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