# NCTB Class 8 Math Chapter Four Exercise 4.4 Solution

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)+y2=

= 25+10y+y2

2> which one of the following is the product of (x-2) and (4x+3)

(x-2)(4x+3)

= 4x2+3x-8x-6

= 4x2-5x-6

Ans: (a) (d)

3> what is the H.C.F of x2-2x-3 and x2+2x-3

x2-2x-3

= x2-3x+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+52)-(5)2+2.5x.3x+(3x)2}

= (9x2-30x+25)-(25+30x+9x2)

= 9x2-30x+25-25+30x-9x2

= -60x

8> which one of the following expressions in the factor of x2-x-30 ?

x2-x-30

= x2-6x+5x-30

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

9>  If x2-10x+21 and x2-6x-7 are two algebraic expressions –

(a) x2-10x+21

= x2-7x-3x+21

=x(x-7)-3(x-7)

(b) x2-6x-7

= x2+7x-x-7

= x(x+7)-1(x+7)

= (x+7)(x-1)

Product = (x2-10x+21)( x2-6x-7)

= x4+6x3-7x2-103-60x2+70x+21x2+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=32

or, (x+y)2-4xy=9

or, (5)2-4xy=9

or, -4xy = 9-25

= -16

Or, xy=4

X2+y2

= (x+y)2-2.xy

= (5)2-2.4

= 25-8

= 17 (c)

(2) The value of xy is = 4

(3) The value of (x2-y2)

= (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) x3+1/x3

= (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 :

36a2b2c4d5, 54a5c2d4, 90a4b3c2

Ans is = 18a2c2

14> 20x3y2a3b4, 15x4y3a4b3 and 35x2y4a3b2

Ans is = 5x2y2a3b2

15> 15x2y3z4a3, 12x3y2z3a4 and 27x3y4z5a7

Ans is = 3x2y2z3a3

16> 18a3b4c5, 42a4c3d4, 60b3c4d5 and 78a2b4d3

Ans is = 6

17> x2-3x= x(x-3)

X2 – 9 = (x)2-(3)2

= (x+3)(x-3)

X2-4x+3

= x2-3x-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(x2-y2)

1st expression , 18(x+y)3=

2nd expression , 24(x+y)2

3rd expression, 32(x2-y2)

Therefore H.C.F is = 2(x+y).

20>

1st, a3-3a2-10a

= a(a2-3a-10)

= a(a2-5a+2a-10)

= a {a(a-5)+2(a-5)}

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

2nd, a3+6a2+8a

= a(a2+6a+8)

= a(a2+4a+2a+8)

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

= a (a+4)(a+2)

3rd, a4-5a3-14a2

= a2(a2-5a-14)

= a2(a2-7a+2a-14)

= a2{a(a+7)+2(a-7)}

= a2(a-7)(a+2)

H.C.F is = a(a+2)

21> 1st, a5b2c

2nd, ab3c2

3rd, a7b4c3

Therefore, L.C.M is = a7b4c3

22> 1st, 5a2b3c2

2nd, 10ab2c3

3rd, 15ab3c

Therefore, L.C.M is = 30a2b3c3

23> 1st, 3x3y2

2nd, 4xy3z

3rd, 5x4y2z2

4th, 12xy4z2

Therefore, L.C.M is = 120x4y4z2

24> 1st, 3a2d3; 2nd, 9d2b2; 3rd, 12c3d2; 4th, 24a3b2; 5th, 35c3d2

Therefore, L.C.M is 72a3b2c3d3

25> 1st, x2+3x+2

= x2+2x+x+2

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

= (x+2)(x+1)

2nd, x2-1

= (x+1)(x-1)

3rd, x2+x-2

= x2+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> 1st term, x2-4= x2-22=(x+2)(x-2)

2nd, x2+4x+4

= x2+2x+2x+4

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

= (x+2)2

3rd, x3-8 = x3-23

= (x-2)(x2+2x+4)

Therefore, L.C.M is (x+2)2(x-2)(x2+2x+4)

27> 1st, 6x2-x-1

= 6x2-3x+2x-1

= 3x (2x-1)+1(2x-1)

= (3x+1)(2x-1)

2nd, 3x2+7x+2

= 3x2+6x+x+2

= 3x(x+2)+1(x+2)

= (3x+1)(x+2)

3rd, 2x2+3x-2

= 2x2+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> x2+ 1/x2=3

Or, (x+1/x)2-2.x.1/x =3

Or, (x+1/x)2-2 =3

(b) x6+1/x3

= x3+1/x3

= (x+1/x)3-3.x.1/x (x+1/x)

= – 3

= 2

Updated: March 25, 2021 — 1:22 pm