Chapter 13 Solution West Bengal Board : Class 8 Mathematics
West Bengal Board Class 8 Math Solution full Exercise 13.1, 13.2 by Experts. Here in this page WB Board Class 8 Student can find Ganit Prabha Class 8 Math Chapter 13 বীজগাণিতিক সংখ্যামালার উৎপাদকে বিশ্লেষণ Exercise 13.1, 13.2 Solution.
Board | WB Board |
Class | 8 (Eight) |
Subject | Mathematics |
Book Name | Ganit Prabha |
Chapter | 13 |
Chapter Page Number on Book | 127 and 131 |
13.1 Solution
(1) (x2 – 40x – 12y)
= x2 – 43x + 3x – 129
= x (x – 43) + 3 (x – 43)
= (x + 3) (x – 43)
∴ P = 3 q = – 43
m2 + 19m + 60
= m2 + 150m + 4m + 60
= m (m + 15) + 4 (m + 15)
= (m + 4) (m + 15)
x2 – x – 6
= x2 – 3x + 2x – 6
= x (x – 3) + 2 (x – 3)
= (x – 3) (x + 3)
p = – 3; q = 2
(a + b)2 – 4 (a + b) – 12
a + b = x Let,
x2 – 4x – 12
= x2 – 6x + 2x – 12
= x (x – 6) + 2(x – 6)
= (x – 6) (x + 2)
= (a + b – 6) (a + b + 2)
p = -6; q = 2
(2) (i) (a + b) – 5 (a + b) – 6
Let, a + b = x
x2 – 5x – 6
= x2 – 6x + x – 6
= x (x – 6) + 1(x – 6)
= (x + 1) (x – 6)
= (a + b + 1) (a + b – 6)
(ii) (x2 – 2x)2 + 5 (x2 – 2x) – 36
Let, x2 – 2x = a
a2 – 5a – 36
= a2 – 9a + 4a – 36
= a (a – 9) + 4 (a – 9)
= (a + 4) (a – 9)
= (x2 – 2x + 4) (x2 – 2x – 9)
(iii) (p2 – 3q2)2 – 16 (p2 – 3q2) + 63
Let, (p2 – 3q2) = a
a2 – 16a + 63
= a2 – 9a – 7a + 63
= a (a – 9) – 7 (a – 9)
= (a – 7) (a – 9)
= (p2 – 3q2 – 7) (p2 – 3q2 – 9)
(iv) a4 + 4a2 – 5
=a2 (a2 + 4 – 5)
= a2 (a2 – 1)
= a4 – a2
(v) x2y2 + 23xy – 420
= (xy)2 + (35 – 12) xy – 420
= (xy)2 + 35xy – 12xy – 420
= xy (xy + 35) – 12 (xy + 35)
= (xy – 12) (xy + 35)
(vi) x4 – 7x2 + 12
= x4 – 4x2 – 3x2 + 12
=x2 (x2 – 4) – 3 (x2 – 4)
= (x2 – 4) (x2 – 3)
= (x + 2) (x – 2) ((x2 – 3)
(vii) a2 + ab – 12b2
= a2 + 4ab – 3ab + 12b2
= a (a + 4b) – 3b (a + 4b)
= (a + 4b) (a – 3b)
(viii) p2 + 31pq + 108q2
= p2 + 27pq + 4pq + 108pq2
= p (p + 27q) + 4q (p + 27q)
= (p + 27q) (p + 4q)
(ix) a6 + 3a3b3 – 40b6
= a6 + 8a3b3 – 5a3b3 – 40b6
= a3 (a3 + 8b3) – 5b3 (a3 + 8b3)
= (a3 + 8b3) (a3 – 5b3)
= {a3 + (2b)3} (a3 – 5b3)
= (a + 2b) (a2 – 2ab + 4b2) (a3 – 5b3)
(x) (x + 1) (x + 3) (x – 4) (x – 6) + 24
= (x + 1) (x – 4) (x + 3) (x – 6) + 24
= (x2 – 4x + x – 4) (x2 – 6x + 3x – 18) + 24
= (x2 – 3x – 4) (x2 – 3x – 18) + 24
Let, x2 – 3x = p
(p – 4) (p – 18) + 24
= p2 – 18p – 4p + 72 + 24
= p2 – 22p + 96
= p2 – 16p – 6p + 96
= p (p – 16) – 6 (p – 16)
= (p – 16) (p – 6)
= (x2 – 3x – 16) (x2 – 3x – 6)
(xi) (x + 1) (x + 9) (x + 5)2 + 63
= (x2 + 9x + x + 9) (x2 + 10x + 25) + 63
= (x2 + 10x + 9) (x2 + 10x + 25) + 63
[x2 + 10x = p]
(p + 9) (p + 25) + 63
= p2 + 25p + 9p + 225 + 63
= p2 + 34p + 288
= p2 + 18p + 16p + 288
= p (p + 18) + 16 (p + 18)
= (p + 18) (p + 16)
∴ (x2 + 10x + 18) (x2 + 10x + 16)
= (x2 + 10x + 18) (x2 + 8x + 2x + 16)
= (x2 + 10x + 18) {x (x + 8) + 2 (x +8)}
= (x2 + 10x + 18) (x + 8) (x + 2)
(xii) x (x + 3) (x + 6) (x + 9) + 56
= x (x + 9) (x + 3) (x + 6) + 56
= (x2 + 9x) (x2 + 6x + 3x + 18) + 56
=(x2 + 9x) (x2 + 9x + 18 + 56)
Let, x2 + 9x = p
p (p + 18) + 56
= p2 + 18p + 56
=p2 + 14p + 4p + 56
=p (p + 14) + 4 (p + 14)
= (p + 14) (p + 4)
= (x2 + 9x + 14) (x2 + 9x + 4)
=(x2 + 7x + 2x + 14) (x2 + 9x + 4)
= x (x + 7) + 2 (x + 7) (x2 + 9x + 4)
= (x + 7) (x + 2) (x2 + 9x + 4)
(xiii) x2 – 2ax + (a + b) (a – b)
= x2 – {(a + b) + (a – b) } x + (a + b) (a – b)
= x2 – (a + b) x – (a – b) x + (a + b) (a – b)
= x {x – (a + b)} – (a – b) x – (a + b)
= [x – (a + b)} x – (a – b)}
= (x – a – b) (x – a + b)
(xiv) x2 – bx – (a + 3b) (a + 2b)
= x2 – {(a + 3b) – (a + 2b)} x – (a + 3b) (a + 2b)
= x2 – (a + 3b) x + (a + 2b) x – (a + 3b) (a + 2b)
= x (x – a – 3b) + (a + 2b) {x – a – 3b}
=(x – a – 3b) (x + a + 2b)
(xv) (a + b)2 – 5a – 5b + 6
= (a + b)2 – 5 (a + b) + 6
Let, a + b = p
P2 – 5p + 6
= p2 – 3p – 2p + 6
= p (p – 3) – 2 (p – 3)
= (P – 3) (p – 2)
= (a + b – 3) (a + b – 2)
(xvi) x2 + 4abx – (a2 – b2)2
= x2 + 4abx – (a + b)2 (a – b)2
= x2 + {(a + b)2 – (a – b)2} – (a + b)2 (a – b)2
= x2 + (a + b)2x – (a – b)2 x – (a + b)2 (a – b)2
= x {(x + (a + b)2} – (a – b)2 {x + a + b)2}
= {x + (a + b)2} {x – (a – b)2}
= {x + (a2 + 2ab + b2)} {x – (a2 – 2ab + b2)}
= {x + (a2 + 2ab + b2)} {x – (a2 – 2ab + b2)}
= (x + a2 + 2ab + b2) (x – a2 + 2ab – b2)
(xviii) x6y6 – 9x3y3 + 8
= x6y6 – (8 + 1) x3y3 + 8
= x6y6 – 8x3y3 – x3y3 + 8
= x3y3 (x3y3 – 8) – 1 (x3y3 – 8)
= (x3y3 – 8) (x3y3 – 1)
= {(xy)3 – (2)3} {(xy)3 – (1)3}
= (xy – 2) (x2y2 + 2xy + 4) (xy – 1) (x2y2 + xy + 1)
= (xy – 2) (xy – 1) (x2y2 + 2xy + 4) (x2y2 + xy + 1)
13.2 Solution
(1)(i) 2a2 + 5a + 2
= 2a2 + 4a + a + 2
= 2a (a + 2) + 1 (a + 2)
= (a + 2) (2a + 1)
(ii) 3x2 + 14x + 8
= 3x2 + 12x + 2x + 8
= 3x (x + 4) + 2 (x + 4)
= (x + 4) (3x + 2)
(iii) 2m2 + 7m + 6
= 2m2 + 4m + 3m + 6
= 2m (m +2) + 3 (m + 2)
= (2m + 3) (m + 2)
(iv) 6x2 – x – 15
= 6x2 – 10x + 9x – 15
= 2x (3x – 5) + 3 (3x – 5)
= (2x + 3) (3x – 5)
(v) 9r2 + r – 8
= 9r2 + 9r – 8r – 8
= 9r (r + 1) – 8 (r + 1)
= (r + 1) (9r – 8)
(vi) 6m2 – 11mn + 10n2
= 6m2 – 15mn + 4mn – 10n2
= 3m (2m – 5n) + 2n (3m – 5n)
= (3m + 2n) (2m – 5n)
(vii) 7x2 + 48xy – 7y2
= 7x2 + 49xy – xy – 7y2
= 7x (x + 7y) – y (x + 7y)
= (7x – y) (x + 7y)
(viii) 12 + x – 6x2
= 6x2 – x – 12
=6x2 – 9x + 8x – 12
= 3x (2x – 3) + 4 (2x – 3)
= (3x + 4) (2x – 3)
(ix) 6 + 5a – 6a2
= 6a2 – 5a – 6
= 6a2 – 9a + 4a – 6
= 3a (2a – 3) + 2 (2a – 3)
= (2a- ) (3a + 2)
(x) 6x2 – 13x + 6
= 6x2 – 9x – 4x + 6
= 3x (2x – 3) – 2 (2x – 3)
= (3x – 2) (2x – 3)
(xi) 99a2 – 202ab + 99b2
= 99a2 – (121 + 81)ab + 99b2
= 99a2 – 121ab – 81ab + 99b2
= 11a (9a – 11b) – 9b (9b – 11b)
= (11a – 9b) (9a – 11b)
(xii) 2a6 – 13a3 – 24
= 2a6 – 16a3 + 3a3 – 24
= 2a3 (a3 – 8) + 3 (a3 – 8)
= (2a3 + 3) (a – 8)
(xiii) 8a4 + 2a2 – 45
= 8a4 + 20a2 – 18a2 – 45
= 4a2 (2a2 + 5) – 9 (2a2 + 5)
= (4a2 – 9) (2a2 + 5)
= {(2a)2 – (3)2} (2a2 + 5)
= (2a + 3) (2a – 3) (2a2 + 5)
(xiv) 6 (x – y)2 – x + y – 15
= 6 (x – y)2 – (x – y) – 15
x – y = p Let,
6p2 – p – 15
=6p2 – 10p + 9p – 15
= 2p (3p – 5) + 3 (3 (3p – 5)
= (3p – 5) (2p + 3)
= {3 (x – y) – 5} {2(x – y) + 3}
= (3x – 3y – 5) (2x – 2y + 3)
(xv) 3 (a + b) – 2a – 2b – 8
= 3 (a + b)2 – 2 (a + b) – 8
Let, a + b = p
3p – 2p – 8
=3p – 6p + 4p – 8
= 3p (p – 2) + 4 (p – 2)
= (p – 2) (3p + 4)
(2) (i) x2 – 2x – 3
= x2 – 2.x.1 + (1)2 – (1)2 – 3
= (x – 1)2 – 1 – 3
= (x – 1)2 – 4
= (x – 1)2 – (2)2
= (x – 1 + 2) (x – 1 – 2)
= (x + 1) (x – 3)
(iii) 3x2 – 7x – 6
= 3x2 – 9x + 2x – 6
= 3x (x – 3) + 2 (x – 3)
= (x – 3) (3x + 2)
(iv) 3a2 – 2a – 5
= 3a2 + 3a – 5a – 5
= 3a (a + 1) – 5 (a + 1)
= (3a – 5) (a + 1)
(3) (i) ax2 + (a2 + 1)x + a
= ax2 + a2x + x + a
= ax (x + a) + 1 (x + a)
= (ax + 1) (x + a)
(ii) x2 + 2ax + (a + b) (a – b)
= x2 {(a + b) + (a – b)} x + (a + b) (a – b)
= x2 + (a + b) x + (a – b) x + (a + b) (a – b)
= x {x + (a + b)} + (a – b) {x + (a – b)}
= (x + a + b) (x + a – b)
(iii) ax2 – (a2 + 1) x + a
= ax2 – a2x – x + a
= ax (x – a) – 1 (x – a)
= (x – a) (ax – 1)
(iv) ax2 + (a2 – 1) x – a
= ax2 + a2x – x – a
= ax (x + a) – 1 (x + a)
= (ax – 1) (x + a)
(v) ax2 – (a2 – 2)x – 2a
= ax2 – a2x + 2x – 2a
= ax (x- a) + 2 (x – a)
= (ax + 2) (x – a)