NCTB Class 7 Math Chapter Five Exercise 5.4 Solution by Math Expert. Bangladesh Board Class 7 Math Solution Chapter 5.4“Algebraic Formulaic and Applications” Exercise 5.4 Solution.

Board |
NCTB |

Class |
7 |

Subject |
Math |

Chapter |
Five |

Chapter Name |
“Algebraic Formulaic and Applications” |

Exercise |
5.4 Solution |

__Exercise:- 5.4__

__Exercise:- 5.4__

**(1) Solution:-** (a-5)^{2} = a^{2}– 2xax5= (5)^{2}

(b) = a^{2}-10a+25

**(2) Solution:-** (x +y)^{2}+ 2(x +y) (x-y) + (x-y)^{2}

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

= (2x)^{2}

= 4x^{2}

**(3) Solution:-** We know, 4ab = (a +b)^{2} – (a-b)^{2}

=> 4ab = 4^{2}-2^{2}

= 16-4

= 12

=> a b = 12/4 = 3 (a)

**(4) Solution:-** If a quantity is divisible without remainder another quantity is called – (c) Multiple

**(5) Solution:-** Least common multiple of a, a^{2}, a (a +b) is a^{2} (a +b) (d) Answer.

**(6) Solution:-** H.C.F of 2a and 3b is 1 (a)

**(7) Solution:-** If a, b are real numbers – (d) (i), (ii), (iii) are correct.

**(8) Solution:-** (d) x y (x +y) (x-y)

**(9) Solution:-** H.C.F if two algebraic expression, (d) x (x-y)

**(10) Solution:-** The L.C.M is — (d) x y (x +y) (x+2y)

**(11) Solution:-** L.C.M of 9x^{2}– 25y^{2} and 15ax-25xy is (d) 5a (9x^{2}-25y^{2})

**(12) Solution:-** H.C.F of x^{3} y^{5} and a^{2}-b^{2} is (d) 1.

**(13) Solution:-** x-1/x = 0; (d) (i), (ii), (iii) Correct.

**(14) Solution:-** a+1/a= 4 ; a^{2}-4a+1= (d) 0.

**(15) Solution:-** (a+5)^{2} = a^{2}+2.a.5+25

= a^{2}+10a+25

**(16) Solution:-** We know, 4ab= (a +b)^{2} – (a-b)^{2}

=> 4ab = (8)^{2} – (4)^{2}

= 64-16

= 48

=> a b= 48/4

= 12

**(17) Solution:-** 1^{st}, 3a^{2} b^{2} c= 3^{2} x a^{2} x a x a xb^{2}xb^{2}xc^{2}

2^{nd}, 6ab2 c2= 2×3^{2}x a^{2}xb^{2}xb^{2}xc^{2}xc

Therefore, H.C.F= 3a^{2} b^{2}c

**(18)Solution:-** 1^{st}, 5ab^{2} x^{2}= 5^{2}xa^{2}xb^{2}xbxXxX

2^{nd}, 10a^{2 }by^{2} = 2×5^{2}xa^{2}xaxaxb^{2}xyxy

H.C.F = 5ab

**(19) Solution:-** 1^{st}, 3a^{2 }x^{2} = 3^{2}xa^{2}x ax X x

2^{nd}, 6axy^{2} = 2×3^{2}x a^{2}x X x y^{2} x y

3^{rd}, 9 ay^{2}= 3 x3^{2}xa^{2 }xy^{2 }x y

H.C.F = 3a

**(20) Solution:-** 1^{st}, 16a^{3} x^{4} y = 2x2x2x2xa^{3}xX^{4}xy

2^{nd}, 40a^{2} y^{3} x= 2^{2}x 2x 2x5xa^{2}xy^{3}x X

3^{rd}, 28a x^{3}= 2^{2}x2 x7 xa^{2}x X^{3}

H.C.F= 2ax

**(21) Solution:- **1^{st}, a^{2}+ab = a (a +b)

2^{nd}, a^{2}-b^{2}= (a +b) (a-b)

Therefore, a2- b2= (a +b) (a-b)

Therefore, H.C.F = (a +b)

**(22) Solution:-** 1^{st}, x^{3}y – xy^{3}

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

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

2^{nd}, (x-y)^{2 }= (x-y) (x-y)

Therefore, H.C.F = (x-y)

**(23) Solution:-** 1^{st}, x^{2} +7x+12

= x^{2}+4x+3x+12

= x (x+4) +3(x+4)

= (x+4) (x+3)

2nd, x^{2}+9x+20

= x^{2} + 5x+ 4x+20

= x(x+5) +4 (x+5)

= (x+5) (x+4)

**(24) Solution:-** 1^{st}, a^{3}-ab^{2}= a (a^{2}-b^{2})

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

2^{nd}, a^{4}+2a^{3}b + a^{2} b^{2}

= a^{2 }(a^{2}+2ab+b^{2})

= a^{2} (a +b)^{2}

= a. a (a +b) (a +b)

Therefore, H.C.F = a (a +b)

**(25) Solution:-** 1^{st}, a^{2}-16= (a)^{2} – (4)^{2}

= (a+4) (a-4)

2^{nd}, 3a +12 = 3 (a+4)

3^{rd}, a^{2}+ 5a +4

= a^{2}+ 4a + a+ 4

= a (a +4) +1(a +4)

= (a +4) (a +1)

**(27) Solution:-** 1^{st}, 6a^{3} b^{2} c

= 2x 3x a^{3}x b^{2}xc

2^{nd}, 9a^{4} bd^{2}= 3 x3 xa^{4}x b x d^{2}

Therefore, L.C.M = 2x 3x 3xa^{4}xb^{2}xc xd^{2}

= 18a^{4} b^{2} cd^{2}

**(28) Solution:-** 1^{st}, 5x^{2} y^{2}= 5x x^{2}xy^{2}

2^{nd}, 10xz^{3}= 2x 5x X xz^{3}

3^{rd}, 15y^{3} z^{4}= 3x 5 x y^{3}x z4

Therefore, L.C.M= 2x 3 x5x X^{2}x y^{3}xz^{4}

= 30x^{2} y^{3} z^{4}

**(29) Solution:-** 1^{st}, 2p^{2} xy2= 2x p^{2}x X x y^{2}

2^{nd}, 3p q^{2 }= 3x p x q^{2}

3^{rd}, 6pqx^{2 }= 2 x 3 x p x q x X^{2}

Therefore, L.C.M= 6p^{2} q^{2} x^{2} y^{2}

**(30) Solution:-** 1^{st}, (b^{2}-c^{2})= (b +c) (b-c)

2^{nd}, (b +c)^{2} = (b +c)

Therefore, L.C.M= (b +c)^{2} (b-c)

**(31) Solution:-** 1^{st}, x^{2}=2x

= x( x+2)

2^{nd}, x^{2}+3x+2

= x(x+2) +1(x+2)

= (x+2) (x-1)

Therefore, L.C.M= x (x+1) (x+2)

**(32) Solution:-** 1^{st}, 9x^{2}-25y^{2}

= (3x)^{2} – (5y)^{2}

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

2^{nd}, 15ax-25ay = 5a (3x-5y)

**(33) Solution:-** 1^{st}, x^{2}-3x-10

= x^{2}-5x=2x-10

= (x-5) (x+2)

2^{nd}, x^{2}-10x+25

= x^{2}– 5x-5x+25

= (x-5) -5 (x-5)

= (x-5) (x-5)

= (x-5)^{2}

Therefore, L.C.M= (x-5)^{2} (x+2)

**(34) Solution:-** 1^{st}, a^{2}-7a+12

= a^{2}-3a-4a+12

= a (a-3) -4 (a-3)

= (a-3) (a-4)

2^{nd}, a^{2}+a-20

= a^{2}+5a-4a-20

= a(a+5) -4(a+5)

= (a-4) (a+5)

3^{rd}, a^{2}+ 2a-15

= a^{2}+5a-3a-15

= (a-5) -3(a+5)

= (a-3) (a+5)

Therefore, L.C.M, (a-3) (a-4) (a+5)

**(35) Solution:-** 1^{st}, x^{2}-8x+15

= x^{2}-5x-3x+15

= x (x-5) -3(x-5)

= (x-5) (x-3)

2^{nd}, x^{2}-25= (x+5) (x-5)

3^{rd}, x^{2}+2x-25

= x^{2}-3x+5x-15

= x (x-3) +5 (x-3)

= (x-3) (x+5)

Therefore, L.C.M = (x-3) (x2-25)

**(36) Solution:-** 1^{st}, (x+5)

2^{nd}, x^{2}+5x

= x (x+5)

3^{rd}, x^{2}+7x+10

= (x^{2}+5x+2x+10)

= (x+5) (x+2)

Therefore, L.C.M= x (x+5) (x+2)

**(37) (a)** **Solution:-** (a +b)

= 2x-3+2x+5

= (4x+2)

**(b) Solution:- **a^{2}= (2x-3)^{2}

= (2x)^{2}– 2. 2x.3+ (3)2

= 4x^{2}-12x +9

**(c) Solution:-** a b

= (2x-3) (2x=50

= 4x^{2}+10x-6x-15

= 4x^{2}+4x-15

If, x= 2;

4x (2)^{2}+ 4×2-15

= 16+8-15

= 24-15

= 9

**(38) (a) Solution:-** x^{2}+3x-10

= x^{2}+5x-2x-10

= x(x+5)-2(x+5)

= (x+5) (x-2)

**(b) Solution:-** 1^{st}, x^{2}-625

= (x)^{2}– (25)^{2}

= (x+25) (x-25)

= (x+25){(x)^{2} – (5)^{2}}

= (x+25) (x+5) (x-5)

2^{nd}, x^{2}+3x-10

= (x=5) (x-2)

L.C.M = (x+5) (x-2) (x-5) (x+25)

**(c) Solution:-** We get,

From 1^{st} expression, (x+5) (x-2)

2^{nd}, expression (x^{2}+25) (x+5) (x-5)

L.C.M= (x+5) (x-2) (x-5) (x^{2} +25)

**(39) Solution:-** **(a)** (3x-2y+z)^{2}

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

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

= (3x)^{2} – 2.3x.2y+ (2y)^{2}+ 6xz- 4yz +z^{2}

= 9x^{2}+ 4y^{2}+z^{2}– 12xy +6x^{2}– 4yz

**(b) Solution:-** 1^{st}, x^{2}-3x-10

= (x+2)(x-3)

2rd, X^{3}+6x^{2}+8x

= x(x+6x+8)

= x(x^{2}+4x+2x+8)

= x{x (x+4) +2 (x+4)}

= x(x+4) (x+2)

Therefore, H.C.F= (x+2)

**(c) Solution:-** 3^{rd}, expression, x^{4}-5x^{3}-14x^{2}

= x^{2}(x^{2}-5x-14)

= x^{2}(x^{2}-7x+2x-14)

= x^{2}{x(x-7) +2(x-7)}

= x^{2} (x-7) (x+2)