CBSE Class 8 Maths Chapter 14 Factorization NCERT Solutions

Reviewed by NCERT Help Editorial Team · Updated on 29 Jul 2026

NCERT Solutions PDF Class 8 PDF

CBSE Class 8 Mathematics Chapter 14, Factorization, introduces students to a key algebraic concept. This chapter focuses on breaking down algebraic terms and expressions into their simplest multiplicative components, known as factors. The NCERT Solutions for this chapter provide detailed explanations and step-by-step problem-solving for Exercise 14.1. Students will learn to identify common numerical and variable factors within terms and expressions. A significant part of the chapter involves applying the distributive property to factorize expressions, which is a crucial skill for simplifying algebraic equations and solving more complex problems in higher mathematics. Mastering factorization not only strengthens algebraic manipulation abilities but also builds confidence in tackling mathematical challenges, ensuring students are well-prepared for their examinations.

Home

Quick info

Board	CBSE
Class	Class 8
Subject	Mathematics
Session	2026
Language	English
Type	NCERT Solutions
Chapter	Chapter 14: Factorization

Chapter summary

Chapter 14 on Factorization for Class 8 Maths focuses on the core concepts of identifying common factors in algebraic terms and expressions. The NCERT Solutions cover finding the greatest common divisor (GCD) for numerical coefficients and common variables. It then progresses to factorizing expressions by taking out the common factor, illustrating the reverse of the distributive property. The exercise provides ample practice in handling various forms of algebraic expressions, including those with negative terms and multiple variables.

Learning outcomes

Understand the concept of common factors in algebraic terms.
Identify and calculate the greatest common factor (GCF) for given terms.
Apply the concept of common factors to factorize algebraic expressions.
Factorize expressions by taking out the common factor using the distributive property.
Solve problems involving factorization of polynomials with up to three terms.

Topics covered

Paper topics

Common Factors
Greatest Common Factor (GCF)
Prime Factorization
Factorization of Algebraic Expressions
Distributive Property in Factorization
Factoring out common terms
Identifying common numerical factors
Identifying common variable factors

Important topics

Finding the Greatest Common Factor (GCF)
Factorizing expressions by taking out the GCF
Understanding common factors in terms
Applying factorization to algebraic expressions

PDF preview

Read page by page below. PDF is streamed from the official NCERT website — no download button on this page.

Loading document …

Page of

Loading page …

Questions and Solutions

Question 1

Find the common factors of the given terms:

(i) 12x, 36

(ii) 2y, 22xy

(iii) 14pq, 28p²q²

(iv) 2x, 3x², 4

(v) 6abc, 24ab², 12a²b

(vi) 16x³, -4x², 32x

(vii) 10pq, 20qr, 30rp

(viii) 3x²y³, 10x³y², 6x²y²z

Solution:

To find the common factors, we first find the prime factorization of each term and then identify the factors that are common to all terms.

(i) For 12x and 36:

Prime factorization of 12x: $12x = 2 \times 2 \times 3 \times x$
Prime factorization of 36: $36 = 2 \times 2 \times 3 \times 3$
The common factors are 2, 2, and 3.
The greatest common factor (GCF) is $2 \times 2 \times 3 = 12$ .

(ii) For 2y and 22xy:

Prime factorization of 2y: $2y = 2 \times y$
Prime factorization of 22xy: $22xy = 2 \times 11 \times x \times y$
The common factors are 2 and y.
The GCF is $2 \times y = 2y$ .

(iii) For 14pq and 28p²q²:

Prime factorization of 14pq: $14pq = 2 \times 7 \times p \times q$
Prime factorization of 28p²q²: $28p^2q^2 = 2 \times 2 \times 7 \times p \times p \times q \times q$
The common factors are 2, 7, p, and q.
The GCF is $2 \times 7 \times p \times q = 14pq$ .

(iv) For 2x, 3x², and 4:

Prime factorization of 2x: $2x = 2 \times x \times 1$
Prime factorization of 3x²: $3x^2 = 3 \times x \times x \times 1$
Prime factorization of 4: $4 = 2 \times 2 \times 1$
The only common factor among all three terms is 1.
The GCF is 1.

(v) For 6abc, 24ab², and 12a²b:

Prime factorization of 6abc: $6abc = 2 \times 3 \times a \times b \times c$
Prime factorization of 24ab²: $24ab^2 = 2 \times 2 \times 2 \times 3 \times a \times b \times b$
Prime factorization of 12a²b: $12a^2b = 2 \times 2 \times 3 \times a \times a \times b$
The common factors are 2, 3, a, and b.
The GCF is $2 \times 3 \times a \times b = 6ab$ .

(vi) For 16x³, -4x², and 32x:

Prime factorization of 16x³: $16x^3 = 2 \times 2 \times 2 \times 2 \times x \times x \times x$
Prime factorization of -4x²: $-4x^2 = (-1) \times 2 \times 2 \times x \times x$
Prime factorization of 32x: $32x = 2 \times 2 \times 2 \times 2 \times 2 \times x$
The common factors are 2, 2, and x.
The GCF is $2 \times 2 \times x = 4x$ . (Note: The sign is usually taken as positive for GCF unless specified otherwise).

(vii) For 10pq, 20qr, and 30rp:

Prime factorization of 10pq: $10pq = 2 \times 5 \times p \times q$
Prime factorization of 20qr: $20qr = 2 \times 2 \times 5 \times q \times r$
Prime factorization of 30rp: $30rp = 2 \times 3 \times 5 \times r \times p$
The common factors are 2 and 5.
The GCF is $2 \times 5 = 10$ .

(viii) For 3x²y³, 10x³y², and 6x²y²z:

Prime factorization of 3x²y³: $3x^2y^3 = 3 \times x \times x \times y \times y \times y$
Prime factorization of 10x³y²: $10x^3y^2 = 2 \times 5 \times x \times x \times x \times y \times y$
Prime factorization of 6x²y²z: $6x^2y^2z = 2 \times 3 \times x \times x \times y \times y \times z$
The common factors are x, x, and y, y.
The GCF is $x \times x \times y \times y = x^2 y^2$ .

Question 2

Factorize the following expressions:

(i) 7x - 42

(ii) 6p - 12q

(iii) 7a² + 14a

(iv) -16z + 20z³

(v) 20l²m + 30alm

(vi) 5x²y - 15xy²

(vii) 10a² - 15b² + 20c²

(viii) -4a² + 4ab - 4ca

(ix) x²yz + xy²z + xyz²

(x) ax²y + bxy² + cxyz

Solution:

To factorize these expressions, we identify the greatest common factor (GCF) of the terms and then use the distributive property in reverse, i.e., $a(b+c) = ab + ac$ .

(i) For 7x - 42:

The terms are 7x and -42.
The prime factorization of 7x is $7 \times x$ .
The prime factorization of 42 is $2 \times 3 \times 7$ .
The GCF of 7x and 42 is 7.
Factoring out 7: $7x - 42 = 7(x) - 7(6)$
Using the distributive property: $7(x - 6)$ .

(ii) For 6p - 12q:

The terms are 6p and -12q.
The prime factorization of 6p is $2 \times 3 \times p$ .
The prime factorization of 12q is $2 \times 2 \times 3 \times q$ .
The GCF of 6p and 12q is $2 \times 3 = 6$ .
Factoring out 6: $6p - 12q = 6(p) - 6(2q)$
Using the distributive property: $6(p - 2q)$ .

(iii) For 7a² + 14a:

The terms are 7a² and 14a.
The prime factorization of 7a² is $7 \times a \times a$ .
The prime factorization of 14a is $2 \times 7 \times a$ .
The GCF of 7a² and 14a is $7 \times a = 7a$ .
Factoring out 7a: $7a^2 + 14a = 7a(a) + 7a(2)$
Using the distributive property: $7a(a + 2)$ .

(iv) For -16z + 20z³:

The terms are -16z and 20z³.
The prime factorization of 16z is $2 \times 2 \times 2 \times 2 \times z$ .
The prime factorization of 20z³ is $2 \times 2 \times 5 \times z \times z \times z$ .
The GCF of 16z and 20z³ is $2 \times 2 \times z = 4z$ .
We can factor out either 4z or -4z. Let's factor out -4z to make the first term in the parenthesis positive.
Factoring out -4z: $-16z + 20z^3 = (-4z)(4) + (-4z)(-5z^2)$
Using the distributive property: $-4z(4 - 5z^2)$ .

(v) For 20l²m + 30alm:

The terms are 20l²m and 30alm.
Prime factorization of 20l²m: $2 \times 2 \times 5 \times l \times l \times m$
Prime factorization of 30alm: $2 \times 3 \times 5 \times a \times l \times m$
The common factors are 2, 5, l, and m.
The GCF is $2 \times 5 \times l \times m = 10lm$ .
Factoring out 10lm: $20l^2m + 30alm = 10lm(2l) + 10lm(3a)$
Using the distributive property: $10lm(2l + 3a)$ .

(vi) For 5x²y - 15xy²:

The terms are 5x²y and -15xy².
Prime factorization of 5x²y: $5 \times x \times x \times y$
Prime factorization of 15xy²: $3 \times 5 \times x \times y \times y$
The GCF is $5 \times x \times y = 5xy$ .
Factoring out 5xy: $5x^2y - 15xy^2 = 5xy(x) - 5xy(3y)$
Using the distributive property: $5xy(x - 3y)$ .

(vii) For 10a² - 15b² + 20c²:

The terms are 10a², -15b², and 20c².
Prime factorization of 10a²: $2 \times 5 \times a \times a$
Prime factorization of 15b²: $3 \times 5 \times b \times b$
Prime factorization of 20c²: $2 \times 2 \times 5 \times c \times c$
The GCF of the coefficients 10, 15, and 20 is 5. There are no common variables.
Factoring out 5: $10a^2 - 15b^2 + 20c^2 = 5(2a^2) - 5(3b^2) + 5(4c^2)$
Using the distributive property: $5(2a^2 - 3b^2 + 4c^2)$ .

(viii) For -4a² + 4ab - 4ca:

The terms are -4a², 4ab, and -4ca.
Prime factorization of 4a²: $2 \times 2 \times a \times a$
Prime factorization of 4ab: $2 \times 2 \times a \times b$
Prime factorization of 4ca: $2 \times 2 \times c \times a$
The GCF of the coefficients is 4. The common variable is 'a'.
The GCF of all terms is $4a$ .
Let's factor out -4a to make the first term positive: $-4a^2 + 4ab - 4ca = (-4a)(a) + (-4a)(-b) + (-4a)(c)$
Using the distributive property: $-4a(a - b + c)$ .

(ix) For x²yz + xy²z + xyz²:

The terms are x²yz, xy²z, and xyz².
The common factors are x, y, and z.
The GCF is $xyz$ .
Factoring out xyz: $x^2yz + xy^2z + xyz^2 = xyz(x) + xyz(y) + xyz(z)$
Using the distributive property: $xyz(x + y + z)$ .

(x) For ax²y + bxy² + cxyz:

The terms are ax²y, bxy², and cxyz.
The common variable factors are x and y.
The GCF is $xy$ .
Factoring out xy: $ax^2y + bxy^2 + cxyz = xy(ax) + xy(by) + xy(cz)$
Using the distributive property: $xy(ax + by + cz)$ .

Common mistakes

Incorrectly identifying the greatest common factor (GCF) of coefficients.
Missing common variable factors or their highest powers.
Errors in signs when factoring out negative common factors.
Forgetting to factor out all common factors, leaving a non-factorized expression.
Mistakes in applying the distributive property in reverse.

Revision tips

Review the prime factorization of numbers to easily find the GCF of coefficients.
Systematically check for common factors in both the numerical coefficients and the variables of each term.
Practice taking out common factors from expressions with negative terms.
Verify your factorization by multiplying the factors back together to see if you get the original expression.
Work through examples with multiple terms to build confidence in handling complex expressions.

Practice MCQs

Q1. What is the greatest common factor of 12x and 36?

Q2. Which of the following is a common factor of 2y and 22xy?

Q3. What is the common factor in the expression 7 $a^2$ + 14a?

Q4. Factorize 6p - 12q.

Q5. What is the common factor of 10pq, 20qr, and 30rp?

Q6. $Factorize -16z + 20z^3.$

Frequently asked questions

What is factorization in Class 8 Maths?

Factorization is the process of expressing an algebraic expression as a product of its factors. In Class 8, this primarily involves finding common factors among terms and using them to rewrite the expression.

How do I find the common factors of algebraic terms?

To find common factors, first find the greatest common factor (GCF) of the numerical coefficients. Then, identify the common variables and take the lowest power of each common variable present in all terms.

What is the main goal of Exercise 14.1?

Exercise 14.1 focuses on two main skills: finding the common factors of given terms and factorizing algebraic expressions by taking out the common factor.

How can these NCERT solutions help me prepare for exams?

These solutions provide clear, step-by-step explanations for each problem, helping you understand the methods. Practicing with them reinforces your understanding of factorization, a key topic for exams.

What does it mean to 'factorize' an expression?

Factorizing an expression means rewriting it as a multiplication of simpler expressions (its factors). For example, factorizing 7x - 42 means writing it as 7(x - 6).

Content reviewed by the NCERT Help team. Editorial Team and update policy

NCERT Solutions PDF PDF on NCERT Help. URL unchanged for search indexing.