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:-** x^{2}+x y +z x +y z

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

= (x +y) (x +z)

**(2) Solution:-** a^{2}+bc+ ca+ a b

= a^{2}+ a b +ca +b c

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

= (a +c) (a +b)

**(3) Solution:-** a b(p x+ q y)+a^{2}q x +b^{2}x + b^{2} p y

= a b p x + a b q y + a^{2} q x+ b^{2} p y

= a b p x+ a^{2} q x+ b^{2} 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:-** 4x^{2}-y^{2}

= (2x)^{2} – (y)^{2}

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

**(5) Solution:-** 9a^{2} – 4b^{2}

= (3b)^{2} – (2b)^{2}

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

**(6) Solution:-** a^{2} b^{2} – 49y^{2}

= (a b)^{2} – (7y)^{2}

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

**(7) Solution:-** 16x^{4} – 81y^{4}

= (4x^{2})^{2} – (9y^{2})^{2}

= (4x^{2}+9y^{2}) (4x^{2}-9y^{2})

= (4x^{2}+9y^{2}) {(2x)^{2} – (3y)^{2}}

= (4x^{2}+9y^{2}) (2x+3y) (2x- 3y)

**(8) Solution:-** a^{2} – (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+ 8a^{2}+9a^{4})

= (2)^{2}+2x 2x 3a^{2}+ (3a^{2})^{2} – 4a^{2}

= (2+3a^{2})^{2} – (2a)^{2}

= (2+ 3a^{2}+2a) (2+ 3a^{2}-2a)

**(11) Solution:- **2a^{2}+6a-80

= 2 (a^{2}+3a-40)

= 2 (a^{2}+8a-5a-40)

= 2{a (a+8) – 5 (a+8)}

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

**(12) Solution:-** y^{2}-6y- 91

= y^{2} – 13y+7y- 91

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

= (y-13) (y+7)

**(13)** **Solution:-** p^{2}– 15p+ 56

= p^{2}– 8p-7p+56

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

= (p-8) (p-7)

**(14) Solution:-** 45a^{8} – a^{4}x^{4}

= 5a^{4} (9a^{4}-x^{4})

= 5a^{4 }{(3a^{2})^{2} – (x^{2})^{2}}

= 5a^{4} (3a^{2}+x^{2}) (3a^{2}-x^{2})

**(15) Solution:-** a^{2}+3a-40

= a^{2}+8a-5a-40

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

= (a+8) (a-5)

**(16) Solution:–** (x^{2}+1)^{2} – (y^{2}+1)^{2}

= {(x^{2}+1) + (y^{2}+1) }{(x^{2}+1) – (y^{2}+1)}

= (x^{2}+1+y^{2}+1) (x^{2}+1-y^{2}-1)

= (x^{2}+y^{2}+2) (x +y) (x-y)

= (x^{2}+y^{2}+2) (x +y) (x-y)

**(17) Solution:-** x^{2}+11x+30

= x^{2}+5x+6x+30

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

= (x+6) (x+5)

**(18) Solution:-** a^{2}-b^{2}+2bc-c^{2}

= a^{2}-(b^{2}-2bc+c^{2})

= (a)^{2} – (b-c)^{2}

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

**(19)** **Solution:-** 144x^{7}– 25x^{3} a^{4}

= x^{3} (144x^{4}– 25a^{4})

= x^{3}{(12x^{2})^{2} – (5a^{2})^{2}}

= x^{3} (12x^{2}+5a^{2}) (12x^{2}-5a^{2})

**(20) Solution:-** 4x^{2}+12xy+9y^{2}-16a^{2}

= (2x)^{2} + 2X 2 x X 3y+ (3y)^{2} – 16a^{2}

= (2x+3y)^{2} – (4a)^{2}

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

Thanks for this solutions

BRO!! Which book you are using?

where is the solution of 2021 year?

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

Book link Send Please

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

Like no 14

Like num 14

New curriculum for 2023 please

New curriculum for 2023 please