NCERT Solutions for Class 8 Maths Chapter 14 Factorisation (EX 14.1) Exercise 14.1

VSAT 2022

NCERT Solutions for Class 8 Maths Chapter 14 Factorisation Exercise 14.1

Factorisation can be defined as breaking down a number into smaller numbers, and multiplying the same will provide you with the original numeral. It is one of the essential concepts in maths, and every learner must be well accustomed to knowing more about equations. NCERT Solutions For Class 8 Maths Chapter 14 Exercise 14.1 prepared by Vedantu contain various problems on factorisation. Besides the first exercise, the remaining ones are also affixed in the PDF. So, you can refer to this Class 8 Maths Chapter 14 Exercise 14.1 to get familiar with the NCERT solution steps and also clear your concepts. Students can also download the NCERT Solutions for Class 8 Science curated by our Master Teachers really Helpful.

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Exercise 14.1

1. It is required to find common factors of the following terms.

i. Find common factors of terms $12x$, 36.

Ans: To find common factors of terms $12x$ and 36 write these terms as a multiple of different numbers.

$12x$ can be written as $2 \times 2 \times 3 \times x$.

36 can be written as $2 \times 2 \times 3 \times 3$.

On comparing both the terms $2 \times 2 \times 3 \times x$ and $2 \times 2 \times 3 \times 3$the common factors obtained is $2 \times 2 \times 3$.

Thus, $2 \times 2 \times 3$ can be simplified as 12.

Therefore, the common factor of terms $12x$and 36 is 12.

 

ii. Find common factors of terms $2y$, $22xy$.

Ans: To find common factors of terms $2y$ and $22xy$ write these terms as a multiple of different numbers.

$2y$ can be written as $2 \times y$.

$22xy$ can be written as $2 \times 11 \times x \times y$.

On comparing both the terms $2 \times y$ and $2 \times 11 \times x \times y$ the common factors obtained is $2 \times y$.

Thus, $2 \times y$ can be simplified as $2y$.

Therefore, the common factor of terms $2y$ and $22xy$ is $2y$.

 

iii. Find common factors of terms $14pq$, $28{p^2}{q^2}$.

Ans: To find common factors of terms $14pq$ and $28{p^2}{q^2}$ write these terms as a multiple of different numbers.

$14pq$ can be written as $2 \times 7 \times p \times q$.

$28{p^2}{q^2}$ can be written as $2 \times 2 \times 7 \times p \times q$.

On comparing both the terms $2 \times 7 \times p \times q$ and $2 \times 2 \times 7 \times p \times q$ the common factors obtained is $2 \times 7 \times p \times q$.

Thus, $2 \times 7 \times p \times q$ can be simplified as $14pq$.

Therefore, the common factor of terms $14pq$ and $28{p^2}{q^2}$ is $14pq$.

 

iv. Find common factors of terms $2x$, $3{x^2}$, 4.

Ans: To find common factors of terms $2x$, $3{x^2}$ and 4 write these terms as a multiple of different numbers.

$2x$ can be written as $2 \times x$.

$3{x^2}$ can be written as $3 \times x \times x$.

4 can be written as $2 \times 2$.

On comparing the terms $2 \times x$, $3 \times x \times x$ and $2 \times 2$ the common factors obtained is 1.

Therefore, the common factor of terms $2x$, $3{x^2}$ and 4 is 1.

 

v. Find common factors of terms $6abc$, $24a{b^2}$, $12{a^2}b$.

Ans: To find common factors of terms $6abc$, $24a{b^2}$ and $12{a^2}b$ write these terms as a multiple of different numbers.

$6abc$can be written as $2 \times 3 \times a \times b \times c$.

$24a{b^2}$ can be written as $2 \times 2 \times 2 \times 3 \times a \times b \times b$.

$12{a^2}b$ can be written as $2 \times 2 \times 3 \times a \times a \times b$.

On comparing the terms $2 \times 3 \times a \times b \times c$, $2 \times 2 \times 2 \times 3 \times a \times b \times b$ and $2 \times 2 \times 3 \times a \times a \times b$ the common factors obtained is $2 \times 3 \times a \times b$.

Thus, $2 \times 3 \times a \times b$ can be simplified as $6ab$.

Therefore, the common factor of terms $6abc$, $24a{b^2}$ and $12{a^2}b$is $6ab$.

 

vi. Find common factors of terms $16{x^3}$, $ - 4{x^2}$, $32x$.

Ans: To find common factors of terms $16{x^3}$, $ - 4{x^2}$ and $32x$ write these terms as a multiple of different numbers.

$16{x^3}$ can be written as $2 \times 2 \times 2 \times 2 \times x \times x \times x$.

$ - 4{x^2}$ can be written as $ - 1 \times 2 \times 2 \times x \times x$.

$32x$ can be written as $2 \times 2 \times 2 \times 2 \times 2 \times x$.

On comparing the terms $2 \times 2 \times 2 \times 2 \times x \times x \times x$, $ - 1 \times 2 \times 2 \times x \times x$ and $2 \times 2 \times 2 \times 2 \times 2 \times x$ the common factors obtained is $2 \times 2 \times x$.

Thus, $2 \times 2 \times x$ can be simplified as $4x$.

Therefore, the common factor of terms $16{x^3}$, $ - 4{x^2}$ and $32x$ is $4x$.

 

vii. Find common factors of terms $10pq$, $20qr$, $30rp$.

Ans: To find common factors of terms $10pq$, $20qr$ and $30rp$ write these terms as a multiple of different numbers.

$10pq$ can be written as $2 \times 5 \times p \times q$.

$20qr$ can be written as $2 \times 2 \times 5 \times q \times r$.

$30rp$ can be written as $2 \times 3 \times 5 \times r \times p$.

On comparing the terms $2 \times 5 \times p \times q$, $2 \times 2 \times 5 \times q \times r$ and $2 \times 3 \times 5 \times r \times p$ the common factors obtained is $2 \times 5$.

Thus, $2 \times 5$ can be simplified as 10.

Therefore, the common factor of terms $10pq$, $20qr$ and $30rp$is 10.

 

viii. Find common factor of terms $3{x^2}{y^3}$, $10{x^3}{y^2}$, $6{x^2}{y^2}z$.

Ans: To find common factors of terms $3{x^2}{y^3}$, $10{x^3}{y^2}$ and $6{x^2}{y^2}z$ write these terms as a multiple of different numbers.

$3{x^2}{y^3}$ can be written as $3 \times x \times x \times y \times y \times y$.

$10{x^3}{y^2}$ can be written as $2 \times 5 \times x \times x \times x \times y \times y$.

$6{x^2}{y^2}z$ can be written as $2 \times 3 \times x \times x \times y \times y \times z$.

On comparing the terms $3 \times x \times x \times y \times y \times y$, $2 \times 5 \times x \times x \times x \times y \times y$ and $2 \times 3 \times x \times x \times y \times y \times z$ the common factors obtained is $x \times x \times y \times y$.

Thus, $x \times x \times y \times y$ can be simplified as ${x^2}{y^2}$.

Therefore, the common factor of terms $3{x^2}{y^3}$, $10{x^3}{y^2}$ and $6{x^2}{y^2}z$ is ${x^2}{y^2}$.

 

2. Factorise the following expressions.

(i) Factorise the expression $7x - 42$.

Ans: To factorise the expression $7x - 42$write $7x$ and 42 as a product of different numbers.

$7x$ can be written as $7 \times x$.

42 can be written as $2 \times 3 \times 7$.

Substitute $7 \times x$ for $7x$ and $2 \times 3 \times 7$ for 42 in expression $7x - 42$.

$7x - 42 = \left( {7 \times x} \right) - \left( {2 \times 3 \times 7} \right)$

7 is the common factor of $7 \times x$ and $2 \times 3 \times 7$ 

Thus, take 7 as a common factor from right hand side of expression $7x - 42 = \left( {7 \times x} \right) - \left( {2 \times 3 \times 7} \right)$.

$7x - 42 = 7\left( {x - 6} \right)$

Therefore, $7x - 42$ can be factorized as $7\left( {x - 6} \right)$.

 

(ii) Factorise the expression $6p - 12q$.

Ans: To factorise the expression $6p - 12q$ write $6p$ and $12q$ as products of different numbers.

$6p$ can be written as $2 \times 3 \times p$.

$12q$ can be written as $2 \times 2 \times 3 \times q$.

Substitute $2 \times 3 \times p$ for $6p$ and $2 \times 2 \times 3 \times q$ for $12q$ in expression $6p - 12q$.

$6p - 12q = \left( {2 \times 3 \times p} \right) - \left( {2 \times 2 \times 3 \times q} \right)$

$\left( {2 \times 3} \right)$ is the common factor of $2 \times 3 \times p$ and $2 \times 2 \times 3 \times q$ 

Thus, take $\left( {2 \times 3} \right)$ as a common factor from right hand side of expression $6p - 12q = \left( {2 \times 3 \times p} \right) - \left( {2 \times 2 \times 3 \times q} \right)$.

$6p - 12q = \left( {2 \times 3} \right)\left[ {p - \left( {2 \times q} \right)} \right]$

$6p - 12q= 6\left( {p - 2q} \right)$

Therefore, $6p - 12q$ can be factorized as $6\left( {p - 2q} \right)$.

 

(iii) Factorise the expression $7{a^2} + 14a$.

Ans: To factorise the expression $7{a^2} + 14a$ write $7{a^2}$ and $14a$ as products of different numbers.

$7{a^2}$ can be written as $7 \times a \times a$.

$14a$ can be written as $7 \times 2 \times a$.

Substitute $7 \times a \times a$ for $7{a^2}$ and $7 \times 2 \times a$ for $14a$ in expression $7{a^2} + 14a$.

$7{a^2} + 14a = \left( {7 \times a \times a} \right) + \left( {7 \times 2 \times a} \right)$

$\left( {7 \times a} \right)$is the common factor of $7 \times a \times a$ and $7 \times 2 \times a$ 

Thus, take $\left( {7 \times a} \right)$ as a common factor from right hand side of expression $7{a^2} + 14a = \left( {7 \times a \times a} \right) + \left( {7 \times 2 \times a} \right)$.

$7{a^2} + 14a = \left( {7 \times a} \right)\left[ {a + 2} \right]$

$= 7a\left( {a + 2} \right)$

Therefore, $7{a^2} + 14a$ can be factorized as $7a\left( {a + 2} \right)$.

 

(iv) Factorise the expression $ - 16z + 20{z^3}$.

Ans: To factorise the expression $ - 16z + 20{z^3}$write $ - 16z$ and $20{z^3}$ as products of different numbers.

$ - 16z$ can be written as $ - 1 \times 2 \times 2 \times 2 \times 2 \times z$.

$20{z^3}$ can be written as $2 \times 2 \times 5 \times z \times z \times z$.

Substitute $ - 1 \times 2 \times 2 \times 2 \times 2 \times z$ for $ - 16z$ and $2 \times 2 \times 5 \times z \times z \times z$ for $20{z^3}$ in expression $ - 16z + 20{z^3}$.

$ - 16z + 20{z^3} = \left( { - 1 \times 2 \times 2 \times 2 \times 2 \times z} \right) + \left( {2 \times 2 \times 5 \times z \times z \times z} \right)$

$\left( {2 \times 2 \times z} \right)$ is the common factor of $ - 1 \times 2 \times 2 \times 2 \times 2 \times z$ and $2 \times 2 \times 5 \times z \times z \times z$ 

Thus, take  $\left( {2 \times 2 \times z} \right)$ as a common factor from right hand side of expression $ - 16z + 20{z^3} = \left( { - 1 \times 2 \times 2 \times 2 \times 2 \times z} \right) + \left( {2 \times 2 \times 5 \times z \times z \times z} \right)$.

$- 16z + 20{z^3} = \left( {2 \times 2 \times z} \right)\left[ {\left( { - 1 \times 2 \times 2} \right) + \left( {5 \times z \times z} \right)} \right]$

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

Therefore, $ - 16z + 20{z^3}$ can be factorized as \[4z\left( { - 4 + 5{z^2}} \right)\].

 

v. Factorise the expression $20{l^2}m + 30alm$.

Ans: To factorise the expression $20{l^2}m + 30alm$ write $20{l^2}m$ and $30alm$ as products of different numbers.

$20{l^2}m$ can be written as $2 \times 2 \times 5 \times l \times l \times m$.

$30alm$ can be written as $2 \times 3 \times 5 \times a \times l \times m$.

Substitute $2 \times 2 \times 5 \times l \times l \times m$ for $20{l^2}m$ and $2 \times 3 \times 5 \times a \times l \times m$ for $30alm$ in expression $20{l^2}m + 30alm$.

$20{l^2}m + 30alm = \left( {2 \times 2 \times 5 \times l \times l \times m} \right) + \left( {2 \times 3 \times 5 \times a \times l \times m} \right)$

$\left( {2 \times 5 \times l \times m} \right)$ is the common factor of $2 \times 2 \times 5 \times l \times l \times m$ and $2 \times 3 \times 5 \times a \times l \times m$ 

Thus, take  $\left( {2 \times 5 \times l \times m} \right)$ as a common factor from right hand side of expression $20{l^2}m + 30alm = \left( {2 \times 2 \times 5 \times l \times l \times m} \right) + \left( {2 \times 3 \times 5 \times a \times l \times m} \right)$.

$20{l^2}m + 30alm = \left( {2 \times 5 \times l \times m} \right)\left[ {\left( {2 \times l} \right) + \left( {3 \times a} \right)} \right]$

$= 10lm\left( {2l + 3a} \right)$

Therefore, $20{l^2}m + 30alm$ can be factored as \[10lm\left( {2l + 3a} \right)\].

 

(vi) Factorise the expression $5{x^2}y - 15x{y^2}$.

Ans: To factorise the expression $5{x^2}y - 15x{y^2}$ write $5{x^2}y$ and $15x{y^2}$ as products of different numbers.

$5{x^2}y$ can be written as $5 \times x \times x \times y$.

$15x{y^2}$ can be written as $3 \times 5 \times x \times y \times y$.

Substitute $5 \times x \times x \times y$ for $5{x^2}y$ and $3 \times 5 \times x \times y \times y$ for $15x{y^2}$ in expression $5{x^2}y - 15x{y^2}$.

$5{x^2}y - 15x{y^2} = \left( {5 \times x \times x \times y} \right) - \left( {3 \times 5 \times x \times y \times y} \right)$

$\left( {5 \times x \times y} \right)$ is the common factor of $5 \times x \times x \times y$ and $3 \times 5 \times x \times y \times y$ 

Thus, take  $\left( {5 \times x \times y} \right)$ as a common factor from right hand side of expression $5{x^2}y - 15x{y^2} = \left( {5 \times x \times x \times y} \right) - \left( {3 \times 5 \times x \times y \times y} \right)$.

$5{x^2}y - 15x{y^2} = \left( {5 \times x \times y} \right)\left[ {\left( x \right) - \left( {3 \times y} \right)} \right]$

$= 5xy\left( {x - 3y} \right)$ 

Therefore, $5{x^2}y - 15x{y^2}$ can be factorized as $5xy\left( {x - 3y} \right)$.

 

(vii) Factorise the expression $10{a^2} - 15{b^2} + 20{c^2}$.

Ans: To factorise the expression $10{a^2} - 15{b^2} + 20{c^2}$ write $10{a^2}$, $15{b^2}$ and $20{c^2}$ as product of different numbers.

$10{a^2}$ can be written as $2 \times 5 \times a \times a$.

$15{b^2}$ can be written as $3 \times 5 \times b \times b$.

$20{c^2}$ can be written as $2 \times 2 \times 5 \times c \times c$.

Substitute $2 \times 5 \times a \times a$for $10{a^2}$, $3 \times 5 \times b \times b$ for $15{b^2}$ and $2 \times 2 \times 5 \times c \times c$ for $20{c^2}$ in expression $10{a^2} - 15{b^2} + 20{c^2}$.

$10{a^2} - 15{b^2} + 20{c^2} = \left( {2 \times 5 \times a \times a} \right) - \left( {3 \times 5 \times b \times b} \right) + \left( {2 \times 2 \times 5 \times c \times c} \right)$

5 is the common factor of $2 \times 5 \times a \times a$, $3 \times 5 \times b \times b$ and $2 \times 2 \times 5 \times c \times c$ 

Thus, take  5 as a common factor from right hand side of expression $10{a^2} - 15{b^2} + 20{c^2} = \left( {2 \times 5 \times a \times a} \right) - \left( {3 \times 5 \times b \times b} \right) + \left( {2 \times 2 \times 5 \times c \times c} \right)$.

$10{a^2} - 15{b^2} + 20{c^2} = 5\left[ {\left( {2 \times a \times a} \right) - \left( {3 \times b \times b} \right) + \left( {2 \times 2 \times c \times c} \right)} \right]$

$= 5\left( {2{a^2} - 3{b^2} + 4{c^2}} \right)$

Therefore, $10{a^2} - 15{b^2} + 20{c^2}$ can be factorized as $5\left( {2{a^2} - 3{b^2} + 4{c^2}} \right)$.

 

(viii) Factorise the expression $ - 4{a^2} + 4ab - 4ca$.

Ans: To factorise the expression $ - 4{a^2} + 4ab - 4ca$ write $4{a^2}$, $4ab$ and $4ca$ as products of different numbers.

$4{a^2}$ can be written as $2 \times 2 \times a \times a$.

$4ab$ can be written as $2 \times 2 \times a \times b$.

$4ca$ can be written as $2 \times 2 \times c \times a$.

Substitute $2 \times 2 \times a \times a$ for $4{a^2}$, $2 \times 2 \times a \times b$ for $4ab$ and $2 \times 2 \times c \times a$ for $4ca$ in expression $ - 4{a^2} + 4ab - 4ca$.

$ - 4{a^2} + 4ab - 4ca =  - \left( {2 \times 2 \times a \times a} \right) + \left( {2 \times 2 \times a \times b} \right) - \left( {2 \times 2 \times c \times a} \right)$

$\left( {2 \times 2 \times a} \right)$ is the common factor of $2 \times 2 \times a \times a$, $2 \times 2 \times a \times b$ and $2 \times 2 \times c \times a$ 

Thus, take  $\left( {2 \times 2 \times a} \right)$ as a common factor from right hand side of expression $ - 4{a^2} + 4ab - 4ca =  - \left( {2 \times 2 \times a \times a} \right) + \left( {2 \times 2 \times a \times b} \right) - \left( {2 \times 2 \times c \times a} \right)$.

$- 4{a^2} + 4ab - 4ca = \left( {2 \times 2 \times a} \right)\left[ { - \left( a \right) + \left( b \right) - \left( c \right)} \right]$

$= 4a\left( { - a + b - c} \right)$

Therefore, $ - 4{a^2} + 4ab - 4ca$ can be factorized as $4a\left( { - a + b - c} \right)$.

 

(ix) Factorise the expression ${x^2}yz + x{y^2}z + xy{z^2}$.

Ans: To factorise the expression ${x^2}yz + x{y^2}z + xy{z^2}$ write ${x^2}yz$, $x{y^2}z$ and $xy{z^2}$ as product of different numbers.

${x^2}yz$ can be written as $x \times x \times y \times z$.

$x{y^2}z$ can be written as $x \times y \times y \times z$.

$xy{z^2}$can be written as $x \times y \times z \times z$.

Substitute $x \times x \times y \times z$for ${x^2}yz$, $x \times y \times y \times z$ for $x{y^2}z$ and $x \times y \times z \times z$ for $xy{z^2}$ in expression ${x^2}yz + x{y^2}z + xy{z^2}$.

${x^2}yz + x{y^2}z + xy{z^2} = \left( {x \times x \times y \times z} \right) + \left( {x \times y \times y \times z} \right) + \left( {x \times y \times z \times z} \right)$

$\left( {x \times y \times z} \right)$ is the common factor of $x \times x \times y \times z$, $x \times y \times y \times z$ and $x \times y \times z \times z$ 

Thus, take  $\left( {x \times y \times z} \right)$ as a common factor from right hand side of expression ${x^2}yz + x{y^2}z + xy{z^2} = \left( {x \times x \times y \times z} \right) + \left( {x \times y \times y \times z} \right) + \left( {x \times y \times z \times z} \right)$.

${x^2}yz + x{y^2}z + xy{z^2} = \left( {x \times y \times z} \right)\left[ {\left( x \right) + \left( y \right) + \left( z \right)} \right]$

$= xyz\left( {x + y + z} \right)$

Therefore, ${x^2}yz + x{y^2}z + xy{z^2}$ can be factorized as $xyz\left( {x + y + z} \right)$.

 

(x) Factorise the expression $a{x^2}y + bx{y^2} + cxyz$.

Ans: To factorise the expression $a{x^2}y + bx{y^2} + cxyz$ write $a{x^2}y$, $bx{y^2}$ and $cxyz$ as product of different numbers.

$a{x^2}y$ can be written as $a \times x \times x \times y$.

$bx{y^2}$ can be written as $b \times x \times y \times y$.

$cxyz$ can be written as $c \times x \times y \times z$.

Substitute $a \times x \times x \times y$ for $a{x^2}y$, $b \times x \times y \times y$ for $bx{y^2}$ and $c \times x \times y \times z$ for $cxyz$ in expression $a{x^2}y + bx{y^2} + cxyz$.

$a{x^2}y + bx{y^2} + cxyz = \left( {a \times x \times x \times y} \right) + \left( {b \times x \times y \times y} \right) + \left( {c \times x \times y \times z} \right)$

$\left( {x \times y} \right)$is the common factor of $a \times x \times x \times y$, $b \times x \times y \times y$ and $c \times x \times y \times z$ 

Thus, take  $\left( {x \times y} \right)$ as a common factor from right hand side of expression $a{x^2}y + bx{y^2} + cxyz = \left( {a \times x \times x \times y} \right) + \left( {b \times x \times y \times y} \right) + \left( {c \times x \times y \times z} \right)$.

$a{x^2}y + bx{y^2} + cxyz = \left( {x \times y} \right)\left[ {\left( {a \times x} \right) + \left( {b \times y} \right) + \left( {c \times z} \right)} \right]$

$= xy\left( {ax + by + cz} \right)$

Therefore, $a{x^2}y + bx{y^2} + cxyz$ can be factorized as $xy\left( {ax + by + cz} \right)$.

 

3. Factorise the following.

i. Factorise the expression ${x^2} + xy + 8x + 8y$.

Ans: To factorise the expression ${x^2} + xy + 8x + 8y$ write ${x^2}$, $xy$, $8x$ and $8y$ as products of different numbers.

${x^2}$ can be written as $x \times x$.

$xy$ can be written as $x \times y$.

$8x$ can be written as $8 \times x$.

$8y$ can be written as $8 \times y$.

Substitute $x \times x$ for ${x^2}$, $x \times y$ for $xy$, $8 \times x$ for $8x$ and $8 \times y$ for $8y$ in expression ${x^2} + xy + 8x + 8y$.

${x^2} + xy + 8x + 8y = \left( {x \times x} \right) + \left( {x \times y} \right) + \left( {8 \times x} \right) + \left( {8 \times y} \right)$

Take $x$ as the common factor from $x \times x$, $x \times y$ and 8 as the common factor from $8 \times x$, $8 \times y$ and simplify.

${x^2} + xy + 8x + 8y = x\left[ {\left( x \right) + \left( y \right)} \right] + 8\left[ {\left( x \right) + \left( y \right)} \right]$

$= x\left( {x + y} \right) + 8\left( {x + y} \right)$

$= \left( {x + y} \right)\left( {x + 8} \right)$

Therefore, ${x^2} + xy + 8x + 8y$ can be factorized as $\left( {x + y} \right)\left( {x + 8} \right)$.

 

ii. Factorise the expression $15xy - 6x + 5y - 2$.

Ans: To factorise the expression $15xy - 6x + 5y - 2$ write $15xy$, $6x$, $5y$ and 2 as products of different numbers.

$15xy$ can be written as $3 \times 5 \times x \times y$.

$6x$ can be written as $2 \times 3 \times x$.

$5y$ can be written as $5 \times y$.

2 can be written as $1 \times 2$.

Substitute $3 \times 5 \times x \times y$ for $15xy$, $2 \times 3 \times x$ for $6x$, $5 \times y$ for $5y$ and $1 \times 2$ for 2 in expression $15xy - 6x + 5y - 2$.

$15xy - 6x + 5y - 2 = \left( {3 \times 5 \times x \times y} \right) - \left( {2 \times 3 \times x} \right) + \left( {5 \times y} \right) - \left( {2 \times 1} \right)$

Take $3 \times x$ as the common factor from $3 \times 5 \times x \times y$, $2 \times 3 \times x$ and 1 as the common factor from $5 \times y$, $1 \times 2$  and simplify.

$15xy - 6x + 5y - 2 = \left( {3 \times x} \right)\left[ {\left( {5 \times y} \right) - \left( 2 \right)} \right] + 1\left[ {\left( {5 \times y} \right) - \left( 2 \right)} \right]$

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

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

Therefore, $15xy - 6x + 5y - 2$ can be factorized as $\left( {5y - 2} \right)\left( {3x + 1} \right)$.

 

iii. Factorise the expression $ax + bx - ay - by$.

Ans: To factorise the expression $ax + bx - ay - by$ write $ax$, $bx$, $ay$ and $by$ as a product of different numbers.

$ax$ can be written as $a \times x$.

$bx$ can be written as $b \times x$.

$ay$can be written as $a \times y$.

$by$ can be written as $b \times y$.

Substitute $a \times x$ for $ax$, $b \times x$ for $bx$, $a \times y$ for $ay$ and $b \times y$ for $by$ in expression $ax + bx - ay - by$.

$ax + bx - ay - by = \left( {a \times x} \right) + \left( {b \times x} \right) - \left( {a \times y} \right) - \left( {b \times y} \right)$

Take $x$ as the common factor from $a \times x$, $b \times x$ and $y$ as the common factor from $a \times y$, $b \times y$ and simplify.

$ax + bx - ay - by = x\left[ {\left( a \right) + \left( b \right)} \right] - y\left[ {\left( a \right) + \left( b \right)} \right]$

$= \left( {x - y} \right)\left( {a + b} \right)$

Therefore, $ax + bx - ay - by$ can be factorized as $\left( {x - y} \right)\left( {a + b} \right)$.

 

iv. Factorise the expression $15pq + 15 + 9q + 25p$.

Ans: To factorise the expression $15pq + 15 + 9q + 25p$ write $15pq$, 15, $9q$ and $25p$ as products of different numbers.

$15pq$ can be written as $3 \times 5 \times p \times q$.

15 can be written as $3 \times 5$.

$9q$ can be written as $3 \times 3 \times q$.

$25p$can be written as $5 \times 5 \times p$.

Substitute $3 \times 5 \times p \times q$ for $15pq$, $3 \times 5$ for 15, $3 \times 3 \times q$ for $9q$ and $5 \times 5 \times p$ for $25p$ in expression $15pq + 15 + 9q + 25p$.

$15pq + 15 + 9q + 25p = \left( {3 \times 5 \times p \times q} \right) + \left( {3 \times 5} \right) + \left( {3 \times 3 \times q} \right) + \left( {5 \times 5 \times p} \right)$

Take $3 \times q$ as the common factor from $3 \times 5 \times p \times q$, $3 \times 3 \times q$ and 5 as the common factor from $3 \times 5$, $5 \times 5 \times p$ and simplify.

$15pq + 15 + 9q + 25p = \left( {3 \times q} \right)\left[ {\left( {5 \times p} \right) + \left( 3 \right)} \right] + \left( 5 \right)\left[ {\left( 3 \right) + \left( {5 \times p} \right)} \right]$

$= 3q\left( {5p + 3} \right) + 5\left( {3 + 5p} \right)$

$= \left( {5p + 3} \right)\left( {3q + 5} \right)$

Therefore, $15pq + 15 + 9q + 25p$ can be factorized as $\left( {5p + 3} \right)\left( {3q + 5} \right)$.

 

v. Factorise the expression $z - 7 + 7xy - xyz$.

Ans: To factorise the expression $z - 7 + 7xy - xyz$ write $7xy$ and $xyz$ as products of different numbers.

$7xy$ can be written as $7 \times x \times y$.

$xyz$can be written as $x \times y \times z$.

Substitute $7 \times x \times y$ for $7xy$ and $x \times y \times z$ for $xyz$ in expression $z - 7 + 7xy - xyz$.

$z - 7 + 7xy - xyz = z - 7 + \left( {7 \times x \times y} \right) - \left( {x \times y \times z} \right)$

Take $x \times y$ as the common factor from $7 \times x \times y$ and $x \times y \times z$ and simplify.

$z - 7 + 7xy - xyz = z - 7 + \left( {x \times y} \right)\left[ {\left( 7 \right) - \left( z \right)} \right]$

$= 1\left( {z - 7} \right) - xy\left( {z - 7} \right)$

$= \left( {1 - xy} \right)\left( {z - 7} \right)$

Therefore, $z - 7 + 7xy - xyz$ can be factorized as $\left( {1 - xy} \right)\left( {z - 7} \right)$.


Class 8 Maths Chapter 14 Factorisation Exercise 14.1 Free PDF Download

Chapter wise NCERT Solutions for Class 8 Maths


Class 8 Maths Chapter 14 Includes:

Chapter 14 Factorisation All Exercises in PDF Format

Exercise 14.2

5 Questions & Solutions

Exercise 14.3

5 Questions & Solutions

Exercise 14.4

21 Questions & Solutions


Class 8 Maths Ch 14 Ex 14.1 Free PDF Download

We advise you to download NCERT Solutions Class 8 Maths Chapter 14 Exercise 14.1 PDF and follow it throughout your academic year. As said earlier, this material will assist you in understanding the factorisation methods in a better way, enabling you to score better in exams.


NCERT Maths Class 8 Chapter 14 Exercise 14.1 - Overview

Alongside exercise 14.1, the other sections are also provided in this PDF file. Please go through the following section to have a glimpse of what you are going to study in NCERT Solutions for Class 8th Maths Chapter 14 Exercise 14.1.

Important Concepts Covered in Exercise 14.1 of Class 8 Maths NCERT Solutions

Exercise 14.1 of Class 8 Maths NCERT Solutions is mainly based on finding factors and also finding common factors of the given numbers and terms. The solutions present in this exercise will help students to build a solid base of the concepts and they can easily solve the complex algebraic expressions. 


By practising the questions focused on finding factors of terms and numbers, students can score well in exams and it will help them in advanced Mathematics studies.

Benefits of Exercise 14.1 of Class 8 Maths NCERT Solutions

Some of the prime benefits of Exercise 14.1 of Class 8 Maths NCERT Solutions are listed below.

  • Solving the questions present in Exercise 14.1 of Class 8 Maths NCERT Solutions will help students to improve their speed in answering the problems.

  • Students can self-assess their knowledge about the concept of factors easily by solving the questions of Exercise 14.1 of Class 8 Maths NCERT Solutions.

  • The questions are solved by keeping in mind the CBSE format and these solutions are provided by expert teachers having years of experience.

Exercise 14.1: Question 1, 2 and 3

The first question of NCERT Class 8 Maths Chapter 14 Exercise 14.1 is about finding the common factors of some expressions like (i) 12x, 36, (ii) 2y, 22y, etc. These questions are more or less straightforward, but one has to be very careful with the variables.


Exercise 14.2

  • Question 1: Here, you are required to factorise various expressions. For instance, a2 + 8a + 16, p2 – 10p + 25, 25m2 + 30m + 9 and others. All total, there are eight expressions which you need to factorise, and you can go through Maths Class 8 Chapter 14 Exercise 14.1 to understand the solution steps.

  • Question 2: Similar to the previous set of questions, here also you are supposed to factorise the given expressions. 

  • Question 3: Here, the expressions for factorisation are slightly more complicated the previous ones like (i) ax2 + bx (ii) am2 + bm2 + bn2 + an2 (iii) 10ab + 4a + 5b + 2 and others.

  • Question 4 and 5: These two sets of questions contain five and three factorisation problems, respectively. The processes to solve each problem is different, and you must study from maths NCERT Solutions Class 8 Chapter 14 Exercise 14.1 for clear comprehension.


Exercise 14.3

  • Question 1: This section of NCERT Solutions Class 8 Chapter 14 begins with divisions of various expressions like 28x2 / 56x, 66pq2r3 / 11qr2, 12a2b2 / (-6a2b2).

  • Question 2 and 3: Problems in these two questions are related to dividing a polynomial by a monomial. Following are some of the sums you will find the PDF file: a. (5x2 – 6x) / 3x b. 8(x3y2z2 + x2y3z2 + x2y2z3) / 4x2y2z2

  • Question 4 and 5: Here, you need to proceed with the sums as directed. These problems are a bit more complicated than the above-mentioned ones, hence practise them on a regular basis.

At the end of class 8 Maths Chapter 14 Exercise 14.1, you will get to come across some other problems. These questions have some errors in them, and you are supposed to find and correct the same.

Along with various textbooks, prefer to download NCERT Solutions for class 8 maths chapter 14 exercise 14.1, 14.2 and 14.3 offered by Vedantu, and get familiar with the techniques to solve factorisation sums.

FAQs on NCERT Solutions for Class 8 Maths Chapter 14 Factorisation (EX 14.1) Exercise 14.1

1. How to factorise an expression?

If you are given an expression (a+ b) (c + d) and asked to factorise the same, the expanded one can be written as ac + ad + bc + bd. This means that all the variables in the first bracket must undergo multiplication with the second bracket.

2. What are the different techniques of factorisation?

There are various by which you can factorise an expression. They are (i) By taking a common factor (ii) Difference of two squares (iii) Perfect square method and (iv) Grouping of the terms.

3. What are the applications of factorisation in day-to-day life?

Some very general applications of factorisation include exchanging money, diving things into equal parts, travel time calculations, etc. Plus, if you have clear factorisation concepts, you can quickly resolve number relationships without the use of calculators.

4. How many questions are there in each exercise including Exercise 14.1 of Chapter 14 Class 8 Maths?

In Chapter 14 Factorisation of Class 8 Maths, there are four exercises in total. 


In exercise 14.1, there are three questions. These questions are further divided into eight, ten and five questions respectively. 


Talking about Exercise 14.2, there are five questions. These five questions are further divided into eight, eight, nine, five and three questions respectively. 


Exercise 14.3 consists of five questions. Each question further contains five or more than five questions. 


There are 21 questions in Exercise 14.4. 

5. How can I make a study plan for Exercise 14.1 of Chapter 14 of Class 8 Maths?

To make a study plan for Exercise 14.1 of Chapter 14 “Factorisation” of Class 8 Maths, follow the given steps. 

  • Firstly, you have to make changes to your present timetable. Your new timetable should be a perfect balance of your daily activities and your studies. 

  • Study Exercise 14.1 of Chapter 14 of Class 8 Maths thoroughly and its notes. 

  • Try to solve questions within 3 to 4 minutes. This target may be difficult for you to achieve initially but, with constant practice, you’ll be able to accomplish this task easily. 

  • Identify the time in which you are comfortable doing your studies. For instance, you may like studying at night or early in the morning. 

6. What should I do to prepare for Exercise 14.1 of Chapter 14 of Class 8 Maths thoroughly? 

If you are a Class 8 student and want to prepare Exercise 14.1 of Chapter 14 of Class 8 Maths thoroughly then, the first thing you need to do is attend all the classes at your coaching institute as well as in your school. At the same time, you have to keep your concentration on your studies. If you have any queries in your mind while understanding the concepts then, ask them from your mentors. Moreover, you must also revise Chapter 14 and Exercise 14.1 of Class 8 Maths when you reach home. Additionally, referring to study materials like revision notes, important questions and the NCERT Solutions will also help you to understand this exercise in a better way. 

7. Are the NCERT Solutions of Exercise 14.1 Chapter 14 “Factorisation” of Class 8 Maths important from an exam point of view? 

Yes, the NCERT Solutions of Exercise 14.1 of Chapter 14 “Factorisation” of Class 8 Maths are important from an exam point of view. These solutions consist of very short, short and long answer type questions to enhance the problem-solving skills of Class 8 students. These solutions are made by the best subject matter experts of Vedantu that tend to provide authentic information in simple and easy language so that the students may not find any difficulty while comprehending the concepts of the Chapter. These solutions are available free of cost on the Vedantu website and the Vedantu app.

8. How can I get the study materials to prepare Exercise 14.1 of Chapter 14 very well? 

You can get multiple study materials to prepare Exercise 14.1 of Chapter 14 on Vedantu. Here’s how you can download them. 

  • Visit the page NCERT Solutions for Exercise 14.1 of Chapter 14 of Class 8 Maths.

  • The link will land you on the official website of Vedantu where you will discover numerous study stuff like revision notes, textbook questions with their answers, NCERT Solutions, etc. 

  • Choose what type of study material you want and tap on the downloading icon to avail the content for free. Also, you can use them in offline mode too. 

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