NCERT solution class 8 chapter 14 Factorisation exercise 14.3

Exercise 14.3

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Q 1 - Ex 14.3 - Factorization - NCERT Maths Class 8th - Chapter 14

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Question 1 
Carry out the following divisions.
(i) 28x⁴ ÷ 56x
(ii) −36y³ ÷ 9y²
(iii) 66pq²r³ ÷ 11qr²
(iv) 34x³y³z³ ÷ 51xy²z³
(v) 12a⁸b⁸ ÷ (−6a⁶b⁴)
Sol :
(i) 28x⁴ = 2 × 2 × 7 × x × x × x × x
56x = 2 × 2 × 2 × 7 × x

(ii) 36y³ = 2 × 2 × 3 × 3 × y × y × y
9y² = 3 × 3 × y × y

(iii) 66 pq² r³ = 2 × 3 × 11 × p × q × q × r × r × r
11qr² = 11 × q × r × r

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

(v) 12a⁸b⁸ = 2 × 2 × 3 × a⁸ × b⁸
6a⁶b⁴ = 2 × 3 × a⁶ × b⁴
= −2a²b⁴

Q 2 - Ex 14.3 - Factorization - NCERT Maths Class 8th - Chapter 14

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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
Sol :
(i) 5x² − 6x = x(5x − 6)

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

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

(iv) x³ + 2x² + 3x = x(x² + 2x + 3)

(v) p³q⁶− p⁶q³ = p³q³(q³ − p³)

Q 3 - Ex 14.3 - Factorization - NCERT Maths Class 8th - Chapter 14

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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)
Sol :
(i)
(ii)
(iii)

(iv)

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

Q 4 - Ex 14.3 - Factorization - NCERT Maths Class 8th - Chapter 14

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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)
Sol :
(i) = 5(3x + 1)
(ii) = 2y (x + 5)
(iii) 52pqr (p + q) (q + r) (r + p) ÷ 104pq (q + r) (r + p)

(iv) 20(y + 4) (y² + 5y + 3) = 2 × 2 × 5 × (y + 4) (y² + 5y + 3)

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

Q 5 - Ex 14.3 - Factorization - NCERT Maths Class 8th - Chapter 14

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Question 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)
Sol :
(i) (y² + 7y + 10) = y² + 2y + 5y + 10
= y (y + 2) + 5 (y + 2)
= (y + 2) (y + 5)

(ii) m² − 14m − 32 = m² + 2m − 16m − 32
= m (m + 2) − 16 (m + 2)
= (m + 2) (m − 16)

(iii) 5p² − 25p + 20 = 5(p² − 5p + 4)
= 5[p² − p − 4p + 4]
= 5[p(p −1) − 4(p −1)]
= 5(p −1) (p − 4)

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(iv) 4yz(z² + 6z −16) = 4yz [z² − 2z + 8z − 16]
= 4yz [z(z − 2) + 8(z − 2)]
= 4yz(z − 2) (z + 8)

(v) 5pq(p² − q²) = 5pq (p − q) (p + q)

(vi) 12xy(9x² − 16y²) = 12xy[(3x)² − (4y)²] = 12xy(3x − 4y) (3x + 4y)

(vii) 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)