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)