NCERT Solutions Class 8 Maths Chapter 12 Factorisation Exercise 12.1

NCERT Solutions Class 8 Maths Chapter 12 Factorisation Exercise 12.1 introduces students to the basic idea of factorisation, which means writing numbers or algebraic expressions as a product of their factors. In this exercise, students learn how to find common factors of algebraic terms and use them to simplify expressions. It builds the foundation for solving more complex expressions in later exercises.

These NCERT Solutions follow the official NCERT syllabus and are explained in simple steps, making it easier for students to understand and solve each question. With clear methods and useful examples, the solutions help students practice and improve their algebra skills. Practicing NCERT Solutions from Exercise 12.1 strengthens understanding of factorisation and prepares students for higher-level algebra topics.

1.0Download NCERT Solutions Class 8 Maths Chapter 12 Factorisation Exercise 12.1: Free PDF

Download the free NCERT Solutions Class 8 Maths Chapter 12 Factorisation Exercise 12.1 PDF with clear, step-by-step answers to help you understand and solve problems easily.

2.0Key Concepts in Exercise 12.1 of Class 8 Maths Chapter 12

Exercise 12.1 of Chapter 12 – Factorisation introduces the basic concept of finding factors of algebraic expressions, which is the reverse process of multiplication. This exercise helps students develop a strong foundation for simplifying and solving algebraic equations.

• What is Factorisation?
  Factorisation means expressing an algebraic expression as a product of its factors. It’s the opposite of expansion.
• Common Factors
  Students learn how to find the Highest Common Factor (HCF) from algebraic terms and take it out to simplify the expression.
  Example: 6x+9=3(2x+3)
• Factorisation by Taking Common Terms
  The method involves identifying and extracting the common numerical and variable terms from each part of the expression.
• Monomial Common Factors
  Students practice finding common monomial factors from binomial or trinomial expressions.
• Step-by-Step Simplification
  The exercise encourages writing intermediate steps to build clarity and avoid mistakes in longer expressions.

3.0NCERT Class 8 Maths Chapter 12: Other Exercises

4.0NCERT Class 8 Maths Chapter 12 Exercise 12.1: Detailed Solutions

1. Find the common factors of the given terms : (i) $12x,36$ (ii) $2\mathrm{y},22\mathrm{xy}$ (iii) $14pq,28p^2{q}^2$ (iv) $2\mathrm{x},3{\mathrm{x}}^2,4$ (v) $6abc,24a{b}^2,12a^{2}b$ (vi) $16x^3,-4x^2,32x$ (vii)10pq, 20qr, 30 rp (viii) $3x^2{y}^3,10x^3{y}^2,6x^2{y}^{2}z$ Sol. (i) The numerical coefficients in the given monomials are 12 and 36 . The highest common factor of 12 and 36 is 12 . But there is no common literal appearing in the given monomials 12 x and 36. $\therefore$ The highest common factor $=12$ (ii) $2\mathrm{y}=2\times \mathrm{y}$ $22xy=2\times 11\times x\times y$ The common factors are $2,\mathrm{y}$. And, $2\times y=2y$ (iii) The numerical coefficients of the given monomials are 14 and 28. The highest common factor of 14 and 28 is 14 . The common literals appearing in the given monomials are $p$ and $q$. The smallest power of p and q in the two monomials = 1 The monomial of common literals with smallest powers $=\mathrm{pq}$ $\therefore$ The highest common factor $=14\mathrm{pq}$ (iv) $2\mathrm{x}=2\times \mathrm{x}$ $3{\mathrm{x}}^2=3\times \mathrm{x}\times \mathrm{x}$ $4=2\times 2$ The common factor is 1 . (v) The numerical coefficients of the given monomials are 6, 24 and 12. The common literals appearing in the three monomials are a and b . The smallest power of a in the three monomials =1 The smallest power of $b$ in the three monomials $=1$ The monomial of common literals with smallest powers $=ab$ Hence, the highest common factor $=6\mathrm{ab}$ (vi) $16{\mathrm{x}}^3=2\times 2\times 2\times 2\times \mathrm{x}\times \mathrm{x}\times \mathrm{x}$ $-4{\mathrm{x}}^2=-1\times 2\times 2\times \mathrm{x}\times \mathrm{x}$ $32\mathrm{x}=2\times 2\times 2\times 2\times 2\times x$ The common factors are $2,2,x$. And $2\times 2\times x=4x$ (vii) The numerical coefficients of the given monomials are 10,20 and 30. The highest common factor of 10,20 and 30 is 10 . There is no common literal appearing in the three monomials. Hence, the highest common factor $=10$ (viii) $3x^2{y}^3=3\times x\times x\times y\times y\times y$ $10x^3{y}^2=2\times 5\times x\times x\times x\times y\times y$ $6x^2{y}^{2}z=2\times 3\times x\times x\times y\times y\times z$ The common factors are $\mathrm{x},\mathrm{x},\mathrm{y},\mathrm{y}$. and, $x\times x\times y\times y=x^2{y}^2$
2. Factorise the following expressions (i) $7x-42$ (ii) $6p-12q$ (iii) $7a^2+14a$ (iv) $-16z+20z^3$ (v) $20{\ell }^2\mathrm{m}+30\mathrm{a}\ell \mathrm{m}$ (vi) $5x^{2}y-15x{y}^2$ (vii) $10a^2-15b^2+20c^2$ (viii) $-4a^2+4ab-4ca$ (ix) $x^{2}yz+x{y}^{2}z+xy{z}^2$ (x) $a{x}^{2}y+bx{y}^2+cxyz$ Sol. (i) We have, $7\mathrm{x}=7\times \mathrm{x}$ and $42=2\times 3\times 7$ The two terms have 7 as a common factor $7\mathrm{x}-42=(7\times \mathrm{x})-2\times 3\times 7$ $=7\times (x-2\times 3)=7(x-6)$ (ii) $6\mathrm{p}=2\times 3\times \mathrm{p}$ $12q=2\times 2\times 3\times q$ The common factors are 2 and 3 . $\therefore 6\mathrm{p}-12\mathrm{q}=(2\times 3\times \mathrm{p})-(2\times 2\times 3\times \mathrm{q})$ $=2\times 3[p-(2\times q)]$ $=6(p-2q)$ (iii) We have, $7{\mathrm{a}}^2=7\times \mathrm{a}\times \mathrm{a}$ and, $14\mathrm{a}=2\times 7\times \mathrm{a}$ The two terms have 7 and a as common factors $7{\mathrm{a}}^2+14\mathrm{a}=(7\times \mathrm{a}\times \mathrm{a})+(2\times 7\times \mathrm{a})$ $=7\times a\times (a+2)=7a(a+2)$ (iv) $16z=2\times 2\times 2\times 2\times z$ $20{\mathrm{z}}^3=2\times 2\times 5\times \mathrm{z}\times \mathrm{z}\times \mathrm{z}$ The common factors are 2,2 , and z . $\therefore -16z+20z^3=-(2\times 2\times 2\times 2\times z)+$ $(2\times 2\times 5\times \mathrm{z}\times \mathrm{z}\times \mathrm{z})$ $=(2\times 2\times z)[-(2\times 2)+(5\times z\times z)]$ $=4\mathrm{z}\left(-4+5{\mathrm{z}}^2\right)$ (v) We have, $20{\ell }^2\mathrm{m}=2\times 2\times 5\times \ell \times \ell \times \mathrm{m}$ and, $30\mathrm{a}\ell \mathrm{m}=3\times 2\times 5\times \mathrm{a}\times \ell \times \mathrm{m}$ The two terms have $2,5,\ell$ and m as common factors. $\therefore 20{\ell }^2\mathrm{m}+30\mathrm{a}\ell \mathrm{m}=(2\times 2\times 5\times \ell \times \ell$ $\times m)+(3\times 2\times 5\times a\times \ell \times m)$ $=2\times 5\times \ell \times \mathrm{m}\times (2\times \ell +3\times \mathrm{a})$ $=10\ell \mathrm{m}(2\ell +3\mathrm{a})$ (vi) $5x^{2}y=5\times x\times x\times y$ $15x^2=3\times 5\times x\times y\times y$ The common factors are $5,\mathrm{x}$, and y . $\therefore 5x^{2}y-15x{y}^2$ $=(5\times x\times x\times y)-(3\times 5\times x\times y\times y)$ $=5\times x\times y[x-(3\times y)]$ $=5xy(x-3y)$ (vii)We have, $10{\mathrm{a}}^2=2\times 5\times \mathrm{a}\times \mathrm{a}$, $15b^2=3\times 5\times b\times b$ and, $20c^2=2\times 2\times 5\times c\times c$ The three terms have 5 as a common factor $\therefore 10a^2-15b^2+20c^2$ $=(2\times 5\times a\times a)-(3\times 5\times b\times b)$ $+(2\times 2\times 5\times c\times c)$ $=5\times (2\times a\times a-3\times b\times b+4\times c\times c)$ $=5\left(2{\mathrm{a}}^2-3{\mathrm{b}}^2+4{\mathrm{c}}^2\right)$ (viii) We have, $4a^2=2\times 2\times a\times a$, $4\mathrm{ab}=2\times 2\times \mathrm{a}\times \mathrm{b}$ and, $4\mathrm{ca}=2\times 2\times \mathrm{c}\times \mathrm{a}$ The three terms have 2,2 and a as common factors $$\begin{gathered} \therefore -4a^2+4ab-4ca=-(2\times 2\times a\times a) \\ +(2\times 2\times a\times b)-(2\times 2\times c\times a) \\ =2\times 2\times a\times (-a+b-c) \\ =4a(-a+b-c) \end{gathered}$$ (ix) $x^{2}yz=x\times x\times y\times z$ $x{y}^{2}z=x\times y\times y\times z$ $xy{z}^2=x\times y\times z\times z$ The common factors are $\mathrm{x},\mathrm{y}$, and z . $\therefore {\mathrm{x}}^2\mathrm{yz}+{\mathrm{xy}}^2\mathrm{z}+{\mathrm{xyz}}^2=(\mathrm{x}\times \mathrm{x}\times \mathrm{y}\times \mathrm{z})$ $+(x\times y\times y\times z)+(x\times y\times z\times z)$ $=x\times y\times z[x+y+z]$ $=xyz(x+y+z)$ (x) We have, ${\mathrm{ax}}^2\mathrm{y}=\mathrm{a}\times \mathrm{x}\times \mathrm{x}\times \mathrm{y}$ $bx{y}^2=b\times x\times y\times y$ and, $cxyz=c\times x\times y\times z$ The three terms have $x$ and $y$ as common factors $\therefore a^{2}y+bx{y}^2+cxyz=(a\times x\times x\times y)$ $+(b\times x\times y\times y)+(c\times x\times y\times z)$ $=x\times y\times (a\times x+b\times y+c\times z)$ $=xy(ax+by+cz)$
3. Factorise : (i) ${\mathrm{x}}^2+\mathrm{xy}+8\mathrm{x}+8\mathrm{y}$ (ii) $15xy-6x+5y-2$ (iii) $ax+bx-ay-by$ (iv) $15pq+15+9q+25p$ (v) $z-7+7xy-xyz$ Sol. (i) $x^2+xy+8x+8y=\left(x^2+xy\right)+(8x+8y)$ $=x(x+y)+8(x+y)$ $=(x+y)(x+8)$ [Taking $(x+y)$ common] (ii) $15xy-6x+5y-2$ $=3\times 5\times \mathrm{x}\times \mathrm{y}-3\times 2\times \mathrm{x}+5\times \mathrm{y}-2$ $=3x(5y-2)+1(5y-2)$ $=(5y-2)(3x+1)$ (iii) $ax+bx-ay-by=(ax+bx)-(ay+by)$ [Grouping the terms] $=(a+b)x-(a+b)y$ $=(a+b)(x-y)$ [Taking $(a+b)$ common] (iv) $15pq+15+9q+25p$ $=15pq+9q+25p+15$ $=3\times 5\times p\times q+3\times 3\times q+5\times 5\times p+3\times 5$ $=3q(5p+3)+5(5p+3)$ $=(5p+3)(3q+5)$ (v) $z-7+7xy-xyz=z-7-xyz+7xy$ $=1(z-7)-xy(z-7)$ $=(\mathrm{z}-7)(1-\mathrm{xy})$ [Taking (z-7) common]

5.0Key Features and Benefits of Class 8 Maths Chapter 12 Exercise 12.1

• Introduction to Factorisation: Exercise 12.1 helps students learn how to break algebraic expressions into simpler parts called factors.
• Focus on Common Factors: Students practice taking out the common factor from terms, which is an important first step in factorising any expression.
• Strengthens Basic Algebra: This exercise builds a strong foundation for solving more advanced algebraic problems in higher classes.
• Improves Simplification Skills: By learning to factorise correctly, students become better at simplifying expressions and solving equations quickly.
• NCERT-Based Questions: All problems follow the official NCERT syllabus, making it perfect for exams, class tests, and regular practice.