# Ncert Solutions for Class 8 Maths Chapter 14: Factorization CBSE Part 3

In this page we have NCERT book Solutions for Class 8th Maths Chapter 14: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)
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
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)
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)
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)
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)