West Bengal Board Class 7 Math Chapter 19 Solution

Chapter 19 Solution West Bengal Board : Class 7 Mathematics

West Bengal Board Class 7 Math Solution full Chapter 19 by Experts. Here in this page WB Board Class 7 Student can find Ganit Prabha Class 7 Math Chapter 19 উৎপাদকে বিশ্লেষণ (Product analysis) Solution.

Board

WB Board
Class

7 (Seven)

Subject

Mathematics
Book Name

Ganit Prabha

Chapter

19
Chapter Page Number on Book

225 and 228


West Bengal Board Class 7 Math Chapter 19 Solution

Ex – 19.1

(1) (i) 7xy = 7 X x X y

(ii) 9x2 y = 3 X 3 X x X x X y

(iii) 16ab2 c = 2 X 2 X 2 X 2 X a X b X b X c

(iv)  – 25lmn = – 5 X 5 X l X m X n

(v) 12x (2 + x) = 2 X 2 x 3 X x X (2+x)

(vi) -5pq (p2 + 8) = – 5 X p X q X (p2 + 8)

(vii) 21xy2 (3x – 2) = 3 X 7. X x X y X y (3x – 2)

(viii) 121mn (m2 – n) = 11 X 11 X m X n X (m2 – n)

(3) (i) x2

= ax2, x2y

(ii) 2xy

= 2x2y, 6xyz

(iii) 4a2

= 4a2b, 12a2

(iv) (mn + 2)

∴ b2 (mn + 2), 2a (mn + 2)

(v) x (y + 2)

∴ x2 ( y + 2), ax (y + 2)

(4) (i) 5 + 10x

= 5 (1 + 2x)

(ii) 2x – 6

= 2 (x – 3)

(iii) 7m – 14n

= 7 (m – 2n)

(iv) 18xy + 21xz

= 3x (6y + 7z)

(v) 4xy + 6yz

= 2y (2x + 3z)

(vi) 7xyz – 6xy

= xy (7z – 6)

(vii) 7a2 + 14a

= 7a (a + 2)

(viii) – 15m + 20

= 5 (- 3m + 4)

(ix) 6a2b + 8ab2

= 2ab (3a + 4b)

(x) 3a2 – ab2

= a (3a – b2)

(xi) abc – bcd

= bc (a – b)

(xii) 60xy3 + 4xy – 8

= 4 (15xy3 + xy – 2)

(xiii) x2yz + xy2z + xyz2

= xyz (x + y + z)

(xiv) a3 – a2 + a

= a (a2 – a + 1)

(xv) x2 y2 z2 + x2 y2 + x2 y2 q2

= x2 y2 (z2 + 1 + q2) (Ans)

(5) (i) xy + 2x + y + 2

= xy + y + 2x + 2

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

= (y + 2) (x + 1)

(ii) ab – 5b + a – 5

= ab + a – 5b – 5

= a (b + 1) – 5 (b + 1)

= (b + 1) (a – 5) (Ans)

(iii) 6xy – 9y + 4x – 6

= 6xy + 4x – 9y – 6

= 2x (3y + 2) – 3 (3y + 2)

= (3y + 2) (2x – 3) (Ans)

(iv) 15m + 9 – 35mn – 21n

= 3 (5m + 3) – 7n (5m + 3)

= (5m + 3) (3 – 7n)

(v) ax + bx – ay – by

= x (a + b) – y (a + b)

= (a + b) (x – y) (Ans)

 (vi) c -9 + 9ab – abc

= c – abc – 9 + 9ab.

= c (1 – ab) – 9 (1 – ab)

= (1 – ab) (c – 9) (Ans)

Ex – 19.2

(1) (i) x2 + 14x + 49

= x2 + 7x + 7x + 49

= x (x + 7) + 7 (x + 7)

= (x + 7) (x + 7) (Ans).

(ii) 4m2 – 36m + 81

= 4m2 – 18m – 18m + 81

= 2m (2m – 9) -9 (2m – 9)

= (2m – 9) (2m – 9)

(iii) 25x2 + 30x + 9

= 25x2 + 15x + 15x + 9

= 5x (5x + 3) + 3 (5x + 3)

= (5x + 3) (5x + 3) (Ans)

(iv) 121b2 – 88bc + 16

= 121b2 – 44bc – 44bc + 16

= 11b (11b – 4c) – 4 (11b – 4c)

= (11b – 4c) (11b – 4c)

(v) (x2y)2 – 4x2 y2

= x4 y2 – 4x2 y2

= x2 y2 (x2 – 4)

= x2 y2 (x2 – 22)

= (x2 y2) (x + 2) (x – 2)

(vi) a4 + 4a2b2 + 4b4

= (a2)2 + 2.a2.2b2 + (2b2)2

= (a2 + 2b2)2

= (a2 + 2b2) (a2 + 2b2) (Ans)

(vii) 4x2 – 16

= 4 (x2 – 4)

= 4 (x2 – 22)

= 4 (x + 2) (x – 2)

(viii) 121 – 36x2

= (11)2 – (6x)2

= (11 + 6x) (11 – 6x)

(ix) x2 y2 – p2 q2

= (xy)2 – (pq)2

= (xy + pq) (xy – pq)

(x) 80m2 – 125

= 5 (16m2 – 25)

= 5 {(4m)2 – (5)2}

= 5 {(4m + 5) (4m – 5)}

= 5 (4m + 5) (4m – 5) (Ans)

(xi) ax2 – ay2

= a (x2 – y2)

= a (x + y) (x – y) (Ans)

(xii) 1 – (m + n)2

= (1)2 – (m + n)2

= (1 + m + n) (1 – m – n) (Ans)

(xiii) (2a – b – c)2 – (a – 2b – c)2

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

= (3a – 3b – 2c) (a + b) (Ans)

(xiv) x2 – 2xy – 3y2

= x2 – 2xy + y2 – y2 – 3y2

= (x – y)2 – 4y2

= (x – y)2 – (2y)2

= (x – y + 2y) (x – y – 2y)

= (x + y) (x – 3y) (Ans)

(xv) x2 + 9y2 + 6xy – z2

= x2 + 6xy + 9y2 – z2

= x2 + 2x.3y + (3y)2 – z2

= (x + 3y)2 – (z)2

= (x + 3y + z) (x + 3y – z) (Ans)

(xvi) a2 – b2 + 2bc – c2

= a2 – (b2 – 2bc + c2)

= (a)2 – (b – c)2

= (a + b – c) (a – b + c) (Ans)

(xvii) a2 (b – c)2 – b2 (c – a)2

= {a(b – c)}2 – {b(c – a)}2

= (ab – ac)2 – (bc – ba)2

= {(ab – ac) + (bc – ab)} {(ab – ac) – (bc – ab)}

= (ab – ac + bc – ab) (ab – ac – bc + ab) (Cut +ab , -ab fro0m 1st box)

= (bc – ac) (2ab – ac – bc) (Ans)

(xviii) x2 – y2 – 6yz – 9z2

= x2 – (y2 + 6yz + 9z2)

= x2 {(y2 + 2.y.3z + (3z)2}

= x2 – (y + 3z)2

= (x + y + 3z) (x + y – 3z) (Ans)

(xix) x2 – y2 + 4x – 4y

= (x + y) (x – y) +4 (x – y)

= (x – y) (x + y + 4) (Ans)

(xx) a2 – b2 + c2 – d2 – 2 (ac – bd)

= a2 – b2 + c2 – d2 – 2ac + 2ab

= a2 – 2ac + c2 – b2 + 2bd – d2

= (a – c)2 – (b2 – 2bd + d2)

= (a – c)2 – (b – d)2

= (a – c + b – d) (a – c – b + d) (Ans)

(xxi) 2ab – a2 – b2 + c

= c2 – a2 + 2ab – b2

= c2 – (a2 – 2ab + b2)

= c2 – (a – b)2

= (c + a – b) (c – a + b) (Ans)

(xxii) 36x2 – 16a2 – 24ab – 9b2

= 36x2 – (16a2 + 24ab + 9b2)

= (6x)2 – {(4a)2 + 2.4a.3b + (3b)2}

= . (6x)2 – (4a + 3b)2

= (6x + 4a + 3b) (6x – 4a – 3b) (Ans)

(xxiii) a2 – 1 + 2b – b2

= a2 – 1 + 2b – b2

= a2 – (1 – 2b + b2)

= a2 – (1 – b)2

= (a + 1 – b) (a – 1 + b) (Ans)

(xxiv) a2 – 2a – b2 + 2b

= a2 – b2 – 2a + 2b

= (a + b) (a – b) – 2 (a – b)

= (a – b) (a + b – 2) (Ans)

(xxvi) a2 – b2 – 4ac + 4bc

= a2 – b2 – 4ac + 4bc + 4c2 – 4c2

= (a2 – 4ac + 4c2) – b2 + 4bc – 4c2

= {a2 – 2.a. 2c + (2c)2} – (b2 – 4bc + 4c2)

= (a + 2c)2 – (b – 2c)2

= (a + 2c + b – 2c) (a + 2c – b + 2c)

= (a + b) (a – b + 4c) (Ans)

Updated: June 5, 2021 — 11:48 pm

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