NCERT Solutions Class 8 Maths Chapter 12 - Factorisation Exercise 12.1

Exercise 12.1

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

i. $12x$, 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. $2y$, $22xy$.

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. $14pq$, $28{p^2}{q^2}$.

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. $2x$, $3{x^2}$, 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. $6abc$, $24a{b^2}$, $12{a^2}b$.

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. $16{x^3}$, $ - 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. $10pq$, $20qr$, $30rp$.

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. $3{x^2}{y^3}$, $10{x^3}{y^2}$, $6{x^2}{y^2}z$.

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) $7x - 42$.

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) $6p - 12q$.

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) $7{a^2} + 14a$.

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) $ - 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. $20{l^2}m + 30alm$.

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) $5{x^2}y - 15x{y^2}$.

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) $10{a^2} - 15{b^2} + 20{c^2}$.

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) $ - 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) ${x^2}yz + x{y^2}z + xy{z^2}$.

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) $a{x^2}y + bx{y^2} + cxyz$.

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. ${x^2} + xy + 8x + 8y$.

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. $15xy - 6x + 5y - 2$.

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. $ax + bx - ay - by$.

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. $15pq + 15 + 9q + 25p$.

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. $z - 7 + 7xy - xyz$.

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

Conclusion

Class 8 Chapter 12 Exercise 12.1 shows you how to divide integers and expressions into smaller parts. It is important to learn how to identify common elements and group terms to make equations easier to solve. Concentrate on applying these techniques and understand the mathematics. This practice helps you build a solid foundation for solving more complex mathematical problems. By mastering these fundamental skills, you will increase your ability to solve problems and succeed in maths.

Class 8 Maths Chapter 12: Exercises Breakdown

CBSE Class 8 Maths Chapter 12 Other Study Materials

Chapter-Specific NCERT Solutions for Class 8 Maths

Given below are the chapter-wise NCERT Solutions for Class 8 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.