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R S AGGARWAL AND V AGGARWAL Solutions Mathematics Class 9 chapter 3 Factorisation of Polynomials

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

Factorize:

9x2 + 12xy

Solution 1

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

Question 2

Factorize:

x3 + 2x2 + 5x + 10

Solution 2

x3 + 2x2 + 5x + 10

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

=
(x2 + 5) (x + 2)

Question 3

Factorize:

x2
+ xy – 2xz – 2yz

Solution 3

x2
+ xy – 2xz – 2yz

=
x (x + y) – 2z (x + y)

=
(x+ y) (x – 2z)

Question 4

Factorize:

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

Factorize:

8
– 4a – 2a3 + a4

Solution 5

8
– 4a – 2a3 + a4

=
4(2 – a) – a3 (2 – a)

=
(2 – a) (4 – a3)

Question 6

Factorize:

x3
– 2x2y + 3xy2 – 6y3

Solution 6

x3
– 2x2y + 3xy2 – 6y3

=
x2 (x – 2y) + 3y2 (x – 2y)

=
(x – 2y) (x2 + 3y2)

Question 7

Factorize:

px
– 5q + pq – 5x

Solution 7

px
+ pq – 5q – 5x

=
p(x + q) – 5 (q + x)

=
(x + q) (p – 5)

Question 8

Factorize:

x2
+ y – xy – x

Solution 8

x2
– xy + y – x

=
x (x – y) – 1 (x – y)

=
(x – y) (x – 1)

Question 9

Factorize:

(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

Factorize:

(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

Factorize:

a3
+ a – 3a2 – 3

Solution 11

a3
+ a – 3a2 – 3

=
a(a2 + 1) – 3 (a2 + 1)

=
(a – 3) (a2 + 1)

Question 12

Factorize:

18x2y – 24xyz

Solution 12

18x2y – 24xyz = 6xy (3x – 4z)

Question 13

Factorize:

3ax
– 6ay – 8by + 4bx

Solution 13

3ax
– 6ay – 8by + 4bx

=
3a (x – 2y) + 4b (x – 2y)

=
(x – 2y) (3a + 4b)

Question 14

Factorize:

abx2
+ a2x + b2x +ab

Solution 14

abx2
+ a2x + b2x +ab

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

=
(bx + a) (ax + b)

Question 15

Factorize:

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

Factorize:

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

Factorize:

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

Factorize:

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

Factorize:

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

Factorize:

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

Factorize:

(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

Factorize:

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

Factorize:

27a3b3
– 45a4b2

Solution 23

27a3b3
– 45a4b2 = 9a3b2 (3b – 5a)

Question 24

Factorize:

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

Factorize:

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

Factorize:

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

Factorize:

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

Factorize:

Solution 28

Question 29

Factorize:

2a
(x+ y) – 3b (x + y)

Solution 29

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

Question 30

Factorize:

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

Factorize:

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

Factorize:

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

Factorize:

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

Factorize:

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

Factorize:

8ab2
– 18a3

Solution 2

8ab2
– 18a3

=
2a (4b2 – 9a2)

=
2a [(2b)2 – (3a)2]

=
2a (2b + 3a) (2b – 3a)

Question 3

Factorize:

150
– 6x2

Solution 3

150
– 6x2

=
6 (25 – x2)

=
6 (52 – x2)

=
6 (5 + x) (5 – x)

Question 4

Factorize:

2
– 50x2

Solution 4

2
– 50x2

=
2 (1 – 25x2)

=
2 [(1)2 – (5x)2]

=
2 (1 + 5x) (1 – 5x)

Question 5

Factorize:

20x2
– 45

Solution 5

20x2
– 45

=
5(4x2 – 9)

=
5 [(2x)2 – (3)2]

=
5 (2x + 3) (2x – 3)

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

Factorize:

a2
+ 2ab + b2 – 9c2

Solution 11

a2
+ 2ab + b2 – 9c2

=
(a + b)2 – (3c)2

=
(a + b + 3c) (a + b – 3c)

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

Factorize:

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)

Question 14

Factorize:

(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)

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

Factorize:

a
– b – a2 + b2

Solution 18

a
– b – a2 + b2

=
(a – b) – (a2 – b2)

=
(a – b) – (a – b) (a + b)

=
(a – b) (1 – a – b)

Question 19

Factorize:

a2
– 4ac + 4c2 – b2

Solution 19

a2
– 4ac + 4c2 – b2

=
a2 – 4ac + 4c2 – b2

=
a2 – 2 a 2c + (2c)2
– b2

=
(a – 2c)2 – b2

=
(a – 2c + b) (a – 2c – b)

Question 20

Factorize:

9
– a2 + 2ab – b2

Solution 20

9
– a2 + 2ab – b2

=
9 – (a2 – 2ab + b2)

=
32 – (a – b)2

=
(3 + a – b) (3 – a + b)

Question 21

Factorize:

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)

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

Factorize:

x2
– y2 + 6y – 9

Solution 25

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 26

Factorize:

4x2
– 9y2 – 2x – 3y

Solution 26

4x2
– 9y2 – 2x – 3y

=
(2x)2 – (3y)2 – (2x + 3y)

=
(2x + 3y) (2x – 3y) – (2x + 3y)

=
(2x + 3y) (2x – 3y – 1)

Question 27

Factorize:

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)

=
(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

Factorize:

x4
– 625

Solution 35

x4
– 625

=
(x2)2 – (25)2

=
(x2 + 25) (x2 – 25)

=
(x2 + 25) (x2 – 52)

=
(x2 + 25) (x + 5) (x – 5)

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

Factorize:

3a3b
– 243ab3

Solution 37

3a3b
– 243ab3

=
3ab (a2 – 81 b2)

=
3ab [(a)2 – (9b)2]

=
3ab (a + 9b) (a – 9b)

Question 38

Factorize:

3x3
– 48x

Solution 38

3x3
– 48x

=
3x (x2 – 16)

=
3x [(x)2 – (4)2]

=
3x (x + 4) (x – 4)

Question 39

Factorize:

27a2
– 48b2

Solution 39

27a2
– 48b2

=
3 (9a2 – 16b2)

=
3 [(3a)2 – (4b)2]

=
3(3a + 4b) (3a – 4b)

Question 40

Factorize:

x
– 64x3

Solution 40

x
– 64x3

=
x (1 – 64x2)

=
x[(1)2 – (8x)2]

=
x (1 + 8x) (1 – 8x)

Chapter 3 – Factorisation of Polynomials Exercise Ex. 3C

Question 1

Factorize:

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

Factorize:

z2
– 32z – 105

Solution 7

z2
– 32z – 105

=
z2 – 35z + 3z – 105

=
z (z – 35) + 3 (z – 35)

=
(z – 35) (z + 3)

Question 8

Factorize:

x2
– 11x – 80

Solution 8

x2 – 11x – 80

=
x2 – 16x + 5x – 80

=
x (x – 16) + 5 (x – 16)

=
(x – 16) (x + 5).

Question 9

Factorize:

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

Factorize:

40
+ 3x – x2

Solution 11

40
+ 3x – x2

=
40 + 8x – 5x – x2

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

=
(5 + x) (8 – x).

Question 12

Factorize:

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

Factorize:

x2
– x – 156

Solution 18

x2 – x – 156

=
x2 – 13x + 12x – 156

=
x (x – 13) + 12 (x – 13)

=
(x – 13) (x + 12).

Question 19

Factorize:

9x2
+ 18x + 8

Solution 19

9x2
+ 18x + 8

=
9x2 + 12x + 6x + 8

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

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

Question 20

Factorize:

6x2
+ 17x + 12

Solution 20

6x2
+ 17x + 12

=
6x2 + 9x + 8x + 12

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

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

Question 21

Factorize:

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

Factorize:

2x2
+ 11x – 21

Solution 23

2x2
+ 11x – 21

=
2x2 + 14x – 3x – 21

=
2x (x + 7) – 3 (x + 7)

=
(x + 7) (2x – 3).

Question 24

Factorize:

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

Factorize:

24x2
– 41x + 12

Solution 26

24x2
– 41x + 12

=
24x2 – 32x – 9x + 12

=
8x (3x – 4) – 3 (3x – 4)

=
(3x – 4) (8x – 3).

Question 27

Factorize:

3x2
– 14x + 8

Solution 27

3x2
– 14x + 8

=
3x2 – 12x – 2x +8

=
3x (x – 4) – 2(x – 4)

=
(x – 4) (3x – 2).

Question 28

Factorize:

2x2
+ 3x – 90

Solution 28

2x2
+ 3x – 90

=
2x2 – 12x + 15x – 90

=
2x (x – 6) + 15 (x – 6)

=
(x – 6) (2x + 15).

Question 29

Factorize:

Solution 29

Question 30

Factorize:

Solution 30

Question 31

Factorize:

Solution 31

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

Factorize:

Solution 33

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

Factorize:

Solution 35

Question 36

Factorize:

Solution 36

Question 37

Factorize:

15x2
– x – 28

Solution 37

15x2
– x – 28

=
15x2 + 20x – 21x – 28

=
5x (3x + 4) – 7 (3x + 4)

=
(3x + 4) (5x – 7).

Question 38

Factorize:

6x2
– 5x – 21

Solution 38

6x2
– 5x – 21

=
6x2 + 9x – 14x – 21

=
3x (2x + 3) – 7 (2x + 3)

=
(3x – 7) (2x + 3).

Question 39

Factorize:

2x2
– 7x – 15

Solution 39

2x2
– 7x – 15

=
2x2 – 10x + 3x – 15

=
2x (x – 5) + 3 (x – 5)

=
(x – 5) (2x + 3).

Question 40

Factorize:

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

Factorize:

10x2
– 9x – 7

Solution 44

10x2
– 9x – 7

=
10x2 + 5x – 14x – 7

=
5x (2x + 1) – 7 (2x+ 1)

=
(2x + 1) (5x – 7).

Question 45

Factorize:

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

Factorize:

Solution 46

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

Factorize:

Solution 53

Let x + y = z

Then,
2 (x + y)2 – 9 (x + y) – 5

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

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

Factorize:

9
(2a – b)2 – 4 (2a – b) – 13

Solution 55

Let 2a – b = c

Then, 9 (2a – b)2 – 4 (2a – b) -13

Now,
replacing c by (2a – b) , we get

9
(2a – b)2 – 4 (2a – b) – 13

Question 56

Factorize:

7
(x – 2y)2 – 25 (x – 2y) + 12

Solution 56

Let
x – 2y = z

Then,
7 (x – 2y)2 – 25 (x – 2y) + 12

Now
replace z by (x – 2y), we get

7
(x – 2y)2 – 25 (x – 2y) + 12

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

Factorize:

Solution 60

Let
x2 = y

Then,
4x4 + 7x2 – 2

Now
replacing y by x2, we get

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

Factorize:
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

Factorize:
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

Factorize:
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

Factorize:

x3
+ 27

Solution 1

x3
+ 27

=
x3 + 33

= (x + 3) (x2 – 3x + 9)

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

Factorize:

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

= (x – 8) (x2 + 8x + 64).

 

Question 4

Factorize:

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 Polynomials 0.4 + (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

Factorize:

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

Factorize:

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 Polynomials 2y + (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

Factorize:

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 Polynomials 5 + (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

Factorize:

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 Polynomials 3b + (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

x4y4xy

Solution 10

x4y4xy

= 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

Factorize:

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 Polynomials 9 + (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)

= [(x3 – y3)][(x3)2
+ x3y3 + (y3)2]

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

Question 16

Factorize:

(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

Factorize:

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 Polynomials b + (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

Factorize:

a3 + 3a2b + 3ab2 + b3 – 8

Solution 18

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

Question 19

Factorize:

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

Factorize:

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 Polynomials 2b + (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)

= (a2 + b2)[(a2)2
– (a2b2) + (b2)2]

= (a2 + b2)(a4
– a2b2 + b4)

Question 22

Factorise

a12 – b12

Solution 22

a12 – b12

=
(a6)2 – (b6)

= (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)(a2ab + b2)(a2 + b2)(a4
– a2b2 + b4)

= (a – b)(a + b)(a2 + b2)(a2
+ ab + b2)(a2ab + b2)(a4 – a2b2
+ b4)

Question 23

Factorize:

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

Factorize:

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

Factorize:

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

Factorize:

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 Polynomials 2b + (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

Factorize:

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 Polynomials 1 + (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

Factorize:

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 Polynomials 0.2 + (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

Factorize:

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 Polynomials 3x + (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 + xyyz + xz)

Solution 12

(x – y – z)(x2 + y2 + z2 + xyyz + 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 + c2abbcca)

= (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
+ y2xy

(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

 

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