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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) 63x4 y4z3 + 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
= (2x2 – 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)2 – (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)2 – 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)2 – (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 (4x2 – 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)2 – (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 2 – (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
= x2 – 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
= a2 – 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
= a2 (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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