Chapter 3 Factorisation-of-polynomials

Chapter 3

Factorisation-of-polynomials

Exercise 3.1

Page No: 99

Question 1:

Factorize:

9x2 + 12xy

ANSWER:

We have:

9x2+12xy

=3x(3x+4y)

Question 2:

Factorize:

18x2y − 24xyz

ANSWER:

We have:

18x2y−24xyz

=6xy(3y−4z)

Question 3:

Factorize:

27a3b3 − 45a4b2

ANSWER:

We have:

27a3b3−45a4b2

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=9a3b2(3b−5a)

Question 4:

Factorize:

2a(x + y) − 3b(x + y)

ANSWER:

We have:

2a(x+y)−3b(x+y)=(x+y)(2a−3b)

Question 5:

Factorize:

2x(p2 + q2) + 4y(p2 + q2)

ANSWER:

We have:

2x(p2 + q2) + 4y(p2 + q2)

=2[x(p2 + q2)+4y(p2 + q2)]

=2(p2+q2)(x+2y)

Question 6:

Factorize:

x(a − 5) + y(5 − a)

ANSWER:

We have:

x(a−5)+y(5−a)

=x(a−5)−y(a−5)

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=(a−5)(x−y)

Question 7:

Factorize:

4(a + b) − 6(a + b)2

ANSWER:

We have:

4(a+b)−6(a+b)2

=2(a+b)[2−3(a+b)]

=2(a+b)(2−3a−3b)

Question 8:

Factorize:

8(3a − 2b)2 − 10(3a − 2b)

ANSWER:

We have:

8(3a−2b)2−10(3a−2b)

=2(3a−2b)[4(3a−2b)−5]

=2(3a−2b)(12a−8b−5)

Question 9:

Factorize:

x(x + y)3 − 3x2y(x + y)

ANSWER:

We have:

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x(x+y)3−3x2y(x+y)

=x(x+y) [(x+y)2−3xy]

=x(x+y)[ x2+y2+2xy−3xy]

=x(x+y)(x2+y2−xy)

Question 10:

Factorize:

x3 + 2x2 + 5x + 10

ANSWER:

We have:

x3 + 2x2 + 5x + 10

=(x3+2x2)+(5x+10)

=x2(x+2)+5(x+2)

=(x+2)(x2+5)

Question 11:

Factorize:

x2 + xy − 2xz − 2yz

ANSWER:

We have:

x2+xy−2xz−2yz = (x2+xy)−(2xz+2yz)

=x(x+y)−2z(x+y)

=(x+y)(x−2z)

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Question 12:

Factorize:

a3b − a2b + 5ab − 5b

ANSWER:

We have:

a3b − a2b + 5ab − 5b = b(a3−a2+5a−5)

=b[(a3−a2)+(5a−5)]

=b[a2(a−1)+5(a−1)]

=b(a−1)(a2+5)

Question 13:

Factorize:

8 − 4a − 2a3 + a4

ANSWER:

We have:

8 − 4a − 2a3 + a4

= (8−4a)−(2a3−a4)

= 4(2−a)− a3(2−a)

= (2−a) (4 − a3)

Question 14:

Factorize:

x3 − 2x2y + 3xy2 − 6y3

ANSWER:

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We have:

x3 − 2x2y + 3xy2 − 6y3

=( x3 − 2x2y )+( 3xy2 − 6y3)

= x2(x−2y)+3y2 (x−2y)

=(x−2y)(x2+3y2)

Question 15:

Factorize:

px − 5q + pq − 5x

ANSWER:

We have:

px−5q+pq−5x

=(px−5x)+(pq−5q)

= x(p−5)+q(p−5)

=(p−5)(x+q)

Question 16:

Factorize:

x2 + y − xy − x

ANSWER:

We have:

x2+y−xy−x

=(x2−xy)−(x−y)

=x(x−y)−1(x−y)

=(x−y)(x−1)

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Question 17:

Factorize:

(3a − 1)2 − 6a + 2

ANSWER:

We have:

(3a−1)2−6a+2=(3a−1)2−2(3a−1)

=(3a−1)[(3a−1)−2]

=(3a−1)(3a−1−2)

=(3a−1)(3a−3)

=3(3a−1)(a−1)

Question 18:

Factorize:

(2x − 3)2 − 8x + 12

ANSWER:

We have:

(2x−3)2−8x+12 =(2x−3)2−4(2x−3)

=(2x−3)[(2x−3)−4]

=(2x−3)(2x−3−4)

=(2x−3)(2x−7)

Question 19:

Factorize:

a3 + a − 3a2 − 3

ANSWER:

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We have:

a3 + a − 3a2 – 3 = (a3− 3a2 )+(a−3)

=a2(a−3)+1(a−3)

=(a−3)(a2+1)

Question 20:

Factorize:

3ax − 6ay − 8by + 4bx

ANSWER:

We have:

3ax−6ay−8by+4bx

=(3ax−6ay)+(4bx−8by)

=3a(x−2y)+4b(x−2y)

=(x−2y)(3a+4b)

Question 21:

Factorize:

abx2 + a2x + b2x + ab

ANSWER:

We have:

abx2 + a2x + b2x + ab

=(abx2+b2x)+( a2x +ab)

=bx(ax+b)+a(ax+b)

=(ax+b)(bx+a)

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Question 22:

Factorize:

x3 − x2 + ax + x − a − 1

ANSWER:

We have:

x3 − x2 + ax + x − a − 1

=( x3 − x2 )+(ax−a)+(x−1)

= x2 (x−1) + a(x−1)+1(x−1)

=(x−1)( x2 +a+1)

Question 23:

Factorize:

2x + 4y − 8xy − 1

ANSWER:

We have:

2x+4y−8xy−1

=(2x−8xy)−(1−4y)

=2x(1−4y)−1(1−4y)

=(1−4y)(2x−1)

Question 24:

Factorize:

ab(x2 + y2) − xy(a2 + b2)

ANSWER:

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We have:

ab(x2 + y2) − xy(a2 + b2)

= ( 𝑎𝑏𝑥2 − 𝑎2𝑥𝑦) − (𝑏2𝑥𝑦 − 𝑎𝑏𝑦2)

=ax(bx−ay)−by(bx−ay)

=(bx−ay)(ax−by)

Question 25:

Factorize:

a2 + ab(b + 1) + b3

ANSWER:

We have:

a2+ab(b+1)+b3

= 𝑎2 + 𝑎𝑏2 + 𝑎𝑏 + 𝑏3

= 𝑎(𝑎 + 𝑏2) + 𝑏(𝑎 + 𝑏2)

= (𝑎 + 𝑏2)(𝑎 + 𝑏)

Question 26:

Factorize:

a3 + ab(1 − 2a) − 2b2

ANSWER:

We have:

a3 + ab(1 − 2a) − 2b2

= (𝑎3 − 2𝑎2𝑏) + (𝑎𝑏 − 2𝑏2)

= 𝑎2(𝑎 − 2𝑏) + 𝑏(𝑎 − 2𝑏)

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= (𝑎 − 2𝑏)(𝑎2 + 𝑏)

Question 27:

Factorize:

2a2 + bc − 2ab − ac

ANSWER:

We have:

2a2 + bc − 2ab – ac

= (2a2 -2ab) – (ac-bc)

= 2a (a−b)− c(a−b)

=(a−b)(2a−c)

Question 28:

Factorize:

(ax + by)2 + (bx − ay)2

ANSWER:

We have:

(ax + by)2 + (bx − ay)2 = [(𝑎𝑥)2 + 2 × 𝑎𝑥 × 𝑏𝑦 + (𝑏𝑦)2] +

[(𝑏𝑥)2 − 2 × 𝑏𝑥 × 𝑎𝑦 + (𝑎𝑦)2]

= 𝑎2𝑥2 + 2𝑎𝑏𝑥𝑦 + 𝑏2𝑦2 + 𝑏2𝑥2 − 2𝑎𝑏𝑥𝑦 + 𝑎2𝑦2

= (𝑎2𝑥2 + 𝑏2𝑥2) + (𝑎2𝑦2 + 𝑏2𝑦2)

= 𝑥2(𝑎2 + 𝑏2) + 𝑦2(𝑎2 + 𝑏2)

= (𝑎2 + 𝑏2)(𝑥2 + 𝑦2)

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Question 29:

Factorize:

a(a + b − c) − bc

ANSWER:

We have:

a(a+b−c)−bc

= 𝑎2+ab-ac-bc

=(𝑎2−ac)+(ab−bc)

= a(a−c)+b(a−c)

= (a−c) (a+b)

Question 30:

Factorize:

a(a − 2b − c) + 2bc

ANSWER:

We have:

a(a−2b−c)+2bc

= 𝑎2−2ab−ac+2bc

=( 𝑎2−2ab)−(ac−2bc)

=a(a−2b)−c(a−2b)

=(a−2b)(a−c)

Question: 31

Factorize:

a2x2 + (ax2 + 1)x + a

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ANSWER:

We have:

a2x2 + (ax2 + 1)x + a

= (ax2 + 1)x + (𝑎2x2 + a)

= 𝑥(𝑎𝑥2 + 1) + 𝑎(𝑎𝑥2 + 1)

= (𝑎𝑥2 + 1)(𝑥 + 𝑎)

Question 32:

Factorize:

ab(x2 + 1) + x(a2 + b2)

ANSWER:

We have:

ab(x2 + 1) + x(a2 + b2)

= 𝑎𝑏𝑥2 + 𝑎𝑏 + 𝑎2𝑥 + 𝑏2𝑥

=ax(bx+a)+b(bx+a)

=(bx+a)(ax+b)

Question 33:

Factorize:

x2 − (a + b)x + ab

ANSWER:

We have:

x2 − (a + b)x + ab

= x2 – ax - 𝑏𝑥 + 𝑎𝑏

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= 𝑥(𝑥 − 𝑎) − 𝑏(𝑥 − 𝑎)

=(𝑥 − 𝑎)(𝑥 − 𝑏)

Question: 34

Factorize

𝑥2 + 1

𝑥2 – 2 – 3x+ 3

𝑥

Answer:

We have:

𝑥2 + 1

𝑥2 – 2 – 3x+ 3

𝑥

= 𝑥2 - 2 + 1

𝑥2 - 3x + 3

𝑥

= 𝑥2 - 2 × 𝑥 ×1

𝑥 + (

1

𝑥)

2 – 3 (𝑥 −

1

𝑥)

= (𝑥 −1

𝑥)

2− 3 (𝑥 −

1

𝑥)

= (𝑥 −1

𝑥) (𝑥 −

1

𝑥− 3)

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Exercise 3.2

Page No: 105

Question 1:

Factorise:

9x2 – 16y2

ANSWER:

9x2 – 16y2

=(3x)2−(4y)2

=(3x+4y)(3x−3y) [a2−b2=(a+b)(a−b)]

Question 2:

(25

4𝑥2 −

1

9𝑦2)

Answer:

(25

4𝑥2 −

1

9𝑦2)

= (5

2𝑥)

2- (

1

3𝑦)

2

= (5

2𝑥 +

1

3𝑦 ) (

5

2𝑥 −

1

3𝑦 )

[𝑎2 − 𝑏2 = (𝑎 + 𝑏)(𝑎 − 𝑏)]

Question 3:

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Factorise:

81 – 16x2

ANSWER:

81−16x2

=92−(4x)2

=(9+4x)(9−4x) [a2−b2=(a+b)(a−b)]

Question 4:

Factorise:

5 – 20x2

ANSWER:

5−20 x2

=5(1−4x2)

=5[12−(2x)2]

=5(1+2x)(1−2x) [a2−b2=(a+b)(a−b)]

Question 5:

Factorise:

2x4 – 32

ANSWER:

2x4 -32=2( x4-16) =2[ (𝑥2)2 -42 ]

=2(𝑥2+4)( 𝑥2-4) [𝑎2 − 𝑏2 = (𝑎 + 𝑏)(𝑎 − 𝑏)]

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

=2(x2+4)(x+2)(x-2) [𝑎2 − 𝑏2 = (𝑎 + 𝑏)(𝑎 − 𝑏)]

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Question 6:

Factorize:

3a3b − 243ab3

ANSWER:

a3b − 243ab3

= 3ab( 𝑎2−81𝑏2)

=3ab [𝑎2 − (9𝑏)2]

= 3ab ( 𝑎 - 9b)( 𝑎 + 9b)

= (a−9b) (a+9b)

Question 7:

Factorize:

3x3 − 48x

ANSWER:

3x3 − 48x

=3x(x2−16)

=3x(𝑥2 − 42)

=3x (x −4) (x +4)

Question 8:

Factorize:

27a2 − 48b2

ANSWER:

27a2 − 48b2

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=3 (9𝑎2 − 16𝑏2)

=3[(3a)2−(4b)2]

=3(3a−4b)(3a+4b)

Question 9:

Factorize:

x − 64

ANSWER:

x3 = x(1−64 x2)

= x [1 − (8𝑥)2]

= x(1−8x) (1+8x)

Question 10:

Factorize:

8ab2 − 18a3

ANSWER:

8ab2 − 18a3

=2a(4b2−9a2)

=2a[(2b)2−(3a)2]

=2a(2b−3a)(2b+3a)

Question 11:

Factorize:

150 − 6x2

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ANSWER:

150 − 6x2

=6(25− x2)

= 6(52− x2)

= 6(5−x)(5+x)

Question 12:

Factorise:

2 – 50x2

ANSWER:

2−50x2

=2(1− 25𝑥2) =2[12−(5x)2]

=2(1+5x)(1−5x) [𝑎2 − 𝑏2 = (𝑎 + 𝑏)(𝑎 − 𝑏)]

Question 13:

Factorise:

20x2 – 45

ANSWER:

20x2 – 45

=5(4x2 − 9)

=5[(2𝑥)2 − 32]

=5(2x+3)(2x−3) [𝑎2 − 𝑏2 = (𝑎 + 𝑏)(𝑎 − 𝑏)]

Question 14:

Factorise:

(3a + 5b)2 – 4c2

ANSWER:

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(3a + 5b)2 – 4c2

= (3a + 5b)2 – (2c)2

=(3a+5b+2c) (3a+5b−2c) [𝑎2 − 𝑏2 = (𝑎 + 𝑏)(𝑎 − 𝑏)]

Question 15:

Factorise:

a2 – b2 – a – b

ANSWER:

a2 – b2 – a – b

=( a + b)( a – b)−1(a + b)

[𝑎2 − 𝑏2 = (𝑎 + 𝑏)(𝑎 − 𝑏)]

= (a + b) [(𝑎 − 𝑏) − 1]

= (a + b) [𝑎 − 𝑏 − 1]

Question 16:

Factorise:

4a2 – 9b2 – 2a – 3b

ANSWER:

4a2 – 9b2 – 2a – 3b

= (2𝑎)2 − (3𝑏)2 − 1(2𝑎 + 3𝑏)

=(2𝑎 +3𝑏)(2𝑎 − 3𝑏)−1(2𝑎 + 3𝑏)

[𝑎2 − 𝑏2 = (𝑎 + 𝑏)(𝑎 − 𝑏)]

=(2𝑎 + 3𝑏)[( 2𝑎 − 3𝑏)−1]

=(2𝑎 + 3𝑏)( 2𝑎 − 3𝑏−1)

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Question 17:

Factorise:

a2 – b2 + 2bc – c2

ANSWER:

a2 – b2 + 2bc – c2

= a2 −( b2 −2bc+ c2)

= a2 –(𝑏 − 𝑐)2

[ a2 −2ab+ b2 = (𝑎 − 𝑏)2]

=[𝑎 + (𝑏 − 𝑐)][𝑎 − (𝑏 − 𝑐)]

[ a2− b2 =(𝑎 + 𝑏)( 𝑎 − 𝑏)]

=(a+b−c)(a−b+c)

Question 18:

Factorise:

4a2 – 4b2 + 4a + 1

ANSWER:

4a2 – 4b2 + 4a + 1

=(4a2+4a+1)− 4b2

=[(2 a)2+2×2a×1+12]−4b2

=(2a +1)2−(2b)2

[ 𝑎2 + 2𝑎𝑏 + 𝑏2 = (𝑎 + 𝑏)2]

=[(2a+1)+2b][(2a+1)−2b] [ 𝑎2−𝑏2=( 𝑎 + 𝑏)( 𝑎 − 𝑏)]

=(2a +1+2b)(2a +1−2b)

=(2a +2b +1)(2a −2b +1)

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Question 19:

Factorize:

a2 + 2ab + b2 − 9c2

ANSWER:

a2 + 2ab + b2 − 9c2

= (𝑎 + 𝑏)2 − (3𝑐)2

= (a+b−3c)(a+b+3c)

Question 20:

Factorize:

108a2 − 3(b − c)2

ANSWER:

108a2 − 3(b − c)2

= 3[36𝑎2−(b−c)2]

= 3[(6a)2−(b−c)2]

= 3(6a−b+c)(6a+b−c)

Question 21:

Factorize:

(a + b)3 − a − b

ANSWER:

(a + b)3 − a − b

=(a + b)3 −(a+b)

= (a+b) [(𝑎 + 𝑏)2 − 12]

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= (a+b)(𝑎 + 𝑏 − 1)(𝑎 + 𝑏 + 1)

Question 22:

Factorise:

x2 + y2 – z2 – 2xy

ANSWER:

x2 + y2 – z2 – 2xy

= (𝑥2 + 𝑦2 − 2xy) − 𝑧2

= (𝑥 − 𝑦)2 − 𝑧2 [a2−2ab+b2=(a−b)2]

[𝑎2 − 2𝑎𝑏 + 𝑏2 = (𝑎 − 𝑏)2]

= (𝑥 − 𝑦 + 𝑧)(𝑥 − 𝑦 − 𝑧)

[𝑎2 − 𝑏2 = (𝑎 + 𝑏)(𝑎 − 𝑏)]

Question 23:

Factorise:

x2 + 2xy + y2 – a2 + 2ab – b2

ANSWER:

x2 + 2xy + y2 – a2 + 2ab – b2

=( x2 + 2xy + y2 ) - (𝑎2 + 2𝑎𝑏 – 𝑏2)

= (𝑥 + 𝑦)2 − (𝑎 − 𝑏)2

[𝑎2 + 2𝑎𝑏 – 𝑏2 = (𝑎 + 𝑏)2 𝑎𝑛𝑑 𝑎2 − 2𝑎𝑏 + 𝑏2 = (𝑎 − 𝑏)2]

= [(𝑥 + 𝑦) + (𝑎 − 𝑏)][(𝑥 + 𝑦) − (𝑎 − 𝑏)]

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[𝑎2 − 𝑏2 = (𝑎 + 𝑏)(𝑎 − 𝑏)]

= (𝑥 + 𝑦 + 𝑎 − 𝑏)(𝑥 + 𝑦 − 𝑎 + 𝑏)

Question 24:

Factorise:

25x2 – 10x + 1 – 36y2

ANSWER:

25x2 – 10x + 1 – 36y2

= [(5𝑥)2 − 2 × 5𝑥 × 1 + 12] − (6𝑦)2

= (5𝑥 − 1)2 − (6𝑦)2

[𝑎2 − 2𝑎𝑏 + 𝑏2 = (𝑎 − 𝑏)2]

= [(5x−1+6y)][(5x−1−6y)]

[𝑎2 − 𝑏2 = (𝑎 + 𝑏)(𝑎 − 𝑏)]

= [(5𝑥 + 6𝑦 − 1)(5𝑥 − 6𝑦 − 1)]

Question 25:

Factorize:

a − b − a2 + b2

ANSWER:

a−b− a2 + b2=(a−b)−(𝑎2 − 𝑏2)

=(a−b)−(a−b)(a+b)

=(a−b)[1−(a+b)]

=(a−b)(1−a−b)

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Question 26:

Factorize:

a2 − b2 − 4ac + 4c2

ANSWER:

a2 − b2 − 4ac + 4c2

= (𝑎2 − 4𝑎𝑐 + 4𝑐2) − 𝑏2

=𝑎2 −2×2𝑎×c +(2c)2−b2

=(a−2c)2−b2

=(a − 2c + b)(a − 2c –b )

Question 27:

Factorize:

9 − a2 + 2ab − b2

ANSWER:

9 − a2 + 2ab − b2

=9 − ( a2 − 2ab + b2)

= 32− (𝑎 − 𝑏)2

=[3−(a−b)][3+(a−b)]

=(3−a+b)(3+a−b)

Question 28:

Factorize:

x3 − 5x2 − x + 5

ANSWER:

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x3 − 5x2 − x + 5

= x2(𝑥 −5)−1(𝑥 −5)

=(x−5)( x2−1)

=( x−5)( x2−12)

=( x −5)( x −1)( x +1)

Question 29:

Factorise:

1 + 2ab – (a2 + b2)

ANSWER:

1 + 2ab – (a2 + b2)

= 1+2ab− a2 - b2

= 1− a2 + 2ab − b2

= 12 − (a2 + 2ab − b2)

= 12 − (𝑎 − 𝑏)2

[ a2 - 2ab + b2 = (𝑎 − 𝑏)2]

= [1 + (𝑎 − 𝑏)] [1 − (𝑎 − 𝑏)] [𝑎2 − 𝑏2 = (𝑎 + 𝑏)(𝑎 − 𝑏)]

= ( 1 + 𝑎 − 𝑏) (1- 𝑎 + 𝑏 )

Question 30:

Factorise:

9a2 + 6a + 1 – 36b2

ANSWER:

9a2 + 6a + 1 – 36b2

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= [(3𝑎)2 + 2 × 3𝑎 × 1 + 12] − (6𝑏)2

= (3𝑎 + 1)2 − (6𝑏)2

[𝑎2 + 2𝑎𝑏 + 𝑏2 = (𝑎 + 𝑏)2]

= ( 3𝑎 + 1 – 6𝑏) (3𝑎 + 1 + 6𝑏)

[𝑎2 − 𝑏2 = (𝑎 − 𝑏)(𝑎 + 𝑏)]

= ( 3a – 6b + 1) ( 3a + 6b +1)

Question 31:

Factorize:

x2 − y2 + 6y − 9

ANSWER:

x2 − y2 + 6y – 9 = x2 – (y2 - 6y +9)

= x2 – (𝑦2 − 2 × 𝑦 × 3 + 32)

= x2 – (𝑦 − 3)2

= [𝑥 + (𝑦 − 3)] [𝑥 − (𝑦 − 3)]

= (𝑥 + 𝑦 − 3)(𝑥 − 𝑦 + 3)

Question 32:

Factorize:

4x2 − 9y2 − 2x − 3y

ANSWER:

4x2 − 9y2 − 2x − 3y

=(4x2 − 9y2 )− (2x+3y)

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= [(2𝑥)2 − (3𝑦)2] − (2𝑥 + 3𝑦)

= (2𝑥 − 3𝑦) (2𝑥 + 3𝑦) − 1(2𝑥 + 3𝑦)

= (2𝑥 + 3𝑦)(2𝑥 − 3𝑦 − 1)

Question 33:

Factorize:

9a2 + 3a − 8b − 64b2

ANSWER:

9a2 + 3a − 8b − 64b2

= 9a2 − 64b2 + 3a − 8b

= (3𝑎)2 − (8𝑏)2 + ( 3𝑎 − 8𝑏)

= (3𝑎 − 8𝑏)(3𝑎 + 8𝑏) + 1(3𝑎 − 8𝑏)

= (3𝑎 − 8𝑏)(3𝑎 + 8𝑏 + 1)

Question 34:

Factorise:

𝑥2 +1

𝑥2− 3

ANSWER:

𝑥2 +1

𝑥2− 3

= 𝑥2 +1

𝑥2 − 2 − 1

= [𝑥2 + (1

𝑥)

2− 2 × 𝑥 ×

1

𝑥] − 1

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= (𝒙 −𝟏

𝒙)

𝟐− 𝟏𝟐

[𝑎2 − 2𝑎𝑏 + 𝑏2 = (𝑎 − 𝑏)2]

= (𝑥 −1

𝑥+ 1) (𝑥 −

1

𝑥− 1)

[𝑎2 − 𝑏2 = (𝑎 − 𝑏)(𝑎 + 𝑏)]

Question :35

Factorise:

𝑥2 − 2 +1

𝑥2 −𝑦2

= [𝑥2 − 2 × 𝑥 ×1

𝑥+ (

1

𝑥)

2] − 𝑦2

= (𝑥 −1

𝑥)

𝟐− 𝑦2

[𝑎2 − 2𝑎𝑏 + 𝑏2 = (𝑎 − 𝑏)2]

= (𝑥 −1

𝑥+ 𝑦) (𝑥 −

1

𝑥− 𝑦)

[𝑎2 − 𝑏2 = (𝑎 + 𝑏)(𝑎 − 𝑏)]

Disclaimer: The expression of the question should be 𝑥2 − 2 +1

𝑥2 −𝑦2. The same has been done before solving the question.

Question 36:

Factorise:

𝑥4 +4

𝑥4

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ANSWER:

𝑥4 +4

𝑥4

= 𝑥4 +4

𝑥4 + 4 - 4

= [(𝑥2)2 + (2

𝑥2)2

+ 2 × (𝑥2) × (2

𝑥2)] − 22

= (𝑥2 +2

𝑥2)2

− 22

[𝑎2 + 2𝑎𝑏 + 𝑏2 = (𝑎 + 𝑏)2]

= (𝑥2 +2

𝑥2 + 2) (𝑥2 +2

𝑥2 − 2)

[𝑎2 − 𝑏2 = (𝑎 + 𝑏)(𝑎 − 𝑏)]

Question 37:

Factorise:

x8 – 1

ANSWER:

x8 – 1

= (𝑥4)2 − 12

= (𝑥4 + 1)(𝑥4 − 1)

[𝑎2 − 𝑏2 = (𝑎 + 𝑏)(𝑎 − 𝑏)]

= (𝑥4 + 1)[(𝑥2)2 − 12]

= (𝑥4 + 1)(𝑥2 + 1)(𝑥2 − 1)

[𝑎2 − 𝑏2 = (𝑎 + 𝑏)(𝑎 − 𝑏)]

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Question 38:

Factorise:

16x4 – 1

ANSWER:

16x4 – 1

= (4𝑥2)2 − 12

= (4𝑥2 + 1)(4𝑥2 − 1)

[𝑎2 − 𝑏2 = (𝑎 + 𝑏)(𝑎 − 𝑏)]

= (4𝑥2 + 1)[(2𝑥)2 − 12]

= (4𝑥2 + 1)(2𝑥 + 1)(2𝑥 − 1)

[𝑎2 − 𝑏2 = (𝑎 + 𝑏)(𝑎 − 𝑏)]

Question 39:

81x4 – y4

ANSWER:

81x4 – y4

= (9𝑥2)2 − (𝑦2)2

= (9𝑥2 + 𝑦2)(9𝑥2 − 𝑦2)

= [𝑎2 − 𝑏2 = (𝑎 + 𝑏)(𝑎 − 𝑏)]

= (9𝑥2 + 𝑦2)[(3𝑥)2 − 𝑦2]

= (9𝑥2 + 𝑦2)(3𝑥 + 𝑦)(3𝑥 − 𝑦)

[𝑎2 − 𝑏2 = (𝑎 + 𝑏)(𝑎 − 𝑏)]

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Question 40:

x4 – 625

ANSWER:

x4 – 625

= (𝑥2)2 − 252

= (𝑥2 + 25)( 𝑥2 − 25)

[𝑎2 − 𝑏2 = (𝑎 + 𝑏)(𝑎 − 𝑏)]

= (𝑥2 + 25) (𝑥2 − 52)

= (𝑥2 + 25) (𝑥 + 5)(𝑥 − 5)

[𝑎2 − 𝑏2 = (𝑎 + 𝑏)(𝑎 − 𝑏)]

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Exercise 3.3

Page No: 114

Question 1:

Factorize:

x2 + 11x + 30

ANSWER:

We have:

x2 + 11x + 30

We have to split 11 into two numbers such that their sum of is 11 and

their product is 30.

Clearly, 5+6=11 and 5×6=30

∴ 𝑥2+11x+30

= 𝑥2+5𝑥 + 6𝑥 + 30

= (𝑥 + 5)(𝑥 + 6)

Question 2:

Factorize:

x2 + 18x + 32

ANSWER:

We have:

x2 + 18x + 32

We have to split 18 into two numbers such that their sum is 18 and their

product is 32.

Clearly, 16 +2 = 18

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and 16 × 2 = 32.

∴ x2 + 18x + 32 = x2 +16𝑥 + 2𝑥 + 32

= 𝑥(𝑥 + 16) + 2(𝑥 + 16)

= (𝑥 + 16)(𝑥 + 2)

Question 3:

Factorise:

x2 + 20x – 69

ANSWER:

x2 + 20x – 69

= 𝑥2 + 23𝑥 − 3𝑥 − 69

=𝑥(𝑥 + 23) − 3(𝑥 + 23)

= (𝑥 + 23)(𝑥 − 3)

Question 4:

x2 + 19x – 150

ANSWER:

x2 + 19x – 150

= x2 + 25x−6x−150

= 𝑥(𝑥 + 25) − 6(𝑥 + 25)

= (𝑥 + 25)(𝑥 − 6)

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Question 5:

Factorise:

x2 + 7x – 98

ANSWER:

x2 + 7x – 98

= 𝑥2 + 14𝑥 − 7𝑥 − 98

= 𝑥(𝑥 + 14) − 7(𝑥 + 14)

= (𝑥 + 14)(𝑥 − 7)

Question 6:

Factorise:

𝑥2 + 2√3x – 24

ANSWER:

𝑥2 + 2√3x – 24

= 𝑥2 + 4√3x−2√3x − 24

= 𝑥(𝑥 + 4√3) - 2√3 ( 𝑥 + 4√3 )

= (𝑥 + 4√3) (𝑥 − 2√3)

Question 7:

Factorise:

x2 – 21x + 90

ANSWER:

x2 – 21x + 90

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= x2 – 15x – 6x + 90

= 𝑥(𝑥 − 15) − 6(𝑥 − 15)

= (𝑥 − 6)(𝑥 − 15)

Question 8:

Factorise:

x2 – 22x + 120

ANSWER:

x2 – 22x + 120

= 𝑥2 − 12𝑥 − 10𝑥 + 120

= 𝑥(𝑥 − 12) − 10 (𝑥 − 12)

= (𝑥 − 12)(𝑥 − 10)

Question 9:

Factorise:

x2 – 4x + 3

ANSWER:

x2 – 4x + 3

= 𝑥2 − 3𝑥 − 𝑥 + 3

= 𝑥(𝑥 − 3) − 1(𝑥 − 3)

= (𝑥 − 1)(𝑥 − 3)

Question 10:

Factorise:

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𝑥2 + 7√6 𝑥 + 60

ANSWER:

𝑥2 + 7√6 𝑥 + 60

= 𝑥2 + 5√6 𝑥 + 2√6 𝑥+ 60

= 𝑥 ( 𝑥 + 5√6) + 2√6 ( 𝑥 + 5√6)

= (𝑥 + 5√6) ( 𝑥 + 2√6 )

Question 11:

Factorise:

𝑥2 + 3√3𝑥 + 6

ANSWER:

𝑥2 + 3√3𝑥 + 6

= 𝑥2 + 2√3x+ √3x +6

= 𝑥(𝑥 + 2√3) + √3 (𝑥 + 2√3)

= (𝑥 + 2√3)(𝑥 + √3)

Question 12:

Factorise:

𝑥2 + 6√6𝑥 + 48

ANSWER:

𝑥2 + 6√6𝑥 + 48

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= 𝑥2 + 4√6𝑥 + 2√6x + 48

= 𝑥 + (𝑥 + 4√6) + 2√6(𝑥 + 4√6)

= (𝑥 + 4√6) (𝑥 + 2√6)

Question 13:

Factorise:

𝑥2 + 5√5 𝑥 + 30

ANSWER:

𝑥2 + 5√5 𝑥 + 30

= 𝑥2 + √53

𝑥 + √52

x + 30

= 𝑥(𝑥 + 3√5) + 2√5 (𝑥 + 3√5)

= (𝑥 + 3√5)+ 2√5(𝑥 + 3√5)

= (𝑥 + 3√5)(𝑥 + 2√5)

Question 14:

Factorise:

𝑥2 − 24𝑥 − 180

ANSWER:

𝑥2 − 24𝑥 − 180

= 𝑥2 − 30𝑥 + 6𝑥 − 180

= 𝑥(𝑥 − 30)+ 6 ( 𝑥 − 30)

= (𝑥 − 30)(𝑥 + 6)

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Question 15:

Factorise:

x2 – 32x – 105

ANSWER:

x2 – 32x – 105

= x2 – 35x+3x – 105

= 𝑥(𝑥 − 35) + 3(𝑥 − 35)

= (𝑥 − 35)(𝑥 + 3)

Question 16:

Factorise:

x2 – 11x – 80

ANSWER:

x2 – 11x – 80

= x2 – 16x +5x – 80

= 𝑥(𝑥 − 16) + 5(𝑥 − 16)

= (𝑥 − 16)(𝑥 + 5)

Question 17:

Factorise:

6 – x – x2

ANSWER:

– x2 −𝑥 − 6

= −𝑥2 − 3𝑥 + 2𝑥 − 6

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= −𝑥(𝑥 + 3) + 2(𝑥 + 3)

= (𝑥 + 3)(−𝑥 + 2)

= (𝑥 + 3)(2 − 𝑥)

Question 18:

Factorise:

𝑥2 − √3𝑥 − 6

ANSWER:

= 𝑥2 − 2√3𝑥 + √3𝑥 − 6

= 𝑥(𝑥 − 2√3) + √3 (𝑥 − 2√3)

= (𝑥 − 2√3) (𝑥 + √3)

Question 19:

Factorise:

40 + 3x – x2

ANSWER:

– x2 + 3𝑥 + 40

= – x2 + 8x – 5x + 40

= −𝑥 (𝑥 − 8) − 5(𝑥 − 8)

= (𝑥 − 8)(−𝑥 − 5)

= (8 − 𝑥)(𝑥 + 5)

Question 20:

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Factorise:

x2 – 26x + 133

ANSWER:

x2 – 26x + 133

= 𝑥2 − 19𝑥 − 7𝑥 + 133

= 𝑥(𝑥 − 19) − 7(𝑥 − 19)

= (𝑥 − 19)(𝑥 − 7)

Question 21:

Factorise:

𝑥2 − 2√3𝑥 − 24

ANSWER:

𝑥2 − 2√3𝑥 − 24

= 𝑥2 − 4√3x + 2√3𝑥 − 24

= 𝑥 (𝑥 − 4√3) + 2√3(𝑥 − 4√3)

= (𝑥 − 4√3) + (𝑥 + 2√3)

Question 22:

Factorise:

𝑥2 − 3√5𝑥 − 20

ANSWER :

𝑥2 − 3√5𝑥 − 20

= 𝑥2 − 4√5𝑥 + √5𝑥 − 20

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= 𝑥 (𝑥 − 4√5) + √5(𝑥 − 4√5)

= (𝑥 − 4√5) + (𝑥 + √5)

Question 23:

Factorise:

𝑥2 + √2x -24

ANSWER:

𝑥2 + √2x -24

= 𝑥2 + 4√2x - 3√2x – 24

= 𝑥 ( 𝑥 + 4√2) − 3√2 (𝑥 + 4√2 )

= (𝑥 + 4√2 )(𝑥 − 3√2 )

Question 24:

Factorise:

𝑥2 − 2√2 𝑥 − 30

ANSWER:

𝑥2 − 2√2 𝑥 − 30

= 𝑥2 − 5√2 𝑥 + 3 √2 𝑥 − 30

= 𝑥(𝑥 − 5√2) +3 √2 (𝑥 − 5√2)

= (𝑥 − 5√2) (𝑥 + 3√2)

Question 25:

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Factorize:

x2 − x − 156

ANSWER:

We have:

x2 − x − 156

We have to split (−1) into two numbers such that their sum is (−1) and

their product is (−156).

Clearly, −13+12 = −1 and −13×12 = −156

∴ 𝑥2 − 𝑥 − 156 = 𝑥2 − 13𝑥 − 12𝑥 − 156

= 𝑥(𝑥 − 13) + 12 (𝑥 − 13)

= (𝑥 − 13)(𝑥 + 12)

Question 26:

Factorise:

x2 – 32x – 105

ANSWER:

x2 – 32x – 105

= x2 – 35x + 3x −105

= 𝑥(𝑥 − 35) + 3(𝑥 − 35)

= (𝑥 − 35)(𝑥 + 3)

Question 27:

Factorise:

9x2 + 18x + 8

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ANSWER:

9x2 + 18x + 8

= 9𝑥2 + 12𝑥 + 6𝑥 + 8

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

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

Question 28:

Factorise:

6x2 + 17x + 12

ANSWER:

6x2 + 17x + 12

= 6x2 +9𝑥 + 8𝑥 + 12

= 3𝑥(2𝑥 + 3) + 4(2𝑥 + 3)

= (2𝑥 + 3)(3𝑥 + 4)

Question 29:

Factorize:

18x2 + 3x − 10

ANSWER:

We have:

18x2 + 3x − 10


product is (−180), i.e., 18×(−10).

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Clearly, 15+(−12)=3 and 15×(−12)=−180

∴ 18𝑥2 + 3𝑥 − 10 = 18𝑥2 + 15𝑥 − 12𝑥 − 10

= 3𝑥(6𝑥 + 5) − 2(6𝑥 + 5)

=(6x+5)(3x−2)

Question 30:

Factorize:

2x2 + 11x − 21

ANSWER:

We have:

2x2 + 11x − 21


product is (−42), i.e., 2×(−21).

Clearly, 14+ (−3) =11 and 14×(−3)= −42.

∴ 2𝑥2+ 11𝑥 − 21 = 2𝑥2+14𝑥 − 3𝑥 − 21

=2𝑥(𝑥 + 7) − 3(𝑥 + 7)

= (𝑥 + 7)(2𝑥 − 3)

Question 31:

Factorize:

15x2 + 2x − 8

ANSWER:

We have:

15x2 + 2x − 8

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product is (−120), i.e., 15×(−8).

Clearly, 12+(−10)=2 and 12×(−10) = −120

∴15x2 + 2x – 8 = 15x2 + 12𝑥 − 10𝑥 − 8

= 3𝑥(5𝑥 + 4) − 2(5𝑥 + 4)

= (5𝑥 + 4)(3𝑥 − 2)

Question 32:

Factorise:

21x2 + 5x – 6

ANSWER:

21x2 + 5x – 6

= 21x2 + 14𝑥 − 9𝑥 – 6

= 7𝑥(3𝑥 + 2) − 3(3𝑥 + 2)

= (3𝑥 + 2)(7𝑥 − 3)

Question 33:

Factorize:

24x2 − 41x + 12

ANSWER:

We have:

24x2 − 41x + 12

We have to split (−-41) into two numbers such that their sum is (−-41)

and their product is 288, i.e., 24×1224×12.

Clearly, (−32)+(−9)=−41 and (−32)×(−9)=288

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∴ 24𝑥2 - 41𝑥 + 12

= 24𝑥2- 32𝑥 − 9𝑥 + 12

=8x(3x−4)−3(3x−4)

=(3x−4)(8x−3)

Question 34:

Factorise:

3x2 – 14x + 8

ANSWER:

3x2 – 14x + 8

= 3x2 −12x−2x+8

= 3𝑥 (𝑥 − 4) − 2(𝑥 − 4)

= (𝑥 − 4)(3𝑥 − 2)

Hence, factorisation of 3x2 – 14x + 8 is (𝑥 − 4)(3𝑥 − 2).

Question 35:

Factorize:

2x2 + 3x − 90

ANSWER:

We have:

2x2 + 3x − 90


product is (−180), i.e., 2×(−90).

Clearly, −12 + 15 = 3 and −12×15 = −180

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∴ 2𝑥2 + 3𝑥 − 90 = 2𝑥2 − 12𝑥 + 15𝑥 − 90

= 2𝑥(𝑥 − 6) + 15 (𝑥 − 6)

= (𝑥 − 6)(2𝑥 + 15)

Question 36:

Factorize:

√5𝑥2 + 2𝑥 − 3 − 3√5

ANSWER:

We have:

√5𝑥2 + 2𝑥 − 3 − 3√5

We have to split 2 into two numbers such that their sum is 2 and product

is (−15), i.e., √5 ×(−3√5).

Clearly, 5+(−3)=2 and 5×(−3) = −15.

∴ √5𝑥2 + 2x - 3√5 = √5𝑥2 +5𝑥 − 3𝑥 − 3√5

= √5𝑥 (𝑥 + √5) – 3(𝑥 + √5)

= (𝑥 + √5)( √5𝑥 − 3)

Question 37:

Factorize:

2√3𝑥2 + 𝑥 − 5√3

ANSWER:

We have:

2√3𝑥2 + 𝑥 − 5√3

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We have to split 1 into two numbers such that their sum is 1 and product

is 30, i.e., 2√3 ×(−5√3).

Clearly, 6+(−5)=1 and 6×(−5)= −30

∴ 2√3𝑥2 + 𝑥 − 5√3

= 2√3𝑥2 + 6𝑥 − 5𝑥 − 5√3

= 2√3x (𝑥 + √3) -5(𝑥 + √3)

= (𝑥 + √3)(2√3x − 5)

Question 38:

Factorize:

7𝑥2 + 2√14 𝑥 + 2

ANSWER:

We have:

7𝑥2 + 2√14 𝑥 + 2

We have to split 2√14 into two numbers such that their sum

is 2√14 and product is 14.

Clearly,

√14 + √14 = 2√14 and √14 × √14 = 14

∴ 7𝑥2 + 2√14 𝑥 + 2

= 7𝑥2 + √14 𝑥 + √14 𝑥 + 2

= √7𝑥 (√7𝑥 + √2) (√7𝑥 + √2)

= (√7𝑥 + √2)2

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Question 39:

Factorize:

6√3𝑥2 − 47𝑥 + 5√3

ANSWER:

We have:

6√3𝑥2 − 47𝑥 + 5√3

Now, we have to split (−-47) into two numbers such that their sum is (−-

47) and their product is 90.

Clearly, (−45)+(−2)=−47 and (−45)×(−2)=90.

∴ 6√3𝑥2 − 47𝑥 + 5√3 = 6√3𝑥2 − 2𝑥 − 45𝑥 + 5√3

= 2𝑥 (3√3𝑥 − 1) − 5√3 (3√3𝑥 − 1)

= (3√3𝑥 − 1)(2𝑥 − 5√3)

Question 40:

Factorize:

5√5 𝑥2 + 20𝑥 + 3√5

ANSWER:

We have:

5√5 𝑥2 + 20𝑥 + 3√5


product is 75.

Clearly,

15+5 = 20 and 15×5 = 75

∴ 5√5 𝑥2 + 20𝑥 + 3√5

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= 5√5 𝑥2 + 15𝑥 + 5𝑥 + 3√5

= 5𝑥 (√5𝑥 + 3) + √5 (√5𝑥 + 3)

= (√5𝑥 + 3)(5𝑥 + √5 )

Question 41:

Factorise:

√3 𝑥2 + 10𝑥 + 8√3

ANSWER:

√3 𝑥2 + 10𝑥 + 8√3

= √3 𝑥2 + 6𝑥 + 4𝑥 + 8√3

= √3x (𝑥 + 2√3 ) + 4 (𝑥 + 2√3)

= (𝑥 + 2√3 )(√3𝑥 + 4)

Hence, factorisation of √3 𝑥2 + 10𝑥 + 8√3 is (𝑥 + 2√3 )(√3𝑥 +

4).

Question 42:

Factorize:

√2𝑥2 + 3x + √2 .

ANSWER:

We have:

√2𝑥2 + 3x + √2 .


product is 2, i.e., √2 × √2 .

Clearly, 2+1=3 and 2×1= 2.

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∴ √2𝑥2 + 3x + √2 = √2 𝑥2 + 2𝑥 + 𝑥 + √2.

= √2𝑥 (𝑥 + √2 )+1 (𝑥 + √2)

= (𝑥 + √2) (√2𝑥 + 1).

Question 43:

Factorize:

2𝑥2 + 3√3𝑥 + 3

ANSWER:

We have:

2𝑥2 + 3√3𝑥 + 3

We have to split 3√3 into two numbers such that their sum is 3√3 and

their product is 6, i.e., 2×3.

Clearly, 2√3 + √3 = 3√3 and 2√3 × √3 =6.

∴ 2𝑥2 + 3√3𝑥 + 3 = 2𝑥2 + 2√3x + √3𝑥 + 3.

= 2𝑥(𝑥 + √3) + √3 (𝑥 + √3)

= (𝑥 + √3) (2𝑥 + √3).

Question 44:

Factorize:

15x2 − x − 128

ANSWER:

We have:

15x2 − x − 128

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their product is (−420), i.e., 15×(−28).

Clearly, (−21)+20=−1 and (−21)×20=−420.

∴ 15x2 − x − 128

= 15x2 – 21𝑥 + 20𝑥 − 28

=3x(5x−7)+4(5x−7)

=(5x−7)(3x+4)

Question 45:

Factorize:

6x2 − 5x − 21

ANSWER:

We have:

6x2 − 5x − 21



Clearly, 9+(−14)=−5 and 9×(−14)= −126

∴ 6x2 − 5x − 21

= 6x2 −9𝑥 − 14𝑥 − 21

= 3𝑥(2𝑥 + 3) − 7(2𝑥 + 3)

= (2𝑥 + 3)(3𝑥 − 7).

Question 46:

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Factorize:

2x2 − 7x − 15

ANSWER:

We have:

2x2 − 7x − 15



Clearly, (−10) + 3= −7 and (−10)×3 = −30.

∴ 2x2 − 7x − 15

= 2x2 -10𝑥 + 3𝑥 − 15

= 2𝑥(𝑥 − 5) + 3(𝑥 − 5)

= (𝑥 − 5)(2𝑥 + 3)

Question 47:

Factorize:

5x2 − 16x − 21

ANSWER:

We have:

5x2 − 16x − 21



Clearly, (−21) + 5 = −16 and (−21)×5 = −105

∴ 5x2 − 16x − 21

= 5x2 + 5𝑥 − 21𝑥 − 21

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= 5𝑥(𝑥 + 1) − 21 (𝑥 + 1)

= (𝑥 + 1)(5𝑥 − 21)

Question 48:

Factorise:

6x2 – 11x – 35

ANSWER:

6x2 – 11x – 35

= 6x2 – 21x + 10x – 35

= 3𝑥(2𝑥 − 7) + 5(2𝑥 − 7)

= (2𝑥 − 7)(3𝑥 + 5)

Hence, factorisation of 6x2 – 11x – 35 is (2x−7)(3x+5).

Question 49:

Factorise:

9x2 – 3x – 20

ANSWER:

9x2 – 3x – 20

= 9x2 – 15x +12x – 20

= 3𝑥(3𝑥 − 5) + 4(3𝑥 − 5)

= (3𝑥 − 5) + (3𝑥 + 4)

Hence, factorisation of 9x2 – 3x – 20 is (3𝑥 − 5) + (3𝑥 + 4).

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Question 50:

Factorize:

10x2 − 9x − 7

ANSWER:

We have:

10x2 − 9x − 7


their product is (-70), i.e., 10×(−7).

Clearly, (−14)+5= −9 and (−14)×5= −70.

∴ 10x2 − 9x – 7

= 10𝑥2 + 5𝑥 − 14𝑥 − 7

= 5𝑥(2𝑥 + 1) − 7(2𝑥 + 1)

= (2𝑥 + 1)(5𝑥 − 7)

Question 51:

Factorize:

𝑥2 − 2𝑥 + 7

16

ANSWER:

We have:

𝑥2 − 2𝑥 + 7

16

= 16𝑥2−32𝑥+7

16

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= 1

16 (16𝑥2 − 32𝑥 + 7)

Now, we have to split (−32) into two numbers such that their sum is

(−32) and their product is 112, i.e., 16×7.

Clearly, (−4)+(−28)=−32 and (−4)×(−28)=112

∴ 𝑥2 − 2𝑥 + 7

16 =

1

16 (16𝑥2 − 32𝑥 + 7)

= 1

16 (16𝑥2 − 4𝑥 − 28𝑥 + 7)

= 1

16 [4𝑥(𝑥 − 1) − 7(4𝑥 − 1)]

= 1

16 (4x-1) (4x-7)

Question 52:

Factorise:

1

3𝑥2 − 2𝑥 − 9

ANSWER:

1

3𝑥2 − 2𝑥 − 9

= 𝑥2−6𝑥−27

3

= 𝑥2−9𝑥+3𝑥−27

3

= 𝑥(𝑥−9)+3(𝑥−9)

3

= (𝑥−9)(𝑥+3)

3

= (𝑥−9)

3×

(𝑥+3)

1

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= (1

3𝑥 − 3) (𝑥 + 3)

Hence, factorisation of 1

3𝑥2 − 2𝑥 − 9 is (

1

3𝑥 − 3) (𝑥 + 3).

Question 53:

Factorise:

𝑥2 +12

35𝑥 +

1

35

Answer:

𝑥2 +12

35𝑥 +

1

35 =

35𝑥2+12𝑥+1

35

= 35𝑥2+7𝑥+5𝑥+1

35

= 7𝑥(5𝑥+1)+1(5𝑥+1)

35

= (5𝑥+1)(7𝑥+1)

35

= (5𝑥+1)(7𝑥+1)

5×7

= (5𝑥+1)

5 ×

(7𝑥+1)

7

= (𝑥 +1

5) (𝑥 +

1

7)

Hence, factorisation of 𝑥2 +12

35𝑥 +

1

35 is (𝑥 +

1

5) (𝑥 +

1

7).

Question 54:

Factorise:

21𝑥2 − 2𝑥 +1

21

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ANSWER:

21𝑥2 − 2𝑥 +1

21

= 21𝑥2 − 𝑥 − 𝑥 + 1

21

= 21𝑥 (𝑥 − 1

21) − 1 (𝑥 −

1

21)

= (𝑥 − 1

21) (21𝑥 − 1)

Hence, factorisation of

21𝑥2 − 2𝑥 +1

21 is (𝑥 −

1

21) (21𝑥 − 1).

Question 55:

Factorise:

3

2𝑥2 + 16𝑥 + 10.

ANSWER:

3

2𝑥2 + 16𝑥 + 10 =

3

2𝑥2 + 15𝑥 + 𝑥 + 10

= 3𝑥 (1

2𝑥 + 5) + 1(𝑥 + 10)

= 3

2𝑥 (𝑥 + 10) + 1(𝑥 + 10)

= (𝑥 + 10) (3

2𝑥 + 1)


2𝑥2 + 16𝑥 + 10 is (𝑥 + 10) (

3

2𝑥 + 1).

Question 56:

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Factorise:

2

3𝑥2 −

17

3𝑥 − 28

ANSWER:

2

3𝑥2 −

17

3𝑥 − 28

= 2

3𝑥2 − 8𝑥 +

7

3𝑥 − 28

= (1

3𝑥 − 4) (2𝑥 + 7)


3𝑥2 −

17

3𝑥 − 28 is (

1

3𝑥 − 4) (2𝑥 + 7).

Question 57:

Factorise:

3

5𝑥2 −

19

5𝑥 + 4

ANSWER:

3

5𝑥2 −

19

5𝑥 + 4

= 3

5𝑥2-3𝑥 −

4

5𝑥 + 4

= 3𝑥 (1

5𝑥 − 1) − 4 (

1

5𝑥 − 1)

= (1

5𝑥 − 1) (3𝑥 − 4)


5𝑥2 −

19

5𝑥 + 4 is (

1

5𝑥 − 1) (3𝑥 − 4).

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Question 58:

Factorise:

2𝑥2 − 𝑥 +1

8

ANSWER:

2𝑥2 − 𝑥 +1

8

= 2𝑥2 - 1

2𝑥 −

1

2𝑥 +

1

8.

= 2𝑥 (𝑥 −1

4) −

1

2(𝑥 −

1

4)

= (𝑥 −1

4) (2𝑥 −

1

2)

Hence, factorisation of 2𝑥2 − 𝑥 +1

8 is (𝑥 −

1

4) (2𝑥 −

1

2).

Question 59:

Factorize:

2(x + y)2 − 9(x + y) − 5

ANSWER:

We have:

2(x + y)2 − 9(x + y) − 5

Let:(x+y) = u

Thus, the given expression becomes

2𝑢2 − 9𝑢 − 5

Now, we have to split (−9) into two numbers such that their sum is (−9)

and their product is (-10).

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Clearly, −10+1 = −9 and −10×1 = −10

∴ 2𝑢2 − 9𝑢 − 5 = 2𝑢2 − 10𝑢 − 𝑢 − 5

= 2𝑢(𝑢 − 5) + 1(𝑢 − 5)

= (𝑢 − 5)(2𝑢 + 1)

Putting 𝑢 = (𝑥 + 𝑦), we get:

2(𝑥 + 𝑦)2 − 9(𝑥 + 𝑦) − 5 = (𝑥 + 𝑦 − 5)[2(𝑥 + 𝑦) + 1]

= (𝑥 + 𝑦 − 5)(2𝑥 + 2𝑦 + 1)

Question 60:

Factorize:

9(2a − b)2 − 4(2a − b) − 13

ANSWER:

We have:

9(2a − b)2 − 4(2a − b) − 13

Let:(2a−b) = p

Thus, the given expression becomes

9𝑝2 − 4𝑝 − 13

Now, we must split (−4) into two numbers such that their sum is (−4)

and their product is (−117).

Clearly, −13+9 = −4 and −13×9 = −117.

∴9𝑝2 − 4𝑝 − 13 = 9𝑝2 + 9𝑝 − 13𝑝 − 13

= 9𝑝(𝑝 + 1) − 13 (𝑝 + 1)

= (𝑝 + 1)(9𝑝 − 13)

Putting p=(2a−b), we get:

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9(2a − b)2 − 4(2a − b) – 13 = [(2𝑎 − 𝑏) + 1][9(2𝑎 − 𝑏) − 13]

= (2𝑎 − 𝑏 + 1)[18𝑎 − 9𝑏 − 13]

Question 61:

Factorise:

7(𝑥 − 2𝑦)2 − 25(𝑥 − 2𝑦) + 12

ANSWER:

7(𝑥 − 2𝑦)2 − 25(𝑥 − 2𝑦) + 12

= 7(𝑥 − 2𝑦)2 − 21(𝑥 − 2𝑦) − 4(𝑥 − 2𝑦) + 12

= [7(𝑥 − 2𝑦)](𝑥 − 2𝑦 − 3) − 4(𝑥 − 2𝑦 − 3)

= [7(x-2y)-4] (𝑥 − 2𝑦 − 3)

= (7𝑥 − 14𝑦 − 4)(𝑥 − 2𝑦 − 3)

Hence, factorisation of 7(𝑥 − 2𝑦)2 − 25(𝑥 − 2𝑦) + 12 is (7𝑥 − 14𝑦 −

4)(𝑥 − 2𝑦 − 3).

Question 62:

Factorise:

10 (3𝑥 + 1

𝑥)

2

− (3𝑥 +1

𝑥) − 3

Answer:

10 (3𝑥 + 1

𝑥)

2− (3𝑥 +

1

𝑥) -3

= 10 (3𝑥 + 1

𝑥)

2− 6 (3𝑥 +

1

𝑥) + 5 (3𝑥 +

1

𝑥) − 3

= [2 (3𝑥 +1

𝑥)] [5 (3𝑥 +

1

𝑥) − 3] + 1 [5 (3𝑥 +

1

𝑥) − 3]

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= [5 (3𝑥 +1

𝑥) − 3] [2 (3𝑥 +

1

𝑥) + 1]

= (15𝑥 +5

𝑥− 3) (6𝑥 +

2

𝑥+ 1)

Hence, factorisation of 10(3𝑥 + 1

𝑥)

2− (3𝑥 +

1

𝑥)-3 is (15𝑥 +

5

𝑥−

3) (6𝑥 +2

𝑥+ 1)

Question 63:

Factorise:

6 (2𝑥 −3

𝑥)

2

+ 7 (2𝑥 −3

𝑥) − 20

ANSWER:

6 (2𝑥 −3

𝑥)

2

+ 7 (2𝑥 −3

𝑥) − 20

= 6 (2𝑥 −3

𝑥)

2+ 15 (2𝑥 −

3

𝑥) − 8 (2𝑥 −

3

𝑥) − 20

= [3 (2𝑥 −3

𝑥)] [2 (2𝑥 −

3

𝑥) + 5] − 4 [2 (2𝑥 −

3

𝑥) + 5]

= [2(2𝑥 −3

𝑥) + 5][3 (2𝑥 −

3

𝑥) − 4]

= (4𝑥 −6

𝑥+ 5) (6𝑥 −

9

𝑥− 4).

Hence, factorisation of 6 (2𝑥 −3

𝑥)

2+ 7 (2𝑥 −

3

𝑥) − 20 is (4𝑥 −

6

𝑥+

5) (6𝑥 −9

𝑥− 4).

Question 64:

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Factorise:

(𝑎 + 2𝑏)2 + 101(𝑎 + 2𝑏) + 100

ANSWER:

(𝑎 + 2𝑏)2 + 101(𝑎 + 2𝑏) + 100

= (𝑎 + 2𝑏)2 = 100(𝑎 + 2𝑏) + 1(𝑎 + 2𝑏) + 100

= (𝑎 + 2𝑏) + [(𝑎 + 2𝑏) + 100] + 1[(𝑎 + 2𝑏) + 100]

= [(𝑎 + 2𝑏) + 1](𝑎 + 2𝑏 + 100)

Hence, factorisation of (𝑎 + 2𝑏)2 + 101(𝑎 + 2𝑏) + 100 𝑖𝑠 (𝑎 + 2𝑏 +

1) (𝑎 + 2𝑏 + 100).

Question 65:

Factorise:

4x4 + 7x2 – 2

ANSWER:

4x4 + 7x2 – 2

= 4x4 + 8x2 - x2 – 2

= 4𝑥2(𝑥2 + 2) − 1(𝑥2 + 2)

= (4𝑥2 − 1)(𝑥2 + 2)

Hence, factorisation of 4x4 + 7x2 – 2 is (4𝑥2 − 1)(𝑥2 + 2).

Question 66:

Evaluate {(999)2 – 1}.

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ANSWER:

{(999)2−1} = {(999)2−12}

=(999−1)(999+1)

=(998)(1000)

Hence, {(999)2 – 1} = 998000.

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Exercise 3.4

Page No: 119

Question 1:

Expand:

(i) (a + 2b + 5c)2

(ii) (2b − b + c)2

(iii) (a − 2b − 3c)2

ANSWER:

(i) (a + 2b + 5c)2

(a + 2b + 5c)2 =( a)2 + (2b)2 +(5c)2+2(a)(2b) +2(2b)(5c) + 2(5c)( a)

𝑎2 + 4𝑏2 + 25𝑐2 + 4𝑎𝑏 + 20𝑏𝑐 + 10𝑎𝑐

(ii) (2b − b + c)2 = [(2𝑎) − (𝑏) + (𝑐)]2

= (2𝑎)2 + (−𝑏)2 + (𝑐)2 + 2(2𝑎)(−𝑏) + 2(−𝑏)(𝑐) + 4(𝑎)(𝑐)

= 4𝑎2 + 𝑏2 + 𝑐2 − 4𝑎𝑏 − 2𝑏𝑐 + 4𝑎𝑐

(iii) (a − 2b − 3c)2 = [𝑎 + (−2𝑏) + (−3𝑐)]2

= (𝑎)2 + (−2𝑏)2 + (−3𝑐)2 + 2(𝑎)(−2𝑏) + 2(−2𝑏)(−3𝑐)2(𝑎)(−3𝑐)

= 𝑎2 + 4𝑏2 + 9𝑐2 − 4𝑎𝑏 + 12𝑏𝑐 − 6𝑎𝑐

Question 2:

Expand:

(i) (2a − 5b − 7c)2

(ii) (−3a + 4b − 5c)2

(iii) (12a−14a+2)212a-14a+22

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Answer:

(i) (2a − 5b − 7c)2

=[(2a)+(−5b)+(−7c)]2

=(2a)2+(−5b)2+(−7c)2+2(2a)(−5b)+2(−5b)(−7c)+2(2a)(−7c)

=4a2+25b2+49c2−20ab+70bc−28ac

(ii) (−3𝑎 + 4𝑏 − 5𝑐)2

= [(−3𝑎) + (4𝑏) + (−5𝑐)]2

= (−3𝑎)2+ (4𝑏)2 + (−5𝑐)2 + 2(−3𝑎)(−4𝑏) + 2(4𝑏)(−5𝑐) + 2

(−3𝑎)(−5𝑐)

= 9𝑎2 + 16𝑏2 + 25𝑐2 − 24𝑎𝑏 − 40𝑏𝑐 + 30𝑎𝑐

(iii) (1

2𝑎 −

1

4𝑏 + 2)

2 = [(

𝑎

2) + (−

𝑏

4) + (2)]

2

= (𝑎

2)

2+ (−

𝑏

4)

2+ (2)2 + 2 (

𝑎

2) (−

𝑏

4) + 2 (−

𝑏

4) (2) + 2 (

𝑎

2) (2)

= 𝑎2

4 +

𝑏2

16 + 4 -

𝑎𝑏

4− 𝑏 + 2𝑎

Question 3:

Factorize: 4x2 + 9y2 + 16z2 + 12xy − 24yz − 16xz.

ANSWER:

We have:

4x2 + 9y2 + 16z2 + 12xy − 24yz − 16xz.

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= (2𝑥)2 + (3𝑦)2 + (−4𝑧)2 + 2(2𝑥)(3𝑦) + 2(3𝑦)(−4𝑧) + 2

(−4𝑧)(2𝑥)

= [(2𝑥) + (3𝑦) + (−4𝑧)]2

= (2𝑥 + 3𝑦 − 4𝑧)2

Question 4:

Factorize: 9x2 + 16y2 + 4z2 − 24xy + 16yz − 12xz

ANSWER:

We have:

9x2 + 16y2 + 4z2 − 24xy + 16yz − 12xz

= (−3𝑥)2 + (4𝑦)2 + (2𝑧)2 + 2(−3𝑥)(4𝑦) + 2(4𝑦)(2𝑧)(−3𝑥)

= [(−3𝑥) + (4𝑦) + (2𝑧)]2

= (−3𝑥 + 4𝑦 + 2𝑧)2

Question 5:

Factorize: 25x2 + 4y2 + 9z2 − 20xy − 12yz + 30xz.

ANSWER:

We have:

25x2 + 4y2 + 9z2 − 20xy − 12yz + 30xz.

=(5x)2+(−2y)2+(3z)2+2(5x)(−2y)+2(−2y)(3z)+2(3z)(5x)

=[(5x)+(−2y)+(3z)]2

=(5x−2y+3z)2

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Question 6:

16x2 + 4y2 + 9z2 – 16xy – 12yz + 24xz

ANSWER:

16x2 + 4y2 + 9z2 – 16xy – 12yz + 24xz

=(4x)2+(−2y)2+(3z)2+2(4x)(−2y)+2(−2y)(3z)+2(3z)(4x)

=(4x−2y+3z)2

[using 𝑎2 + 𝑏2 + 𝑐2 + 2𝑎𝑏 + 2𝑏𝑐 + 2𝑐𝑎 = (𝑎 + 𝑏 + 𝑐)2]

Hence, 16x2 + 4y2 + 9z2 – 16xy – 12yz + 24xz = (4x−2y+3z)2.

Question 7:

Evaluate

(i) (99)2

(ii) (995)2

(iii) (107)2

ANSWER:

(i) (99)2=(100−1)2

=[(100)+(−1)]2

= (100)2+2×(100)×(−1)+(−1)2

=10000 −200+1

=9801

(ii) (995)2=(1000−5)2

=[(1000)+(−5)]2

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=(1000)2+2×(1000)×(−5)+(−5)2

=1000000−10000+25

=990025

(iii) (107)2=(100+7)2

=(100)2+2×(100)×(7)+(7)2

=10000+1400+49

=11449

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Exercise 3.5

Page No: 123

Question 1:

Expand

(i) (3x + 2)3

(ii) (3𝑎 +1

4𝑏)

3

(iii) (1 +2

3𝑎)

3

ANSWER:

(i) (3x+2)3

= (3𝑥)3 + 3 × (3𝑥)2 𝑥2 + 3 × 3𝑥 × (2)2 + (2)3

= 27𝑥3 + 54𝑥2+36x + 8

(ii) (3𝑎 +1

4𝑏)

3= (3𝑎)3 + (

1

4𝑏)

3+3(3𝑎)2 (

1

4𝑏) + 3 × (3𝑎) × (

1

4𝑏)

2

= 27𝑎3 +1

64𝑏3 +27𝑎2

4𝑏 +

9𝑎

16𝑏2

(iii) (1 +2

3𝑎)

3= (

2

3𝑎)

3+ 3 × (

2

3𝑎)

2× 1 + 3𝑎

2

3𝑎 × (1)2 + (1)3

= 8

27𝑎3+

4

3𝑎2 + 2𝑎 + 1

Question 2:

Expand

(i) (5a – 3b)3

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(ii) (3𝑥 −5

𝑥)

3

(iii) (4

5𝑎 − 2)

3

ANSWER:

(i) (5a – 3b)3 = (5𝑎)3 − (3𝑏)3 − 3(5𝑎)2(3𝑏) + 3(5𝑎)(3𝑏)2

= 125𝑎3 − 27𝑏3 − 225𝑎2𝑏 + 135𝑎𝑏2

(ii) (3𝑥 −5

𝑥)

3= (3𝑥)3 − (

5

𝑥)

3− 3(3𝑥)3 (

5

𝑥) + 3(3𝑥) (

5

𝑥)

2

= 27𝑥3 −125

𝑥3 − 135𝑥 +225

𝑥

(iii) (4

5𝑎 − 2)

3= (

4

5𝑎)

3− (2)3 − 3 (

4

5𝑎)

2(2) + 3 (

4

5𝑎) (2)2

= 64

125𝑎3 − 8 −

96

25𝑎2 +

48

5𝑎

Question 3:

Factorise

8𝑎3 + 27𝑏3 + 36𝑎2𝑏 + 54𝑎𝑏2

ANSWER:

8𝑎3 + 27𝑏3 + 36𝑎2𝑏 + 54𝑎𝑏2

= (2𝑎)3 + (3𝑏)3 + 3(2𝑎)2(3𝑏) + 3(2𝑎)(3𝑏)2

= (2𝑎 + 3𝑏)3

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8𝑎3 + 27𝑏3 + 36𝑎2𝑏 + 54𝑎𝑏2 is (2𝑎 + 3𝑏)3

Question 4:

Factorise

64𝑎3 − 27𝑏3 − 144𝑎2𝑏 + 108𝑎𝑏2

ANSWER:

64𝑎3 − 27𝑏3 − 144𝑎2𝑏 + 108𝑎𝑏2

= (4a)3 − (3b)3 −3(4a)2(3b) + 3(4a)(3b)2

= (4a−3b)3

Hence, factorisation of 64𝑎3 − 27𝑏3 − 144𝑎2𝑏 + 108𝑎𝑏2 is (4a−3b)3.

Question 5:

Factorise

1 +27

125𝑎3 +

9𝑎

5+

27𝑎2

24

ANSWER:

1 +27

125𝑎3 +

9𝑎

5+

27𝑎2

24

= (1)3 + (3

5𝑎)

3+ 3(1)2 (

3

5𝑎) + 3(1) (

3

5𝑎)

3

= (1 +3

5𝑎)

𝟑

Hence, factorisation of 1 + 27

125𝑎3 +

9𝑎

5+

27𝑎2

24 is (1 +

3

5𝑎)

𝟑

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Question 6:

Factorise

125x3−27y3−225x2y+135xy2

ANSWER:

125x3−27y3−225x2y+135xy2

=(5x)3−(3y)3−3(5x)2(3y)+3(5x)(3y)2

= (5𝑥 − 3𝑦)3

Hence, factorisation of 125x3−27y3−225x2y+135xy2 is (5𝑥 − 3𝑦)3.

Question 7:

Factorise

𝑎3𝑥3 − 3𝑎2𝑏𝑥2 + 3𝑎𝑏2𝑥 − 𝑏3

ANSWER:

𝑎3𝑥3 − 3𝑎2𝑏𝑥2 + 3𝑎𝑏2𝑥 − 𝑏3

= (𝑎𝑥)3 − (𝑏)3 − 3(𝑎𝑥)2(𝑏) + 3(𝑎𝑥)(𝑏)2

= (𝑎𝑥 − 𝑏)3


𝑎3𝑥3 − 3𝑎2𝑏𝑥2 + 3𝑎𝑏2𝑥 − 𝑏3 is (𝑎𝑥 − 𝑏)3.

Question 8:

Factorise

64

125𝑎3 −

96

25𝑎2 +

48

5𝑎 − 8

ANSWER:

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64

125𝑎3 −

96

25𝑎2 +

48

5𝑎 − 8

= (4

5𝑎)

3− (2)3 − 3 (

4

5𝑎)

2(2) + 3 (

4

5𝑎) (2)2

= (4

5𝑎 − 2)

𝟑


125𝑎3 −

96

25𝑎2 +

48

5𝑎 − 8 is (

4

5𝑎 − 2)

𝟑.

Question 9:

Factorise

a3 – 12a(a – 4) – 64

ANSWER:

𝑎3 − 12𝑎(𝑎 − 4) − 64 = 𝑎3 − 12𝑎2 + 48𝑎 − 64

= (𝑎)3 − (4)3 − 3(𝑎)2(4) + 3(𝑎)(4)2

= (𝑎 − 4)3

Hence, factorisation of a3 – 12a(a – 4) – 64 is (𝑎 − 4)3.

Question 10:

Evaluate

(i) (103)3

(ii) (99)3

ANSWER:

(i) (103)3=(100+3)3

=(100)3+(3)3+3(100)2(3)+3(100)(3)2

=1000000+27+90000+2700

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=1092727

(ii) (99)3=(100−1)3

=(100)3−(1)3−3(100)2(1)+3(100)(1)2

=1000000−1−30000+300

=1000300−30001

=970299

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Exercise 3.6

Page No: 129

Question 1:

Factorize:

x3 + 27

ANSWER:

𝑥3+27

=(x)3+(3)3

=(x+3)(x2−3x+32)

=(x+3)(x2−3x+9)

Question 2:

Factorise

27a3 + 64b3

ANSWER:

We know that

x3+y3=(x+y)(x2+y2−xy)

Given: 27a3 + 64b3

x = 3a, y = 4b

27a3 + 64b3

=(3a+4b)(9a2+16b2−12ab)

Question 3:

Factorize:

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125𝑎3 +1

8

ANSWER:

125𝑎3 +1

8

=(5a)3+ (1

2)

3

= (5𝑎 +1

2) [(5𝑎)2 − 5𝑎 ×

1

2+ (

1

2)

3]

= (5𝑎 +1

2) (25𝑎2 −

5𝑎

2+

1

4)

Question 4:

Factorize:

216𝑥3 +1

125

ANSWER:

216𝑥3 +1

125

=(6x)3+ (1

5)

3

= (6𝑥 +1

5) [(6𝑥)2 − 6𝑥 ×

1

5+ (

1

5)

2]

= (6𝑥 +1

5) (36𝑥2 −

6𝑥

5+

1

25)

Question 5:

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Factorize:

16x4 + 54x

ANSWER:

16x4 + 54x

= 2𝑥(8𝑥3 + 27)

= 2𝑥(2𝑥 + 3)[(2𝑥)2 − 2𝑥 × 3 + 32]

= 2𝑥(2𝑥 + 3)(4𝑥2 − 6𝑥 + 9)

Question 6:

Factorize:

7a3 + 56b3

ANSWER:

7a3 + 56b3

=7 [(𝑎)3 + (2𝑏)3]

= 7 (a+2b) [𝑎2 − 𝑎 × 2𝑏 + (2𝑏)2]

= 7 (a+2b) (𝑎2 − 2𝑎𝑏 + 4𝑏2)

Question 7:

Factorize:

x5 + x2

ANSWER:

= 𝑥5 + 𝑥2 = 𝑥2 (𝑥3 + 1 )

= 𝑥2(𝑥3 + 13 )

= 𝑥2(𝑥 + 1)(𝑥2 − 𝑥 × 1 + 12)

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= 𝑥2(𝑥 + 1)(𝑥2 − 𝑥 + 1)

Question 8:

Factorize:

a3 + 0.008

ANSWER:

𝑎3 + 0.008 = 𝑎3 + (0.2)3

= (𝑎 + 0.2)[𝑎2 − 𝑎 × (0.2) + (0.2)2]

= (𝑎 + 0.2)(𝑎2 − 0.2𝑎 + 0.04)

Question 9:

Factorise

1 – 27a3

ANSWER:

1 – 27a3

= 13−(3a)3

=(1−3a)[12+1×3x+(3a)2]

=(1−3a)(1+3a+9a2)

Question 10:

Factorize:

64a3 − 343

ANSWER:

64a3 − 343

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=(4a)3−(7)3

=(4a−7)(16a2+4a×7+72)

=(4a−7)(16a2+28a+49)

Question 11:

Factorize:

x3 − 512

ANSWER:

𝑥3 −512

= 𝑥3 − 83

= (𝑥 − 8)(𝑥2 + 8𝑥 + 82)

= (𝑥 − 8) (𝑥2 + 8𝑥 + 64)

Question 12:

Factorize:

a3 − 0.064

ANSWER:

a3−0.064

=( a)3−(0.4)3

=( a −0.4)(𝑎2 × 0.4𝑎 + 0.16)

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Question 13:

Factorize:

8𝑥3 −1

27𝑦3

ANSWER:

8𝑥3 −1

27𝑦3

= (2𝑥)3 − (1

3𝑦)

3

= (2𝑥 −1

3𝑦) [(2𝑥)2 + 2𝑥 ×

1

3𝑦+ (

1

3𝑦)

3]

= (2𝑥 −1

3𝑦) (4𝑥2 +

2𝑥

3𝑦+

1

9𝑦2)

Question 14:

Factorise

𝑥3

216− 8𝑦3

ANSWER:

We know

𝑎3 − 𝑏3 = (𝑎 − 𝑏)(𝑎2 + 𝑏2 + 𝑎𝑏)

We have,

𝑥3

216− 8𝑦3 = (

𝑥

6)

3

− (2𝑦)3

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So, 𝑎 =𝑥

6, 𝑏 = 2𝑦

𝑥3

216 − 8𝑦3 = (

𝑥

6− 2𝑦) ((

𝑥

6)

2+

𝑥

6× 2𝑦 + (2𝑦)2)

= (𝑥

6− 2𝑦) (

𝑥2

36+

𝑥𝑦

3+ (4𝑦)2)

Question 15:

Factorize:

x − 8xy3

ANSWER:

x − 8xy3

=x(1−8y3)

=x[13−(2y)3]

=x(1−2y)(12+1×2y+(2y)2)

=x(1−2y)(1+2y+4y2)

Question 16:

Factorise

32x4 – 500x

ANSWER:

= 32x4 – 500x

= 4x(8x3−125)

=4x((2x)3−53)

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we know

= 𝑎3 − 𝑏3 = (𝑎 − 𝑏)(𝑎2 + 𝑏2 + 𝑎𝑏)

𝑎 = 2𝑥 , 𝑏 = 5

32𝑥4 − 500𝑥 = 4𝑥((2𝑥3) − 53)

= 4𝑥(2𝑥 − 5)(4𝑥2 + 25 + 10𝑥)

Question 17:

Factorize:

3a7b − 81a4b4

ANSWER:

3a7b − 81a4b4

= 3𝑎4𝑏 (𝑎3 − 27𝑏3)

= 3𝑎4𝑏[𝑎3 − (3𝑏)3]

= 3𝑎4𝑏(𝑎 − 3𝑏)[𝑎2 + 𝑎 × 3𝑏 + (3𝑏)2]

= 3𝑎4𝑏(𝑎 − 3𝑏)(𝑎2 + 3𝑎𝑏 + 9𝑏2)

Question 18:

Factorise

x4 y4 – xy

ANSWER:

Using the identity

= 𝑎3 − 𝑏3 = (𝑎 − 𝑏)(𝑎2 + 𝑏2 + 𝑎𝑏)

= x4 y4 – xy = xy (𝑥3𝑦3 − 1)

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= 𝑥𝑦 (𝑥𝑦 − 1)(𝑥2𝑦2 + 1 + 𝑥𝑦)

Question 19:

Factorise

8x2 y3 – x5

ANSWER:

8x2 y3 – x5

= 𝑥2(8𝑦3 − 𝑥3)

= 𝑥2(2𝑦 − 𝑥)(4𝑦2 + 𝑥2 + 2𝑥𝑦)

Question 20:

Factorise

1029 – 3x3

ANSWER:

1029 – 3x3

= 3(343− x3)

=3(73−x3)

=3(7−x)(49+𝑥2+7x)

Question 21:

Factorize:

x6 − 729

ANSWER:

x6 – 729 = (𝑥2)3 − (9 )3

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= [𝑥2 − 9] [(𝑥2)2 + 𝑥2 × 9 + 92]

= [𝑥2 − 32] (𝑥4 + 9𝑥2 + 81)

= (𝑥 + 3)(𝑥 − 3)(𝑥4 + 18𝑥2 + 81 − 9𝑥2)

= (𝑥 + 3)(𝑥 − 3)((𝑥2)2 + 2 × 𝑥2 × 9 + 92 − 9𝑥2)

= (𝑥 + 3)(𝑥 − 3)[(𝑥2 + 9)2 − (3𝑥)2]

= (𝑥 + 3)(𝑥 − 3)[(𝑥2 + 9 + 3𝑥)(𝑥2 + 9 − 3𝑥)]

= (𝑥 + 3)(𝑥 − 3)(𝑥2 + 3𝑥 + 9)(𝑥2 − 3𝑥 + 9)

Question 22:

Factorise

x9 – y9

ANSWER:

x9 – y9 = (𝑥3)3 − (𝑦3)3

We know :

𝑎3 − 𝑏3 = (𝑎 − 𝑏)(𝑎2 + 𝑏2 + 𝑎𝑏)

𝑎 = 𝑥3, 𝑏 = 𝑦3

So, x9 – y9 = (𝑥3)3 − (𝑦3)3 = (𝑥3 − 𝑦3)(𝑥6 + 𝑦6 + 𝑥3𝑦3)

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

Question 23:

Factorize:

(a + b)3 − (a − b)3

ANSWER:

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(a + b)3 − (a − b)3

= [(𝑎 + 𝑏) − (𝑎 − 𝑏)] [(𝑎 + 𝑏)2 + (𝑎 + 𝑏)(𝑎 − 𝑏) + (𝑎 − 𝑏)2]

= (𝑎 + 𝑏 − 𝑎 + 𝑏)[𝑎2 + 2𝑎𝑏 + 𝑏2 + 𝑎2 − 𝑏2 + 𝑎2 − 2𝑎𝑏 + 𝑏2]

= 2𝑏(3𝑎2 + 𝑏2)

Question 24:

Factorize:

8a3 − b3 − 4ax + 2bx

ANSWER:

8a3 − b3 − 4ax + 2bx = [(2𝑎)3 − (𝑏)3] − 2𝑥(2𝑎 − 𝑏)

= (2𝑎 − 𝑏)[(2𝑎)2 + 2𝑎𝑏 + 𝑏2]- 2𝑥 (2𝑎 − 𝑏)

= (2𝑎 − 𝑏)(4𝑎2 + 2𝑎𝑏 + 𝑏2 ) − 2𝑥(2𝑎 − 𝑏)

= (2a-b)(4𝑎2 + 2𝑎𝑏 + 𝑏2 − 2𝑥)

Question 25:

Factorize :

𝑎3 + 3𝑎2𝑏 + 3𝑎𝑏2 + 𝑏3 − 8

Answer:

𝑎3 + 3𝑎2𝑏 + 3𝑎𝑏2 + 𝑏3 − 8

= (𝑎3 + 𝑏3 + 3𝑎2𝑏 + 3𝑎𝑏2) − 8

= (𝑎 + 𝑏)3 − 23

= (𝑎 + 𝑏 − 2)[(𝑎 + 𝑏)2 + 2(𝑎 + 𝑏) + 22 ]

= (𝑎 + 𝑏 − 2) [(𝑎 + 𝑏)2 + 2(𝑎 + 𝑏) + 4]

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Question 26:

Factorize:

𝑎3 − 1

𝑎3− 2𝑎 +

2

𝑎

ANSWER:

𝑎3 − 1

𝑎3− 2𝑎 +

2

𝑎

= (𝑎3 − 1

𝑎3) − 2 (𝑎 −1

𝑎)

= [(𝑎3) − (1

𝑎)

3] − 2 (𝑎 −

1

𝑎)

= (𝑎 −1

𝑎) [𝑎2 × 𝑎 ×

1

𝑎+ (

1

𝑎)

2] − 2 (𝑎 −

1

𝑎)

= (𝑎 −1

𝑎) (𝑎2 + 1 +

1

𝑎2) − 2 (𝑎 −1

𝑎)

= (𝑎 −1

𝑎) (𝑎2 + 1 +

1

𝑎2 − 2)

= (𝑎 −1

𝑎) (𝑎2 − 1 +

1

𝑎2)

Question: 27

Factorize:

2a3 + 16b3 − 5a − 10b

ANSWER:

2a3 + 16b3 − 5a − 10b

= 2[𝑎3 + 8𝑏3] − 5(𝑎 + 2𝑏)

= 2[𝑎3 + (2𝑏)3] − 5(𝑎 + 2𝑏)

= 2(𝑎 + 2𝑏)[𝑎2 − 𝑎 × 2𝑏 + (2𝑏)2] − 5(𝑎 + 2𝑏)

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= 2(𝑎 + 2𝑏)(𝑎2 − 2𝑎𝑏 + 4𝑏2) − 5(𝑎 + 2𝑏)

= (𝑎 + 2𝑏)[2(𝑎2 − 2𝑎𝑏 + 4𝑏2) − 5]

Question 28:

Factorise

a6 + b6

ANSWER:

a6 + b6

= (𝑎2)3 + (𝑏2)3

= (𝑎2 + 𝑏2)[(𝑎2)2 − 𝑎2𝑏2 + (𝑏2)2]

= (𝑎2 + 𝑏2)(𝑎4 − 𝑎2𝑏2 + 𝑏4)

Question 29:

Factorise

a12 – b12

ANSWER:

a12 – b12

= (𝑎6 + 𝑏6) (𝑎6 − 𝑏6)

= [(𝑎2)3 + (𝑏2)3][(𝑎3)2 − (𝑏3)2]

= [(𝑎2 + 𝑏2)(𝑎4 + 𝑏4 − 𝑎2𝑏2)][(𝑎3 − 𝑏3)(𝑎3 + 𝑏3)]

= [(𝑎2 + 𝑏2)(𝑎4 + 𝑏4 − 𝑎2𝑏2)][(𝑎 − 𝑏)(𝑎2 + 𝑏2 + 𝑎𝑏)(𝑎 + 𝑏)(𝑎2 +

𝑏2 − 𝑎𝑏)]

= (𝑎 − 𝑏)(𝑎2 + 𝑏2 + 𝑎𝑏)(𝑎 + 𝑏)(𝑎2 + 𝑏2 − 𝑎𝑏)(𝑎2 + 𝑏2)(𝑎4 +

𝑏4 − 𝑎2𝑏2)

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Question 30:

Factorise

x6 – 7x3 – 8

ANSWER:

Let

𝑥3 = 𝑦

So, the equation becomes

𝑦2 − 7𝑦 − 8 = 𝑦2 − 8𝑦 + 𝑦 − 8

= 𝑦(𝑦 − 8) + (𝑦 − 8)

= (𝑦 − 8)(𝑦 + 1)

= (𝑥3 − 8)( 𝑥3 + 1)

= (𝑥 − 2)(𝑥2 + 4 + 2𝑥)(𝑥 + 1)(𝑥2 + 1 − 𝑥)

Question 31:

Factorise

x3 – 3x2 + 3x + 7

ANSWER:

x3 – 3x2 + 3x + 7

= 𝑥3 − 3𝑥2 + 3𝑥 + 7

= 𝑥3 − 3𝑥2 + 3𝑥 + 8 − 1

= 𝑥3 − 3𝑥2 + 3𝑥 − 1 + 8

= (𝑥3 − 3𝑥2 + 3𝑥 − 1) + 8

= (𝑥 − 1)3 + (2)3

= (𝑥 − 1 + 2)[(𝑥 − 1)2 + 4 − 2(𝑥 − 1)]

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= (𝑥 + 1)[𝑥2 + 1 − 2𝑥 + 4 − 2𝑥 + 2]

= (𝑥 + 1)(𝑥2 − 4𝑥 + 7)

Question 32:

Factorise

(𝑥 + 1)3 + (𝑥 − 1)3

ANSWER:

(𝑥 + 1)3 + (𝑥 − 1)3

= (𝑥 + 1 + 𝑥 − 1)[(𝑥 + 1)2 + (𝑥 − 1)2 − (𝑥 − 1)(𝑥 + 1)]

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

= 2𝑥(𝑥2 + 1 + 2𝑥 + 𝑥2 + 1 − 2𝑥 − 𝑥2 + 1)

= 2𝑥(𝑥2 + 3)

Question 33:

Factorise

(2a +1)3 + (a – 1)3

ANSWER:

(2a +1)3 + (a – 1)3

= (2𝑎 + 1 + 𝑎 − 1)[(2𝑎 + 1)2 + (𝑎 − 1)2 − (2𝑎 + 1)(𝑎 − 1)]

= (3𝑎)[4𝑎2 + 1 + 4𝑎 + 𝑎2 + 1 − 2𝑎 − 2𝑎2 + 2𝑎 − 𝑎 + 1]

= 3𝑎[3𝑎2 + 3𝑎 + 3]

= 9𝑎(𝑎2 + 𝑎 + 1)

Question 34:

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Factorise

8(𝑥 + 𝑦)3 − 27(𝑥 − 𝑦)3

Answer:

8(𝑥 + 𝑦)3 − 27(𝑥 − 𝑦)3

= [2(𝑥 + 𝑦)]3 − [3(𝑥 − 𝑦)]3

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

= (−𝑥 + 5𝑦)[4(𝑥2 + 𝑦2 + 2𝑥𝑦) + 9(𝑥2 + 𝑦2 − 2𝑥𝑦) + 6(𝑥2 − 𝑦2)]

= (−𝑥 + 5𝑦)[4𝑥2 + 4𝑦2 + 8𝑥𝑦 + 9𝑥2 + 9𝑦2 − 18𝑥𝑦 + 6𝑥2 − 6𝑦2]

= (−𝑥 + 5𝑦)(19𝑥2 + 7𝑦2 − 10𝑥𝑦)

Question 35:

Factorise

(𝑥 + 2)3 + (𝑥 − 2)3

Answer:

(𝑥 + 2)3 + (𝑥 − 2)3

= (𝑥 + 2 + 𝑥 − 2)[(𝑥 + 2)2 + (𝑥 − 2)2 − (𝑥2 − 4)]

= 2𝑥(𝑥2 + 4 + 4𝑥 + 𝑥2 + 4 − 4𝑥 − 𝑥2 + 4)

= 2𝑥(𝑥2 + 12)

Question 36:

Factorise

(𝑥 + 2)3 − (𝑥 − 2)3

Answer:

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(𝑥 + 2)3 − (𝑥 − 2)3

= 4((𝑥2 + 2(𝑥)(2) + 22) + 𝑥2 − 22 + (𝑥2 − 2(𝑥)(2) + 22)

= 4(𝑥2 + 4𝑥 + 4 + 𝑥2 − 4 + 𝑥2 − 4𝑥 + 4)

= 4(3𝑥2 + 4)

Question 37:

Prove that

0.85×0.85×0.85+0.15×0.15×0.15

0.85×0.85−0.85×0.15+0.15×0.15 =1.

ANSWER:

L.H.S.

0.85×0.85×0.85+0.15×0.15×0.15

0.85×0.85−0.85×0.15+0.15×0.15

= (0.85)3+(0.15)3

(0.85)2−0.85×0.15+(0.15)2

We know

𝑎3 + 𝑏3 = (𝑎 + 𝑏)(𝑎2 + 𝑏2 − 𝑎𝑏)

Here, 𝑎 = 0.85, 𝑏 = 0.15

= (0.85)3+(0.15)3

(0.85)2−0.85×0.15+(0.15)2

= (0.85+(0.15)((0.85)2−0.85×0.15+(0.15)2)

(0.85)2−0.85×0.15+ (0.15)2

= 0.85 + 0.15 = 1 ∶ 𝑅𝐻𝑆

Thus, LHS = RHS

Question 38:

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Prove that

59×59×59−9×9×9

59×59+59×9+9×9 =50.

ANSWER:

= 59×59×59−9×9×9

59×59+59×9+9×9

= (59)3−(9)3

592+59×9+9×9

We know

𝑎3 + 𝑏3 = (𝑎 + 𝑏)(𝑎2 + 𝑏2 + 𝑎𝑏)

Here, 𝑎 = 59, 𝑏 = 9

So, (59−9)(592+92+59×9)

592+ 92+59×9 = 59 - 9 = 50 : RHS

Thus, LHS = RHS

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Exercise 3.7

Page No: 136

Question 1:

Find the product:

(𝑥 + 𝑦 + 𝑧)(𝑥2 + 𝑦2 + 𝑧2 − 𝑥𝑦 + 𝑦𝑧 + 𝑧𝑥)

ANSWER:

(𝑥 + 𝑦 + 𝑧)(𝑥2 + 𝑦2 + 𝑧2 − 𝑥𝑦 + 𝑦𝑧 + 𝑧𝑥)

= [𝑥 + 𝑦 + (−𝑧)][𝑥2 + 𝑦2 + (−𝑧)2 − 𝑥𝑦 − 𝑦 × (−𝑧) − [−𝑧] × 𝑥]

= 𝑥3 + 𝑦3 + (−𝑧)3 − 3𝑥 × 𝑦 × (−𝑧)

= 𝑥3 + 𝑦3−𝑧3 + 3𝑥𝑦𝑧

Question 2:

Find the product:

(𝑥 − 𝑦 − 𝑧)( 𝑥2 + 𝑦2+𝑧2 + 𝑥𝑦 − 𝑦𝑧 + 𝑥𝑧)

ANSWER:

(𝑥 − 𝑦 − 𝑧)( 𝑥2 + 𝑦2+𝑧2 + 𝑥𝑦 − 𝑦𝑧 + 𝑥𝑧)

= (𝑥 + (−𝑦) + (−𝑧))( 𝑥2 + 𝑦2+𝑧2 + 𝑥𝑦 − 𝑦𝑧 + 𝑥𝑧)

We know

(𝑎 + 𝑏 + 𝑐)(𝑎2 + 𝑏2+𝑐2 − 𝑎𝑏 − 𝑏𝑐 − 𝑐𝑎)

= 𝑎3 + 𝑏3 + 𝑐3 − 3𝑎𝑏𝑐

Here, 𝑎 = 𝑥, 𝑏 = −𝑦, 𝑐 = −𝑧

(𝑥 + (−𝑦) + (−𝑧)) ( 𝑥2 + 𝑦2+𝑧2 + 𝑥𝑦 − 𝑦𝑧 + 𝑥𝑧)

= 𝑥3 + 𝑦3 + 𝑧3 − 3𝑥𝑦𝑧

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Question 3:

Find the product:

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

ANSWER:

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

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

= [𝑥 + (−2𝑦) + 3][𝑥2 + (−2𝑦)2 + (3)2 − 𝑥 × (−2𝑦) − (−2𝑦) × 3 −

3 × 𝑥]

= (𝑥)3 + (−2𝑦)3 + (3)3 − 3(𝑥)(−2𝑦)(3)

= 𝑥3 − 8𝑦3 + 27 + 18𝑥𝑦.

Question 4:

Find the product:

(3x – 5y + 4) (9x2 + 25y2 + 15xy − 20y + 12x + 16)

ANSWER:

(3x – 5y + 4) (9x2 + 25y2 + 15xy − 20y + 12x + 16)

= (3𝑥 + (−5𝑦) + 4) (9𝑥2 + 25𝑦2 + 16 + 15𝑥𝑦 − 20𝑦 + 12𝑥)

(𝑎 + 𝑏 + 𝑐)(𝑎2 + 𝑏2 + 𝑐2 − 𝑎𝑏 − 𝑏𝑐 − 𝑐𝑎)

= 𝑎3 + 𝑏3 + 𝑐3 − 3𝑎𝑏𝑐

Here, 𝑎 = 3𝑥, 𝑏 = −5𝑦, 𝑐 = 4

(3𝑥 + (−5𝑦) + 4)(9𝑥2 + 25𝑦2 + 16 + 15𝑥𝑦 − 20𝑦 + 12𝑥)

= (3𝑥)3 + (−5𝑦)3 + (4)3 − 3 × 3𝑥(−5𝑦)(4)

= 27𝑥3 − 125𝑦3 + 64 + 180𝑥𝑦

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Question 5:

Factorize:

125a3 + b3 + 64c3 − 60abc

ANSWER:

125a3 + b3 + 64c3 − 60abc

= (5𝑎)3 + (𝑏)3 + (4𝑐)3 − 3 × 5𝑎 × 𝑏 × 4𝑐

= (5𝑎 + 𝑏 + 4𝑐)[(5𝑎)2 + (𝑏)2 + (4𝑐)2 − 5𝑎 × 𝑏 − 𝑏 × 4𝑐 − 5𝑎 ×

4𝑐]

= (5𝑎 + 𝑏 + 4𝑐)(25𝑎2 + 𝑏2 + 16𝑐2 − 5𝑎𝑏 × 4𝑏𝑐 − 20𝑎𝑐)

Question 6:

Factorize:

a3 + 8b3 + 64c3 − 24abc

ANSWER:

a3 + 8b3 + 64c3 − 24abc

= 𝑎3 + (2𝑏)3 + (4𝑐)3 − 3 × 𝑎 × 2𝑏 × 4𝑐

= (𝑎 + 2𝑏 + 4𝑐)[𝑎2 + (2𝑏)2 + (4𝑐)2 − 𝑎 × 2𝑏 − 2𝑏 × 4𝑐 − 4𝑐 × 𝑎]

= (𝑎 + 2𝑏 + 4𝑐)[𝑎2 + 4𝑏2 + 16𝑐2 − 2𝑎𝑏 − 8𝑏𝑐 − 4𝑐𝑎]

Question 7:

Factorize:

1 + b3 + 8c3 − 6bc

ANSWER:

1 + b3 + 8c3 − 6bc

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= (1)3 + (𝑏)3 + (2𝑐)3 − 3 × 1 × 𝑏 × 2𝑐

= (1 + 𝑏 + 2𝑐)[12 + 𝑏2 + (2𝑐)2 − 1 × 𝑏 − 𝑏 × 2𝑐 − 1 × 2𝑐]

= (1 + 𝑏 + 2𝑐)(1 + 𝑏2 + 4𝑐2 − 𝑏 − 2𝑏𝑐 − 2𝑐)

Question 8:

Factorize:

216 + 27b3 + 8c3 − 108abc

ANSWER:

216 + 27b3 + 8c3 − 108abc

= (6)3 + (3𝑏)3 + (2𝑐)3 − 3 × 6 × 3𝑏 × 2𝑐

= (6 + 3𝑏 + 2𝑐)[62 + (3𝑏)2 + (2𝑐)2 − 6 × 3𝑏 − 3𝑏 × 2𝑐 − 2𝑐 × 6]

= (6 + 3𝑏 + 2𝑐)(36 + 9𝑏2 + 4𝑐2 − 18𝑏 − 6𝑏𝑐 − 12𝑐

Question 9:

Factorize:

27a3 − b3 + 8c3 + 18abc

ANSWER:

27a3 − b3 + 8c3 + 18abc

= (3𝑎)3 + (−𝑏)3 + (2𝑐)3 − 3 × (3𝑎) × (−𝑏) × (2𝑐)

= [3𝑎 + (−𝑏) + 2𝑐][(3𝑎)2 + (−𝑏)2 + (2𝑐)2 − 3𝑎(−𝑏) − (−𝑏)2𝑐 −

3𝑎 × 2𝑐]

= (3𝑎 − 𝑏 + 2𝑐)(9𝑎2 + 𝑏2 + 4𝑐2 + 3𝑎𝑏 + 2𝑏𝑐 − 6𝑎𝑐)

Question 10:

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Factorize:

8a3 + 125b3 − 64c3 + 120abc

ANSWER:

8a3 + 125b3 − 64c3 + 120abc

= (2𝑎)3 + (5𝑏)3 + (−4𝑐)3 − 3 × (2𝑎) × (5𝑏) × (−4𝑐)

= (2𝑎 + 5𝑏 − 4𝑐)[(2𝑎)2 + (5𝑏)2 + (−4𝑐)2 − (2𝑎)(5𝑏) −(5𝑏)(−4𝑐) − (2𝑎) × (−4𝑐)]

= (2𝑎 + 5𝑏 − 4𝑐)(4𝑎2 + 25𝑏2 + 16𝑐2 − 10𝑎𝑏 + 20𝑏𝑐 + 8𝑎𝑐)

Question 11:

Factorize:

8 − 27b3 − 343c3 − 126bc

ANSWER:

8 − 27b3 − 343c3 − 126bc

= (2)3+(−3𝑏)3 + (−7𝑐)3 − 3 × (2) × (−3𝑏) × (−7𝑐)

= [2 + (−3𝑏) + (−7𝑐)][(2)2 + (−3𝑏)2 + (−7𝑐)2 − (2)(−3𝑏) −(−3𝑏)(−7𝑐) − (2)(−7𝑐)]

= (2 − 3𝑏 − 7𝑐)( 4 + 9𝑏2+49𝑐2 + 6𝑏 − 21𝑏𝑐 + 14𝑐)

Question 12:

Factorize:

125 − 8x3 − 27y3 − 90xy

ANSWER:

125 − 8x3 − 27y3 − 90xy

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= 53 + (−2𝑥)3 + (−3𝑦)3 − 3 × 5 × (−2𝑥) × (−3𝑦)

= [5 + (−2𝑥) + (−3𝑦)][52 + (−2𝑥)2 + (−3𝑦)2 − 5 × (−2𝑥) −(−2𝑥) − (−2𝑥)9 − 3𝑦) − 5 × (−3𝑦)]

= (5 − 2𝑥 − 3𝑦) (25 + 4𝑥2 + 9𝑦2 + 10𝑥 − 6𝑥𝑦 + 15𝑦)

Question 13:

Factorize:

2√2𝑎3 + 16√2𝑏3 + 𝑐3 − 12𝑎𝑏𝑐

ANSWER:

2√2𝑎3 + 16√2𝑏3 + 𝑐3 − 12𝑎𝑏𝑐

= (√2𝑎)3

+ (2√2𝑏)3

+ 𝑐3 − 3 × (√2𝑎) × (2√2𝑏) × (𝑐)

= (√2𝑎 + 2√2𝑏 + 𝑐)(√2𝑎)2

+ (2√2𝑏)2

+ 𝑐2 − (√2𝑎) × (2√2𝑏) −

(2√2𝑏) × (𝑐) − (√2𝑎) × (𝑐)

= (√2𝑎 + 2√2𝑏 + 𝑐)(2𝑎2 + 8𝑏2 + 𝑐2 − 4𝑎𝑏 − 2√2𝑏𝑐 − √2𝑎𝑐 )

Question 14:

Factorise:

27x3 – y3 – z3 – 9xyz

ANSWER:

27x3 – y3 – z3 – 9xyz

= (3𝑥)3 − 𝑦3 − 𝑧3 − 3 × (3𝑥) × (−𝑦) × (−𝑧)

We know,

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𝑎3 + 𝑏3 + 𝑐3 − 3𝑎𝑏𝑐 = ( 𝑎 + 𝑏 + 𝑐)(𝑎2 + 𝑏2 + 𝑐2 − 𝑎𝑏 − 𝑏𝑐 − 𝑐𝑎)

𝑎 = 3𝑥, 𝑏 = −𝑦, 𝑐 = −𝑧

(3𝑥)3 − 𝑦3 − 𝑧3 − 3 × (3𝑥) × (−𝑦) × (−𝑧)

= (3𝑥 − 𝑦 − 𝑧)(9𝑥2 + 𝑦2 + 𝑧2 + 3𝑥𝑦 − 𝑦𝑧 + 3𝑥𝑧)

Question 15:

Factorise:

2√2𝑎3 + 3√3𝑏3 + 𝑐3 − 3√6𝑎𝑏𝑐

ANSWER:

2√2𝑎3 + 3√3𝑏3 + 𝑐3 − 3√6𝑎𝑏𝑐

= (√2𝑎)3+(√3𝑏)3 + 𝑐3 − 3(√2𝑎)(√3𝑏)𝑐

We know:

𝑥3 + 𝑦3 + 𝑧3-3𝑥𝑦𝑧

= (𝑥 + 𝑦 + 𝑧)(𝑥2 + 𝑦2 + 𝑧2 − 𝑥𝑦 − 𝑦𝑧 − 𝑧𝑥)

𝑥 = √2𝑎, 𝑦 = √3𝑏, 𝑧 = 𝑐

= (√2𝑎)3+(√3𝑏)3 + 𝑐3 − 3(√2𝑎)(√3𝑏)𝑐

= (√2𝑎 + √3𝑏 + 𝑐 )(2𝑎2 + 3𝑏2 + 𝑐2 − √6𝑎𝑏 − √3𝑏𝑐 − √2 𝑎𝑐)

Question 16:

Factorise:

3√3 𝑎3 − 𝑏3 − 5√5 𝑐3 − 3√15 𝑎𝑏𝑐

ANSWER:

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3√3 𝑎3 − 𝑏3 − 5√5 𝑐3 − 3√15 𝑎𝑏𝑐

= (√3𝑎)3+(−𝑏)3 + (√5𝑐)3 − 3(√3𝑎)(−𝑏)(−√5 𝑐)

We know

𝑥3 + 𝑦3 + 𝑧3-3𝑥𝑦𝑧

= (𝑥 + 𝑦 + 𝑧)(𝑥2 + 𝑦2 + 𝑧2 − 𝑥𝑦 − 𝑦𝑧 − 𝑧𝑥)

Here, 𝑥 = (√3𝑎), 𝑦 = (−𝑏), 𝑧 = (−√5 c)

3√3 𝑎3 − 𝑏3 − 5√5 𝑐3 − 3√15 𝑎𝑏𝑐

= (√3𝑎)3+(−𝑏)3 + (√5𝑐)3 − 3(√3𝑎)(−𝑏)(−√5 𝑐)

= (√3𝑎 − 𝑏 − √5𝑐)(3𝑎2 + 𝑏2 + 5𝑐2 + √3𝑎𝑏 − √5𝑏𝑐 + √15𝑐)

Question 17:

Factorize:

(a − b)3 + (b − c)3 + (c − a)3

ANSWER:

(a − b)3 + (b − c)3 + (c − a)3

Putting (a−b)=x, (b−c)=y and (c−a)=z, we get:

(a − b)3 + (b − c)3 + (c − a)3

[Where (x+y+z)=(a−b)+(b−c)+(c−a)=0]

=3xyz

[(x+y+z)=0 ⇒x3+y3+z3=3xyz]

=3(a−b)(b−c)(c−a)

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Question 18:

Factorise:

(𝑎 − 3𝑏)3 + (3𝑏 − 𝑐)3 + (𝑐 − 𝑎)3

ANSWER:

We know

𝑥3 + 𝑦3 + 𝑧3-3𝑥𝑦𝑧

= (𝑥 + 𝑦 + 𝑧)(𝑥2 + 𝑦2 + 𝑧2 − 𝑥𝑦 − 𝑦𝑧 − 𝑧𝑥)

𝑥3 + 𝑦3 + 𝑧3 = (𝑥 + 𝑦 + 𝑧)(𝑥2 + 𝑦2 + 𝑧2 − 𝑥𝑦 − 𝑦𝑧 − 𝑧𝑥) + 3𝑥𝑦𝑧

Here, 𝑥 = (𝑎 − 3𝑏), 𝑦 = (3𝑏 − 𝑐), 𝑧 = (𝑐 − 𝑎)

(𝑎 − 3𝑏)3 + (3𝑏 − 𝑐)3 + (𝑐 − 𝑎)3

= (𝑎 − 3𝑏 + 3𝑏 − 𝑐 + 𝑐 − 𝑎)[(𝑎 − 3𝑏)2 + (3𝑏 − 𝑐)2 + (𝑐 − 𝑎)2 −(𝑎 − 3𝑏)(3𝑏 − 𝑐) − (3𝑏 − 𝑐)(𝑐 − 𝑎) − (𝑐 − 𝑎)(𝑎 − 3𝑏)] +

3(𝑎 − 3𝑏)(3𝑏 − 𝑐)(𝑐 − 𝑎)

= 0 + 3(𝑎 − 3𝑏)(3𝑏 − 𝑐)(𝑐 − 𝑎)

= 3(𝑎 − 3𝑏)(3𝑏 − 𝑐)(𝑐 − 𝑎)

Question 19:

Factorize:

(3a − 2b)3 + (2b − 5c)3 + (5c − 3a)3

ANSWER:

Put (3a−2b)=x, (2b−5c)=y and (5c−3a)=z.

We have:

𝑥 + 𝑦 + 𝑧 = 3𝑎 − 2𝑏 + 2𝑏 − 5𝑐 + 5𝑐 − 3𝑎 = 0

Now, (3a − 2b)3 + (2b − 5c)3 + (5c − 3a)3 = 𝑥3 + 𝑦3 + 𝑧3

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=3𝑥𝑦𝑧

[Here, 𝑥 + 𝑦 + 𝑧 = 0. So, 𝑥3 + 𝑦3 + 𝑧3= 3xyz]

=3(3a−2b)(2b−5c)(5c−3a)

Question 20:

Factorize:

(5a − 7b)3 + (9c − 5a)3 + (7b − 9c)3

ANSWER:

Put

(5𝑎 − 7𝑏) = 𝑥, (9𝑐 − 5𝑎) = 𝑧 𝑎𝑛𝑑 (7𝑏 − 9𝑐) = 𝑦.

Here,

𝑥 + 𝑦 + 𝑧 = 5𝑎 − 7𝑏 + 9𝑐 − 5𝑎 + 7𝑏 − 9𝑐 = 0

Thus, we have:

(5a − 7b)3 + (9c − 5a)3 + (7b − 9c)3 = 𝑥3 + 𝑧3 + 𝑦3

=3xzy [When 𝑥 + 𝑦 + 𝑧 =0, 𝑥3 + 𝑧3 + 𝑦3 = 3xyz.]

=3 (5a−7b)(9c−5a)(7b−9c)

Question 21:

Factorize:

a3(b − c)3 + b3(c − a)3 + c3(a − b)3

ANSWER:

We have:

a3(b − c)3 + b3(c − a)3 + c3(a − b)3 = [𝑎(𝑏 − 𝑐)]3 + [𝑏(𝑐 − 𝑎)]3 +

[𝑐(𝑎 − 𝑏)]3

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Put

a(b−c) = x

b(c−a) = y

c(a−b) = z

Here,

𝑥 + 𝑦 + 𝑧 = 𝑎(𝑏 − 𝑐) + 𝑏(𝑐 − 𝑎) + 𝑐(𝑎 − 𝑏)

= 𝑎𝑏 − 𝑎𝑐 + 𝑏𝑐 − 𝑎𝑏 + 𝑎𝑐 − 𝑏𝑐 = 0

Thus, we have:

a3(b − c)3 + b3(c − a)3 + c3(a − b)3 = 𝑥3 + 𝑦3 + 𝑧3 = 3𝑥𝑦𝑧

[When 𝑥 + 𝑦 + 𝑧 =0, 𝑥3 + 𝑦3 + 𝑧3 =3xyz.]

=3a(b−c)b(c−a)c(a−b)

=3abc(a−b)(b−c)(c−a)

Question 22:

Evaluate

(i) (–12)3 + 73 + 53

(ii) (28)3 + (–15)3 + (–13)3

ANSWER:

(i) (–12)3 + 73 + 53

We know

𝑥3 + 𝑦3 + 𝑧3-3𝑥𝑦𝑧 = (𝑥 + 𝑦 + 𝑧)(𝑥2 + 𝑦2 + 𝑧2 − 𝑥𝑦 − 𝑦𝑧 − 𝑧𝑥)


Here,

𝑥 = (−12), 𝑦 = 7, 𝑧 = 5

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(–12)3 + 73 + 53

= (−12 + 7 + 5)[(−12)2 + (7)2 + (5)2 − 7(−12) − 35 + 60] + 3 ×(−12) × 35

= 0 – 1260 = -1260

(ii) (28)3 + (–15)3 + (–13)3

We know

𝑥3 + 𝑦3 + 𝑧3-3𝑥𝑦𝑧 = (𝑥 + 𝑦 + 𝑧)(𝑥2 + 𝑦2 + 𝑧2 − 𝑥𝑦 − 𝑦𝑧 − 𝑧𝑥)


Here,

𝑥 = (−28) , 𝑦 = −15, 𝑧 = −13

(–28)3 + (-15)3 + (-13)3 = (28 − 15 − 13)[(28)2 + (−15)2 + (−13)2 −

28 (−15)— 15)(−13) − 28 (−13) + 3 × 28(−15)(−13)

= 0 + 16380 = 16380

Question 23:

Prove that

(𝑎 + 𝑏 + 𝑐)3 − 𝑎3 − 𝑏3 − 𝑐3 = 3(𝑎 + 𝑏)(𝑏 + 𝑐)(𝑐 + 𝑎)

ANSWER:

(𝑎 + 𝑏 + 𝑐)3 = [(𝑎 + 𝑏) + 𝑐]3= (𝑎 + 𝑏)3+ 𝑐3 + 3(𝑎 + 𝑏)𝑐(𝑎 + 𝑏 + 𝑐)

⟹ (𝑎 + 𝑏 + 𝑐)3 = 𝑎3 + 𝑏3 + 3𝑎𝑏(𝑎 + 𝑏) + 𝑐3 + 3(𝑎 + 𝑏)𝑐(𝑎 + 𝑏 +

𝐶)

⟹ (𝑎 + 𝑏 + 𝑐)3 − 𝑎3 + 𝑏3 + 𝑐3 = 3𝑎𝑏(𝑎 + 𝑏) + 3(𝑎 + 𝑏)𝑐(𝑎 + 𝑏 +

𝐶)

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⟹ (𝑎 + 𝑏 + 𝑐)3 − 𝑎3 + 𝑏3 + 𝑐3 = 3(𝑎 + 𝑏)[𝑎𝑏 + 𝑐𝑎 + 𝑐𝑏 + 𝑐2]

⟹ (𝑎 + 𝑏 + 𝑐)3 − 𝑎3 + 𝑏3 + 𝑐3 = 3(𝑎 + 𝑏)(𝑏 + 𝑐)(𝑎 + 𝑐)

Question 24:

If a, b, c are all nonzero and a + b + c = 0, prove that 𝑎2

𝑏𝑐 +

𝑏2

𝑐𝑎 +

𝑐2

𝑎𝑏 =3.

ANSWER:

𝑎 + 𝑏 + 𝑐 = 𝑎3 + 𝑏3 + 𝑐3 = 3𝑎𝑏𝑐

Thus, we have:

𝑎2

𝑏𝑐 +

𝑏2

𝑐𝑎 +

𝑐2

𝑎𝑏 =

𝑎3+𝑏3+𝑐3

𝑎𝑏𝑐

= 3𝑎𝑏𝑐

𝑎𝑏𝑐

=3

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CCE Test Paper – 3

Question 1:

If a + b + c = 9 and a2 + b2 + c2 = 35, find the value of (a3 + b3 + c3 –

3abc).

ANSWER:

a + b + c = 9

⇒ (𝑎 + 𝑏 + 𝑐)2 =92=81

⇒ a2 + b2 + c2 +2(𝑎𝑏 + 𝑏𝑐 + 𝑐𝑎) = 81

⇒ 35 + 2𝑎𝑏 + 𝑏𝑐 + 𝑐𝑎 = 81

⇒ (𝑎𝑏 + 𝑏𝑐 + 𝑐𝑎) = 23

We know,

(a3 + b3 + c3 – 3abc) = (𝑎 + 𝑏 + 𝑐)(𝑎2 + 𝑏2 + 𝑐2 − 𝑎𝑏 − 𝑏𝑐 − 𝑐𝑎)

=(9)(35−23)

=108

Question 2:

If (x + 1) is factor of the polynomial (2x2 + kx) then the value of k is

(a) –2

(b) –3

(c) 2

(d) 3

ANSWER:

(c) 2

(x+1) is a factor of 2x2 + kx.

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So, −1 is a zero of 2x2 + kx.

Thus, we have:

2×(−1)2+k×(−1)=0

⇒2−k=0

⇒k=2

Question 3:

The value of (249)2 – (248)2 is

(a) 12

(b) 477

(c) 487

(d) 497

ANSWER:

(249)2 – (248)2

We know

𝑎2 − 𝑏2 =(𝑎 + 𝑏)(𝑎 − 𝑏)

So,

(249)2 – (248)2

= (249 − 248)(249 + 248)

= 497

Hence, the correct answer is option (d).

Question 4:

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If 𝑥

𝑦+

𝑦

𝑥 =−1 Where, xy+yx=-1, where x ≠ 0 and y ≠ 0, then the value

of (x3 − y3) is

(a) 1

(b) −1

(c) 0

(d) 1

2

ANSWER:

(c) 0 𝑥

𝑦+

𝑦

𝑥 =−1

⇒ 𝑥2+𝑦2

𝑥𝑦 = -1

⇒x2 + y2 = −xy

⇒ x2 + y2 + xy = 0

Thus, we have:

(x3 − y3) =(x−y)(𝑥2+ 𝑦2 +xy)

=(x−y)×0

=0

Question 5:

If a + b + c = 0, then a3 + b3 + c3 = ?

(a) 0

(b) abc

(c) 2abc

(d) 3abc

ANSWER:

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(d) 3abc

𝑎 + 𝑏 + 𝑐 = 0

⟹ 𝑎 + 𝑏 = −𝑐

⟹ (𝑎 + 𝑏)3 = (−𝑐)3

⟹ 𝑎3+𝑏3 + 3𝑎𝑏(𝑎 + 𝑏) = (−𝑐)3

⟹ 𝑎3+𝑏3 + 3𝑎𝑏(-c) = (−𝑐)3

⟹ 𝑎3+𝑏3 + 𝑐3 = 3𝑎𝑏𝑐

Question 6:

If (3𝑥 +1

2) (3𝑥 −

1

2) = 9𝑥2 − 𝑝 then the value of p is

(a) 0

(b) −1

4

(c) 1

4

(d) 1

2

ANSWER:

(3𝑥 +1

2) (3𝑥 −

1

2) = 9𝑥2 − 𝑝

= 9𝑥2 − 1

4 = 9𝑥2 − 𝑝

(∵𝑎2+𝑏2) = (𝑎 + 𝑏)(𝑎 − 𝑏)

⇒ p = 1

4

Hence, the correct answer is option (c).

Question 7:

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The coefficient of x in the expansion of (x + 3)3 is

(a) 1

(b) 9

(c) 18

(d) 27

ANSWER:

(x + 3)3

= 𝑥3 + 33 + 9𝑥(𝑥 + 3)

= 𝑥3 + 27 + 9𝑥2 + 27

So, the coefficient of x in (x + 3)3 is 27.


Question 8:

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

(a) x2 + y2 + 2xy

(b) x2 + y2 – xy

(c) xy2

(d) 3xy

ANSWER:

(x + y)3 – (x3 + y3)

= 𝑥3 + 𝑦3 + 3𝑥𝑦(𝑥 + 𝑦) − (𝑥3 + 𝑦3)

= 3𝑥𝑦 (𝑥 + 𝑦)

Thus, the factors of (x + y)3 – (x3 + y3) are 3xy and (x + y).


Question 9:

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

(a) 5 + x

(b) 5 – x

(c) 5x – 1

(d) 10x

ANSWER:

(25𝑥2 − 1) + (1 + 5𝑥)2

= (5x−1)(5x+1)+ (1 + 5𝑥)2

= (5𝑥 + 1)(5𝑥 − 1)+(1+5x)

= (5𝑥 + 1)(10𝑥)

So, the factors of (25𝑥2 − 1) + (1 + 5𝑥)2 are (5x + 1) and 10x


Question 10:

If (x + 5) is a factor of p(x) = x3 − 20x + 5k, then k = ?

(a) −5

(b) 5

(c) 3

(d) −3

ANSWER:

(b) 5

(x+5) is a factor of p(x)= x3 − 20x + 5k.

∴ p(−5)=0

⇒(−5)3−20×(−5)+5k=0

⇒−125+100+5k=0

⇒5k=25

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⇒k=5

Question 11:

If (x + 2) and (x − 1) are factors of (x3 + 10x2 + mx + n), then

(a) m = 5, n = −3

(b) m = 7, n = −18

(c) m = 17, n = −8

(d) m = 23, n = −19

ANSWER:

(b) m = 7, n = −18

Let:

p(x)= x3 + 10x2 + mx + n,

Now,

x+2 =0

⟹ 𝑥 = −2,

(x + 2) is a factor of p(x).

So, we have p(-2)=0

⇒(−2)3+10×(−2)2+m×(−2)+n=0

⇒ −8 + 40 −2m + n=0

⇒32 −2m + n=0

⇒2m – n =32 .....(i)

Now,

𝑥 − 1 = 0

⟹ 𝑥 = 1

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Also,

(x − 1) is a factor of p(x).

We have:

p(1) = 0

⇒ 13 + 10 × 12 + 𝑚 × 1 + 𝑛 = 0

⇒ 1 + 10 + 𝑚 + 𝑛 = 0

⇒ 11 + 𝑚 + 𝑛 = 0

⇒ 𝑚 + 𝑛 = −11 .....(ii)

From (i) and (ii),

we get:

3m=21

⇒m=7

By substituting the value of m in (i),

we get n = −18.

∴ m = 7 and n = −18

Question 12:

104 × 96 = ?

(a) 9864

(b) 9984

(c) 9684

(d) 9884

ANSWER:

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(b) 9984

104×96=(100+4)(100−4)

=1002−42

=(10000−16)

=9984

Question 13:

305 × 308 = ?

(a) 94940

(b) 93840

(c) 93940

(d) 94840

ANSWER:

(c) 93940

305×308 =(300+5)×(300+8)

=(300)2+300×(5+8)+5×8

=90000+3900+40

=93940

Question 14:

207 × 193 = ?

(a) 39851

(b) 39951

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(c) 39961

(d) 38951

ANSWER:

(b) 39951

207×193

=(200+7)(200−7)

=(200)2−(7)2

=40000−49

=39951

Question 15:

4a2 + b2 + 4ab + 8a + 4b + 4 = ?

(a) (2a + b + 2)2

(b) (2a − b + 2)2

(c) (a + 2b + 2)2

(d) none of these

ANSWER:

(a) (2a + b + 2)2

4a2 + b2 + 4ab + 8a + 4b + 4

= 4a2 + b2 +4+ 4ab + 4b + 8a

= (2𝑎)2+ b2 + 22 + 2 × 2𝑎 × 𝑏 + 2 × 𝑏 × 2 + 2 × 2𝑎 × 2

= (2𝑎 + 𝑏 + 2)2

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Question 16:

𝑥2 − 4𝑥 − 21 = ?

(a) (x − 7)(x − 3)

(b) (x + 7)(x − 3)

(c) (x − 7)(x + 3)

(d) none of these

ANSWER:

(c) (x − 7)(x + 3)

𝑥2 − 4𝑥 − 21

= 𝑥2 − 7𝑥 + 3𝑥 − 21

= 𝑥(𝑥 − 7) + 3(𝑥 − 7)

= (𝑥 − 7)(𝑥 + 3)

Question 17:

(4x2 + 4x − 3) = ?

(a) (2x − 1) (2x − 3)

(b) (2x + 1) (2x − 3)

(c) (2x + 3) (2x − 1)

(d) none of these

ANSWER:

(c) (2x + 3) (2x − 1)

4x2 + 4x – 3 = 4x2 +6𝑥 − 2𝑥 − 3

= 2𝑥(2𝑥 + 3) − 1(2𝑥 + 3)

= (2𝑥 + 3)(2𝑥 − 1)

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Question 18:

6x2 + 17x + 5 = ?

(a) (2x + 1)(3x + 5)

(b) (2x + 5)(3x + 1)

(c) (6x + 5)(x + 1)

(d) none of these

ANSWER:

(b) (2x + 5)(3x + 1)

6x2 + 17x + 5 = 6𝑥2 + 5x + 2x + 5

= 3𝑥(2𝑥 + 5) + 1(2𝑥 + 5)

= (2𝑥 + 5)(3𝑥 + 1)

Question 19:

(x + 1) is a factor of the polynomial

(a) x3 − 2x2 + x + 2

(b) x3 + 2x2 + x − 2

(c) x3 − 2x2 − x − 2

(d) x3 − 2x2 − x + 2

ANSWER:

(c) x3 − 2x2 − x − 2

Let:

f(x)= x3 − 2x2 − x − 2

By the factor theorem, (x + 1) will be a factor of f (x) if f (−1) = 0.

We have:

f(−1)= (−1)3 − 2 × (−1)2 + (−1) + 2

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= −1 − 2 − 1 + 2

= -2≠ 0

Hence, (x + 1) is not a factor of

f(x)= x3 − 2x2 − x − 2

Now,

Let:

f(x)= x3 − 2x2 − x − 2


We have:

f(−1)=(−1)3+2×(−1)2+(−1)−2

=−1+2−1−2

=−2 ≠ 0

Hence, (x + 1) is not a factor of f(x)= x3 − 2x2 + x − 2

Now,

Let:

f(x) = x3 − 2x2 − x − 2


We have:

f(−1)=(−1)3+2×(−1)2−(−1)−2

= −1+2+1−2

=0

Hence, (x + 1) is a factor of f(x)= x3 − 2x2 − x – 2.

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Now,

Let:

f(x)= x3 − 2x2 − x + 2


We have:

f(−1)=(−1)3 +2×(−1)2 −(−1)−2

= −1+2+1−2

=0

Hence, (x + 1) is a factor of f(x)= x3 − 2x2 − x + 2.

Question 20:

3x3 + 2x2 + 3x + 2 = ?

(a) (3x − 2)(x2 − 1)

(b) (3x − 2)(x2 + 1)

(c) (3x + 2)(x2 − 1)

(d) (3x + 2)(x2 + 1)

ANSWER:

(d) (3x + 2)(x2 + 1)

3x3 + 2x2 + 3x + 2

= 𝑥2(3x + 2) + 1(3x + 2)

=(3x+2)(𝑥2+1)

Question 21:

If a + b + c = 0, then (𝑎2

𝑏𝑐+

𝑏2

𝑐𝑎+

𝑎2

𝑎𝑏 ) = ?

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ANSWER:

(d) 3

𝑎 + 𝑏 + 𝑐 = 0

⇒ 𝑎3 + 𝑏3 + 𝑐3 = 3𝑎𝑏𝑐

Thus, we have:

(𝑎2

𝑏𝑐+

𝑏2

𝑐𝑎+

𝑎2

𝑎𝑏 ) =

𝑎3+𝑏3+𝑐3

3𝑎𝑏𝑐

= 3𝑎𝑏𝑐

𝑎𝑏𝑐

= 3

Question 22:

If x + y + z = 9 and xy + yz + zx = 23, then the value of (x3 + y3 + z3 −

3xyz) = ?

(a) 108

(b) 207

(c) 669

(d) 729

ANSWER:

(a) 108

𝑥3 + 𝑦3 + 𝑧3 = 3𝑥𝑦𝑧

= ( 𝑥2 + 𝑦2 + 𝑧2 − 𝑥𝑦 − 𝑦𝑧 − 𝑧𝑥)

= (𝑥 + 𝑦 + 𝑧)[(𝑥 + 𝑦 + 𝑧)2 − 3(𝑥𝑦 + 𝑦𝑧 + 𝑧𝑥)]

=9×(81−3×23)

=9×12

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=108

Question 23:

If 𝑎

𝑏+

𝑏

𝑎 = −1 then (a3 − b3) = ?

(a) −3

(b) −2

(c) −1

(d) 0

ANSWER:

𝑎

𝑏+

𝑏

𝑎 = −1

⇒ 𝑎2+𝑏2

𝑎𝑏 =-1

⇒a2 + b2 = − ab

⇒ a2 + b2 + ab = 0

Thus, we have:

(a3 − b3) = (𝑎 − 𝑏)(𝑎2 + 𝑏2 + 𝑎𝑏)

=(𝑎 − 𝑏) × 0

= 0

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Chapter 3 Factorisation-of-polynomials

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