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

10 Comments

Add a Comment
  1. Thanks for this solutions

    1. BRO!! Which book you are using?

  2. where is the solution of 2021 year?

  3. Plz give the solution of year 2021. I need it urgent!!!!!

    1. Book link Send Please

  4. there are also some wrong maths. where is the correct?

    1. Cadet SM Alif bin Ashraf

      Like no 14

    2. Cadet SM Alif bin Ashraf

      Like num 14

  5. Cadet SM Alif bin Ashraf

    New curriculum for 2023 please

  6. Cadet Alif bin Ashraf

    New curriculum for 2023 please

Leave a Reply

Your email address will not be published. Required fields are marked *