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

## 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

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.

### 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:

1. (5x2 – 6x) / 3x

2. 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.

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. SHARE TWEET SHARE SUBSCRIBE