NCTB Class 8 Math Chapter Four Exercise 4.4 Solutions by Math Expert. Bangladesh Board Class 8 Math Solution Chapter Four Algebraic Formulae and Applications Exercise 4.4 Solution.
Board |
NCTB |
Class |
8 |
Subject |
Mathematics |
Chapter |
4 |
Chapter Name |
Algebraic Formulae and Applications |
Exercise |
4.4 Solution |
Exercise: 4.4
1> Which of the following is the square of (-5-y)?
= (-5)^{2}-2.(-5).(y)+y^{2=}
= 25+10y+y^{2}
2> which one of the following is the product of (x-2) and (4x+3)
(x-2)(4x+3)
= 4x^{2}+3x-8x-6
= 4x^{2}-5x-6
Ans: (a) (d)
3> what is the H.C.F of x^{2}-2x-3 and x^{2}+2x-3
x^{2}-2x-3
= x^{2}-3x+1x-3
= x(x-3)+1(x-3)
= (x-3) (x+1)
4> which of the following will be right if we express (3x-5)(5+3x) in the form of difference between two square:
(3x-5)^{2}-(5+3x)^{2}
= {(3x)^{2}-2.3x.5+5^{2})-(5)^{2}+2.5x.3x+(3x)^{2}}
= (9x^{2}-30x+25)-(25+30x+9x^{2})
= 9x^{2}-30x+25-25+30x-9x^{2}
= -60x
8> which one of the following expressions in the factor of x^{2}-x-30 ?
x^{2}-x-30
= x^{2}-6x+5x-30
= x (x-6)+5(x-6)
= (x+5)(x-6) (b) Answer.
9> If x^{2}-10x+21 and x^{2}-6x-7 are two algebraic expressions –
(a) x^{2}-10x+21
= x^{2}-7x-3x+21
=x(x-7)-3(x-7)
(b) x^{2}-6x-7
= x^{2}+7x-x-7
= x(x+7)-1(x+7)
= (x+7)(x-1)
Product = (x^{2}-10x+21)( x^{2}-6x-7)
= x^{4}+6x^{3}-7x^{2}-10^{3}-60x^{2}+70x+21x^{2}+126x-147
None of the following are correct
10> In algebraic formulae:
According to the question the correct answer is (d)
11> If x+y=5 and x-y=3, then
x+y=5;
x-y=3
or, (x-y)^{2}=3^{2}
or, (x+y)^{2}-4xy=9
or, (5)^{2}-4xy=9
or, -4xy = 9-25
= -16
Or, xy=4
X^{2}+y^{2}
= (x+y)^{2}-2.xy
= (5)^{2}-2.4
= 25-8
= 17 (c)
(2) The value of xy is = 4
(3) The value of (x^{2}-y^{2})
= (x+y)(x-y)
= 5 x 3
= 15 (c)
12> If x+1/x=2, then ,
(1) (X-1/x)^{2}
= (x+1/x)^{2}-4.x.1/x
= (2)^{2}-4
= 0
(2) x^{3}+1/x^{3}
= (x+1/x)^{3}-3.x.1/x (x+1/x)
= (2)^{3}-3.2
= 8-6
= 2
13> Find the H.C.F of the following :
36a^{2}b^{2}c^{4}d^{5}, 54a^{5}c^{2}d^{4}, 90a^{4}b^{3}c^{2}
Ans is = 18a^{2}c^{2}
14> 20x^{3}y^{2}a^{3}b^{4}, 15x^{4}y^{3}a^{4}b^{3} and 35x^{2}y^{4}a^{3}b^{2}
Ans is = 5x^{2}y^{2}a^{3}b^{2}
15> 15x^{2}y^{3}z^{4}a^{3}, 12x^{3}y^{2}z^{3}a^{4} and 27x^{3}y^{4}z^{5}a^{7}
Ans is = 3x^{2}y^{2}z^{3}a^{3}
16> 18a^{3}b^{4}c^{5}, 42a^{4}c^{3}d^{4}, 60b^{3}c^{4}d^{5} and 78a^{2}b^{4}d^{3}
Ans is = 6
17> x^{2}-3x= x(x-3)
X^{2 }– 9 = (x)^{2}-(3)^{2}
= (x+3)(x-3)
X^{2}-4x+3
= x^{2}-3x-x+3
= x(x-3)-1(x-3)
=(x-1)(x-3)
H.C.F of this expressions = (x-3)
18> 18 (x+y)^{3}, 24(x+y)^{2}, 32(x^{2}-y^{2})
1^{st} expression , 18(x+y)^{3=}
2^{nd} expression , 24(x+y)^{2}
3^{rd} expression, 32(x^{2}-y^{2})
Therefore H.C.F is = 2(x+y).
20>
1^{st}, a^{3}-3a^{2}-10a
= a(a^{2}-3a-10)
= a(a^{2}-5a+2a-10)
= a {a(a-5)+2(a-5)}
= a (a-5)(a+2)
2^{nd}, a^{3}+6a^{2}+8a
= a(a^{2}+6a+8)
= a(a^{2}+4a+2a+8)
= a {a(a+4)+2(a+4)}
= a (a+4)(a+2)
3^{rd}, a^{4}-5a^{3}-14a^{2}
= a^{2}(a^{2}-5a-14)
= a^{2}(a^{2}-7a+2a-14)
= a^{2}{a(a+7)+2(a-7)}
= a^{2}(a-7)(a+2)
H.C.F is = a(a+2)
21> 1^{st}, a^{5}b^{2}c
2^{nd}, ab^{3}c^{2}
3^{rd}, a^{7}b^{4}c^{3}
Therefore, L.C.M is = a^{7}b^{4}c^{3}
22> 1^{st}, 5a^{2}b^{3}c^{2}
2^{nd}, 10ab^{2}c^{3}
3^{rd}, 15ab^{3}c
Therefore, L.C.M is = 30a^{2}b^{3}c^{3}
23> 1^{st}, 3x^{3}y^{2}
2^{nd}, 4xy^{3}z
3^{rd}, 5x^{4}y^{2}z^{2}
4^{th}, 12xy^{4}z^{2}
Therefore, L.C.M is = 120x^{4}y^{4}z^{2}
24> 1^{st}, 3a^{2}d^{3}; 2^{nd}, 9d^{2}b^{2}; 3^{rd}, 12c^{3}d^{2}; 4^{th}, 24a^{3}b^{2}; 5^{th}, 35c^{3}d^{2}
Therefore, L.C.M is 72a^{3}b^{2}c^{3}d^{3}
25> 1^{st}, x^{2}+3x+2
= x^{2}+2x+x+2
= x(x+2)+1(x+2)
= (x+2)(x+1)
2^{nd}, x^{2}-1
= (x+1)(x-1)
3^{rd}, x^{2}+x-2
= x^{2}+2x-x-2
= x(x+2)-1(x+2)
= (x-1)(x+2)
Therefore, L.C.M is (x+2)(x+1)(x-1)
26> 1^{st} term, x^{2}-4= x^{2}-2^{2}=(x+2)(x-2)
2^{nd}, x^{2}+4x+4
= x^{2}+2x+2x+4
= x(x+2)+2(x+2)
= (x+2)^{2}
3^{rd}, x^{3}-8 = x^{3}-2^{3}
= (x-2)(x^{2}+2x+4)
Therefore, L.C.M is (x+2)^{2}(x-2)(x^{2}+2x+4)
27> 1^{st}, 6x^{2}-x-1
= 6x^{2}-3x+2x-1
= 3x (2x-1)+1(2x-1)
= (3x+1)(2x-1)
2^{nd}, 3x^{2}+7x+2
= 3x^{2}+6x+x+2
= 3x(x+2)+1(x+2)
= (3x+1)(x+2)
3^{rd}, 2x^{2}+3x-2
= 2x^{2}+4x-x-2
= 2x (x+2)-1(x+2)
= (2x-1)(x+2)
Therefore, L.C.M is (3x+1)(2x-1)(x+2)
29> x^{2}+ 1/x^{2}=3
Or, (x+1/x)^{2}-2.x.1/x =3
Or, (x+1/x)^{2}-2 =3
(b) x^{6}+1/x^{3}
= x^{3}+1/x^{3}
= (x+1/x)^{3}-3.x.1/x (x+1/x)
= – 3
= 2
In math number 23, the LCM is wrong. It will be 60.