NCERT Solutions Class 8 Maths Chapter 14 Factorisation Ex 14.3

NCERT Solutions Class 8 Mathematics 
Chapter – 14 (Factorisation) 

The NCERT Solutions in English Language for Class 8 Mathematics Chapter – 14 Factorisation  Exercise 14.3 has been provided here to help the students in solving the questions from this exercise. 

Chapter 14: Factorisation

Exercise – 14.3

1. Carry out the following divisions.
(i) 28x4 ÷ 56x
(ii) –36y3 ÷ 9y2
(iii) 66pq2r3 ÷ 11qr2
(iv) 34x3y3z3 ÷ 51xy2z3
(v) 12a8b8 ÷ (– 6a6b4)

Solution –

(i) 28x4 ÷ 56x
28x4 = 2×2×7×x×x×x×x
56x = 2×2×2×7×x
28x4 ÷ 56x = \frac{2\times 2\times 7\times x\times x\times x\times x}{2\times 2\times 2\times 7\times x}

= \frac{x^3}{2} = \frac{1}{2}x^3

(ii) –36y3 ÷ 9y2
-36y3 = – 2 × 2 × 3 × 3 × y × y × y
9y2 = 3 × 3 × y × y
-36y3 ÷ 9y2 = \frac{- 2 \times 2 \times 3 \times 3 \times y \times y \times y}{3\times 3\times y\times y}

= -4y

(iii) 66 pq2r3 ÷ 11qr2
66pq2r3 = 2 × 3 × 11 × p × q × q × r × r × r 
11qr2 = 11 × q × r × r
66 pq2r3 ÷ 11qr2 = \frac{2\times3\times 11 \times p\times q \times q \times r \times r \times r }{11\times q\times r\times r}

= 6pqr

(iv) 34x3y3z3 ÷ 51xy2z3
34x3y3z3 = 2 × 17 × x × x × x × y × y × y × z × z × z 
51xy2z3 = 3 × 17 × x × y × y × z × z × z

34x3y3z3 ÷ 51xy2z3 = \frac{2 \times 17 \times x \times x \times x \times y \times y \times y \times z \times z \times z}{3\times 17\times x\times y\times y\times z\times z\times z}

= \frac{2x^2y}{3} 

(v) 12a8b8 ÷ (-6a6b4)
12a8b8 = 2 × 2 × 3 × a8 × b8
-6a6b4 = -2 × 3 × a6 × b4

12a8b8 ÷ (-6a6b4) = \frac{2\times 2 \times 3 \times a^8 \times b^8}{-2\times 3\times a^6\times b^4}
= -2a2b4

2. Divide the given polynomial by the given monomial.
(i) (5x2–6x) ÷ 3x
(ii) (3y8–4y6+5y4) ÷ y4
(iii) 8(x3y2z2+x2y3z2+x2y2z3)÷ 4x2 y2 z2
(iv) (x3+2x2+3x) ÷2x
(v) (p3q6–p6q3) ÷ p3q3

Solution –

(i) (5x2 – 6x) ÷ 3x
(5x2 – 6x) = x(5x – 6)

(5x2 – 6x) ÷ 3x = \frac{x(5x - 6)}{3x} 

= \frac{(5x - 6)}{3}

(ii) (3y8 – 4y6 + 5y4) ÷ y4
(3y8 – 4y6 + 5y4) = y4(3y4 – 4y2 + 5)

(3y8 – 4y6 + 5y4) ÷ y4 = \frac{y^4(3y^4 - 4y^2 + 5)}{y^4}
= 3y4 – 4y2 + 5

(iii) 8(x3y2z2 + x2y3z2 + x2y2z3) ÷ 4x2y2z2
8(x3y2z2 + x2y3z2 + x2y2z3) = 8x2y2z2(x + y + z)

8(x3y2z2 + x2y3z2 + x2y2z3) ÷ 4x2y2z2 = \frac{8x^2y^2z^2(x + y + z)}{4x^2y^2z^2}
= 2(x + y + z)

(iv) (x3 + 2x2 + 3x) ÷ 2x
(x3 + 2x2 + 3x) = x(x2 + 2x + 3)
(x3 + 2x2 + 3x) ÷ 2x = \frac{x(x^2 + 2x + 3)}{2x}

= \frac{(x^2 + 2x + 3)}{2}

(v) (p3q6 – p6q3) ÷ p3q3

(p3q6 – p6q3) = p3q3 (q3 – p3)

(p3q6 – p6q3) ÷ p3q3 = \frac{p^3q^3 (q^3 - p^3)}{p^3q^3}
= q3 – p3

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
(10x – 25) = 5× 2 × x – 5 × 5
= 5(2x – 5)
(10x – 25) ÷ 5 = \frac{5(2x - 5)}{5}
= 2x – 5

(ii) (10x – 25) ÷ (2x – 5)
(10x – 25) = 5 × 2 × x – 5× 5
= 5(2x – 5)
(10x – 25) ÷ (2x – 5) = \frac{5(2x - 5)}{(2x - 5)} 
= 5

(iii) 10y(6y + 21) ÷ 5(2y + 7)
10y(6y + 21) = 5 × 2 × y ×(2 × 3 × y + 3 × 7)
= 5 × 2 × y × 3(2 × y + 7)
= 30y(2y + 7)
10y(6y + 21) ÷ 5(2y + 7) = \frac{30y(2y + 7)}{5(2y + 7)}
= 6y

(iv) 9x2y2(3z – 24) ÷ 27xy(z – 8)
9x2y2(3z – 24) = 3 × 3 × x × x × y × y ×(3 × z – 2 × 2 × 2 × 3)
= 3 × 3 × x × x × y × y × 3( z – 2 × 2 × 2)
= 27x2y2(z – 8)
9x2y2(3z – 24) ÷ 27xy(z – 8) = \frac{27x^2y^2(z - 8)}{27xy(z - 8)}
= xy

(v) 96abc(3a – 12)(5b – 30) ÷ 144(a – 4)(b – 6)
96abc(3a – 12)(5b – 30) = 96abc ×(3 × a – 2 × 2 × 3) × (5 × b – 5 × 2 × 3)
= 96abc × 3(a – 2 × 2)× 5(b – 2 × 3)
= 1440abc(a – 4)(b – 6)
96abc(3a – 12)(5b – 30) ÷ 144(a – 4)(b – 6) = \frac{1440abc(a - 4)(b - 6)}{144(a - 4)(b - 6)}
= 10abc

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 –

(i) 5(2x+1)(3x+5) ÷ (2x+1)

= \frac{5(2x +1)(3x + 5) }{(2x +1)}

= 5(3x + 5)

(ii) 26xy(x + 5)(y – 4) ÷ 13x(y – 4)

= \frac{2 \times 13 \times xy(x + 5)(y - 4) }{13x(y - 4)}

= 2y(x + 5)

(iii) 52pqr (p + q)(q + r)(r + p) ÷ 104pq(q + r)(r + p)
= \frac{2\times 2 \times 13 \times p \times q \times r \times ( p + q) \times (q + r) \times (r + p)}{2 \times 2 \times 2 \times 13 \times p \times q \times (q + r) \times (r + p)}

= \frac{r(p + q)}{2}

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

= \frac{2 \times 2 \times 5 \times (y + 4) \times (y^2 + 5 y + 3)}{5 \times (y + 4)}

= 4(y2 + 5 y + 3)

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

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(9x2–16y2)÷4xy(3x+4y)
(vii) 39y3(50y2–98) ÷ 26y2(5y+7)

Solution –

(i) (y2+7y+10)÷(y+5)
(y2+7y+10) = y2+2y+5y+10
= y(y+2)+5(y+2)
= (y+2)(y+5)
(y2+7y+10)÷(y+5) = \frac{(y+2)(y+5)}{(y+5)}
= y+2

(ii) (m2–14m–32)÷ (m+2)
m2–14m–32
= m2+2m-16m–32
= m(m+2)–16(m+2)
= (m–16)(m+2)
(m2–14m–32)÷(m+2) = \frac{(m - 16)(m+2)}{(m+2)}
= m – 16

(iii) (5p2–25p+20)÷(p–1)
5p2–25p+20 = 5(p2 – 5p + 4)
= 5(p2 – p – 4p + 4)
= 5[p(p – 1) – 4(p – 1)]
= 5(p – 1)(p – 4)
(5p2–25p+20)÷(p–1) = \frac{5(p-1)(p-4)}{(p-1)}
= 5(p–4)

(iv) 4yz(z2 + 6z–16)÷ 2y(z+8)

4yz(z2 + 6z -16) = 4 yz (z2 – 2z + 8z -16)
= 4 yz [z(z – 2) + 8(z – 2)]
= 4 yz(z – 2)(z + 8)
4yz(z2+6z–16) ÷ 2y(z+8) = \frac{4yz(z-2)(z+8)}{2y(z+8)}
= 2z(z-2)

(v) 5pq(p2–q2) ÷ 2p(p+q)
5pq(p2 – q2) = 5pq(p–q)(p+q)
5pq(p2–q2) ÷ 2p(p+q) = \frac{5pq(p-q)(p+q)}{2p(p+q)}

= \frac{5q(p-q)}{2}

(vi) 12xy(9x2–16y2) ÷ 4xy(3x+4y)
12xy(9x2 -16y2) = 12xy[(3x)2 – (4y)2]
= 12xy(3x – 4y)(3x + 4y) 
12xy(9x2–16y2) ÷ 4xy(3x+4y) = \frac{12xy(3x+4y)(3x-4y)}{4xy(3x+4y)}
= 3(3x – 4y)

(vii) 39y3(50y2–98) ÷ 26y2(5y+7)
39y3(50y2 – 98) = 3 × 13 × y × y × y × [2 × (25y2 – 49)]
= 3 × 13 × 2 × y × y × y × [(5y)2 – (7)2]
= 3 × 13 × 2 × y × y × y(5y – 7)(5y + 7)

26y2(5y + 7) =  2 × 13 × y × y × (5y + 7)

39y3(50y2–98) ÷ 26y2(5y+7) = \frac{3\times 13 \times 2 \times y \times y \times y(5y - 7)(5y + 7)}{2\times13\times y \times y \times (5y+7)}
= 3y (5y – 7) 

 

NCERT Class 8th Solution 
NCERT Solutions Class 8 English
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NCERT Solutions Class 8 Mathematics 
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