NCERT Solutions for Class 8 Maths Chapter 14 - Exercise

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

• Chapter 1 - Rational Numbers

• Chapter 2 - Linear Equations in One Variable

• Chapter 3 - Understanding Quadrilaterals

• Chapter 4 - Practical Geometry

• Chapter 5 - Data Handling

• Chapter 6 - Squares and Square Roots

• Chapter 7 - Cubes and Cube Roots

• Chapter 8 - Comparing Quantities

• Chapter 9 - Algebraic Expressions and Identities

• Chapter 10 - Visualising Solid Shapes

• Chapter 11 - Mensuration

• Chapter 12 - Exponents and Powers

• Chapter 13 - Direct and Inverse Proportions

• Chapter 14 - Factorisation

• Chapter 15 - Introduction to Graphs

• Chapter 16 - Playing with Numbers

Class 8 Maths Chapter 14 Includes:

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.