NCERT Solutions for Class 8 Maths Chapter 14 Factorisation (EX 14.3) Exercise 14.3

NCERT Solutions for Class 8 Maths Chapter 14 Factorisation (EX 14.3) Exercise 14.3

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Access NCERT Solutions for Class 8 Maths Chapter 14 - Factorization part-1

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.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 8 students who are thorough with all the concepts from the Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 8 Maths Chapter 14 Exercise 14.3 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 8 Maths Chapter 14 Exercise 14.3, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it.

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