Evaluate the following by using factors:

(99.9)² - (0.1)²

Show SolutionHide Solution↓

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.

1(i)Exercise 4.4

Factorise the following:

x² + 5x + 6

Show SolutionHide Solution↓

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).

1(ii)Exercise 4.4

Factorise the following:

x² - 8x + 7

Show SolutionHide Solution↓

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).

2(i)Exercise 4.4

Factorise the following:

x² + 6x - 7

Show SolutionHide Solution↓

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).

2(ii)Exercise 4.4

Factorise the following:

y² + 7y - 18

Show SolutionHide Solution↓

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).

3(i)Exercise 4.4

Factorise the following:

y² - 7y - 18

Show SolutionHide Solution↓

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).

3(ii)Exercise 4.4

Factorise the following:

a² - 3a - 54

Show SolutionHide Solution↓

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).

4(i)Exercise 4.4

Factorise the following:

2x² - 7x + 6

Show SolutionHide Solution↓

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).

4(ii)Exercise 4.4

Factorise the following:

6x² + 13x - 5

Show SolutionHide Solution↓

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).

5(i)Exercise 4.4

Factorise the following:

6x² + 11x - 10

Show SolutionHide Solution↓

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).

5(ii)Exercise 4.4

Factorise the following:

6x² - 7x - 3

Show SolutionHide Solution↓

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).

6(i)Exercise 4.4

Factorise the following:

2x² - x - 6

Show SolutionHide Solution↓

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).

6(ii)Exercise 4.4

Factorise the following:

1 - 18y - 63y²

Show SolutionHide Solution↓

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).

7(i)Exercise 4.4

Factorise the following:

2y² + y - 45

Show SolutionHide Solution↓

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).

7(ii)Exercise 4.4

Factorise the following:

5 - 4x - 12x²

Show SolutionHide Solution↓

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).

8(i)Exercise 4.4

Factorise the following:

x(12x + 7) - 10

Show SolutionHide Solution↓

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).

8(ii)Exercise 4.4

Factorise the following:

(4 - x)² - 2x

Show SolutionHide Solution↓

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).

9(i)Exercise 4.4

Factorise the following:

60x² - 70x - 30

Show SolutionHide Solution↓

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).

9(ii)Exercise 4.4

Factorise the following:

x² - 6xy - 7y²

Show SolutionHide Solution↓

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).

10(i)Exercise 4.4

Factorise the following:

2x² + 13xy - 24y²

Show SolutionHide Solution↓

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).

10(ii)Exercise 4.4

Factorise the following:

6x² - 5xy - 6y²

Show SolutionHide Solution↓

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).

11(i)Exercise 4.4

Factorise the following:

5x² + 17xy - 12y²

Show SolutionHide Solution↓

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).

11(ii)Exercise 4.4

Factorise the following:

x²y² - 8xy - 48

Show SolutionHide Solution↓

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).

12(i)Exercise 4.4

Factorise the following:

2a²b² - 7ab - 30

Show SolutionHide Solution↓

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).

12(ii)Exercise 4.4

Factorise the following:

a(2a - b) - b²

Show SolutionHide Solution↓

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).

13(i)Exercise 4.4

Factorise the following:

(x - y)² - 6(x - y) + 5

Show SolutionHide Solution↓

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).

13(ii)Exercise 4.4

Factorise the following:

(2x - y)² - 11(2x - y) + 28

Show SolutionHide Solution↓

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).

14(i)Exercise 4.4

Factorise the following:

4(a - 1)² - 4(a - 1) - 3

Show SolutionHide Solution↓

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).

14(ii)Exercise 4.4

Factorise the following:

1 - 2a - 2b - 3(a + b)².

Show SolutionHide Solution↓

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).

15(i)Exercise 4.4

Factorise the following:

3 - 5a - 5b - 12(a + b)²

Show SolutionHide Solution↓

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).

15(ii)Exercise 4.4

Factorise the following:

a⁴ - 11a² + 10

Show SolutionHide Solution↓

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).

16(i)Exercise 4.4

Factorise the following:

(x + 4)² - 5xy - 20y - 6y²

Show SolutionHide Solution↓

(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).

16(ii)Exercise 4.4

Factorise the following:

(x² - 2x)² - 23(x² - 2x) + 120

Show SolutionHide Solution↓

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).

17Exercise 4.4

Factorise the following:

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

Show SolutionHide Solution↓

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).

18Exercise 4.4

Factorise the following:

(2x² + 5x)(2x² + 5x - 19) + 84

Show SolutionHide Solution↓

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).

1(i)Exercise 4.5

Factorise the following:

8x³ + y³

Show SolutionHide Solution↓

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).

1(ii)Exercise 4.5

Factorise the following:

64x³ - 125y³

Show SolutionHide Solution↓

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).

2(i)Exercise 4.5

Factorise the following:

64x³ + 1

Show SolutionHide Solution↓

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).

2(ii)Exercise 4.5

Factorise the following:

7a³ + 56b³

Show SolutionHide Solution↓

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).

3(i)Exercise 4.5

Factorise the following:

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

Show SolutionHide Solution↓

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

We know that,

a³ + b³ = (a + b)(a² - ab + b²)

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

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

3(ii)Exercise 4.5

Factorise the following:

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

Show SolutionHide Solution↓

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

We know that,

a³ - b³ = (a - b)(a² + ab + b²)

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

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

4(i)Exercise 4.5

Factorise the following:

x² + x⁵

Show SolutionHide Solution↓

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).

4(ii)Exercise 4.5

Factorise the following:

32x⁴ - 500x

Show SolutionHide Solution↓

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).

5(i)Exercise 4.5

Factorise the following:

27x³y³ - 8

Show SolutionHide Solution↓

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).

5(ii)Exercise 4.5

Factorise the following:

27(x + y)³ + 8(2x - y)³

Show SolutionHide Solution↓

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).

6(i)Exercise 4.5

Factorise the following:

a³ + b³ + a + b

Show SolutionHide Solution↓

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).

6(ii)Exercise 4.5

Factorise the following:

a³ - b³ - a + b

Show SolutionHide Solution↓

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).

7(i)Exercise 4.5

Factorise the following:

x³ + x + 2

Show SolutionHide Solution↓

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).

7(ii)Exercise 4.5

Factorise the following:

a³ - a - 120

Show SolutionHide Solution↓

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).

7(iii)Exercise 4.5

a³ - 2a - 115

Show SolutionHide Solution↓

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).

8(i)Exercise 4.5

Factorise the following:

x³ + 6x² + 12x + 16

Show SolutionHide Solution↓

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).

8(ii)Exercise 4.5

Factorise the following:

a³ - 3a²b + 3ab² - 2b³

Show SolutionHide Solution↓

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).

9(i)Exercise 4.5

Factorise the following:

2a³ + 16b³ - 5a - 10b

Show SolutionHide Solution↓

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).

9(ii)Exercise 4.5

Factorise the following:

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

Show SolutionHide Solution↓

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

We know that,

a³ - b³ = (a - b)(a² + ab + b²).

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

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

10(i)Exercise 4.5

Factorise the following:

a⁶ - b⁶

Show SolutionHide Solution↓

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).

10(ii)Exercise 4.5

Factorise the following:

x⁶ - 1

Show SolutionHide Solution↓

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).

11(i)Exercise 4.5

Factorise the following:

64x⁶ - 729y⁶

Show SolutionHide Solution↓

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).

11(ii)Exercise 4.5

Factorise the following:

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

Show SolutionHide Solution↓

Above terms can be written as,

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

We know that,

a³ - b³ = (a - b)(a² + ab + b²).

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

We know that,

a³ - b³ = (a - b)(a² + ab + b²).

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

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

12(i)Exercise 4.5

Factorise the following:

250(a - b)³ + 2

Show SolutionHide Solution↓

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).

12(ii)Exercise 4.5

Factorise the following:

32a²x³ - 8b²x³ - 4a²y³ + b²y³.

Show SolutionHide Solution↓

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).

13(i)Exercise 4.5

Factorise the following:

x⁹ + y⁹

Show SolutionHide Solution↓

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).

13(ii)Exercise 4.5

Factorise the following:

x⁶ - 7x³ - 8

Show SolutionHide Solution↓

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).

11 Mark · Multiple Choice Questions

Factorisation of 12a²b + 15ab² is

(a)3a(4ab + 5b²)
(b)3b(4a² + 5ab)
(c)3ab(4a + 5b)
(d)none of these

Show SolutionHide Solution↓

H.C.F. of 12a²b and 15ab² is 3ab.

∴ 12a²b + 15ab² = 3ab(4a + 5b).

Hence, Option 3 is the correct option.

21 Mark · Multiple Choice Questions

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

(a)(3y - 2)(2x - 3)
(b)(3x - 2)(2y - 3)
(c)(2y - 3)(2 - 3x)
(d)none of these

Show SolutionHide Solution↓

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.

31 Mark · Multiple Choice Questions

Factorisation of 49p³q - 36pq is

(a)p(7p + 6q)(7p - 6q)
(b)q(7p - 6)(7p + 6)
(c)pq(7p + 6)(7p - 6)
(d)none of these

Show SolutionHide Solution↓

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.

41 Mark · Multiple Choice Questions

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

(a)(y - z)(y + 9)
(b)(y - z)(y - 9)
(c)(z - y)(y + 9)
(d)none of these

Show SolutionHide Solution↓

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.

51 Mark · Multiple Choice Questions

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

(a)(lm + 1)(m + l)
(b)(lm + m)(l + 1)
(c)l(m + 1)
(d)(l + 1)(m + 1)

Show SolutionHide Solution↓

(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.

61 Mark · Multiple Choice Questions

Factorisation of 63x² - 112y² is

(a)63(x - 2y)(x + 2y)
(b)7(3x + 2y)(3x - 2y)
(c)7(3x + 4y)(3x - 4y)
(d)none of these

Show SolutionHide Solution↓

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.

71 Mark · Multiple Choice Questions

Factorisation of p⁴ - 81 is

(a)(p² - 9)(p² + 9)
(b)(p - 3)(p + 3)(p² + 9)
(c)(p - 3)²(p + 3)²
(d)none of these

Show SolutionHide Solution↓

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.

81 Mark · Multiple Choice Questions

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

(a)5 + x
(b)5 - x
(c)5x - 1
(d)10x

Show SolutionHide Solution↓

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

= 50x² + 10x

= 10x(5x + 1).

Hence, Option 4 is the correct option.

91 Mark · Multiple Choice Questions

Factorisation of x² - 4x - 12 is

(a)(x + 6)(x - 2)
(b)(x - 6)(x + 2)
(c)(x - 6)(x - 2)
(d)(x + 6)(x + 2)

Show SolutionHide Solution↓

x² - 4x - 12 = x² - 6x + 2x - 12

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

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

Hence, Option 2 is the correct option.

101 Mark · Multiple Choice Questions

Factorisation of 3x² + 7x - 6

(a)(3x - 2)(x + 3)
(b)(3x + 2)(x - 3)
(c)(3x - 2)(x - 3)
(d)(3x + 2)(x + 3)

Show SolutionHide Solution↓

3x² + 7x - 6 = 3x² + 9x - 2x - 6

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

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

Hence, Option 1 is the correct option.

111 Mark · Multiple Choice Questions

Factorisation of 4x² + 8x + 3 is

(a)(x + 1)(x + 3)
(b)(2x + 1)(2x + 3)
(c)(2x + 2)(2x + 5)
(d)(2x - 1)(2x - 3)

Show SolutionHide Solution↓

4x² + 8x + 3 = 4x² + 6x + 2x + 3

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

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

Hence, Option 2 is the correct option.

121 Mark · Multiple Choice Questions

Factorisation of 16x² + 40x + 25 is

(a)(4x + 5)(4x + 5)
(b)(4x + 5)(4x - 5)
(c)(4x - 5)(4x - 5)
(d)(4x + 5)(4x + 7)

Show SolutionHide Solution↓

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

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

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

Hence, Option 1 is the correct option.

131 Mark · Multiple Choice Questions

Factorisation of x² - 4xy + 4y² is

(a)(x + 2y)(x - 2y)
(b)(x + 2y)(x + 2y)
(c)(x - 2y)(x - 2y)
(d)(2x - y)(2x + y)

Show SolutionHide Solution↓

x² - 4xy + 4y² = x² - 2xy - 2xy + 4y²

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

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

Hence, Option 3 is the correct option.

141 Mark · Multiple Choice Questions

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

(a)x² + xy + 2xy
(b)x² + y² - xy
(c)xy²
(d)3xy

Show SolutionHide Solution↓

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.

151 Mark · Multiple Choice Questions

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

(a)1
(b)-1
(c)0
(d)$\frac{1}{2}$

Show SolutionHide Solution↓

Given,

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

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.

161 Mark · Multiple Choice Questions

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

(a)0
(b)abc
(c)2abc
(d)3abc

Show SolutionHide Solution↓

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.

171 Mark · Multiple Choice Questions

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

(a)x³ + y³ + z³ = 0
(b)x³ + y³ + z³ = 27xyz
(c)(x + y + z)³ = 27xyz
(d)x + y + z = 3xyz

Show SolutionHide Solution↓

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,

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

Cubing both sides we get :

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

Hence, option 3 is the correct option.

181 Mark · Multiple Choice Questions

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?

(a)Both the statements are true.
(b)Both the statements are false.
(c)Statement 1 is true, and Statement 2 is false.
(d)Statement 1 is false, and Statement 2 is true.

Show SolutionHide Solution↓

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.

1Assertion Reason Type Questions

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.

(a)Assertion (A) is true, Reason (R) is false.
(b)Assertion (A) is false, Reason (R) is true.
(c)Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).
(d)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason (or explanation) for Assertion (A).

Show SolutionHide Solution↓

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

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.

2Assertion Reason Type Questions

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

Reason (R): a² - b² = (a - b)(a + b)

(a)Assertion (A) is true, Reason (R) is false.
(b)Assertion (A) is false, Reason (R) is true.
(c)Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).
(d)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason (or explanation) for Assertion (A).

Show SolutionHide Solution↓

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.

3Assertion Reason Type Questions

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

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

(a)Assertion (A) is true, Reason (R) is false.
(b)Assertion (A) is false, Reason (R) is true.
(c)Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).
(d)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason (or explanation) for Assertion (A).

Show SolutionHide Solution↓

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.

1(i)Chapter Test

Factorise the following:

15(2x - 3)³ - 10(2x - 3)

Show SolutionHide Solution↓

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].

1(ii)Chapter Test

Factorise the following:

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

Show SolutionHide Solution↓

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].

2(i)Chapter Test

Factorise the following:

2a²x - bx + 2a² - b

Show SolutionHide Solution↓

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).

2(ii)Chapter Test

Factorise the following:

p² - (a + 2b)p + 2ab

Show SolutionHide Solution↓

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).

3(i)Chapter Test

Factorise the following:

(x² - y²)z + (y² - z²)x

Show SolutionHide Solution↓

(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).

3(ii)Chapter Test

Factorise the following:

5a⁴ - 5a³ + 30a² - 30a

Show SolutionHide Solution↓

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).

4(i)Chapter Test

Factorise the following:

b(c - d)² + a(d - c) + 3c - 3d

Show SolutionHide Solution↓

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).

4(ii)Chapter Test

Factorise the following:

x³ - x² - xy + x + y - 1

Show SolutionHide Solution↓

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).

5(i)Chapter Test

Factorise the following:

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

Show SolutionHide Solution↓

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).

5(ii)Chapter Test

Factorise the following:

a¹²x⁴ - a⁴x¹²

Show SolutionHide Solution↓

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).

6(i)Chapter Test

Factorise the following:

9x² + 12x + 4 - 16y²

Show SolutionHide Solution↓

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).

6(ii)Chapter Test

Factorise the following:

x⁴ + 3x² + 4

Show SolutionHide Solution↓

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).

7(i)Chapter Test

Factorise the following:

21x² - 59xy + 40y²

Show SolutionHide Solution↓

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).

7(ii)Chapter Test

Factorise the following:

4x³y - 44x²y + 112xy

Show SolutionHide Solution↓

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).

8(i)Chapter Test

Factorise the following:

x²y² - xy - 72

Show SolutionHide Solution↓

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).

8(ii)Chapter Test

Factorise the following:

9x³y + 41x²y² + 20xy³

Show SolutionHide Solution↓

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).

9(i)Chapter Test

Factorise the following:

(3a - 2b)² + 3(3a - 2b) - 10

Show SolutionHide Solution↓

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).

9(ii)Chapter Test

Factorise the following:

(x² - 3x)(x² - 3x + 7) + 10

Show SolutionHide Solution↓

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).

10(i)Chapter Test

Factorise the following:

(x² - x)(4x² - 4x - 5) - 6

Show SolutionHide Solution↓

(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).

10(ii)Chapter Test

Factorise the following:

x⁴ + 9x²y² + 81y⁴

Show SolutionHide Solution↓

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).

11(i)Chapter Test

Factorise the following:

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

Show SolutionHide Solution↓

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

We know that,

a³ - b³ = (a - b)(a² + ab + b²)

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

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

11(ii)Chapter Test

Factorise the following:

x⁶ + 63x³ - 64

Show SolutionHide Solution↓

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).

12(i)Chapter Test

Factorise the following:

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

Show SolutionHide Solution↓

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

We know that,

a² - b² = (a + b)(a - b)

a³ + b³ = (a + b)(a² - ab + b²)

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

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

12(ii)Chapter Test

Factorise the following:

(x + 1)⁶ - (x - 1)⁶

Show SolutionHide Solution↓

(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).

13Chapter Test

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

Show SolutionHide Solution↓

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).

14Chapter Test

Show that 97³ + 14³ is divisible by 111.

Show SolutionHide Solution↓

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.

15Chapter Test

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

Show SolutionHide Solution↓

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.

Press ↓ to reveal a solution and move to the next question. Back to ML Aggarwal Solutions Understanding ICSE Mathematics Class 9.