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# NCERT Solutions for Class 8 Maths Chapter 14: Factorisation - Exercise 14.3

Last updated date: 07th Aug 2024
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## NCERT Solutions for Class 8 Maths Chapter 14 (EX 14.3)

Free PDF download of NCERT Solutions for Class 8 Maths Chapter 14 Exercise 14.3 (EX 14.3) and all chapter exercises at one place prepared by an expert teacher as per NCERT (CBSE) books guidelines. Class 8 Maths Chapter 14 Factorisation Exercise 14.3 Questions with Solutions to help you to revise complete Syllabus and Score More marks. Register and get all exercise solutions in your emails.

 Class: NCERT Solutions for Class 8 Subject: Class 8 Maths Chapter Name: Chapter 14 - Factorisation Exercise: Exercise - 14.3 Content-Type: Text, Videos, Images and PDF Format Academic Year: 2024-25 Medium: English and Hindi Available Materials: Chapter WiseExercise Wise Other Materials Important QuestionsRevision Notes

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## Access NCERT Solutions for Class 8 Maths Chapter 14 - Factorization

Refer to page 10 for exercise 14.3 in the PDF.

1.  Carry out the following divisions.

i. Find the division $28{x^4} \div 56x$.

Ans: To find the division $28{x^4} \div 56x$, expand both the expressions.

The expansion of $28{x^4}$ is $28{x^4} = 2 \times 2 \times 7 \times x \times x \times x \times x$.

The expansion of $56x$ is $56x = 2 \times 2 \times 2 \times 7 \times x$.

To find the division $28{x^4} \div 56x$, divide the expansion of $28{x^4}$ by $56x$.

$28{x^4} \div 56x = \dfrac{{2 \times 2 \times 7 \times x \times x \times x \times x}}{{2 \times 2 \times 2 \times 7 \times x}}$

$28{x^4} \div 56x = \dfrac{{{x^3}}}{2}$

The value of $28{x^4} \div 56x$ is $\dfrac{{{x^3}}}{2}$.

ii. Find the division $- 36{y^3} \div 9{y^2}$.

Ans: To find the division $- 36{y^3} \div 9{y^2}$, expand both the expressions.

The expansion of $- 36{y^3}$ is $- 36{y^3} = - 2 \times 2 \times 3 \times 3 \times y \times y \times y$.

The expansion of $9{y^2}$ is $9{y^2} = 3 \times 3 \times y \times y$.

To find the division $- 36{y^3} \div 9{y^2}$, divide the expansion of $- 36{y^3}$ by $9{y^2}$.

$- 36{y^3} \div 9{y^2} = \dfrac{{ - 2 \times 2 \times 3 \times 3 \times y \times y \times y}}{{3 \times 3 \times y \times y}}$

$- 36{y^3} \div 9{y^2} = - 4y$

The value of $- 36{y^3} \div 9{y^2}$ is $- 4y$.

iii. Find the division $66p{q^2}{r^3} \div 11q{r^2}$.

Ans: To find the division $66p{q^2}{r^3} \div 11q{r^2}$, expand both the expressions.

The expansion of $66p{q^2}{r^3}$ is $66p{q^2}{r^3} = 2 \times 3 \times 11 \times p \times q \times q \times r \times r \times r$.

The expansion of $11q{r^2}$ is $11q{r^2} = 11 \times q \times r \times r$.

To find the division $66p{q^2}{r^3} \div 11q{r^2}$, divide the expansion of $66p{q^2}{r^3}$ by $11q{r^2}$.

$66p{q^2}{r^3} \div 11q{r^2} = \dfrac{{2 \times 3 \times 11 \times p \times q \times q \times r \times r \times r}}{{11 \times q \times r \times r}}$

$66p{q^2}{r^3} \div 11q{r^2} = 6pqr$

The value of $66p{q^2}{r^3} \div 11q{r^2}$ is $6pqr$.

iv. Find the division $34{x^3}{y^3}{z^3} \div 51x{y^2}{z^3}$.

Ans: To find the division $34{x^3}{y^3}{z^3} \div 51x{y^2}{z^3}$, expand both the expressions.

The expansion of $34{x^3}{y^3}{z^3}$ is $34{x^3}{y^3}{z^3} = 2 \times 17 \times x \times x \times x \times y \times y \times y \times z \times z \times z$.

The expansion of $51x{y^2}{z^3}$ is $51x{y^2}{z^3} = 3 \times 17 \times x \times y \times y \times z \times z \times z$.

To find the division $34{x^3}{y^3}{z^3} \div 51x{y^2}{z^3}$, divide the expansion of $34{x^3}{y^3}{z^3}$ by $51x{y^2}{z^3}$.

$34{x^3}{y^3}{z^3} \div 51x{y^2}{z^3} = \dfrac{{2 \times 17 \times x \times x \times x \times y \times y \times y \times z \times z \times z}}{{3 \times 17 \times x \times y \times y \times z \times z \times z}}$

$34{x^3}{y^3}{z^3} \div 51x{y^2}{z^3} = \dfrac{2}{3}{x^2}y$

The value of $34{x^3}{y^3}{z^3} \div 51x{y^2}{z^3}$ is $\dfrac{2}{3}{x^2}y$.

v. Find the division $12{a^8}{b^8} \div 6{a^6}{b^4}$.

Ans: To find the division $12{a^8}{b^8} \div \left( { - 6{a^6}{b^4}} \right)$, expand both the expressions.

The expansion of $12{a^8}{b^8}$ is $12{a^8}{b^8} = 2 \times 2 \times 3 \times {a^8} \times {b^8}$.

The expansion of $- 6{a^6}{b^4}$ is $- 6{a^6}{b^4} = - 2 \times 3 \times {a^6} \times {b^4}$.

To find the division $12{a^8}{b^8} \div \left( { - 6{a^6}{b^4}} \right)$, divide the expansion of $12{a^8}{b^8}$ by $6{a^6}{b^4}$.

$12{a^8}{b^8} \div \left( { - 6{a^6}{b^4}} \right) = \dfrac{{2 \times 2 \times 3 \times {a^8} \times {b^8}}}{{ - 2 \times 3 \times {a^6} \times {b^4}}}$

$12{a^8}{b^8} \div \left( { - 6{a^6}{b^4}} \right) = - 2{a^2}{b^4}$

The value of $12{a^8}{b^8} \div \left( { - 6{a^6}{b^4}} \right)$ is $- 2{a^2}{b^4}$.

2. Divide the given polynomial by the given monomial.

i. Find the division $\left( {5{x^2} - 6x} \right) \div 3x$.

Ans: The expression $5{x^2} - 6x$ can be written as $x\left( {5x - 6} \right)$.

Substitute $x\left( {5x - 6} \right)$ for $5{x^2} - 6x$ in $\left( {5{x^2} - 6x} \right) \div 3x$ and simplify using division.

$\left( {5{x^2} - 6x} \right) \div 3x = \dfrac{{x\left( {5x - 6} \right)}}{{3x}}$

$\left( {5{x^2} - 6x} \right) \div 3x = \dfrac{1}{3}\left( {5x - 6} \right)$

The value of $\left( {5{x^2} - 6x} \right) \div 3x$ is $\dfrac{1}{3}\left( {5x - 6} \right)$.

ii. Find the division $\left( {3{y^8} - 4{y^6} + 5{y^4}} \right) \div {y^4}$.

Ans: The expression $3{y^8} - 4{y^6} + 5{y^4}$ can be written as ${y^4}\left( {3{y^4} - 4{y^2} + 5} \right)$.

Substitute ${y^4}\left( {3{y^4} - 4{y^2} + 5} \right)$ for $3{y^8} - 4{y^6} + 5{y^4}$ in $\left( {3{y^8} - 4{y^6} + 5{y^4}} \right) \div {y^4}$ and simplify using division.

$\left( {3{y^8} - 4{y^6} + 5{y^4}} \right) \div {y^4} = \dfrac{{{y^4}\left( {3{y^4} - 4{y^2} + 5} \right)}}{{{y^4}}}$

$\left( {3{y^8} - 4{y^6} + 5{y^4}} \right) \div {y^4} = 3{y^4} - 4{y^2} + 5$

The value of $\left( {3{y^8} - 4{y^6} + 5{y^4}} \right) \div {y^4}$ is $3{y^4} - 4{y^2} + 5$.

iii. Find the division $8\left( {{x^3}{y^2}{z^2} + {x^2}{y^3}{z^2} + {x^2}{y^2}{z^3}} \right) \div 4{x^2}{y^2}{z^2}$.

Ans: The expression $8\left( {{x^3}{y^2}{z^2} + {x^2}{y^3}{z^2} + {x^2}{y^2}{z^3}} \right)$ can be written as $8{x^2}{y^2}{z^2}\left( {x + y + z} \right)$.

Substitute $8{x^2}{y^2}{z^2}\left( {x + y + z} \right)$ for $8\left( {{x^3}{y^2}{z^2} + {x^2}{y^3}{z^2} + {x^2}{y^2}{z^3}} \right)$ in $8\left( {{x^3}{y^2}{z^2} + {x^2}{y^3}{z^2} + {x^2}{y^2}{z^3}} \right) \div 4{x^2}{y^2}{z^2}$ and simplify using division.

$8\left( {{x^3}{y^2}{z^2} + {x^2}{y^3}{z^2} + {x^2}{y^2}{z^3}} \right) \div 4{x^2}{y^2}{z^2} = \dfrac{{8{x^2}{y^2}{z^2}\left( {x + y + z} \right)}}{{4{x^2}{y^2}{z^2}}}$

$8\left( {{x^3}{y^2}{z^2} + {x^2}{y^3}{z^2} + {x^2}{y^2}{z^3}} \right) \div 4{x^2}{y^2}{z^2} = 2\left( {x + y + z} \right)$

The value of $8\left( {{x^3}{y^2}{z^2} + {x^2}{y^3}{z^2} + {x^2}{y^2}{z^3}} \right) \div 4{x^2}{y^2}{z^2}$is $2\left( {x + y + z} \right)$.

iv. Find the division $\left( {{x^3} + 2{x^2} + 3x} \right) \div 2x$.

Ans: The expression $\left( {{x^3} + 2{x^2} + 3x} \right)$ can be written as $x\left( {{x^2} + 2x + 3} \right)$.

Substitute $x\left( {{x^2} + 2x + 3} \right)$ for $\left( {{x^3} + 2{x^2} + 3x} \right)$ in $\left( {{x^3} + 2{x^2} + 3x} \right) \div 2x$ and simplify using division.

$\left( {{x^3} + 2{x^2} + 3x} \right) \div 2x = \dfrac{{x\left( {{x^2} + 2x + 3} \right)}}{{2x}}$

$\left( {{x^3} + 2{x^2} + 3x} \right) \div 2x = \dfrac{1}{2}\left( {{x^2} + 2x + 3} \right)$

The value of $\left( {{x^3} + 2{x^2} + 3x} \right) \div 2x$is $\dfrac{1}{2}\left( {{x^2} + 2x + 3} \right)$.

v. Find the division $\left( {{p^3}{q^6} - {p^6}{q^3}} \right) \div {p^3}{q^3}$.

Ans: The expression $\left( {{p^3}{q^6} - {p^6}{q^3}} \right)$ can be written as ${p^3}{q^3}\left( {{q^3} - {p^3}} \right)$.

Substitute ${p^3}{q^3}\left( {{q^3} - {p^3}} \right)$ for $\left( {{p^3}{q^6} - {p^6}{q^3}} \right)$ in $\left( {{p^3}{q^6} - {p^6}{q^3}} \right) \div {p^3}{q^3}$ and simplify using division.

$\left( {{p^3}{q^6} - {p^6}{q^3}} \right) \div {p^3}{q^3} = \dfrac{{{p^3}{q^3}\left( {{q^3} - {p^3}} \right)}}{{{p^3}{q^3}}}$

$\left( {{p^3}{q^6} - {p^6}{q^3}} \right) \div {p^3}{q^3} = {q^3} - {p^3}$

The value of $\left( {{p^3}{q^6} - {p^6}{q^3}} \right) \div {p^3}{q^3}$is ${q^3} - {p^3}$.

3. Work out the following divisions.

i. Find the division $\left( {10x - 25} \right) \div 5$.

Ans: The expression $\left( {10x - 25} \right)$ can be written as $5\left( {2x - 5} \right)$.

Substitute $5\left( {2x - 5} \right)$ for $\left( {10x - 25} \right)$ in $\left( {10x - 25} \right) \div 5$ and simplify using division.

$\left( {10x - 25} \right) \div 5 = \dfrac{{5\left( {2x - 5} \right)}}{5}$

$\left( {10x - 25} \right) \div 5 = 2x - 5$

The value of $\left( {10x - 25} \right) \div 5$ is $2x - 5$.

ii. Find the division $\left( {10x - 25} \right) \div \left( {2x - 5} \right)$.

Ans: The expression $\left( {10x - 25} \right)$ can be written as $5\left( {2x - 5} \right)$.

Substitute $5\left( {2x - 5} \right)$ for $\left( {10x - 25} \right)$ in $\left( {10x - 25} \right) \div \left( {2x - 5} \right)$ and simplify using division.

$\left( {10x - 25} \right) \div \left( {2x - 5} \right) = \dfrac{{5\left( {2x - 5} \right)}}{{\left( {2x - 5} \right)}}$

$\left( {10x - 25} \right) \div \left( {2x - 5} \right) = 5$

The value of $\left( {10x - 25} \right) \div \left( {2x - 5} \right)$ is 5.

iii. Find the division $10y\left( {6y + 21} \right) \div 5\left( {2y + 7} \right)$.

Ans: The expression $10y\left( {6y + 21} \right)$ can be written as $2 \times 5 \times 3y\left( {2y + 7} \right)$.

Substitute $2 \times 5 \times 3y\left( {2y + 7} \right)$ for $10y\left( {6y + 21} \right)$ in $10y\left( {6y + 21} \right) \div 5\left( {2y + 7} \right)$ and simplify using division.

$10y\left( {6y + 21} \right) \div 5\left( {2y + 7} \right) = \dfrac{{2 \times 5 \times 3y\left( {2y + 7} \right)}}{{5\left( {2y + 7} \right)}}$

$10y\left( {6y + 21} \right) \div 5\left( {2y + 7} \right) = 6y$

The value of $10y\left( {6y + 21} \right) \div 5\left( {2y + 7} \right)$is $6y$.

iv. Find the division $9{x^2}{y^2}\left( {3z - 24} \right) \div 27xy\left( {z - 8} \right)$.

Ans: The expression $9{x^2}{y^2}\left( {3z - 24} \right)$ can be written as $9\left( 3 \right){x^2}{y^2}\left( {z - 8} \right)$.

Substitute $9\left( 3 \right){x^2}{y^2}\left( {z - 8} \right)$ for $9{x^2}{y^2}\left( {3z - 24} \right)$ in $9{x^2}{y^2}\left( {3z - 24} \right) \div 27xy\left( {z - 8} \right)$ and simplify using division.

$9{x^2}{y^2}\left( {3z - 24} \right) \div 27xy\left( {z - 8} \right) = \dfrac{{9\left( 3 \right){x^2}{y^2}\left( {z - 8} \right)}}{{27xy\left( {z - 8} \right)}}$

$9{x^2}{y^2}\left( {3z - 24} \right) \div 27xy\left( {z - 8} \right) = xy$

The value of $9{x^2}{y^2}\left( {3z - 24} \right) \div 27xy\left( {z - 8} \right)$is $xy$.

v. Find the division $96abc\left( {3a - 12} \right)\left( {5b - 30} \right) \div 144\left( {a - 4} \right)\left( {b - 6} \right)$.

Ans: The expression $96abc\left( {3a - 12} \right)\left( {5b - 30} \right)$ can be written as $96 \times 3 \times 5abc\left( {a - 4} \right)\left( {b - 6} \right)$.

Substitute $96 \times 3 \times 5abc\left( {a - 4} \right)\left( {b - 6} \right)$ for $96abc\left( {3a - 12} \right)\left( {5b - 30} \right)$ in $96abc\left( {3a - 12} \right)\left( {5b - 30} \right) \div 144\left( {a - 4} \right)\left( {b - 6} \right)$ and simplify using division.

$96abc\left( {3a - 12} \right)\left( {5b - 30} \right) \div 144\left( {a - 4} \right)\left( {b - 6} \right) = \dfrac{{96 \times 3 \times 5abc\left( {a - 4} \right)\left( {b - 6} \right)}}{{144\left( {a - 4} \right)\left( {b - 6} \right)}}$

$96abc\left( {3a - 12} \right)\left( {5b - 30} \right) \div 144\left( {a - 4} \right)\left( {b - 6} \right) = 10abc$

The value of $96abc\left( {3a - 12} \right)\left( {5b - 30} \right) \div 144\left( {a - 4} \right)\left( {b - 6} \right)$is $10abc$.

4. Divide as directed.

i. Find the division $5\left( {2x + 1} \right)\left( {3x + 5} \right) \div \left( {2x + 1} \right)$.

Ans: To divide $5\left( {2x + 1} \right)\left( {3x + 5} \right) \div \left( {2x + 1} \right)$, cut the like terms and the remaining will be the answer.

$5\left( {2x + 1} \right)\left( {3x + 5} \right) \div \left( {2x + 1} \right) = \dfrac{{5\left( {2x + 1} \right)\left( {3x + 5} \right)}}{{\left( {2x + 1} \right)}}$

$5\left( {2x + 1} \right)\left( {3x + 5} \right) \div \left( {2x + 1} \right) = 5\left( {3x + 5} \right)$

The value of $5\left( {2x + 1} \right)\left( {3x + 5} \right) \div \left( {2x + 1} \right)$is $5\left( {3x + 5} \right)$.

ii. Find the division $26xy\left( {x + 5} \right)\left( {y - 4} \right) \div 13x\left( {y - 4} \right)$.

Ans: To divide $26xy\left( {x + 5} \right)\left( {y - 4} \right) \div 13x\left( {y - 4} \right)$, cut the like terms and the remaining will be the answer.

$26xy\left( {x + 5} \right)\left( {y - 4} \right) \div 13x\left( {y - 4} \right) = \dfrac{{26xy\left( {x + 5} \right)\left( {y - 4} \right)}}{{13x\left( {y - 4} \right)}}$

$26xy\left( {x + 5} \right)\left( {y - 4} \right) \div 13x\left( {y - 4} \right) = \dfrac{{2\left( {13} \right)xy\left( {x + 5} \right)\left( {y - 4} \right)}}{{13x\left( {y - 4} \right)}}$

$26xy\left( {x + 5} \right)\left( {y - 4} \right) \div 13x\left( {y - 4} \right) = 2y\left( {x + 5} \right)$

The value of $26xy\left( {x + 5} \right)\left( {y - 4} \right) \div 13x\left( {y - 4} \right)$is $2y\left( {x + 5} \right)$.

iii. Find the division $52pqr\left( {p + q} \right)\left( {q + r} \right)\left( {r + p} \right) \div 104pq\left( {q + r} \right)\left( {r + p} \right)$.

Ans: To divide $52pqr\left( {p + q} \right)\left( {q + r} \right)\left( {r + p} \right) \div 104pq\left( {q + r} \right)\left( {r + p} \right)$, cut the like terms and the remaining will be the answer.

$52pqr\left( {p + q} \right)\left( {q + r} \right)\left( {r + p} \right) \div 104pq\left( {q + r} \right)\left( {r + p} \right) = \dfrac{{52pqr\left( {p + q} \right)\left( {q + r} \right)\left( {r + p} \right)}}{{104pq\left( {q + r} \right)\left( {r + p} \right)}}$

$52pqr\left( {p + q} \right)\left( {q + r} \right)\left( {r + p} \right) \div 104pq\left( {q + r} \right)\left( {r + p} \right) = \dfrac{{52pqr\left( {p + q} \right)\left( {q + r} \right)\left( {r + p} \right)}}{{2\left( {52} \right)pq\left( {q + r} \right)\left( {r + p} \right)}}$$52pqr\left( {p + q} \right)\left( {q + r} \right)\left( {r + p} \right) \div 104pq\left( {q + r} \right)\left( {r + p} \right) = \dfrac{1}{2}r\left( {p + q} \right)$

The value of $52pqr\left( {p + q} \right)\left( {q + r} \right)\left( {r + p} \right) \div 104pq\left( {q + r} \right)\left( {r + p} \right)$is $\dfrac{1}{2}r\left( {p + q} \right)$.

iv. Find the division $20\left( {y + 4} \right)\left( {{y^2} + 5y + 3} \right) \div 5\left( {y + 4} \right)$.

Ans: To divide $20\left( {y + 4} \right)\left( {{y^2} + 5y + 3} \right) \div 5\left( {y + 4} \right)$, cut the like terms and the remaining will be the answer.

$20\left( {y + 4} \right)\left( {{y^2} + 5y + 3} \right) \div 5\left( {y + 4} \right) = \dfrac{{20\left( {y + 4} \right)\left( {{y^2} + 5y + 3} \right)}}{{5\left( {y + 4} \right)}}$

$20\left( {y + 4} \right)\left( {{y^2} + 5y + 3} \right) \div 5\left( {y + 4} \right) = \dfrac{{5 \times 4\left( {y + 4} \right)\left( {{y^2} + 5y + 3} \right)}}{{5\left( {y + 4} \right)}}$

$20\left( {y + 4} \right)\left( {{y^2} + 5y + 3} \right) \div 5\left( {y + 4} \right) = 4\left( {{y^2} + 5y + 3} \right)$

The value of $20\left( {y + 4} \right)\left( {{y^2} + 5y + 3} \right) \div 5\left( {y + 4} \right)$is $4\left( {{y^2} + 5y + 3} \right)$.

v. Find the division $x\left( {x + 1} \right)\left( {x + 2} \right)\left( {x + 3} \right) \div x\left( {x + 1} \right)$.

Ans: To divide $x\left( {x + 1} \right)\left( {x + 2} \right)\left( {x + 3} \right) \div x\left( {x + 1} \right)$, cut the like terms and the remaining will be the answer.

$x\left( {x + 1} \right)\left( {x + 2} \right)\left( {x + 3} \right) \div x\left( {x + 1} \right) = \dfrac{{x\left( {x + 1} \right)\left( {x + 2} \right)\left( {x + 3} \right)}}{{x\left( {x + 1} \right)}}$

$x\left( {x + 1} \right)\left( {x + 2} \right)\left( {x + 3} \right) \div x\left( {x + 1} \right) = \left( {x + 2} \right)\left( {x + 3} \right)$

The value of $x\left( {x + 1} \right)\left( {x + 2} \right)\left( {x + 3} \right) \div x\left( {x + 1} \right)$is $\left( {x + 2} \right)\left( {x + 3} \right)$.

5. Factorise the expressions and divide them as directed.

i. Factorise the expression and divide$\left( {{y^2} + 7y + 10} \right) \div \left( {y + 5} \right)$.

Ans:  Factorise $\left( {{y^2} + 7y + 10} \right)$ by splitting the middle term.

${y^2} + 7y + 10 = {y^2} + 2y + 5y + 10$

${y^2} + 7y + 10 = y\left( {y + 2} \right) + 5\left( {y + 2} \right)$

${y^2} + 7y + 10 = \left( {y + 2} \right)\left( {y + 5} \right)$

Substitute $\left( {y + 2} \right)\left( {y + 5} \right)$ for $\left( {{y^2} + 7y + 10} \right)$ in $\left( {{y^2} + 7y + 10} \right) \div \left( {y + 5} \right)$ and simplify.

$\left( {{y^2} + 7y + 10} \right) \div \left( {y + 5} \right) = \dfrac{{\left( {y + 2} \right)\left( {y + 5} \right)}}{{\left( {y + 5} \right)}}$

$\left( {{y^2} + 7y + 10} \right) \div \left( {y + 5} \right) = y + 2$

The value of $\left( {{y^2} + 7y + 10} \right) \div \left( {y + 5} \right)$is $y + 2$.

ii. Factorise the expression and divide $\left( {{m^2} - 14m - 32} \right) \div \left( {m + 2} \right)$.

Ans:  Factorise ${m^2} - 14m - 32$ by splitting the middle term.

${m^2} - 14m - 32 = {m^2} + 2m - 16m - 32$

${m^2} - 14m - 32 = m\left( {m + 2} \right) - 16\left( {m + 2} \right)$

${m^2} - 14m - 32 = \left( {m + 2} \right)\left( {m - 16} \right)$

Substitute $\left( {m + 2} \right)\left( {m - 16} \right)$ for ${m^2} - 14m - 32$ in $\left( {{m^2} - 14m - 32} \right) \div \left( {m + 2} \right)$ and simplify.

$\left( {{m^2} - 14m - 32} \right) \div \left( {m + 2} \right) = \dfrac{{\left( {m + 2} \right)\left( {m - 16} \right)}}{{\left( {m + 2} \right)}}$

$\left( {{m^2} - 14m - 32} \right) \div \left( {m + 2} \right) = m - 16$

The value of $\left( {{m^2} - 14m - 32} \right) \div \left( {m + 2} \right)$ is $m - 16$.

iii. Factorise the expression and divide $\left( {5{p^2} - 25p + 20} \right) \div \left( {p - 1} \right)$.

Ans: The expression $5{p^2} - 25p + 20$ can be written as $5\left( {{p^2} - 5p + 4} \right)$.

Factorise $5\left( {{p^2} - 5p + 4} \right)$ by splitting the middle term.

$5\left( {{p^2} - 5p + 4} \right) = 5\left( {{p^2} - 1p - 4p + 4} \right)$

$5\left( {{p^2} - 5p + 4} \right) = 5\left( {p\left( {p - 1} \right) - 4\left( {p - 1} \right)} \right)$

$5\left( {{p^2} - 5p + 4} \right) = 5\left( {p - 1} \right)\left( {p - 4} \right)$

Substitute $5\left( {p - 1} \right)\left( {p - 4} \right)$ for $5{p^2} - 25p + 20$ in $\left( {5{p^2} - 25p + 20} \right) \div \left( {p - 1} \right)$ and simplify.

$\left( {5{p^2} - 25p + 20} \right) \div \left( {p - 1} \right) = \dfrac{{5\left( {p - 1} \right)\left( {p - 4} \right)}}{{\left( {p - 1} \right)}}$

$\left( {5{p^2} - 25p + 20} \right) \div \left( {p - 1} \right) = 5\left( {p - 4} \right)$

The value of $\left( {5{p^2} - 25p + 20} \right) \div \left( {p - 1} \right)$ is $5\left( {p - 4} \right)$.

iv. Factorise the expression and divide $4yz\left( {{z^2} + 6z - 16} \right) \div 2y\left( {z + 8} \right)$.

Ans:  Factorise $4yz\left( {{z^2} + 6z - 16} \right)$ by splitting the middle term.

$4yz\left( {{z^2} + 6z - 16} \right) = 4yz\left( {{z^2} - 2z + 8z - 16} \right)$

$4yz\left( {{z^2} + 6z - 16} \right) = 4yz\left( {z\left( {z - 2} \right) + 8\left( {z - 2} \right)} \right)$

$4yz\left( {{z^2} + 6z - 16} \right) = 4yz\left( {z - 2} \right)\left( {z + 8} \right)$

Substitute $4yz\left( {z - 2} \right)\left( {z + 8} \right)$ for $4yz\left( {{z^2} + 6z - 16} \right)$ in $4yz\left( {{z^2} + 6z - 16} \right) \div 2y\left( {z + 8} \right)$ and simplify.

$4yz\left( {{z^2} + 6z - 16} \right) \div 2y\left( {z + 8} \right) = \dfrac{{4yz\left( {z - 2} \right)\left( {z + 8} \right)}}{{2y\left( {z + 8} \right)}}$

$4yz\left( {{z^2} + 6z - 16} \right) \div 2y\left( {z + 8} \right) = 2z\left( {z - 2} \right)$

The value of $4yz\left( {{z^2} + 6z - 16} \right) \div 2y\left( {z + 8} \right)$ is $2z\left( {z - 2} \right)$.

v. Factorise the expression and divide $5pq\left( {{p^2} - {q^2}} \right) \div 2p\left( {p + q} \right)$.

Ans: Factorise $5pq\left( {{p^2} - {q^2}} \right)$ by using the formula ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$.

$5pq\left( {{p^2} - {q^2}} \right) = 5pq\left( {p - q} \right)\left( {p + q} \right)$

Substitute $5pq\left( {p - q} \right)\left( {p + q} \right)$ for $5pq\left( {{p^2} - {q^2}} \right)$ in $5pq\left( {{p^2} - {q^2}} \right) \div 2p\left( {p + q} \right)$ and simplify.

$5pq\left( {{p^2} - {q^2}} \right) \div 2p\left( {p + q} \right) = \dfrac{{5pq\left( {p - q} \right)\left( {p + q} \right)}}{{2p\left( {p + q} \right)}}$

$5pq\left( {{p^2} - {q^2}} \right) \div 2p\left( {p + q} \right) = \dfrac{5}{2}q\left( {p - q} \right)$

The value of $5pq\left( {{p^2} - {q^2}} \right) \div 2p\left( {p + q} \right)$ is $\dfrac{5}{2}q\left( {p - q} \right)$.

vi. Factorise the expression and divide $12xy\left( {9{x^2} - 16{y^2}} \right) \div 4xy\left( {3x + 4y} \right)$.

Ans: Factorise $12xy\left( {9{x^2} - 16{y^2}} \right)$ by using the formula ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$.

$12xy\left( {9{x^2} - 16{y^2}} \right) = 12xy\left( {3x - 4y} \right)\left( {3x + 4y} \right)$

Substitute $12xy\left( {3x - 4y} \right)\left( {3x + 4y} \right)$ for $12xy\left( {9{x^2} - 16{y^2}} \right)$ in $12xy\left( {9{x^2} - 16{y^2}} \right) \div 4xy\left( {3x + 4y} \right)$ and simplify.

$12xy\left( {9{x^2} - 16{y^2}} \right) \div 4xy\left( {3x + 4y} \right) = \dfrac{{12xy\left( {3x - 4y} \right)\left( {3x + 4y} \right)}}{{4xy\left( {3x + 4y} \right)}}$

$12xy\left( {9{x^2} - 16{y^2}} \right) \div 4xy\left( {3x + 4y} \right) = 3\left( {3x - 4y} \right)$

The value of $12xy\left( {9{x^2} - 16{y^2}} \right) \div 4xy\left( {3x + 4y} \right)$ is $3\left( {3x - 4y} \right)$.

vii. Factorise the expression and divide $39{y^3}\left( {50{y^2} - 98} \right) \div 26{y^2}\left( {5y + 7} \right)$.

Ans: The expression $39{y^3}\left( {50{y^2} - 98} \right)$ can be written as $39\left( 2 \right){y^3}\left( {{{\left( {5y} \right)}^2} - {{\left( 7 \right)}^2}} \right)$.

Factorise $39\left( 2 \right){y^3}\left( {{{\left( {5y} \right)}^2} - {{\left( 7 \right)}^2}} \right)$ by using the formula ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$.

$39\left( 2 \right){y^3}\left( {{{\left( {5y} \right)}^2} - {{\left( 7 \right)}^2}} \right) = 39\left( 2 \right){y^3}\left( {5y - 7} \right)\left( {5y + 7} \right)$

Substitute $39\left( 2 \right){y^3}\left( {5y - 7} \right)\left( {5y + 7} \right)$ for $39{y^3}\left( {50{y^2} - 98} \right)$ in $39{y^3}\left( {50{y^2} - 98} \right) \div 26{y^2}\left( {5y + 7} \right)$ and simplify.

$39{y^3}\left( {50{y^2} - 98} \right) \div 26{y^2}\left( {5y + 7} \right) = \dfrac{{39\left( 2 \right){y^3}\left( {5y - 7} \right)\left( {5y + 7} \right)}}{{26{y^2}\left( {5y + 7} \right)}}$

$39{y^3}\left( {50{y^2} - 98} \right) \div 26{y^2}\left( {5y + 7} \right) = 3y\left( {5y - 7} \right)$

The value of $39{y^3}\left( {50{y^2} - 98} \right) \div 26{y^2}\left( {5y + 7} \right)$ is $3y\left( {5y - 7} \right)$.

## NCERT Solutions For Class 8 Maths Chapter 14 Factorisation Exercise 14.3

Opting for the NCERT solutions for Ex 14.3 Class 8 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 14.3 Class 8 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

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4. How can we solve Class 8 Maths Chapter 14 factorization exercise 14.3?

The questions that are included in the first exercise of class 8 Maths chapter 14 factorization exercise 14.3 are quite simple and can easily be done if the students have good conceptual knowledge about the factors. For better conceptual understanding, students can look for the NCERT solution for class 7 math chapter 14 factorization exercise 14.3 on the Vedantu website.

5. What is the importance of the syllabus?

A syllabus is very essential, as it works as a skeleton and gives a framework to both the pupils and the teachers. It also helps the student develop effective learning strategies. Besides this, it also helps the teachers organize their upcoming classes and plan accordingly.