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)