NCERT Solutions Class 8th Maths Chapter 14 Factorisation Exercise 14.3

NCERT Solutions Class 8th Maths Chapter 14 Factorisation Exercise 14.3

NCERT Solutions Class 8th Maths Chapter 14 Factorisation Exercise 14.3

?Chapter – 14?

✍Factorisation✍

?Exercise 14.3?

1. Carry out the following divisions.

(i) 28×4 ÷ 56x
(ii) –36y3 ÷ 9y2
(iii) 66pq2r3 ÷ 11qr2
(iv) 34x3y3z3 ÷ 51xy2z3
(v) 12a8b8 ÷ (– 6a6b4)

Solution:  (i)28×4 = 2×2×7×x×x×x×x
56x = 2×2×2×7×x

2. Divide the given polynomial by the given monomial.

(i)(5×2–6x) ÷ 3x
(ii)(3y8–4y6+5y4) ÷ y4
(iii) 8(x3y2z2+x2y3z2+x2y2z3)÷ 4×2 y2 z2
(iv)(x3+2×2+3x) ÷2x
(v) (p3q6–p6q3) ÷ p3q3

Solution: 

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)

Solution: 

(i) (10x–25) ÷ 5 = 5(2x-5)/5 = 2x-5
(ii) (10x–25) ÷ (2x–5) = 5(2x-5)/( 2x-5) = 5
(iii) 10y(6y+21) ÷ 5(2y+7) = 10y×3(2y+7)/5(2y+7) = 6y
(iv) 9x2y2(3z–24) ÷ 27xy(z–8) = 9x2y2×3(z-8)/27xy(z-8) = xy

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)

Solution: 

5. Factorise 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(9×2–16y2)÷4xy(3x+4y)
(vii) 39y3(50y2–98) ÷ 26y2(5y+7)

Solution: 

(i) (y2+7y+10)÷(y+5)
First solve for equation, (y2+7y+10)
(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)
Solve for m2–14m–32, we have
m2–14m–32 = m2+2m-16m–32 = m(m+2)–16(m+2) = (m–16)(m+2)
Now, (m2–14m–32)÷(m+2) = (m–16)(m+2)/(m+2) = m-16

(iii) (5p2–25p+20)÷(p–1)
Step 1: Take 5 common from the equation, 5p2–25p+20, we get
5p2–25p+20 = 5(p2–5p+4)
Step 2: Factorize p2–5p+4
p2–5p+4 = p2–p-4p+4 = (p–1)(p–4)
Step 3: Solve original equation
(5p2–25p+20)÷(p–1) = 5(p–1)(p–4)/(p-1) = 5(p–4)

(iv) 4yz(z2 + 6z–16)÷ 2y(z+8)
Factorize z2+6z–16,
z2+6z–16 = z2-2z+8z–16 = (z–2)(z+8)
Now, 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)
p2–q2 can be written as (p–q)(p+q) using identity.
5pq(p2–q2) ÷ 2p(p+q) = 5pq(p–q)(p+q)/2p(p+q) = 5q(p–q)/2

(vi) 12xy(9×2–16y2) ÷ 4xy(3x+4y)
Factorize 9×2–16y2 , we have
9×2–16y2 = (3x)2–(4y)2 = (3x+4y)(3x-4y) using identity: p2–q2 = (p–q)(p+q)
Now, 12xy(9×2–16y2) ÷ 4xy(3x+4y) = 12xy(3x+4y)(3x-4y) /4xy(3x+4y) = 3(3x-4y)

(vii) 39y3(50y2–98) ÷ 26y2(5y+7)
st solve for 50y2–98, we have
50y2–98 = 2(25y2–49) = 2((5y)2–72) = 2(5y–7)(5y+7)
Now, 39y3(50y2–98) ÷ 26y2(5y+7) =