Class 8 NCERT Solutions- Chapter 14 Factorisation – Exercise 14.3

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Question 1. Carry out the following divisions.

(i) 28x⁴ ÷ 56x

Solution:

28x⁴ = 2 × 2 ×7 × x × x × x × x

56x = 2 × 2 × 2 × 7 × x

28x⁴÷ 56x = $\frac{(2 * 2 *7 * x * x * x * x)}{(2 * 2 * 2 * 7 * x)}$ (grouping 28x to cancel)

= ½ × x × x × x

= ½ x³

(ii) -36y³ ÷ 9y²

Solution:

-36y³= -2 × 2 × 3 × 3 × y × y × y

9y²= 3 × 3 × y × y

-36y³ ÷ 9y² = $\frac{(-2 * 2 * 3 * 3 * y * y * y)}{(3 * 3 * y * y) }$ (grouping 9y² to cancel)

= -(2 × 2 × y)

= -4y

(iii) 66pq²r³ ÷ 11qr²

Solution:

66pq²r³ = 2 × 3 × 11 × p × q × q × r × r × r

11qr² = 11 × q × r × r

66pq²r³ ÷ 11qr² = $\frac{(2 * 3 * 11 * p * q * q * r * r * r)}{(11 * q * r * r)}$ (grouping 11qr²to cancel)

= (2 × 3 × p × q × r)

= 6pqr

(iv) 34x³y³z³ ÷ 51xy²z³

Solution:

34x³y³z³ = 2 × 17 × x × x × x × y × y × y × z × z × z

51xy²z³ = 3 × 17 × x × y × y × z × z × z

34x³y³z³ ÷ 51xy²z³ = $\frac{(2 * 17 * x * x * x * y * y * y * z * z * z)}{(3 * 17 * x * y * y * z * z * z)}$

= $\frac{(2 * x * x * y )}{(3)}$ (grouping 17xy²z³ to cancel)

= $\mathbf{\frac{2}{3}}$ x²y

(v) 12a⁸b⁸ ÷ (-6a⁶b⁴)

Solution:

12a⁸b⁸ = 2 × 2 × 3 × a × a × a × a × a × a × a × a × b × b × b × b × b × b × b × b

-6a⁶b⁴= -2 × 3 × a × a × a × a × a × a × b × b × b × b

12a⁸b⁸ ÷ (-6a⁶b⁴) = $\frac{(2 * 2 * 3 * a * a * a * a * a * a * a * a * b * b * b * b * b * b * b * b)}{(-2 * 3 * a * a * a * a * a * a * b * b * b * b)}$

= – (2 × a × a × b × b × b × b) (grouping 6a⁶b⁴ to cancel)

= -2a²b⁴

Question 2. Divide the given polynomial by the given monomial.

(i) (5x² – 6x) ÷ 3x

Solution:

5x²– 6x = (5 × x × x) – (2 × 3 × x)

= 5x × (x) – 6 × (x)

= x(5x – 6)

3x = 3 × (x)

(5x²– 6x) ÷ 3x = $\frac{x * (5x - 6)}{3 * (x)}$ (grouping x to cancel)

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

(ii) (3y⁸ – 4y⁶ + 5y⁴) ÷ y⁴

Solution:

3y⁸-4y⁶+5y⁴ = y⁴ [(3 × y × y × y × y) – (2× 2 × y × y) + (5)]

y⁴ = (y × y × y × y)

(3y⁸-4y⁶+5y⁴) ÷ y⁴= $\frac{y^4(3y^4-4y^2+5)}{y^4}$ (grouping y⁴ to cancel)

= (3x⁴-4y²+5 )

(iii) 8(x³y²z²+ x²y³z² + x²y²z³) ÷ 4x²y²z²

Solution:

8 (x³y²z² + x²y³z² + x²y²z³ ) = 2 × 2 × 2 × x²y²z² (x + y + z)

4 x x²y²z² = 2 × 2 × x²y²z²

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

= 2(x+y+z)

(iv) (x³+2x²+3x) ÷ 2x

Solution:

x³+2x²+3x = x × (x²+2x+3 )

(x³+2x²+3x) ÷ 2x = $\frac{x(x^2+2x+3)}{ 2 × x}$ (grouping x to cancel)

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

(v) (p³q⁶-p⁶q³) ÷ p³q³

Solution:

p³q⁶-p⁶q³ = p³q³(q³-p³)

(p³q⁶-p⁶q³) ÷ p³q³ = $\frac{p^3q^3(q^3-p^3)}{p^3q^3}$ (grouping p³q³ to cancel)

= q³– p³

Question 3. Work out the following divisions.

(i) (10x – 25) ÷ 5

Solution:

10x-25 = (5 × 2 × x) – (5 × 5)

= 5(2x-5)

(10x-25) ÷ 5 = $\frac{5(2x-5)}{5}$ (grouping 5 to cancel)

= (2x – 5)

(ii) (10x – 25) ÷ (2x – 5)

Solution:

10x-25 = 5(2x-5)

(10x-25)÷(2x-5) = $\frac{5(2x-5)}{(2x-5)}$ (grouping (2x-5) to cancel)

= 5

(iii) 10y(6y+21) ÷ 5(2y+7)

Solution:

10y(6y+21) = 5 × 2 × y × 3 × (2y+7)

10y(6y+21) ÷ 5(2y+7) = $\frac{5 * 2 * y * 3 * (2y+7)}{5 * (2y+7)}$ (grouping 5(2y+7) to cancel)

= 2 × 3 × y

= 6y

(iv) 9x²y²(3z-24) ÷ 27xy(z-8)

Solution:

9x²y²(3z-24) = 3 × 3 × x² × y² × 3 × (z-8)

27xy(z-8) = 3 × 3 × 3 × x × y × (z-8)

9x²y²(3z-24)÷27xy(z-8)= $\frac{3 * 3 * 3 * x^2 * y^2 * (z-8)}{3 * 3 * 3 * x * y * (z-8)}$ (grouping (27xy(z-8)) to cancel)

= xy

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

Solution:

96abc(3a-12)(5b-30) = 2 × 2 × 2 × 2 × 2 × 3 × a × b × c × 3 × (a-4) × 5 × (b-6)

144(a-4)(b-6) = 2 × 2 × 2 × 2 × 3 × 3 × (a-4) × (b-6)

96abc(3a-12)(5b-30) ÷ 144(a-4)(b-6) = $\frac{(2 * 2 * 2 * 2 * 2 * 3 * a * b * c * 3 * (a-4) * 5 * (b-6)) }{ (2 * 2 * 2 *× 2 * 3 * 3 * (a-4) * (b-6))}$

= (2 × 5 × a × b × c) (grouping (144(a-4)(b-6)) to cancel)

= 10abc

Question 4. Divide as directed.

(i) 5(2x+1)(3x+5) ÷ (2x+1)

Solution:

= 5(3x+1) (grouping (2x+1) to cancel)

(ii) 26xy(x+5)(y-4)÷13x(y-4)

Solution:

26xy(x+5)(y-4) = 2 × 13 × x × y × (x+5) × (y-4)

26xy(x+5)(y-4)÷13x(y-4) = $\frac{(2 * 13 * x * y * (x+5) * (y-4))}{ (13 * x * (y-4))}$ (grouping 13x(y-4) to cancel)

= (2 × y × (x+5))

= 2y(x+5)

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

Solution:

52pqr(p+q)(q+r)(r+p) = 13 × 2 × 2 × pqr(p+q)(q+r)(r+p)

104pq(q+r)(r+p) = 13 × 2 × 2 × 2 × pq(q+r)(r+p)

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

= $\mathbf{\frac{r(p+q)}{ 2}}$ (grouping (52pq(q+r)(r+p)) to cancel)

(iv) 20(y+4)(y²+5y+3)÷5(y+4)

Solution:

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

20(y+4)(y²+5y+3)÷5(y+4) = $\frac{(2 * 2 * 5 * (y+4) * (y^2+5y+3))}{(5(y+4))}$ (grouping (5(y+4)) to cancel)

= 2 × 2 × (y²+5y+3)

= 4(y²+5y+3)

(v) x(x+1)(x+2)(x+3) ÷ x(x+1)

Solution:

= (x+2)(x+3) (grouping x(x+1) to cancel)

Question 5. Factorise the expressions and divide them as directed.

(i) (y²+7y+10) ÷ (y+5)

Solution:

(y²+7y+10) = (y²+5y+2y+10)

= (y(y+5) + 2(y+5)) (2 + 5 = 7 & 2 × 5 = 10)

= (y+5) (y+2)

(y²+7y+10) ÷ (y+5) = $\frac{(y+5) (y+2) }{(y+5)}$ (grouping (y+5) to cancel)

= (y+2)

(ii) (m²-14m-32)÷(m+2)

Solution:

(m²-14m-32) = (m²-16m+2m-32 )

= (m(m-16) + 2(m-16)) (-16 + 2 = -14 & -16 × 2 = -32)

= (m+2) (m-16)

(m²-14m-32)÷(m+2) = $\frac{(m+2) (m-16)}{ (m+2)}$ (grouping (m+2) to cancel)

= (m-16)

(iii) (5p²-25p+20) ÷ (p-1)

Solution:

(5p²-25p+20) = (5p²-20p-5p+20)

=(5p(p-4)-5(p-4)) (-20 – 5 = -25 )

=(5p-5) (p-4)

=5 (p-1) (p-4)

(5p²-25p+20)÷(p-1) = $\frac{5(p-1)(p-4)}{(p-1)}$ (grouping (p-1) to cancel)

= 5(p-4)

(iv) 4yz(z²+6z-16)÷2y(z+8)

Solution:

4yz(z²+6z-16) = 2 × 2 × y × z × (z²+8z-2z-16)

= 2 × 2 × y × z × (z(z+8)-2(z+8)) (8 + (-2) = 6 & 8 × (-2) = -16)

= 2 × 2 × y × z × (z+8) (z-2))

4yz(z²+6z-16) ÷ 2y(z+8) = $\frac{(2 * 2 * y * z * (z+8) (z-2))}{(2y(z+8))}$ (grouping 2y(z+8) to cancel)

= 2 × z × (z-2)

= 2z(z-2)

(v) 5pq(p²-q²)÷2p(p+q)

Solution:

(p²-q²) = (p+q) (p-q) (IDENTITY a²-b² = (a+b)(a-b) )

5pq(p²-q²)÷2p(p+q) = $\frac{5pq(p+q)(p-q)}{2p(p+q)}$ (grouping p(p+q) to cancel)

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

(vi) 12xy(9x²-16y²) ÷ 4xy(3x+4y)

Soln.

12xy(9x²-16y²) = 2 × 2 × 3 × ((3x)²-(4y)²)

12xy(9x²-16y²) = 2 × 2 × 3 × (3x+4y) (3x-4y) (IDENTITY a²-b² = (a+b)(a-b) )

12xy(9x²-16y²) ÷ 4xy(3x+4y) = $\frac{2 * 2 * 3 * (3x+4y) (3x-4y)}{2 * 2 * xy (3x+4y)}$ (grouping 4xy(3x+4y) to cancel)

= 3 (3x-4y)

(vii) 39y³(50y²-98) ÷ 26y²(5y+7)

Solution:

39y³(50y²-98) = 3 × 13 × y³ × 2 × (25y²-49)

= 3 × 13 × y³ × 2 × ((5y)²-(7)²) (IDENTITY a²-b² = (a+b)(a-b) )

= 3 × 13 × y³ × 2 × (5y+7) (5y-7)

26y²(5y+7) = 2 × 13 × y² × (5y+7)

39y³(50y²-98)÷26y²(5y+7) = $\frac{(3 * 13 * y^3 * 2 * (5y+7) (5y-7))}{(2 * 13 * y^2 * (5y+7))}$ (grouping 26y²(5y+7) to cancel)

= (3 × y × (5y-7))

= 3y(5y-7)

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Last Updated : 09 Mar, 2021

Related Articles

Class 8 NCERT Solutions- Chapter 14 Factorisation – Exercise 14.3

Question 1. Carry out the following divisions.

(i) 28x⁴ ÷ 56x

(ii) -36y³ ÷ 9y²

(iii) 66pq²r³ ÷ 11qr²

(iv) 34x³y³z³ ÷ 51xy²z³

(v) 12a⁸b⁸ ÷ (-6a⁶b⁴)

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

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)

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)

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)

CBSE Class 12 Computer Science

GATE CS & IT 2024

Mastering System Design: From Low-Level to High-Level Solutions

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Class 8 NCERT Solutions- Chapter 14 Factorisation – Exercise 14.3

Question 1. Carry out the following divisions.

(i) 28x4 ÷ 56x

(ii) -36y3 ÷ 9y2

(iii) 66pq2r3 ÷ 11qr2

(iv) 34x3y3z3 ÷ 51xy2z3

(v) 12a8b8 ÷ (-6a6b4)

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

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)

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)

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)

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CBSE Class 12 Computer Science

GATE CS & IT 2024

Mastering System Design: From Low-Level to High-Level Solutions

(i) 28x⁴ ÷ 56x

(ii) -36y³ ÷ 9y²

(iii) 66pq²r³ ÷ 11qr²

(iv) 34x³y³z³ ÷ 51xy²z³

(v) 12a⁸b⁸ ÷ (-6a⁶b⁴)

(i) (5x² – 6x) ÷ 3x

(ii) (3y⁸ – 4y⁶ + 5y⁴) ÷ y⁴

(iii) 8(x³y²z²+ x²y³z² + x²y²z³) ÷ 4x²y²z²

(iv) (x³+2x²+3x) ÷ 2x

(v) (p³q⁶-p⁶q³) ÷ p³q³

(iv) 9x²y²(3z-24) ÷ 27xy(z-8)

(iv) 20(y+4)(y²+5y+3)÷5(y+4)

(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)