R S AGGARWAL AND V AGGARWAL Solutions for Class 9 Maths Chapter 3 - Factorisation of Polynomials

Chapter 3 - Factorisation of Polynomials Exercise Ex. 3A

Question 1

Factorise:

9x2 + 12xy

Solution 1

9x2 + 12xy = 3x (3x + 4y)

Question 2

Factorise:

x3+ 2x2 + 5x + 10

Solution 2

x3+ 2x2 + 5x + 10

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

= (x2 + 5) (x + 2)

Question 3

Factorise:

x2 + xy - 2xz - 2yz

Solution 3

x2 + xy - 2xz - 2yz

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

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

Question 4

Factorise:

a3b - a2b + 5ab - 5b

Solution 4

a3b - a2b + 5ab - 5b

= a2b (a - 1) + 5b (a - 1)

= (a - 1) (a2b + 5b)

= (a - 1) b (a2 + 5)

= b (a - 1) (a2 + 5)

Question 5

Factorise:

8 - 4a - 2a3 + a4

Solution 5

8 - 4a - 2a3 + a4

= 4(2 - a) - a3 (2 - a)

= (2 - a) (4 - a3)

Question 6

Factorise:

x3 - 2x2y + 3xy2 - 6y3

Solution 6

x3 - 2x2y + 3xy2 - 6y3

= x2 (x - 2y) + 3y2 (x - 2y)

= (x - 2y) (x2 + 3y2)

Question 7

Factorise:

px - 5q + pq - 5x

Solution 7

px + pq - 5q - 5x

= p(x + q) - 5 (q + x)

= (x + q) (p - 5)

Question 8

Factorise:

x2 + y - xy - x

Solution 8

x2 - xy + y - x

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

= (x - y) (x - 1)

Question 9

Factorise:

(3a - 1)2 - 6a + 2

Solution 9

(3a - 1)2 - 6a + 2

= (3a - 1)2 - 2 (3a - 1)

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

= (3a - 1) (3a - 3)

= 3(3a - 1) (a - 1)

Question 10

Factorise:

(2x - 3)2 - 8x + 12

Solution 10

(2x - 3)2 - 8x + 12

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

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

= (2x - 3) (2x - 7)

Question 11

Factorise:

a3 + a - 3a2 - 3

Solution 11

a3 + a - 3a2 - 3

= a(a2 + 1) - 3 (a2 + 1)

= (a - 3) (a2 + 1)

Question 12

Factorise:

18x2y - 24xyz

Solution 12

18x2y - 24xyz = 6xy (3x - 4z)

Question 13

Factorise:

3ax - 6ay - 8by + 4bx

Solution 13

3ax - 6ay - 8by + 4bx

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

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

Question 14

Factorise:

abx2 + a2x + b2x +ab

Solution 14

abx2 + a2x + b2x +ab

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

= (bx + a) (ax + b)

Question 15

Factorise:

x3 - x2 + ax + x - a - 1

Solution 15

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 16

Factorise:

2x + 4y - 8xy - 1

Solution 16

2x + 4y - 8xy - 1

= 2x - 1 - 8xy + 4y

= (2x - 1) - 4y (2x - 1)

= (2x - 1) (1 - 4y)

Question 17

Factorise:

ab (x2 + y2) - xy (a2 + b2)

Solution 17

ab (x2 + y2) - xy (a2 + b2)

= abx2 + aby2 - a2xy - b2xy

= abx2 - a2xy + aby2 - b2xy

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

= (bx - ay) (ax - by)

Question 18

Factorise:

a2 + ab (b + 1) + b3

Solution 18

a2 + ab (b + 1) + b3

= a2 + ab2 + ab + b3

= a2 + ab + ab2 + b3

= a (a + b) + b2 (a + b)

= (a + b) (a + b2)

Question 19

Factorise:

a3 + ab (1 - 2a) - 2b2

Solution 19

a3 + ab (1 - 2a) - 2b2

= a3 + ab - 2a2b - 2b2

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

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

Question 20

Factorise:

2a2 + bc - 2ab - ac

Solution 20

2a2 + bc - 2ab - ac

= 2a2 - 2ab - ac + bc

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

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

Question 21

Factorise:

(ax + by)2 + (bx - ay)2

Solution 21

(ax + by)2 + (bx - ay)2

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

= a2x2 + b2y2 + b2x2 + a2y2

= a2x2 + b2x2 + b2y2 + a2y2

= x2 (a2 + b2) + y2(a2 + b2)

= (a2 + b2) (x2 + y2)

Question 22

Factorise:

a (a + b - c) - bc

Solution 22

a (a + b - c) - bc

= a2 + ab - ac - bc

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

= (a - c) (a + b)

Question 23

Factorise:

27a3b3 - 45a4b2

Solution 23

27a3b3 - 45a4b2 = 9a3b2 (3b - 5a)

Question 24

Factorise:

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

Solution 24

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

= a2 - 2ab - ac + 2bc

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

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

Question 25

Factorise:

a2x2 + (ax2 + 1)x + a

Solution 25

a2x2 + (ax2 + 1)x + a

= a2x2 + ax3 + x + a

= ax2 (a + x) + 1 (x + a)

= (ax2 + 1) (a + x)

Question 26

Factorise:

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

Solution 26

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

= abx2 + ab + a2x + b2x

= abx2 + a2x + ab + b2x

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

= (bx + a) (ax + b)

Question 27

Factorise:

x2 - (a + b) x+ ab

Solution 27

x2 - (a + b) x+ ab

= x2 - ax - bx + ab

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

= (x - a) (x - b)

Question 28

Factorise:

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Solution 28

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Question 29

Factorise:

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

Solution 29

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

Question 30

Factorise:

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

Solution 30

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

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

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

Question 31

Factorise:

x (a - 5) + y (5 - a)

Solution 31

x (a - 5) + y (5 - a)

= x (a - 5) + y (-1) (a - 5)

= (x - y) (a - 5)

Question 32

Factorise:

4 (a + b) - 6 (a + b)2

Solution 32

4 (a + b) - 6 (a + b)2

= (a + b) [4 - 6 (a + b)]

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

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

Question 33

Factorise:

8 (3a - 2b)2 - 10 (3a - 2b)

Solution 33

8 (3a - 2b)2 - 10 (3a - 2b)

= (3a - 2b) [8(3a - 2b) - 10]

= (3a - 2b) 2[4 (3a - 2b) - 5]

= 2 (3a - 2b) (12 a - 8b - 5)

Question 34

Factorise:

x (x + y)3 - 3x2y (x + y)

Solution 34

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)

Chapter 3 - Factorisation of Polynomials Exercise Ex. 3B

Question 1

Factorise:

9x2 - 16y2

Solution 1

9x2 - 16y2

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

= (3x + 4y)(3x - 4y) 

Question 2

Factorise:

8ab2 - 18a3

Solution 2

8ab2 - 18a3

= 2a (4b2 - 9a2)

= 2a [(2b)2 - (3a)2]

= 2a (2b + 3a) (2b - 3a)R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Question 3

Factorise:

150 - 6x2

Solution 3

150 - 6x2

= 6 (25 - x2)

= 6 (52 - x2)

= 6 (5 + x) (5 - x)R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Question 4

Factorise:

2 - 50x2

Solution 4

2 - 50x2

= 2 (1 - 25x2)

= 2 [(1)2 - (5x)2]

= 2 (1 + 5x) (1 - 5x)R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Question 5

Factorise:

20x2 - 45

Solution 5

20x2 - 45

= 5(4x2 - 9)

= 5 [(2x)2 - (3)2]

= 5 (2x + 3) (2x - 3)R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Question 6

Factorise:

(3a + 5b)2 - 4c2

Solution 6

(3a + 5b)2 - 4c2

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

= (3a + 5b - 2c)(3a + 5b + 2c) 

Question 7

Factorise:

a2 - b2 - a - b

Solution 7

a2 - b2 - a - b

= a2 - b2 - (a + b)

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

= (a + b)(a - b - 1) 

Question 8

Factorise:

4a2 - 9b2 - 2a - 3b

Solution 8

4a2 - 9b2 - 2a - 3b

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

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

= (2a + 3b)(2a - 3b - 1) 

Question 9

Factorise:

a2 - b2 + 2bc - c2

Solution 9

a2 - b2 + 2bc - c2

= a2 - (b2 - 2bc + c2)

= a2 - (b - c)2

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

= (a - b + c)(a + b - c) 

Question 10

Factorise:

4a2 - 4b2 + 4a + 1

Solution 10

4a2 - 4b2 + 4a + 1

= (4a2 + 4a + 1) - 4b2

= [(2a)2 + 2 × 2a × 1 + (1)2] - (2b)2

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

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

= (2a - 2b + 1)(2a + 2b + 1) 

Question 11

Factorise:

a2 + 2ab + b2 - 9c2

Solution 11

a2 + 2ab + b2 - 9c2

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

= (a + b + 3c) (a + b - 3c)R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Question 12

Factorise:

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Solution 12

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials 

Question 13

Factorise:

108a2 - 3(b - c)2

Solution 13

108a2 - 3(b - c)2

= 3 [(36a2 - (b -c)2]

= 3 [(6a)2 - (b - c)2]

= 3 (6a + b - c) (6a - b + c)R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Question 14

Factorise:

(a + b)3 - a - b

Solution 14

(a + b)3 - a - b

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

= (a + b) [(a + b)2 - 12]

= (a + b) (a + b + 1) (a + b - 1)R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Question 15

Factorise:

x2 + y2 - z2 - 2xy

Solution 15

x2 + y2 - z2 - 2xy

= (x2 + y2 - 2xy) - z2

= (x - y)2 - z2

= (x - y - z)(x - y + z) 

Question 16

Factorise:

x2 + 2xy + y2 - a2 + 2ab - b2

Solution 16

x2 + 2xy + y2 - a2 + 2ab - b2

= (x2 + 2xy + y2) - (a2 - 2ab + b2)

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

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

= (x + y - a + b)(x + y + a - b) 

Question 17

Factorise:

25x2 - 10x + 1 - 36y2

Solution 17

25x2 - 10x + 1 - 36y2

= (25x2 - 10x + 1) - 36y2

= [(5x)2 - 2(5x)(1) + (1)2] - (6y)2

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

= (5x - 1 - 6y)(5x - 1 + 6y) 

Question 18

Factorise:

a - b - a2 + b2

Solution 18

a - b - a2 + b2

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

= (a - b) - (a - b) (a + b)R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

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

Question 19

Factorise:

a2 - 4ac + 4c2 - b2

Solution 19

a2 - 4ac + 4c2 - b2

= a2 - 4ac + 4c2 - b2

= a2 - 2 R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomialsa R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials2c + (2c)2 - b2

= (a - 2c)2 - b2

= (a - 2c + b) (a - 2c - b)R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Question 20

Factorise:

9 - a2 + 2ab - b2

Solution 20

9 - a2 + 2ab - b2

= 9 - (a2 - 2ab + b2)

= 32 - (a - b)2

= (3 + a - b) (3 - a + b)R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Question 21

Factorise:

x3 - 5x2 - x + 5

Solution 21

x3 - 5x2 - x + 5

= x2 (x - 5) - 1 (x - 5)

= (x - 5) (x2 - 1)

= (x - 5) (x + 1) (x - 1)R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Question 22

Factorise:

1 + 2ab - (a2 + b2)

Solution 22

1 + 2ab - (a2 + b2)

= 1 - (a2 + b2 - 2ab)

= (1)2 - (a - b)2

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

= (1 - a + b)(1 + a - b) 

Question 23

Factorise:

81 - 16x2

Solution 23

81 - 16x2

= (9)2 - (4x)2

= (9 - 4x)(9 + 4x) 

Question 24

Factorise:

9a2 + 6a + 1 - 36b2

Solution 24

9a2 + 6a + 1 - 36b2

= (9a2 + 6a + 1) - 36b2

= [(3a)2 + 2(3a)(1) + (1)2] - (6b)2

= (3a + 1)2 - (6b)2

= (3a + 1 - 6b)(3a + 1 + 6b) 

Question 25

Factorise:

x2 - y2 + 6y - 9

Solution 25

x2 - y2 + 6y - 9

= x2 - (y2 - 6y + 9)

= x2 - (y2 - 2 R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomialsy R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials3 + 32)

= x2 - (y - 3)2

= [x + (y - 3)] [x - (y - 3)] R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

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

Question 26

Factorise:

4x2 - 9y2 - 2x - 3y

Solution 26

4x2 - 9y2 - 2x - 3y

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

= (2x + 3y) (2x - 3y) - (2x + 3y) R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

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

Question 27

Factorise:

9a2 + 3a - 8b - 64b2

Solution 27

9a2 + 3a - 8b - 64b2

= 9a2 - 64b2 + 3a - 8b

= (3a)2 - (8b)2 + (3a - 8b)

= (3a + 8b) (3a - 8b) + (3a - 8b)R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

= (3a - 8b) (3a + 8b + 1)

Question 28

Factorise:

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Solution 28

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials 

Question 29

Factorise:

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Solution 29

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials 

Question 30

Factorise:

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Solution 30

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials 

Question 31

Factorise:

x8 - 1

Solution 31

x8 - 1

= (x4)2 - (1)2

= (x4 - 1)(x4 + 1)

= [(x2)2 - (1)2)(x4 + 1)

= (x2 - 1)(x2 + 1)(x4 + 1)

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

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

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Question 32

Factorise:

16x4 - 1

Solution 32

16x4 - 1

= (4x2)2 - (1)2

= (4x2 - 1)(4x2 + 1)

= [(2x)2 - (1)2](4x2 + 1)

= (2x - 1)(2x + 1)(4x2 + 1) 

Question 33

Factorise:

81x4 - y4

Solution 33

81x4 - y4

= (9x2)2 - (y2)2

= (9x2 - y2)(9x2 + y2)

= [(3x)2 - y2](9x2 + y2)

= (3x - y)(3x + y)(9x2 + y2) 

Question 34

Factorise:

5 - 20x2

Solution 34

5 - 20x2

= 5(1 - 4x2)

= 5[(1)2 - (2x)2]

= 5[(1 - 2x)(1 + 2x)]

= 5(1 - 2x)(1 + 2x) 

Question 35

Factorise:

x4 - 625

Solution 35

x4 - 625

= (x2)2 - (25)2

= (x2 + 25) (x2 - 25)R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

= (x2 + 25) (x2 - 52)

= (x2 + 25) (x + 5) (x - 5)R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Question 36

Factorise:

2x4 - 32

Solution 36

2x4 - 32

= 2(x4 - 16)

= 2[(x2)2 - (4)2]

= 2[(x2 - 4)(x2 + 4)]

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

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

= 2(x - 2)(x + 2)(x2 + 4) 

Question 37

Factorise:

3a3b - 243ab3

Solution 37

3a3b - 243ab3

= 3ab (a2 - 81 b2)

= 3ab [(a)2 - (9b)2]

= 3ab (a + 9b) (a - 9b)R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Question 38

Factorise:

3x3 - 48x

Solution 38

3x3 - 48x

= 3x (x2 - 16)

= 3x [(x)2 - (4)2]

= 3x (x + 4) (x - 4)R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Question 39

Factorise:

27a2 - 48b2

Solution 39

27a2 - 48b2

= 3 (9a2 - 16b2)

= 3 [(3a)2 - (4b)2]

= 3(3a + 4b) (3a - 4b)R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Question 40

Factorise:

x - 64x3

Solution 40

x - 64x3

= x (1 - 64x2)

= x[(1)2 - (8x)2]

= x (1 + 8x) (1 - 8x)R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Chapter 3 - Factorisation of Polynomials Exercise Ex. 3C

Question 1

Factorise:

x2 + 11x + 30

Solution 1

x2 + 11x + 30

= x2 + 6x + 5x + 30

= x (x + 6) + 5 (x + 6)

= (x + 6) (x + 5).

Question 2

Factorise:

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Solution 2

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials 

Question 3

Factorise:

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Solution 3

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials 

Question 4

Factorise:

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Solution 4

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials 

Question 5

Factorise:

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Solution 5

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials 

Question 6

Factorise:

x2 - 24x - 180

Solution 6

x2 - 24x - 180

= x2 - 30x + 6x - 180

= x(x - 30) + 6(x - 30)

= (x - 30)(x + 6) 

Question 7

Factorise:

z2 - 32z - 105

Solution 7

z2 - 32z - 105

= z2 - 35z + 3z - 105

= z (z - 35) + 3 (z - 35)

= (z - 35) (z + 3)

Question 8

Factorise:

x2 - 11x - 80

Solution 8

x2 - 11x - 80

= x2 - 16x + 5x - 80

= x (x - 16) + 5 (x - 16)

= (x - 16) (x + 5).

Question 9

Factorise:

6 - x - x2

Solution 9

6 - x - x2

= 6 + 2x - 3x - x2

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

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

Question 10

Factorise:

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Solution 10

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials 

Question 11

Factorise:

40 + 3x - x2

Solution 11

40 + 3x - x2

= 40 + 8x - 5x - x2

= 8 (5 + x) -x (5 + x)

= (5 + x) (8 - x).

Question 12

Factorise:

x2 + 18x + 32

Solution 12

x2 + 18x + 32

= x2 + 16x + 2x + 32

= x (x + 16) + 2 (x + 16)

= (x + 16) (x + 2).

Question 13

Factorise:

x2 - 26x + 133

Solution 13

x2 - 26x + 133

= x2 - 19x - 7x + 133

= x(x - 19) - 7(x - 19)

= (x - 19)(x - 7) 

Question 14

Factorise:

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Solution 14

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials 

Question 15

Factorise:

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Solution 15

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials 

Question 16

Factorise:

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Solution 16

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials 

Question 17

Factorise:

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Solution 17

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials 

Question 18

Factorise:

x2 - x - 156

Solution 18

x2 - x - 156

= x2 - 13x + 12x - 156

= x (x - 13) + 12 (x - 13)

= (x - 13) (x + 12).

Question 19

Factorise:

9x2 + 18x + 8

Solution 19

9x2 + 18x + 8

= 9x2 + 12x + 6x + 8

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

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

Question 20

Factorise:

6x2 + 17x + 12

Solution 20

6x2 + 17x + 12

= 6x2 + 9x + 8x + 12

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

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

Question 21

Factorise:

18x2 + 3x - 10

Solution 21

18x2 + 3x - 10

= 18x2 - 12x + 15x - 10

= 6x (3x - 2) + 5 (3x - 2)

= (6x + 5) (3x - 2).

Question 22

Factorise:

x2 + 20x - 69

Solution 22

x2 + 20x - 69

= x2 + 23x - 3x - 69

= x(x + 23) - 3(x + 23)

= (x + 23)(x - 3) 

Question 23

Factorise:

2x2 + 11x - 21

Solution 23

2x2 + 11x - 21

= 2x2 + 14x - 3x - 21

= 2x (x + 7) - 3 (x + 7)

= (x + 7) (2x - 3).

Question 24

Factorise:

15x2 + 2x - 8

Solution 24

15x2 + 2x - 8

= 15x2 - 10x + 12x - 8

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

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

Question 25

Factorise:

21x2 + 5x - 6

Solution 25

21x2 + 5x - 6

= 21x2 + 14x - 9x - 6

= 7x(3x + 2) - 3(3x + 2)

= (3x + 2)(7x - 3) 

Question 26

Factorise:

24x2 - 41x + 12

Solution 26

24x2 - 41x + 12

= 24x2 - 32x - 9x + 12

= 8x (3x - 4) - 3 (3x - 4)

= (3x - 4) (8x - 3).

Question 27

Factorise:

3x2 - 14x + 8

Solution 27

3x2 - 14x + 8

= 3x2 - 12x - 2x +8

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

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

Question 28

Factorise:

2x2 + 3x - 90

Solution 28

2x2 + 3x - 90

= 2x2 - 12x + 15x - 90

= 2x (x - 6) + 15 (x - 6)

= (x - 6) (2x + 15).

Question 29

Factorise:

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Solution 29

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Question 30

Factorise:

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Solution 30

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Question 31

Factorise:

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Solution 31

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Question 32

Factorise:

x2 + 19x - 150

Solution 32

x2 + 19x - 150

= x2 + 25x - 6x - 150

= x(x + 25) - 6(x + 25)

= (x + 25)(x - 6) 

Question 33

Factorise:

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Solution 33

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Question 34

Factorise:

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Solution 34

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials 

Question 35

Factorise:

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Solution 35

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Question 36

Factorise:

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Solution 36

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Question 37

Factorise:

15x2 - x - 28

Solution 37

15x2 - x - 28

= 15x2 + 20x - 21x - 28

= 5x (3x + 4) - 7 (3x + 4)

= (3x + 4) (5x - 7).

Question 38

Factorise:

6x2 - 5x - 21

Solution 38

6x2 - 5x - 21

= 6x2 + 9x - 14x - 21

= 3x (2x + 3) - 7 (2x + 3)

= (3x - 7) (2x + 3).

Question 39

Factorise:

2x2 - 7x - 15

Solution 39

2x2 - 7x - 15

= 2x2 - 10x + 3x - 15

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

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

Question 40

Factorise:

5x2 - 16x - 21

Solution 40

5x2 - 16x - 21

= 5x2 + 5x - 21x - 21

= 5x (x + 1) -21 (x + 1)

= (x + 1) (5x - 21).

Question 41

Factorise:

6x2 - 11x - 35

Solution 41

6x2 - 11x - 35

= 6x2 - 21x + 10x - 35

= 3x(2x - 7) + 5(2x - 7)

= (2x - 7)(3x + 5) 

Question 42

Factorise:

9x2 - 3x - 20

Solution 42

9x2 - 3x - 20

= 9x2 - 15x + 12x - 20

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

= (3x - 5)(3x + 4) 

Question 43

Factorise:

x2 + 7x - 98

Solution 43

x2 + 7x - 98

= x2 + 14x - 7x - 98

= x(x + 14) - 7(x + 14)

= (x + 14)(x - 7) 

Question 44

Factorise:

10x2 - 9x - 7

Solution 44

10x2 - 9x - 7

= 10x2 + 5x - 14x - 7

= 5x (2x + 1) - 7 (2x+ 1)

= (2x + 1) (5x - 7).

Question 45

Factorise:

x2 - 2x + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Solution 45

x2 - 2x + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Question 46

Factorise:

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Solution 46

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Question 47

Factorise:

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Solution 47

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials 

Question 48

Factorise:

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Solution 48

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials 

Question 49

Factorise:

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Solution 49

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials 

Question 50

Factorise:

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Solution 50

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials 

Question 51

Factorise:

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Solution 51

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials 

Question 52

Factorise:

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Solution 52

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials 

Question 53

Factorise:

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Solution 53

Letx + y = z

Then, 2 (x+ y)2 - 9 (x + y) - 5

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Now, replacing z by (x + y), we get

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Question 54

Factorise:

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Solution 54

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials 

Question 55

Factorise:

9 (2a - b)2 - 4 (2a - b) - 13

Solution 55

Let2a - b = c

Then,9 (2a - b)2 - 4 (2a - b) -13

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Now, replacing c by (2a - b) , we get

9 (2a - b)2 - 4 (2a - b) - 13

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Question 56

Factorise:

7 (x - 2y)2 - 25 (x - 2y) + 12

Solution 56

Let x - 2y = z

Then, 7 (x - 2y)2 - 25 (x - 2y) + 12

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Now replace z by (x - 2y), we get

7 (x - 2y)2 - 25 (x - 2y) + 12

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Question 57

Factorise:

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Solution 57

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials 

Question 58

Factorise:

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Solution 58

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials 

Question 59

Factorise:

(a + 2b)2 + 101(a + 2b) + 100

Solution 59

Given equation: (a + 2b)2 + 101(a + 2b) + 100

Let (a + 2b) = x

Then, we have

x2 + 101x + 100

= x2 + 100x + x + 100

= x(x + 100) + 1(x + 100)

= (x + 100)(x + 1)

= (a + 2b + 100)(a + 2b + 1) 

Question 60

Factorise:

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Solution 60

Let x2 = y

Then, 4x4 + 7x2 - 2

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Now replacing y by x2, we get

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Question 61

Evaluate {(999)2 - 1}.

Solution 61

{(999)2 - 1}

= {(999)2 - (1)2}

= {(999 - 1)(999 + 1)}

= 998 × 1000

= 998000 

Question 62

Factorise:

x2 - 21x + 90

Solution 62

x2 - 21x + 90

= x2 - 6x - 15x + 90

= x(x - 6) - 15(x - 6)

= (x - 6)(x - 15) 

Question 63

Factorise:

x2 - 22x + 120

Solution 63

x2 - 22x + 120

= x2 - 10x - 12x + 120

= x(x - 10) - 12(x - 10)

= (x - 10)(x - 12) 

Question 64

Factorise:

x2 - 4x + 3

Solution 64

x2 - 4x + 3

= x2 - 3x - x + 3

= x(x - 3) - 1(x - 3)

= (x - 3)(x - 1) 

Chapter 3 - Factorisation of Polynomials Exercise Ex. 3D

Question 1

Expand:

(a + 2b + 5c)2

Solution 1

We know:

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

(i) (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

Question 2

Expand:

(2a - b + c)2

Solution 2

We know:

(2a - b + c)2

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

= 4a2 + b2 + c2 - 4ab - 2bc + 4ac.

Question 3

Expand:

(a - 2b - 3c)2

Solution 3

We know:

(a - 2b - 3c)2

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

= a2 + 4b2 + 9c2 - 4ab + 12bc - 6ac.

Question 4

Expand:

(2a - 5b - 7c)2

Solution 4

We know:

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

(2a - 5b - 7c)2

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

= 4a2 + 25b2 + 49c2 - 20ab + 70bc - 28ac.

Question 5

Expand:

(ii) (-3a + 4b - 5c)2

Solution 5

(-3a + 4b - 5c)2

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

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

Question 6

Expand:

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Solution 6

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials= R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Question 7

Factorise: 4x2 + 9y2 + 16z2 + 12xy - 24yz - 16xz.

Solution 7

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

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

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

Question 8

Factorise: 9x2 + 16y2 + 4z2 - 24xy + 16yz - 12xz

Solution 8

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

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

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

Question 9

Factorise: 25x2 + 4y2 + 9z2 - 20xy - 12yz + 30xz.

Solution 9

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

Question 10

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

Solution 10

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

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

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

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

= (4x - 2y + 3z)(4x - 2y + 3z) 

Question 11

Evaluate

(99)2

Solution 11

(99)2

= (100 - 1)2R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

= (100)2 - 2(100) (1) + (1)2

= 10000 - 200 + 1

= 9801.

Question 12

Evaluate

(995)2

Solution 12

(995)2

= (1000 - 5)2

= (1000)2 + (5)2 - 2(1000)(5)

= 1000000 + 25 - 10000

= 990025 

Question 13

Evaluate

(107)2

Solution 13

(107)2

= (100 + 7)2

= (100)2 + (7)2 + 2(100)(7)

= 10000 + 49 + 1400

= 11449 

Chapter 3 - Factorisation of Polynomials Exercise Ex. 3E

Question 1

Expand:

(3x + 2)3

Solution 1

(3x + 2)3

= (3x)3 + (2)3 + 3 R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials3x R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials2 (3x + 2)

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

= 27x3 + 8 + 18x (3x + 2)

= 27x3 + 8 + 54x2 + 36x.

Question 2

Expand:

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Solution 2

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Question 3

Expand

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Solution 3

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials 

Question 4

Evaluate

(103)3

Solution 4

(103)3

= (100 + 3)3

= (100)3 + (3)3 + 3 × 100 × 3 × (100 + 3)

= 1000000 + 27 + 900(103)

= 1000000 + 27 + 92700

= 1092727 

Question 5

Evaluate

(99)3

Solution 5

(99)3

= (100 - 1)3

= (100)3 - (1)3 - 3 × 100 × 1 × (100 - 1)

= 1000000 - 1 - 300(99)

= 1000000 - 1 - 29700

= 970299  

Question 6

Expand

(5a - 3b)3

Solution 6

(5a - 3b)3

= (5a)3 - (3b)3 - 3(5a)(3b)(5a - 3b)

= 125a3 - 27b3 - 45ab(5a - 3b)

= 125a3 - 27b3 - 225a2b + 135ab2 

Question 7

Expand

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Solution 7

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials 

Question 8

Expand

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Solution 8

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials 

Question 9

Factorise 

8a3 + 27b3 + 36a2b + 54ab2

Solution 9

8a3 + 27b3 + 36a2b + 54ab2

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

= (2a + 3b)3

= (2a + 3b)(2a + 3b)(2a + 3b) 

Question 10

Factorise 

64a3 - 27b3 - 144a2b + 108ab2

Solution 10

64a3 - 27b3 - 144a2b + 108ab2

= (4a)3 - (3b)3 - 3 × (4a)2 × (3b) + 3 × (4a) × (3b)2

= (4a - 3b)3

= (4a - 3b)(4a - 3b)(4a - 3b) 

Question 11

Factorise 

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Solution 11

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials 

Question 12

Factorise 

125x3 - 27y3 - 225x2y + 135xy2

Solution 12

125x3 - 27y3 - 225x2y + 135xy2

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

= (5x - 3y)3

= (5x - 3y)(5x - 3y)(5x - 3y)  

Question 13

Factorise 

a3x3 - 3a2bx2 + 3ab2x - b3

Solution 13

a3x3 - 3a2bx2 + 3ab2x - b3

= (ax)3 - 3 × (ax)2 × b + 3 × (ax) × (b)2 - (b)3

= (ax - b)3

= (ax - b)(ax - b)(ax - b)  

Question 14

Factorise 

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Solution 14

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials 

Question 15

Factorise 

a3 - 12a(a - 4) - 64

Solution 15

a3 - 12a(a - 4) - 64

= a3 - 3 × a × 4(a - 4) - (4)3

= (a - 4)3

= (a - 4)(a - 4)(a - 4) 

Chapter 3 - Factorisation of Polynomials Exercise Ex. 3F

Question 1

Factorise:

x3 + 27

Solution 1

x3 + 27

= x3 + 33

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

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Question 2

Factorise 

64a3 - 343

Solution 2

64a3 - 343

= (4a)3 - (7)3

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

= (4a - 7)(16a2 + 28a + 49) 

Question 3

Factorise:

x3 - 512

Solution 3

x3 - 512

= (x)3 - (8)3

= (x - 8) [(x)2 + x R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials8 + (8)2]       R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

= (x - 8) (x2 + 8x + 64).

Question 4

Factorise:

a3 - 0.064

Solution 4

a3 - 0.064

= (a)3 - (0.4)3

= (a- 0.4) [(a)2 + a R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials0.4 + (0.4)2]     R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

= (a - 0.4) (a2 + 0.4 a + 0.16).

Question 5

Factorise:

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Solution 5

We know that R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Let us rewrite

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Question 6

Factorise 

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Solution 6

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials 

Question 7

Factorise:

x - 8xy3

Solution 7

x - 8xy3

= x (1 - 8y3)

= x [(1)3 - (2y)3]

= x (1 - 2y) [(1)2 + 1 R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials2y + (2y)2]    R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

= x (1 - 2y) (1 + 2y + 4y2).

Question 8

Factorise:

32x4 - 500x

Solution 8

32x4 - 500x

= 4x (8x3 - 125)

= 4x [(2x)3 - (5)3]

= 4x [(2x - 5) [(2x)2 + 2x R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials5 + (5)2]      R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

= 4x (2x - 5) (4x2 + 10x + 25).

Question 9

Factorise:

3a7b - 81a4b4

Solution 9

3a7b - 81a4b4

= 3a4b (a3 - 27b3)

= 3a4b [(a)3 - (3b)3]

= 3a4b (a - 3b) [(a)2 + a R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials3b + (3b)2]      R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

= 3a4b (a - 3b) (a2 + 3ab + 9b2).

Question 10

Factorise 

x4y4 - xy

Solution 10

x4y4 - xy

= xy(x3y3 - 1)

= xy[(xy)3 - (1)3]

= xy{(xy - 1)[(xy)2 + (xy)(1) + (1)2]}

= xy(xy - 1)(x2y2 + xy + 1) 

Question 11

Factorise 

8x2y3 - x5

Solution 11

8x2y3 - x5

= x2 (8y3 - x3)

= x2 [(2y)3 - x3]

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

= x2 (2y - x)(4y2 + 2xy + x2) 

Question 12

Factorise 

27a3 + 64b3

Solution 12

27a3 + 64b3

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

= (3a + 4b)[(3a)2 - (3a)(4b) + (4b)2]

= (3a + 4b)(9a2 - 12ab + 16b2) 

Question 13

Factorise 

1029 - 3x3

Solution 13

1029 - 3x3

= 3(343 - x3)

= 3[(7)3 - x3]

= 3[(7 - x)(72 + 7x + x2)]

= 3(7 - x)(49 + 7x + x2) 

Question 14

Factorise:

x6 - 729

Solution 14

x6 - 729

= (x2)3 - (9)3

= (x2 - 9) [(x2)2 + x2 R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials9 + (9)2]      R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

= (x2 - 9) (x4 + 9x2 + 81)

= (x + 3) (x - 3) [(x2 + 9)2 - (3x)2]

= (x + 3) (x - 3) (x2 + 3x + 9) (x2 - 3x + 9).

Question 15

Factorise 

x9 - y9

Solution 15

x9 - y9

= (x3)3 - (y3)3 

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

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

Question 16

Factorise:

(a + b)3 - (a - b)3

Solution 16

We know that, R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Therefore,

(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 + b2 + 2ab + a2 - b2 + a2 + b2 - 2ab]

= 2b (3a2 + b2).

Question 17

Factorise:

8a3 - b3 - 4ax + 2bx

Solution 17

8a3 - b3 - 4ax + 2bx

= 8a3 - b3 - 2x (2a - b)

= (2a)3 - (b)3 - 2x (2a - b)

= (2a - b) [(2a)2 + 2a R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomialsb + (b)2] - 2x (2a - b)    R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

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

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

Question 18

Factorise:

a3 + 3a2b + 3ab2 + b3 - 8

Solution 18

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Question 19

Factorise:

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Solution 19

We know that R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Question 20

Factorise:

2a3 + 16b3 - 5a - 10b

Solution 20

2a3 + 16b3 - 5a - 10b

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

= 2 [(a)3 + (2b)3] - 5 (a + 2b)

= 2 (a + 2b) [(a)2 - a R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials2b + (2b)2 ] - 5 (a + 2b)     R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

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

Question 21

Factorise 

a6 + b6

Solution 21

a6 + b6

= (a2)3 + (b2)3 

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

= (a2 + b2)(a4 - a2b2 + b4) 

Question 22

Factorise 

a12 - b12

Solution 22

a12 - b12

= (a6)2 - (b6)2 

= (a6 - b6)(a6 + b6)

= [(a3)2 - (b3)2][(a2)3 + (b2)3]

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

= (a - b)(a2 + ab + b2)(a + b)(a2 - ab + b2)(a2 + b2)(a4 - a2b2 + b4)

= (a - b)(a + b)(a2 + b2)(a2 + ab + b2)(a2 - ab + b2)(a4 - a2b2 + b4)

Question 23

Factorise:

125a3 + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Solution 23

125a3 + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

We know that R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Let us rewrite

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Question 24

Factorise 

x6 - 7x3 - 8

Solution 24

Given equation is x6 - 7x3 - 8.

Putting x3 = y, we get

y2 - 7y - 8

= y2 - 8y + y - 8

= y(y - 8) + 1(y - 8)

= (y - 8)(y + 1)

= (x3 - 8)(x3 + 1)

= (x3 - 23)(x3 + 13)

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

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

Question 25

Factorise 

x3 - 3x2 + 3x + 7

Solution 25

x3 - 3x2 + 3x + 7

= x3 - 3x2 + 3x - 1 + 8

= (x3 - 3x2 + 3x - 1) + 8

= (x - 1)3 + 23

= (x - 1 + 2)[(x - 1)2 - (x - 1)(2) + 22]

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

= (x + 1)(x2 - 4x + 7) 

Question 26

Factorise 

(x + 1)3 + (x - 1)3

Solution 26

(x + 1)3 + (x - 1)3

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

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

= 2x(x2 + 3)  

Question 27

Factorise 

(2a + 1)3 + (a - 1)3

Solution 27

(2a + 1)3 + (a - 1)3

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

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

= 3a(3a2 + 3a + 3)

= 9a(a2 + a + 1)  

Question 28

Factorise 

8(x + y)3 - 27(x - y)3

Solution 28

8(x + y)3 - 27(x - y)3

= [23 (x + y)3] - [33 (x - y)3]

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

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

= (-x + 5y){4x2 + 4y2 + 8xy + 6x2 - 6y2 + 9x2 + 9y2 - 18xy}

= (-x + 5y)(19x2 + 7y2 - 10xy) 

Question 29

Factorise 

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

Solution 29

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

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

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

= 2x(x2 + 12) 

Question 30

Factorise 

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

Solution 30

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

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

= 4(x2 + 4x + 4 + x2 - 4 + x2 - 4x + 4)

= 4(3x2 + 4) 

Question 31

Prove that

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Solution 31

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials 

Question 32

Prove that

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Solution 32

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials 

Question 33

Factorise:

216x3 + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Solution 33

216x3 + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

We know that R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Let us rewrite

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Question 34

Factorise:

16x4 + 54x

Solution 34

16x 4 + 54x

= 2x (8x 3 + 27)

= 2x [(2x)3 + (3)3]

= 2x (2x + 3) [(2x)2 - 2xR-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials3 + 32]    R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

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

Question 35

Factorise:

7a3 + 56b3

Solution 35

7a3 + 56b3

= 7(a3 + 8b3)

= 7 [(a)3 + (2b)3]

= 7 (a + 2b) [a2 - a R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials2b + (2b)2]    R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

= 7 (a + 2b) (a2 - 2ab + 4b2).

Question 36

Factorise:

x5 + x2

Solution 36

x5 + x2

= x2(x3 + 1)

= x2 (x + 1) [(x)2 - x R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials1 + (1)2]     R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

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

Question 37

Factorise:

a3 + 0.008

Solution 37

a3 + 0.008

= (a)3 + (0.2)3

= (a + 0.2) [(a)2 - a R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials0.2 + (0.2)2]     R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

= (a + 0.2) (a2 - 0.2a + 0.04).

Question 38

Factorise:

1 - 27x3

Solution 38

1 - 27x3

= (1)3 - (3x)3

= (1 - 3x) [(1)2 + 1 R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials3x + (3x)2]       R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

= (1 - 3x) (1 + 3x + 9x2).

Chapter 3 - Factorisation of Polynomials Exercise Ex. 3G

Question 1

Find the product:

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

Solution 1

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

= [x + y + (-z)] [(x)2 + (y)2 + (-z)2 - (x) (y) - (y) (-z) - (-z) (x)]

= x3 + y3 - z3 + 3xyz.

Question 2

Factorize:

8a3 + 125b3 - 64c3 + 120abc

Solution 2

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) - (-4c) (2a)]

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

Question 3

Factorize:

8 - 27b3 - 343c3 - 126bc

Solution 3

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) - (-7c) (2)]

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

Question 4

Factorize:

125 - 8x3 - 27y3 - 90xy

Solution 4

125 - 8x3 - 27y3 - 90xy

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

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

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

Question 5

Factorize:

Solution 5

Question 6

Factorise:

27x3 - y3 - z3 - 9xyz

Solution 6

27x3 - y3 - z3 - 9xyz

= (3x)3 - y3 - z3 - 3(3x)(y)(z)

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

= [3x + (-y) + (-z)][(3x)2 + (-y)2 + (-z)2 - (3x)(-y) - (-y)(-z) - (3x)(-z)]

= (3x - y - z)(9x2 + y2 + z2 + 3xy - yz + 3zx) 

Question 7

Factorise:

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Solution 7

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials 

Question 8

Factorise:

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Solution 8

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials 

Question 9

Factorize:

(a - b)3 + (b - c)3 + (c - a)3

Solution 9

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

(a - b)3 + (b - c)3 + (c - a)3

= x3 + y3 + z3, 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).

Question 10

Factorise:

(a - 3b)3 + (3b - c)3 + (c - a)3

Solution 10

Given equation is (a - 3b)3 + (3b - c)3 + (c - a)3

Now,

(a - 3b) + (3b - c) + (c - a)

= a - a - 3b + 3b - c + c

= 0

We know that if x + y + z = 0, then x3 + y3 + z3 = 3xyz

Hence,

(a - 3b)3 + (3b - c)3 + (c - a)3

= 3(a - 3b)(3b - c)(c - a)

Question 11

Factorize:

(3a - 2b)3 + (2b - 5c)3 + (5c - 3a)3

Solution 11

We have:

(3a - 2b) + (2b - 5c) + (5c - 3a) = 0

So, (3a - 2b)3 + (2b - 5c)3 + (5c - 3a)3

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

Question 12

Find the product.

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

Solution 12

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

= x3 + xy2 + xz2 + x2y  - xyz + x2z - x2y - y3 - yz2 - xy2 + y2z - xyz - x2z - y2z - z3 - xyz + yz2 - xz2

= x3 - y3 - z3 - xyz - xyz - xyz

= x3 - y3 - z3 - 3xyz 

Question 13

Factorise:

(5a - 7b)3 + (7b - 9c)3 + (9c - 5a)3

Solution 13

Given equation is (5a - 7b)3 + (7b - 9c)3 + (9c - 5a)3

Now,

(5a - 7b) + (7b - 9c) + (9c - 5a)

= 5a - 7b + 7b - 9c + 9c - 5a

= 0

We know that if x + y + z = 0, then x3 + y3 + z3 = 3xyz

Hence,

(5a - 7b)3 + (7b - 9c)3 + (9c - 5a)3

= 3(5a - 7b)(7b - 9c)(9c - 5a) 

Question 14

Factorize:

a3 (b - c)3 + b3 (c - a)3 + c3 (a - b)3

Solution 14

a3 (b - c)3 + b3 (c - a)3 + c3 (a - b)3

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

Now, since, a (b - c) + b (c -a) + c (a - b)

= ab - ac + bc - ba + ca - bc = 0

So, a3 (b - c)3 + b3 (c - a)3 + c3 (a - b)3

= 3a (b - c) b (c - a) c (a - b)

= 3abc (a - b) (b - c) (c - a).

Question 15

Evaluate

(-12)3 + 73 + 53

Solution 15

Given equation is (-12)3 + 73 + 53

Now,

-12 + 7 + 5 = -12 + 12 = 0

We know that if x + y + z = 0, then x3 + y3 + z3 = 3xyz

Hence,

(-12)3 + 73 + 53

= 3(-12)(7)(5)

= -1260 

Question 16

Evaluate

(28)3 + (-15)3 + (-13)3

Solution 16

Given equation is (28)3 + (-15)3 + (-13)3

Now,

28 + (-15) + (-13) = 28 - 28 = 0

We know that if x + y + z = 0, then x3 +  y3 + z3 = 3xyz

Hence,

(28)3 + (-15)3 + (-13)3

= 3(28)(-15)(-13)

= 16380 

Question 17

Prove that (a + b + c)3 - a3 - b3 - c3 = 3(a + b)(b + c)(c + a)

Solution 17

L.H.S. = (a + b + c)3 - a3 - b3 - c3

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

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

= a3 + b3 + 3ab(a + b) + c3 + 3(a + b)c × [(a + b) + c] - a3 - b3 - c3

= 3ab(a + b) + 3(a + b)c × [(a + b) + c]

= 3(a + b)[ab + c(a + b) + c2]

= 3(a + b)[ab + ac + bc + c2]

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

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

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

= R.H.S.

Question 18

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

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials.

Solution 18

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials 

Question 19

If a + b + c = 9 and a2 + b2 + c2 = 35, find the value of (a3 + b3 + c3 - 3abc).

Solution 19

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

= (a + b + c)[(a2 + b2 + c2) - (ab + bc + ca)] ….(i)

Now,

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

(9)2 = 35 + 2(ab + bc + ca)

81 = 35 + 2(ab + bc + ca)

2(ab + bc + ca) = 46

ab + bc + ca = 23

Substituting in (i), we get

a3 + b3 + c3 - 3abc = (9)[35 - 23] = 9 × 12 = 108 

Question 20

Find the product:

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

Solution 20

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

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

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

= a3 + b3 + c3 - 3abc

Where, x = a, (-2y) = b and 3 = c

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

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

= x3 - 8y3 + 27 + 18xy.

Question 21

Find the product.

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

*Modified the question

Solution 21

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

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

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

= 27x3 - 125y3 + 64 + 180xy 

Question 22

Factorize:

125a3 + b3 + 64c3 - 60abc

Solution 22

125a3 + b3 + 64c3 - 60abc

= (5a)3 + (b)3 + (4c)3 - 3 (5a) (b) (4c)

= (5a + b + 4c) [(5a)2 + b2 + (4c)2 - (5a) (b) - (b) (4c) - (5a) (4c)]

[R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials a3 + b3 + c3 - 3abc = (a+ b + c) (a2 + b2 + c2 - ab - bc - ca)]

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

Question 23

Factorize:

a3 + 8b3 + 64c3 - 24abc

Solution 23

a3 + 8b3 + 64c3 - 24abc

= (a)3 + (2b)3 + (4c)3 - 3 a 2b 4c

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

Question 24

Factorize:

1 + b3 + 8c3 - 6bc

Solution 24

1 + b3 + 8c3 - 6bc

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

= (1 + b + 2c) [1 + b2 + (2c)2 - b - b 2c - 2c]

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

Question 25

Factorize:

216 + 27b3 + 8c3 - 108bc

Solution 25

216 + 27b3 + 8c3 - 108bc

= (6)3 + (3b)3 + (2c)2 - 3 6 3b 2c

= (6 + 3b + 2c) [(6)2 + (3b)2 + (2c)2 - 6 3b - 3b 2c - 2c 6]

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

Question 26

Factorize:

27a3 - b3 + 8c3 + 18abc

Solution 26

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) - (2c) (3a)]

= (3a - b + 2c) (9a2 + b2 + 4c2 + 3ab + 2bc - 6ca).

Chapter 3 - Factorisation of Polynomials Exercise MCQ

Question 1

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

(a) -2

(b) -3

(c) 2

(d) 3

Solution 1

Correct option: (c)

Let p(x) = 2x2 + kx

Since (x + 1) is a factor of p(x),

P(-1) = 0

2(-1)2 + k(-1) = 0

2 - k = 0

k = 2

Question 2

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

Solution 2

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials 

Question 3

104 96 =?

(a) 9894

(b) 9984

(c) 9684

(d)9884

Solution 3

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials 

Question 4

305 308 = ?

(a) 94940

(b) 93840

(c) 93940

(d)94840

Solution 4

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials 

Question 5

207 193 = ?

(a) 39851

(b) 39951

(c) 39961

(d)38951

Solution 5

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials 

Question 6

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

Solution 6

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials 

Question 7

(x2 - 4x - 21) =?

(a) (x - 7)(x - 3)

(b) (x + 7)(x - 3)

(c) (x - 7)(x + 3)

(d)none of these

Solution 7

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Question 8

(4x2 + 4x -3) = ?

(a) (2x - 1)(2x - 3)

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

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

(d)none of these

Solution 8

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Question 9

6x2 + 17x + 5 =?

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

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

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

(d)none of these

Solution 9

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Question 10

(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

Solution 10

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Question 11

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

(a) (3x - 2)(x2 - 1)

(b) (3x - 2)(x2 + 1)

(c) (3x + 2)(x2 - 1)

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

Solution 11

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Question 12

The value of (249)2 - (248)2 is

(a) 12

(b) 477

(c) 487

(d) 497

Solution 12

Correct option: (d)

(249)2 - (248)2

= (249 + 248)(249 - 248)

= 497 × 1

= 497

Question 13

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

(a) 1

(b) 0

(c) -1

(d)3

 

Solution 13

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Question 14

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

Solution 14

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials 

Question 15

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomialsthen (a3 - b3) = ?

(a) -3

(b) -2

(c) -1

(d) 0

Solution 15

Correct option: (d)

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials 

Question 16

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Solution 16

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials 

Question 17

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

(a) 0

(b) abc 

(c) 2abc

(d)3abc

Solution 17

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Question 18

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

(a) 0

(b) R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

(c) R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

(d) R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Solution 18

Correct option: (c)

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials 

Question 19

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

(a) 1

(b) 9

(c) 18

(d)27

Solution 19

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials 

Question 20

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

Solution 20

Correct option: (d)

(x + y)3 - (x3 + y3)

= (x + y)3 - [(x + y)(x2 - xy + y2)]

= (x + y)[(x + y)2 - (x2 - xy + y2)]

= (x + y)[x2 + y2 + 2xy - x2 + xy - y2]

= (x + y)(3xy) 

Question 21

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

(a) 5 +x

(b) 5 - x

(c) 5x - 1

(d) 10x

Solution 21

Correct option: (d)

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

= [(5x)2 - (1)2] + (1 + 5x)2

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

= (1 + 5x)[(5x - 1) + (1 + 5x)]

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

= (1 + 5x)(10x)

Question 22

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

(a) -5

(b) 5

(c) 3

(d) -3

Solution 22

Correct option: (b)

Since (x + 5) is a factor of p(x) = x3 - 20x + 5k,

p(-5) = 0

(-5)3 - 20(-5) + 5k = 0

-125 + 100 + 5k = 0

-25 + 5k = 0

5k = 25

k = 5