# Joy of Mathematics Class 8 Solutions Chapter 13

## Joy of Mathematics Class 8 Solutions Chapter 13 Factorisation

Welcome to NCTB Solution. Here with this post we are going to help 8th class students for the Solutions of Joy of Mathematics Class 8 Book, Chapter 13 Factorisation. Here students can easily find step by step solutions of all the problems for Factorisation. Exercise wise proper solutions for every problems. All the problem are solved with easily understandable methods so that all the students can understand easily. Here students will find solutions for Exercise 13.1, 13.2, 13.3 and 13.4

Factorisation Exercise 13.1 Solution :

Question no – (1)

Solution :

(a) 3x – 9a

= Here, 3x is the common factor

3x – 9a = 3 (x – 3a)

(b) 7x3 – 14x4

= Here, 7x3 is the common factor

7x3 – 14x4

= 7x3 (1 – 2x)

(c) 4xy + y2

= Here, y is the common factor

4xy + y2 = y (4x + y)

(d) 24x4 + 18 x3

= Here, 6x3 is the common factor

24x4 + 18x3

= 6x3 (4x + 3)

(e) x5  + x3 – x

= Here, x is the common factor

x5 + x3 – x

= x (x4 + x2 – 1)

(f) 5xyz + 15xy3z – 25xyz3

= Here, 5xyz is the common factor

5xyz (5x2 + 3y – 5z2)

Question no – (2)

Solution :

(a) 63xy4z3 + 27x3y3z3

= Here, 9x3y3z3 is the common factor

9x3y3z3(7x3y2 + 3)

(b) xy + yz + xyz

= Here, y is the common factor

xy  + yz + xyz

= y(x + z + xz)

(c) 30x3 – 25x2 + 10x

= Here, 16x is the common factor

30x3 – 60x + 90x2

= 15x (62x2 – 4 + 6x)

(d) 5x3 – 25x2 + 10x

= Here, 5x is the common factor

5x (x2 – 5x + 2)

(e) 9a2b4 – 6a5c6 – 12a3d

= Here, 3a2 is the common factor

9a2b4 – 6a5c6 – 12a3d

= 3a2 (3b4 – 2a3c6 – 12ad)

(f) 10a2 – 20a + 5

= Here, 5 is the common factor

10a2 – 20a + 5

= 5 (2a2 – 4a + 1)

Question no – (3)

Solution :

(a) 3×2 (1 – y) + 6x (1 – y)l – 12 (1 – y)

= 3 (1 – y) {(x2 + 2x -4)}  [∵ 3 (1 – y) common factor]

= 3 (1-y) (x2 + 2. X . – 4)

(b) ab (x – 2y)2 + a2b (x – 2y)2 – bc (x – 2y)2

=  b(x – 2y)2 {a + a2 – c} [∵b (x – 2y)2 is common factor]

= b(x – 2y)2 (a + a2 – c)

(c) 3yz (a – b – c) – 3z ( a – b – c) + 21z2 (a – b – c)

= 3yz (a – b – c) {(y – 1 + 7z)} [∵ 3z (a – b – c) is the common factor]

= 3z (a – b – c) (y + 7z – 1)

(d) 15x (lu – mv) + 25y (lu – mv) + 30z (lu – mv)

= 5 (lu – mv) {(3x + 5y + 6z)} [∵ 5 (lu – mv) is the common factor]

= 5 (lu – mv) (3x  + 5y + 6z)

Question no – (4)

Solution :

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

= (b – c) is the common factor

= (b – c) { a (b – c) + b + c (b – c) }

= (b – c) (ab – ac +  b +  cb – 2b)

(b) a3 (a – 2b) + a2 (a – 2b)

= Here, a2 (a – 2b) is the common factor

= a2 (a – 2b) {(a +1)}

= a2 (a – 2b) (a + 1)

(c) 14 (a – 3b)2 – 21x (a – 3b)

= Here, 7 (a – 3b) is the common factor

= 7 (a – 3b) {2(a – 3b) – 3x}

= 7 (a – 3b) {2 (a – 3b) – 3x}

= 7 (a – 3b) (2a – 6b – 3x)

(d) 10a (2x + y)3 – 15b (2x + y)2 + 35 (2x + y)

= 5 (2x + y) is the common factor

= 5 (2x + y) {2a (2x + y)2 – 3b (2x + y) + 7)}

= 5 (2x + y) [2a (2x + y)2 – 3b (2x + y) + 7)]

#### Factorisation Exercise 13.2 Solution :

Question no – (1)

Solution :

Given, xy + 4x + 2y + 8

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

= (y + 4) (x + 2)

Question no – (2)

Solution :

ab – 3a + 2b – 6

= a (b – 3) + 2 (b – 3)

= (b – 3) (a + 2)

Question no – (3)

Solution :

Given, ac – ad + bc – bd

= a (c –d) + b (c – d)

= (c – d) (a + b)

Question no – (4)

Solution :

6 + 2y + 3x + xy

= 6 + 3x + 2y + xy

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

= (2 +x) (3 +y)

Question no – (5)

Solution :

ab – ad – bc  +cd

= a (b –d) – c (b –d)

= (b – d) (a – c)

Question no – (6)

Solution :

a – ab + b – b2

= a (1 – b) + b (1 0b)

=(1 – b) (a + b)

Question no – (7)

Solution :

2x – y – 2xy + y2

= 2x – 2xy – y + y2

= 2x (1 – y) – 1 (1 – y)

= (1 – y) (2x – 1)

Question no – (8)

Solution :

4p – 6q – 6p2 +9pq

= 2 (2p – 3q) – 3q (2p – 3q)

= (2p – 3q) (2 – 3p)

Question no – (9)

Solution :

2x – xy + 4y – 8

= x (2 – y) – (2 – y)

= (2- y) (x – 4)

Question no – (10)

Solution :

3x – xy + 4y – 12

= x (3 –y) – 4 (3 – y)

= (3 – y) (x – 4)

Question no – (11)

Solution :

5p – pq + 2q – 10

= 5p – pq – 10 + 2q

= p ( 5 –q) – 2 (5 –q)

= (5 – q) ( p – 2)

Question no –( 12)

Solution :

6 – 2y  +xy – 3x

= 6 – 2y – 3x + xy

= 2 (3 – y) – x (3 – y)

= (3 – y) (2 – x)

Question no – (13)

Solution :

2x2 + 4x3 + 1 + 2x

= 2x2 (2x + 1) + 1 (2x – 1)

= (2x – 1) (2x2 + 1)

Question no – (14)

Solution :

x3y + xy3 + x2 + y2

= xy (x2 + y2) + 1(x2 + y2)

= (x2 + y2) (xy + 1)

Question no – (15)

Solution :

1 – 2x – 5x2 + 10x3

= 1 (1 – 2x) – 5x2 (1 – 2x)

= (1 -2x) (1 -5x2)

Question no – (16)

Solution :

3 (x + y)2 + 4x + 4y

= 3 (x + y)2 + 4 (x + y)

= (x + y) {3 (3 (x + y) + 4}

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

Question no – (17)

Solution :

3x2 + 3y2 – 6 (x2 + y2)

= 3 (x2 + y2) – 6 (x2 + y2)2

= 3 (x2 + y2) {1 – 2 (x2 + y2)}

= 3 (x2 + y2) (1 – 2x2 + 2y2)

Question no – (18)

Solution :

x (y – z) – p (z – y)

= x (y – z) – { – p (y – z)}

= x (y – z) + P (y – z)

= (y – z) {x + p}

= (y – z) (x + p)

Question no – (19)

Solution :

(2x2 – 2y2) a + (2y2 – 2x2)b

= (2x2 – 2y2) a + {- (2x2 – 2y2)}b

= (2x2 – 2y2) a – (2x2 – 2y2)b

= (2x– 2y2) (a – b)

Question no – (20)

Solution :

7 (x + y) – 14x – 14y

= 7 (x + y) – 14 (x + y)

= (x + y) (7 – 14)

= (x + y) (-7)

= -7 (x + y)

Question no – (21)

Solution :

Ax2 + 3az + bx2y + by + 3byz + a

= ax2 + bn2y + 3az + 3byz + by + a

= x2 (a + by) + 3z (a + by) + 3z (a + by) + 1 (a + by)

= (a + by) (x2 + 3z H)

Question no – (22)

Solution :

px + p2x + pqy – + qy – (px + qy)2

= px (1 +p) qy (1 + p) – (px + qy)2

= (1 + p) (px + qy) – (px + qy)2

= (px + qy) {1 + p – (px + qy)}

= (px + qy) (1 + p – px – qy)

#### Factorisation Exercise 13.3 Solution :

Question no – (1)

Solution :

x2 – 4

= (x)2 – (2)2

= (x – 2) (x + 2)

Question no – (2)

Solution :

Given, x2 – 16

= (x)2 –(4)2

= (x -4) (x + 4)

Question no – (3)

Solution :

x2 – 81

= (x)– (9)2

= (x – 9) (x + 9)

Question no – (4)

Solution :

Given, y4 – 1

= (y2)2 –(1)2

= (y2 + 1) (y2 – 1)

= (y2 + 1) (y2 – 12)

= (y2 + 1) (y + 1) (y – 1)

Question no – (5)

Solution :

64x2 – 1

= (8x)2 – (1)2

= (8x + 1) (8x – 1)

Question no – (6)

Solution :

X24 – 2.25

= x24 – 225/100

= (x12)2 – (15/10)2

= (x122 + 15/10) (x12 – 15/10)

= (x12 + 1.5) (x12 – 1.5)

Question no – (7)

Solution :

0.04 – x4y6

= 4/100 – z4y6

= (2/10)2 – (x2y3)2

= (2/10 + x2y3) (2/10 – x3y3)

= (0.2 + x2y3) (0.2 – x2y3)

Question no – (8)

Solution :

16a2 – 0.64b2

= (4a)– 64/100 b2

= (4a)2 – (8/10b)2

= (4a + 8/10b) (4a – 8/10b)

= 4a + 0.8b) (4a – 0.8b)

Question no – (9)

Solution :

64x – x3

= x(64 – x2)

= x {(8)2 – (x)2}

= x (8 + x) (8 – x)

Question no – (10)

Solution :

25 – x8

= (5)2 – (x4)2

= (5 + x4) (5 – x4)

Question no – (11)

Solution :

108 – 3x2

= 3(36 –x2)

= 3 {(6)2 – (x)2}

= 3 (+ x) (6 – x)

Question no – (12)

Solution :

25/36 x2 – y2

= (5/6x)– (y)2

= (5/63x + y) (5/6 – y)

Question no – (13)

Solution :

25 (5x – 3y)2 – 16 (3x – 2y)2

= { 5 (5x – 3y)}2 – {4(3x – 2y)}2

= { 5 (5x -3y) + 4 (3x – 2y)} { 5 (5x – 3y) – 4 (3x – 2y)

= (256x – 15y) + 12x – 8y) (25x – 15y – 12x + 8y)

= (37x – 23y) (13x p- 7y)

Question no – (14)

Solution :

(5p + 7q)2 – (7p +5q)2

= (5p + 7q + 7p + 5q) (5p + 7q – 7p – 5q)

= (12p + 12q) (-2p + 2q)

= 12 (p q) – 2 (q – p)

= 24 (p + q) (q – p)

Question no – (15)

Solution :

28x3 – 7x

= 7x (4x– 1)

= 7x { (2x)2 – (1)2}

= 7x (2x + 1) (2x – 1)

Question no – (16)

Solution :

(4 3/4)2 – (7 1/2 x)2

= (19/4)2 – (15/2x)2

= (19/4) + 15/2x) (19/4 – 15/2x)

Question no – (17)

Solution :

(4p – 3q)– (5p – q)2

= (4p – 3q + 5p – q) (4p – 3q – 5p + q)

= (9p – 4q) (-4p – 2q)

= (9p – 4q) { – 2(p + q) }

= -2 (9p – 4q) ( p + q)

Question no – (18)

Solution :

94q4/121 – x8y8/256

= (p2q2/11)2 – (x4y4/16)2

= (p2q2/11 + x4y4/16) (p2q2/11 – (x4y4/16)

Question no – (19)

Solution :

0.0289y2 – 0.81

= 289/10000 y2 – 81/100

= (14/100y)2 – (9/10)2

= (17/100y + 9/10) (17/100y – 9/10)

= (0.17y + 0.9) (0.17y – 0.9)

Question no – (20)

Solution :

x2 – 6x – 16

= x2 – (8 – 2)x – 8 × 2

= x2 – 8x + 2x – 8 × 2

= x (x – 8) + 2 (x – 8)

= (x – 8) (x + 2)

Question no – (21)

Solution :

9x2 – 30x – 11

= 9x2 – (33 – 3) x – 11

= 9x2 – 33x + 3x – 11

= 3x (3x – 11) + 1 (3x – 11)

= (3x – 11) (3x + 1)

#### Factorisation Exercise 13.4 Solution :

Question no – (1)

Solution :

x2 + 3x – 88

= x2 + (11 – 8) x – 8 × 11

= x2 + 11x – 8x – 8 × 11

= x (x + 11) – 8 (x + 11)

= (x + 11) (x – 8)

Question no – (2)

Solution :

x2 + 16x + 64

= x2 + (8 + 8)x + 8 × 8

= x2 + 8x + 8x + 8 × 8

= (x + 8) (x + 8)

Question no – (3)

Solution :

x2 + 8x + 16)

= x2 + (4 – 4) x + 4 × 4

= x2 + 4x + 4x + 4 × 4

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

= (x + 4) ( x + 4)

Question no – (4)

Solution :

x2 – 14x + 45

= x2 = -(9  + 5) x + 45

= x2 – 9x – 5x + 5 × 9

= x (x – 9) – 5 (x – 9)

= (x – 9) ( x – 5)

Question no – (5)

Solution :

x2 + 16x + 48

= x2 + (12 + 4) x + 48

= x2 + 12x + 4x + 48

= x2 + 12x + 4x + 4 × 12

= x (x + 12) + 4 (x + 12)

= x ) x + 12) + 4 ( x + 12)

= (x + 12) (x + 12)

Question no – (6)

Solution :

x2  – 8x + 16

= x2 – (4 + 4) x + 16

= x2 – 4x  – 4x + 4 × 4

= x ( x – 4) – 4 (x – 4)

= x – 4) (x – 4)

Question no – (7)

Solution :

x2 + 11x – 42

= x2 + (14 – 3) x – 42)

= x2 + 14x – 3x – 42

= x2 + 14x  – 3x – 3 × 14

= x (x + 14) – 3 3 (x + 14)

= (x + 14) (x – 3)

Question no – (8)

Solution :

x2 + 16x – 80

= x2 + (20 – 4) x – 80

= x2 + 20x – 4x – 80

= x2 + 20x – 4x – 4 × 20

= x (x + 20) – 4 (x + 20)

= (x + 20) (x – 4)

Question no – (9)

Solution :

x2 + 18x + 81

= x2 + (9 + 9)x + 9 × 9

= x2 + 9x + 9x + 9 × 9

= x (x + 9) + 9 ( x + 9)

= x + 9) ( x + 9)

Question no – (10)

Solution :

x2 – 12x + 35

= x2 – (7 + 5) x + 35

= x2 – 7x – 5x + 35

= x (x – 7) – 5 (x – 7)

= (x – 7) ( x – 5)

Question no – (11)

Solution :

x2 – 2x – 35

= x2 – (7 – 5)x – 35

= x2 – 7x + 5x – 35

= x2 – 7x + 5x – 5 × 7

= x ( x -7) + 5 (x – 7)

= (x – 7) 9x + 5)

Question no – (12)

Solution :

x2 – 12x + 36

= x2 – (6 + 6) x + 6 × 6

= x2 – 6x – 6x + 6 × 6

= x ((x – 6) – 6 (x – 6)

= (x – 6) )x – 6)

Question no – (13)

Solution :

x4 – 18×2 + 72

= x4 – (12 + 6) x2 + 72

= x2 – 12x2 – 6x2 + 12 × 6

= x2 (x2 – 12) – 6 (x2 – 12)

= (x2 – 12) (x2 – 6)

Question no – (14)

Solution :

X4 + 42x2 + 80

= x4 (40 + 2)x2 + 80

= (x4 + 40x2 + 2x2 + 40 × 2

= x2 (x2 + 40) + 2 (x2 + 40)

= (x2 + 40) (x2 + 2)

Question no – (15)

Solution :

3x2 – 18x + 27

= 3x2 – (9 + 9) x + 3 × 9

= 3x2 – 9x – 9x + 3 × 9

= 3x (x – 3) – 9 (x – 3)

= (x – 3) (3x – 9)

Question no – (16)

Solution :

4x2 +y2 + 4xy

= 4x2 + 4xy + y2

= 4x2 + (2 + 2) xy + y2

= 4x2 + 2xy + 2xy + y2

= 2x (2xl + y) + 2y (2x + y)

= (2x + y) (2x – y)

Question no – (17)

Solution :

x2/4 + y2/9 + xy/3

= 9x2 + 4y2 + 12xy/36

= 9x2 + 12xy + 4y2/36

= 1/36 {9x2 + (6 + 6) xy + 4y2}

= 1/36 { 9x2 + 6xy + 6xy + 4y2}

= 1/36 {3x (3x + 2y) + 2y (3x + 2y) }

= 1/36 {(3x + 2y) (3x + 2y)}

Question no – (18)

Solution :

9x2 – 16x2 – 24xy

= 9x2 – 24xy + 16y2

= 9x2 (12 + 12) xy + 16y2

= 9x2 – 12xy – 12xy + 3 × 4y2

= 3x (3x – 4y) – 4y (3x – 4y)

= (3x – 4y) (3x – 4y)

Question no – (19)

Solution :

X2 – 4x/y + 4/y2

= x2y2 – 4xy + 4/y2

= x2y2 – (2 +2) xy + 4/y2

= x2y2 – 2xy – 2xy + 4/y2

= xy (xy – 2) + 2 (xy – 2) /y2

= (xy – 2) (xy – 2)/y2

Question no – (20)

Solution :

5x2 + x – 18

= 5x2 + (10 – 9) x – 18

= 5x2 + 10x – 9x – 18

= 5x (x + 2) – 9 (x + 2)

= (x+ 2) (5x – 9)

Question no – (21)

Solution :

x2 – 6x – 40

= x2 – (10 *- 4) x – 40

= x2 – 10x + 4x  – 40

= x (x – 10) + 4( x – 10)

= (x – 10) p( x + 4)

Question no – (22)

Solution :

-8a2 – 18b4 + 24ab2

= – (18b4 – 24ab2 + 8a2)

= {18b4 – (12 + 12) ab2 + 8a2}

= 18b4 – 12ab2 – 12ab2 + 8a2}

= {3b2 (6b – 4a) – 2a (6b 2 – 4a)}

= { (6b2 – 4a) (3b2 – 2a()}

Question no – (23)

Solution :

6x2 + 7x – 3

= 6x2 + (9 – 2) x – 3

= 6x2 + 9x – 2x – 3

= 3x (2x + 3) – 1 (2x + 3)

= (2x + 3) (3x – 1)

Question no – (24)

Solution :

-7x2 – 5x + 12

= – (7x2  + 5x – 12)

= –  {7x2 + (12 – 7) x – 12)}

= – { 7x2 + 12x – 7x – 12}

= – { 7x2 + 12x – 7x – 3 × 4}

= – { x (7x + 12) – 1 (7x + 12)}

= – {(7x + 12) ( x- 1)}

Question no – (25)

Solution :

3a2 + 12b2 – 12ab

= 3a2 – 12ab + 12b2

= 3a2 – ( 6+ 6) b = 12b2

= 3a2 – 6ab – 6ab + 12b32

= 3a (a -2b) – 6 (a – 2b)

= (a – 2b)  (3a – 6b)

Question no – (26)

Solution :

16x2 + 14x -15

= 16x2 + (24 – 10) x – 15

= 16x2 + 247x – 10x – 15

= 8x (2x + 3) – 5( 2x + 3)

= (2x + 3) (8x – 5)

Question no – (27)

Solution :

12x2 – x – 6

= 12x2 – (9 – 8) x – 6

= 12x2 – 9x + 8x – 6

= 3x (4x; – 3) + 2 (4x – 3)

= (4x – 3) (3x + 2)

Question no – (28)

Solution :

-6a3b – 18a3 + 24a3b2

= -6a3 (b + 3 – 4b2)

= – 6a3 (-4b2 + b + 3)

= – 6a2 {- 4b2 + (4 – 3) b +9 3}

= – 6a2 {-4b2 + (4 – 3) b + 3}

= -6a2 { -4b2 + 4b – 3b + 3}

= -6a2 {4b (b – 1) – 3( b – 1) }

= + 6a2 {(b – 1) (4b- 3)}

Question no – (29)

Solution :

(a + 2b)2 – 9 + 16 – 8/3 (a + 2b)

Let,  a + 2b = x

x2/9 + 16 – 8/3x

= x2 + 144 – 24x/9

= x2 – 247x + 144/9

= 1/9 (x2 – 24x + 144)

= 1/9 {x – (12 + 12) x + 144}

= 1/9 {x2 – 12x – 12x + 12  × 12}

= 1/9 { x (x – 12) – 12 (x – 12) }

= 1/9 { (x – 12) (x – 12) }

Now, putting the value of x,

= 1/9 { (a + 2b – 12) (a + 2b – 12)}

Question no – (30)

Solution :

8 (1 +x/3)2 – 6 (1 +x/3) – 35

= Let, 1 +x/3 = a

98a2 – 6a – 35

= 8a2 – (20a – 14) a – 35

= 8a2 – 20a +9 14a – 35

= 4a (2a – 5) + 7 (2a – 5)

= (2a – 5) (4a + 7)

Now, putting the value of a,

{2 (1 + 4/3)) – 5} { 4 ( 1+ 4/3) + 7}

= (2 + 2x/3  – 5)  (4 + 4/3x + 7)

= (2x/3 – 3) p(11 + 4/3 x)

Question no – (31)

Solution :

x2 – 10x + 21

= x2 – (7 + 3) x + 21

= x2 – 7x – 3x + 21

= x2 – 7x – 3x + 21

= x– 7x – 3x + 3 × 7

= x (x – 7) – 3 (x – 7)

= (x – 7) (x – 3)

Question no – (32)

Solution :

25/x2 + y2/4 – 5y/x

= 100 +x2y2 – 20xy/4x2

= 1/4x2 (x2y2 – 20xy + 100)

= 1/4x2 {x2y2 – (10 + 10) xy + 100}

= 1/4x2 { x2y2 – 10xy – 10xy}

= 1/4x2 {xy (xy – 10) – 10  (xy – 10)}

= 1/4x2 (xy – 10) (xy + 10)

Question no – (33)

Solution :

(3x – 2y)2 – 4 (3x – 2y) – 32

= Let,  3x – 2y = a

= a– 4a – 32

= a2 – (8 – 4) a – 32

= a2 – 8a + 4a – 4 × 8

= a (a – 8) + 4(a – 8)

= (a – 8) (a + 4)

Now, putting the value of a

= (3x – 2y – 8) (3x – 2y + 4)

Question no – (34)

Solution :

(5x + 2)2 + 16 (5x + 2) + 60

= Let,  5x + 2 = a

a2 + 16 a + 60

= a(10 + 6) a + 60

= a2 + 10a + 6a + 60

= a2 + 10a + 6a + 6 × 10

= a (a + 10) + 6 (a  + 10)

= (a + 10) (a + 6)

Now, putting the value of a

= (5x +2 + 10) (5x + 2 + 6)

= (5x + 12) (5 + 8)

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Updated: May 30, 2023 — 9:46 am