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Ncert Solutions for Factorization Class 8 CBSE Part 3



In this page we have NCERT book Solutions for Class 8th Maths:Factorization for EXERCISE 3 . Hope you like them and do not forget to like , social share and comment at the end of the page.

Question 1

Carry out the following divisions.

(i) 28x4 ÷ 56x

 (ii) –36y³ ÷ 9y²

 (iii) 66pq²r3 ÷ 11qr²

 (iv) 34x3y3z3 ÷ 51xy²z3

(v) 12a8b8 ÷ (– 6a6b4)

Answer:

1) 28x4 ÷ 56x

= (2×2×7×x×x3  ) /( 2×2×2×7×x)

=x3/2

 

2) –36y³ ÷ 9y²

= (-2×2×3×3×y×y2 ) /( 3×3×y2)

=-4y

3) 66pq²r3 ÷ 11qr²

= 6pqr

 

4) 34x3y3z3 ÷ 51xy²z3

=(2/3) x2y

 

5) 12a8b8 ÷ (– 6a6b4)

=-2a²b4

 

Question 2

 Divide the given polynomial by the given monomial.

(i) (5x² – 6x) ÷ 3x

(ii) (3y8 – 4y6 + 5y4) ÷ y4

 (iii) 8(x3y²z² + x²y3z² + x²y²z3) ÷ 4x²y²z²

 (iv) (x3 + 2x² + 3x) ÷ 2x

(v) (p3q6 – p6q3) ÷ p3q3

Answer:

  1. (5x² – 6x) ÷ 3x

    =[x(5x-6)]  / 3x

    Cancelling x

    =(5x-6)/3

  2. (3y8 – 4y6 + 5y4) ÷ y4

    = y4(3y4 -4y2 +5)/ y4

    =3y4-4y²+5

  3. 8(x3y²z² + x²y3z² + x²y²z3) ÷ 4x²y²z²

    =8x2 y²z²(x+y+z)/ 4x²y²z²

     

    = 2(x+y+z)

  4. (x3 + 2x² + 3x) ÷ 2x

    =x(x2 +2x+3)/2x

    =( x2 +2x+3)/2

  5. (p3q6 – p6q3) ÷ p3q3

    = p3 q3(q3 - p3) / p3q3

=(q3 - p3)

 

Question 3

 Work out the following divisions.

(i) (10x – 25) ÷ 5

 (ii) (10x – 25) ÷ (2x – 5)

 (iii) 10y(6y + 21) ÷ 5(2y + 7)

 (iv) 9x²y² (3z – 24) ÷ 27xy(z – 8)

(v) 96abc(3a – 12) (5b – 30) ÷ 144(a – 4) (b – 6)

 

Answer

1) (10x – 25) ÷ 5

=5(2x-5)/5

=(2x-5)

2) (10x – 25) ÷ (2x – 5)

= 5(2x-5) /(2x – 5)

=-5

3) 10y(6y + 21) ÷ 5(2y + 7)

=30y(2y+7)/ 5(2y + 7)

= 6y

4) 9x²y² (3z – 24) ÷ 27xy(z – 8)

=27 x²y²(z-8)/ 27xy(z – 8)

=xy

 

5) 96abc(3a – 12) (5b – 30) ÷ 144(a – 4) (b – 6)

=(96×3×5)abc(a-4)(b-6)/ 144(a – 4) (b – 6)

=10abc

 

 

Question 4

Divide as directed.

(i) 5(2x + 1) (3x + 5) ÷ (2x + 1)

(ii) 26xy(x + 5) (y - 4) ÷ 13x(y - 4)

(iii) 52pqr (p + q) (q + r) (r + p) ÷ 104pq(q + r) (r + p)

(iv) 20(y + 4) (y2 + 5y + 3) ÷ 5(y + 4)

(v) x(x + 1) (x + 2) (x + 3) ÷ x(x + 1)

Answer

i) 5(2x + 1) (3x + 5) ÷ (2x + 1)

=5(2x + 1) (3x + 5) / (2x + 1)

=5(3x+5)

ii) 26xy(x + 5) (y - 4) ÷ 13x(y - 4)

=26xy(x + 5) (y - 4) / 13x(y - 4)

Cancelling 13x(y-4)

=2y(x+5)

 

iii) 52pqr (p + q) (q + r) (r + p) ÷ 104pq(q + r) (r + p)

=52pqr (p + q) (q + r) (r + p)/ 104pq(q + r) (r + p)

Cancelling 52pq(q + r) (r + p)

=r(p+q)/2

iv)   20(y + 4) (y2 + 5y + 3) ÷ 5(y + 4)

=4(y2 + 5y + 3)

v) x(x + 1) (x + 2) (x + 3) ÷ x(x + 1)

= x(x + 1) (x + 2) (x + 3) / x(x + 1)

Cancelling x(x + 1)

=(x + 2) (x + 3)

 

Question 5

Factorize the expressions and divide them as directed.

(i) (y2 + 7y + 10) ÷  (y + 5)

(ii) (m2 - 14m - 32) ÷  (m + 2)

(iii) (5p2 - 25p + 20) ÷  (p - 1)

(iv) 4yz(z2 + 6z - 16) ÷ 2y(z + 8)

(v) 5pq(p2 - q2) ÷ 2p(p + q)

(vi) 12xy(9x2 - 16y2) ÷ 4xy(3x + 4y)

(vii) 39y3(50y2- 98) ÷ 26y2(5y+ 7)

 

Answer

i) (y2 + 7y + 10) ÷  (y + 5)

=( y2 + 5y+2y + 10)/(y+5)

=[y(y+5)+2(y+5)] /(y+5)

=(y+1)(y+5) /(y+5)

=(y+1)

 

ii) (m2 - 14m - 32) ÷  (m + 2)

= (m2 - 16m+2m - 32) / (m + 2)

=[m(m-16)+2(m-16)] /(m+2)

=(m+2)(m-16) /(m+2

=(m-16)

 

 

iii) (5p2 - 25p + 20) ÷  (p - 1)

=5(p2-5p+4)/(p-1)

=5[p(p-1) -4(p-1)] /(p-1)

=5(p-1)(p-4)/(p-1)

=5(p-4)

 

iv) 4yz(z2 + 6z - 16) ÷ 2y(z + 8)

=4yz[z2 -2z+8z-16]/ 2y(z + 8)

=4yz[z(z-2)+8(z-2)] / 2y(z + 8)

=4yz(z-2)(z+8) / 2y(z + 8)

=2z(z-2)

v) 5pq(p2 - q2) ÷ 2p(p + q)

=5pq(p2 - q2) / 2p(p + q)

=5pq(p-q)(p+q) / 2p(p + q)

 

=5q(p-q)/2

vi) 12xy(9x2 - 16y2) ÷ 4xy(3x + 4y)

 

=12xy(9x2 - 16y2) / 4xy(3x + 4y)

=12xy(3x+4y)(3x-4y) / 4xy(3x + 4y)

=3(3x-4y)

 

vii) 39y3(50y2- 98) ÷ 26y2(5y+ 7)

=39y3(50y2- 98) /26y2(5y+ 7)

=78y3(25y2-49)/ 26y2(5y+ 7)

=3y(5y+7)(5y-7)/ (5y+ 7)

=3y(5y-7)

 

 


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