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NCERT Solutions Class 8 Maths Chapter 12 - Factorisation Exercise 12.1

Last updated date: 17th Sep 2024
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NCERT Solutions for Maths Class 8 Chapter 12 Factorisation Exercise 12.1 - FREE PDF Download

NCERT Solutions for Class 8 Maths Chapter 12 Exercise 12.1 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. Class 8 Maths NCERT Solutions are prepared by Vedantu which contains various problems with factorisation.

Table of Content
1. NCERT Solutions for Maths Class 8 Chapter 12 Factorisation Exercise 12.1 - FREE PDF Download
2. Glance on NCERT Solutions Maths Chapter 12 Exercise 12.1 Class 8 | Vedantu
3. Access NCERT Solutions for Maths Class 8 Chapter 12 - Factorisation
4. Class 8 Maths Chapter 12: Exercises Breakdown
5. CBSE Class 8 Maths Chapter 12 Other Study Materials
6. Chapter-Specific NCERT Solutions for Class 8 Maths
FAQs

Besides the first exercise, the following ones are included in this article. So, use this Class 8 Exercise 12.1 to understand the NCERT solution methods and clarify your concepts. Students can also find the latest CBSE Class 8 Maths Syllabus, curated by our Master Teachers.

Glance on NCERT Solutions Maths Chapter 12 Exercise 12.1 Class 8 | Vedantu

• Class 8 Maths Chapter 12 Exercise 12.1 Solutions covers identifying and extracting common factors from each term in an expression (Method of Common Factors) and rearranging and grouping terms to simplify factorisation (Factorisation by Regrouping Terms).

• Factors of natural numbers are numbers that can be multiplied together to get a given number. For example, 2 and 3 are factors of 6.

• Factors of algebraic expressions are terms that can be multiplied together to form an algebraic expression.

• Breaking down a number or expression into its simplest components that, when multiplied together, give the original number or expression is known as Factorisation.

• Identifying and taking out the common factors from each term in an expression is known as the Method of Common Factors.

• Rearranging and grouping terms in an expression to make it easier to factorise is called Factorisation by Regrouping Terms.

• Rearranging terms in an expression to create groups that have common factors, making the expression easier to factorise, is known as regrouping.

• There are 3 fully solved questions in Chapter 12 Exercise 12.1 Factorisation.

Access NCERT Solutions for Maths Class 8 Chapter 12 - Factorisation

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

 Exercise Number of Questions Exercise 12.2 5 Questions & Solutions Exercise 12.3 5 Questions & Solutions

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.

FAQs on NCERT Solutions Class 8 Maths Chapter 12 - Factorisation Exercise 12.1

1. How to factorise an expression in NCERT Class 8 Maths Chapter 12 Exercise 12.1 Solutions?

According to NCERT Class 8 Maths Chapter 12 Exercise 12.1 Solutions, 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 in NCERT Class 8 Maths Chapter 12 Exercise 12.1 Solutions?

NCERT Class 8 Maths Chapter 12 Exercise 12.1 Solutions, there are various by which you can factorise an expression. They are (i) By taking a common factor (ii) the Difference between two squares (iii) the Perfect square method and (iv) the grouping of the terms.

3. What are the applications of factorisation in day-to-day life in Class 8 Maths Exercise 12.1?

In Class 8 Maths Exercise 12.1, some very general applications of factorisation include exchanging money, dividing 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 in Class 8 Maths Exercise 12.1?

In Class 8 Maths Exercise 12.1, there are four exercises in total.

In exercise 12.1, there are three questions. These questions are further divided into eight, ten and five questions respectively.

Talking about Exercise 12.2, there are five questions. These five questions are further divided into eight, eight, nine, five and three questions respectively.

Exercise 12.3 consists of five questions. Each question further contains five or more than five questions.

5. How can I make a study plan for Class 8 Chapter 12 Maths Exercise 12.1?

To make a study plan for Class 8 Chapter 12 Maths Exercise 12.1 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 12.1 of Chapter 12 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 Class 8 Chapter 12 Maths Exercise 12.1 thoroughly?

If you are a Class 8 student and want to prepare Maths Class 8 Chapter 12 Exercise 12.1 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 Maths Class 8 Chapter 12 Exercise 12.1 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 Maths Class 8 Chapter 12 Exercise 12.1 “Factorisation” important from an exam point of view?

Yes, the NCERT Solutions of Maths Class 8 Chapter 12 Exercise 12.1 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 who 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.

8. How can I get the study materials to prepare Maths Class 8 Chapter 12 Exercise 12.1 very well?

You can get multiple study materials to prepare Maths Class 8 Chapter 12 Exercise 12.1 on Vedantu. Here’s how you can download them.

• Visit the page NCERT Solutions for Exercise 12.1 of Chapter 12 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.

9. What is the main concept of Class 8 Chapter 12 Maths Exercise 12.1 Factorisation?

Class 8 Chapter 12 Maths Exercise 12.1 focuses on factorising numbers and algebraic expressions. It teaches you how to break down expressions into simpler parts by finding common factors and using different methods to simplify them. This is a key skill for solving more complex algebra problems.

10. How do you find common factors in an expression in Class 8 Maths Chapter 12 Factorisation Exercise 12.1?

In Class 8 Maths Chapter 12 Factorisation Exercise 12.1, to find common factors, look at each term in the expression and identify the factors that are the same.

11. What are the steps to factorise by regrouping terms in Class 8 Maths Chapter 12 Factorisation Exercise 12.1?

First refer Class 8 Maths Chapter 12 Factorisation Exercise 12.1, and identify terms that can be grouped with a common factor. Next, rearrange the terms to form groups with these common factors. Factor out the common factor from each group, and finally combine the groups into a simpler expression. This makes the expression easier to solve.

12. Why is learning factorisation important in Exercise 12.1 Class 8 Maths NCERT Solutions?

Learning factorisation is important because it simplifies complex expressions, making them easier to work with. This skill is essential for solving algebraic equations, simplifying fractions, and understanding higher-level maths concepts. It also improves problem-solving skills in Exercise 12.1 Class 8 Maths NCERT Solutions.

13. What are some common mistakes to avoid while factorising in Exercise 12.1 Class 8 Maths NCERT Solutions?

Common mistakes include not identifying all common factors, incorrect grouping of terms, and missing factors in the final simplified expression. To avoid these, double-check each step, ensure you have factored out all common elements, and practice regularly to improve accuracy.