NCTB Class 7 Math Chapter Five Exercise 5.3 Solution

NCTB Class 7 Math Chapter Five Exercise 5.3 Solution by Math Expert. Bangladesh Board Class 7 Math Solution Chapter 5.3 “Algebraic Formulaic and Applications” Exercise 5.3 Solution.

Board

NCTB
Class

7

Subject

Math
Chapter

5

Chapter Name

“Algebraic Formulaic and Applications”
Exercise

5.3 Solution

Exercise:- 5.3

(1) Solution:- x2+x y +z x +y z

= x( x +y) + z (x +y)

= (x +y) (x +z)

(2) Solution:- a2+bc+ ca+ a b

= a2+ a b +ca +b c

= a (a +b) + c(a +b)

= (a +c) (a +b)

(3) Solution:- a b(p x+ q y)+a2q x +b2x + b2 p y

= a b p x + a b q y + a2 q x+ b2 p y

= a b p x+ a2 q x+ b2 p y + a b q y

= a x (b p +a q) +by (b p +a q)

= (a x +by) (b p +a q)

(4) Solution:- 4x2-y2

= (2x)2 – (y)2

= (2x+y) (2x-y)

(5) Solution:- 9a2 – 4b2

= (3b)2 – (2b)2

= (3a= 2b) (3a-2b)

(6) Solution:-  a2 b2 – 49y2

= (a b)2 – (7y)2

= (ab+7y) (ab-7y)

(7) Solution:- 16x4 – 81y4

= (4x2)2 – (9y2)2

= (4x2+9y2) (4x2-9y2)

= (4x2+9y2) {(2x)2 – (3y)2}

= (4x2+9y2) (2x+3y) (2x- 3y)

(8) Solution:- a2 – (x +y)2

= (a+ x+ y) (a-x-y)

(9) Solution:- (2x-3y+5z)2 – (x-2y+3z)2

= (2x-3y+5z+x-2y+3z) (2x-3y+5z-x+2y-3z)

= (3x-5y+8z) (x-y+2z)

(10) Solution:- (4a+ 8a2+9a4)

= (2)2+2x 2x 3a2+ (3a2)2 – 4a2

= (2+3a2)2 – (2a)2

= (2+ 3a2+2a) (2+  3a2-2a)

(11) Solution:- 2a2+6a-80

= 2 (a2+3a-40)

= 2 (a2+8a-5a-40)

= 2{a (a+8) – 5 (a+8)}

= 2 (a+8) (a-5)

(12) Solution:- y2-6y- 91

= y2 – 13y+7y- 91

= y(y-13) + 7(y-13)

= (y-13) (y+7)

(13) Solution:- p2– 15p+ 56

= p2– 8p-7p+56

= p (p-8) -7(p-8)

= (p-8) (p-7)

(14) Solution:- 45a8 – a4x4

= 5a4 (9a4-x4)

= 5a4 {(3a2)2 – (x2)2}

= 5a4 (3a2+x2) (3a2-x2)

(15) Solution:- a2+3a-40

= a2+8a-5a-40

= a (a+8) -5 (a+8)

= (a+8) (a-5)

(16) Solution: (x2+1)2 – (y2+1)2

= {(x2+1) + (y2+1) }{(x2+1) – (y2+1)}

= (x2+1+y2+1) (x2+1-y2-1)

= (x2+y2+2) (x +y) (x-y)

= (x2+y2+2) (x +y) (x-y)

(17) Solution:- x2+11x+30

= x2+5x+6x+30

= x (x+5) 6 (x=5)

= (x+6) (x+5)

(18) Solution:- a2-b2+2bc-c2

= a2-(b2-2bc+c2)

= (a)2 – (b-c)2

= (a+ b-c) (a-b +c)

(19) Solution:- 144x7– 25x3 a4

= x3 (144x4– 25a4)

= x3{(12x2)2 – (5a2)2}

= x3 (12x2+5a2) (12x2-5a2)

(20)  Solution:- 4x2+12xy+9y2-16a2

= (2x)2 + 2X 2 x X 3y+ (3y)2 – 16a2

= (2x+3y)2 – (4a)2

= (2x+3y+4a) (2x+3y-4a)

Updated: December 17, 2020 — 7:23 am

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  1. Thanks for this solutions

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