Ncert Solutions For Class 8 Maths Ex 14.3

Ncert Solutions For Class 8 Maths Chapter 14 Ex 14.3

 Question 1:

Solve the following:

i) 18a4 ¸ 36a

ii) -45a3 ¸ 9a2

iii) 77xy2z3 ¸ 11xy2

iv) 51a3b3c3 ¸ 68ab2c3

v) 18x8y8 ¸ (-6x6y4)

 

sol.

i) 18a4 ¸ 36a

= 18a436a=1836a4a=12a3

[Since, xm ¸ xn = xm-n]

 

ii) -45a3 ¸ 9a2

= 45a39a2=459a3a2=5a

 

iii) 77xy2z3 ¸ 11xy2

= 77xy2z311xy2=7711xy2z3xy2=7z3

 

Iv) 51a3b3c3 ¸ 68ab2c3

= 51a3b3c368ab2c3=5168a3b3c3ab2c3=34a2b

 

v) 18x8y8 ¸ (-6x6y4)

= 18x8y86x6y4=186x8y8x6y4=3x2y4

 

Question 2:

Divide the following polynomial by the following monomial:

i) (6x2 – 3x) ¸ 3x

ii) (4y8 – 3y6 + 6y4) ¸ y4

iii) 8(a3b2c2 + a2b3c2 + a2b2c3) ¸ 4a2b2c2

iv) (a3 + 2a2 + 3a) ¸ 2a

 

sol.

i) (6x2 – 3x) ¸ 3x = 6x23x3x

= 6x23x3x3x=2x1

 

ii) (4y8 – 3y6 + 6y4) ¸ y4 = 4y83y6+6y4y4

= 4y8y43y6y4+6y4y4=4y43y2+6

 

iii) 8(a3b2c2 + a2b3c2 + a2b2c3) ¸ 4a2b2c2 = 8(a3b2c2+a2b3c2+a2b2c3)4a2b2c2

=8a3b2c24a2b2c2+8a2b3c24a2b2c2+8a2b2c34a2b2c2=2a+2b+2c=2(a+b+c)

 

iv) (a3 + 2a2 + 3a) ¸ 2a = a3+2a2+3a2a

= a32a+2a22a+3a2a=12a2+22a+32=12(a2+2a+3)

 

Question 3:

Solve:

i) (10a – 25) ¸ 5

ii) (10a – 25) ¸ (2a – 5)

iii) 10a(6a + 21) ¸ 5(2a +7)

iv) 9a2b2(3c – 24) ¸ 27ab(c – 8)

sol.

i) (10a – 25) ¸ 5 = 10a255

= 5(2a5)5=2a5

 

ii) (10a – 25) ¸ (2a – 5) = 10a252a5

= 5(2a5)(2a5)=5

 

iii) 10a(6a + 21) ¸ 5(2a +7) = 10a(6a+21)5(2a+7)

= 25a3(2a+7)5(2a+7)=6a

 

iv) 9a2b2(3c – 24) ¸ 27ab(c – 8) = 9a2b2(3c24)27ac(c8)

= 927abab3(c8)ab(c8)=ab

 

Question 4:

Divide:

i) 6(3a + 2)(3a + 5) ¸ (3a + 2)

ii) 26ab(a + 5)(b – 4) ¸ 13a(b – 4)

iii) 52xyz(x + y)(y + z)(z + x) ¸ 104xy(y + z)(z + x)

iv) 20(b + 4)(b2 + 5b + 3) ¸ 5(b + 4)

sol.

i) 6(3a + 2)(3a + 5) ¸ (3a + 2) = 6(3a+2)(3a+5)(3a+2)

=6(3a + 5)

 

ii) 26ab(a + 5)(b – 4) ¸ 13a(b – 4) = 26ab(a+5)(b4)13a(b4)

= 132ab(a+5)(b4)13a(b4)=2b(a+5)

 

iii) 52xyz(x + y)(y + z)(z + x) ¸ 104xy(y + z)(z + x) = 52xyz(x+y)(y+z)(z+x)104xy(y+z)(z+x)

= 52xyz(x+y)(y+z)(z+x)522xy(y+z)(z+x)=12z(x+y)

 

iv) 20(b + 4)(b2 + 5b + 3) ¸ 5(b + 4) = 20(b+4)(b2+5b+3)5(b+4)

= 54(b+4)(b2+5b+3)5(b+4)=4(b2+5b+3)

 

Question 5:

Factorize the following expression and divide them:

i) (p2 + 7p + 10 ) ¸ (p + 5)

ii) (x2 – 14x – 32) ¸ (x+2)

iii) 5ab(a2 – b2) ¸ 2a(a + b)

iv) 4bc(c2 + 6c – 16) ¸ 2b(c + 8)

v) 12ab(9a2 – 16b2) ¸ 4ab(3a + 4b)

sol.

i) (p2 + 7p + 10 ) ¸ (p + 5) = (p2+7p+10)(p+5)

= p2+(2+5)p+10(p+5)

= p2+2p+5p+10(p+5)=(p+2)(p+5)(p+5)

= (p + 2)

 

ii) (x2 – 14x – 32) ¸ (x+2) = (x214x32)(x+2)

= (x2+(16+2)x+(16)2)(x+2)

= (x16)(x+2)(x+2) = (x – 16)

 

iii) 5ab(a2 – b2) ¸ 2a(a + b) = 5ab(a2b2)2a(a+b)

= 5ab(ab)(a+b)2a(a+b)=52b(ab)

 

iv) 4bc(c2 + 6c – 16) ¸ 2b(c + 8) = 4bc(c2+6c16)2b(c+8)

= 4bc[c2+(82)c+8(2)]2b(c+8)

= 4bc(c2)(c+8)2b(c+8)

= 2c(c2)

 

v) 12ab(9a2 – 16b2) ¸ 4ab(3a + 4b) = 12ab(9a216b2)4ab(3a+4b)

= 12ab[(3a)2(4b)2]4ab(3a+4b)

= 12ab(3a4b)(3a+4b)4ab(3a+4b)                                   [Since, a2 – b2 = (a – b)(a + b)]

= 3(3a – 4b)

 

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