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# Mathematics – Class 8 – Chapter 14 – Factorisation – Exercise 14.3 – NCERT Exercise Solution

1.Carry out the following divisions.

(i) 28x4 ÷ 56x

Solution:

(ii) –36y3 ÷ 9y2

Solution:

(iii) 66pq2r3 ÷ 11qr2

Solution:

(iv) 34x3y3z3 ÷ 51xy2z3

Solution:

(v) 12a8b8 ÷ (– 6a6b4)

Solution:

2. Divide the given polynomial by the given monomial.

(i)(5x2–6x) ÷ 3x

Solution:

(ii)(3y8–4y6+5y4) ÷ y4

Solution:

(iii) 8(x3y2z2+x2y3z2+x2y2z3) ÷ 4x2 y2 z2

Solution:

(iv)(x3+2x2+3x) ÷2x

Solution:

(v) (p3q6–p6q3) ÷ p3q3

Solution:

3.Work out the following divisions.

(i)(10x–25) ÷ 5

Solution:

(ii) (10x–25) ÷ (2x–5)

Solution:

(iii) 10y(6y+21) ÷ 5(2y+7)

Solution:

(iv) 9x2y2(3z–24) ÷ 27xy(z–8)

Solution:

= 9x2y2×3(z-8)/27xy(z-8)

= xy

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

Solution:

4. Divide as directed.

(i) 5(2x+1)(3x+5)÷ (2x+1)

Solution:

(ii) 26xy(x+5)(y–4)÷13x(y–4)

Solution:

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

Solution:

(iv) 20(y+4) (y2+5y+3) ÷ 5(y+4)

Solution:

(v) x(x+1) (x+2)(x+3) ÷ x(x+1)

Solution:

5. Factorise the expressions and divide them as directed.

(i) (y2+7y+10) ÷ (y+5)

Solution:

Firstly, we solve equation, (y2+7y+10)

= y2+2y+5y+10

= y(y+2)+5(y+2)

= (y+2)(y+5)

Now,

(y2+7y+10) ÷ (y+5)

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

= y+2

(ii) (m2–14m–32) ÷ (m+2)

Solution:

Firstly, we solve equation: m2–14m–32

m2–14m–32

= m2+2m – 16m–32

= m(m+2) – 16(m+2)

= (m–16) (m+2)

Now,

= (m–16) (m+2)/(m+2)

= m-16

(iii) (5p2–25p+20) ÷ (p–1)

Solution:

Firstly, we solve equation: (5p2–25p+20)

Taking 5 common

= 5(p2–5p+4)

Now, factorize equation:

= p2–5p+4

= p2–p – 4p+4

= (p–1) (p–4)

Now, we solve the original equation by substituting the value we get

(5p2–25p+20) ÷ (p–1)

= 5(p–1) (p–4)/(p-1)

= 5(p–4)

(iv) 4yz (z2 + 6z–16) ÷ 2y(z+8)

Solution:

First, we factorize equation: z2+6z–16,

= z2-2z+8z–16

= (z–2) (z+8)

Now, solve the original equation

= 4yz(z2+6z–16) ÷ 2y(z+8)

= 4yz(z–2) (z+8)/2y(z+8)

= 2z(z-2)

(v) 5pq(p2–q2) ÷ 2p (p + q)

Solution:

[By using identity (a2 – b2) = (a + b) (a – b)]

We can write p2–q2 as (p–q) (p+q)

Now,

= 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)

Solution:

Firstly, we factorize 9x2–16y2

[By using identity (a2 – b2) = (a + b) (a – b)]

= (3x)2–(4y)2

= (3x+4y) (3x-4y)

Now,

= 12xy(9x2–16y2) ÷ 4xy(3x+4y)

= 12xy(3x+4y) (3x-4y) /4xy(3x+4y)

= 3(3x-4y)

(vii) 39y3(50y2–98) ÷ 26y2(5y+7)

Solution:

First, we solve equation 50y2–98

= 2(25y2–49)

= 2((5y)2–72)

= 2(5y–7) (5y+7)

Now,

= 39y3(50y2–98) ÷ 26y2(5y+7)

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