# West Bengal Board Class 8 Math Chapter 4 Solution

## Chapter 4 Solution West Bengal Board : Class 8 Mathematics

West Bengal Board Class 8 Math Solution full Exercise 4.1 and 4.2 by Experts. Here in this page WB Board Class 8 Student can find Ganit Prabha Class 8 Math Chapter 4 (বীজগাণিতিক  সংখ্যামালার গুণ ও ভাগ) Multiplication and Division of Algebraic Numbers Exercise 4.1, 4.2 Solution.

 Board WB Board Class 8 (Eight) Subject Mathematics Book Name Ganit Prabha Chapter 4

### 4.1 Solution

(1)

(b) (x2+12-7y) (2x-y)

= (2x – x y + 24x – 12y – 14xy + 7y)

= 2 x (-2)3 – (-2)3 x 2 + 24 x (-2) – 12 x 2 – 14 x (-2) (2) + 7 x (2)2

= – 16 – 4 x 2 – 48 – 24 + 56 + 28

= -16 – 8 – 48 – 24 + 56 + 28

= – 24 – 48 – 24 + 56 + 28

= – 12

(d) (6a + 5b + 2) (a – b + 6)

= (6a2 – 6ab + 36a + 5ab – 5b2 + 30b +2a – 2b + 12)

= (6a2 – ab + 38a – 5b2 + 28b + 12)

= (6 x 02 – 0 x (-1) + 38 x 0 – 5 (-1)2 + 28 x (-1) + 12)

= – 5 – 28 + 12

= – 21

(e) (p3 – p2q2 + q3) (p2 + pq + q2)

= (p5 + p4q + p3q2 – p4q2 – p3q3 – p2q4 + p2q3 + pq4 + q5)

= 25 + 24 (-2) + 23 (-2)2 – 24 (-2)2 – (2)3 (-2)3 – 22 (-2)4 + (2)2 (-2)3 + 2 (-2)4 + (-2)5

= – 64

(2)

(i) (x5 + 1) (3 – x4) (4 + x3 + x6)

= (3x5 – x9 + 3 – x4) (4 + x3 + x6)

= (12x5 + 3x8 + 3x11 – 4x9 – x12 – x15 + 12 + 3x3 + 3x6 – 4x4 – x7 – x10)

= – x15 – x12 + 3x11 – x10 – 4x9 + 3x8 – x7 + 3x6 + 12x5 – 4x4 + 3x3 + 12

(ii) (2a3 – 3b5) (23a3 + 3b5) (2a4 – 3a2b2 + b4)

= [(2a3)2 – (3b5)2] (2a4 – 3a2b2 + b4)

= (4a6 – 9b10) (2a4 – 3a2b2 + b4)

= (8a10 – 12a8b2 + 4a6b4 – 18a4b10 + 27a2b12 – 9b14)

(iii) (ax + by) (ax – by) (a4x4 + a2b2x2y2 + b4y4)

= [(ax)2 – (by)2] (a4x4 + a2b2x2y2 + b4y4)

= (a2x2 – b2y2) (a4x4 + a2b2x2y2 + b4y4)

= a6x6 +a4b2x4y2 + a2b4x2y4 – a4x4b2y2 – a2b4x2y4 – b6y6

= a6x6– b6y6

(iv) (a + b + c) (a – b + c) (a + b – c)

= (a2 – ab + ac + ab – b2 + bc + ac – bc + c2) (a + b + c)

= (a2 + 2ac – b2 + c2) (a + b + c)

= (a3 + a2b + a2c + 2a2c + 2abc + 2ac2 – ab2 – b3 – b2c + ac2 + bc2 + c3)

= a3 – b3 + c3 + 2abc – ab2 + a2c + a2b + c2b – b2c + 2a2c

(v) (2p2/q2 + 5q2/p2) (2p2/q2 – 5q2/p2)

= (2p2/q2)2 – (5q2/p2)2

= 4p4/q4 – 25q4/p4

(3)

(i) (x + y) (x2 – xy + y2) + (x – y) (x2 + xy + y2)

= (x3 + y3) + (x3 – y3)

= x3 + y3 + x3 – y3

= 2x3

(ii) a2(b2 – c2) + b2 (c2 – a2) + c2 (a2 – b2)

= a2b2 – a2c2 + b2c2 – b2a2 + a2c2 – c2b2

= 0

(4)

(i) a = x2 + xy + y2

b = y2 + yz + z2

c = z2 + xz + x2

(x – y) a + (y – z) b + (z – x) c

= (x – y) (x2 + xy + y2) + (y – z) (y2 + yz + z2) + (z – x) (z2 + xz + x2)

= (x3 + x3y + xy2 – x2y – xy2 – y3 + y3 + y2z + yz2 – zy2 – yz2 – z3 + z3 + xz3 + x2z – xz2 – x2z – x3)

= 0

(ii) a (m+n) + b (n + l) + c (l + m)

= (lm + my +n) (m + n) + (mx + ny + l) (n + l) + (nx + ly + m) (l + m)

= lmn + lnx + m2y + mny + mn + n2 + mnx + mlx + n2y + nly + ln + l2 + nlx + mnx + l2y + mly + ml + m2

= y (l2 + m2 + n2 + lm +mn +nl) + 2x (lm + ln + mn) + m2 + n2 + l2 + lm + mn + ln

### Exercise 4.2 Solution

Updated: September 11, 2020 — 3:13 pm