Factorisation Class 8 PDF: Chapter 12 Explained Simply

This page gives you the official NCERT Class 8 Maths Chapter 12 Factorisation PDF straight from ncert.nic.in, in the current rationalised edition of the book. The factorisation class 8 chapter runs about 12 printed pages: four methods of breaking expressions into factors, then division that cancels those factors.

The file is right below, and every explanation on this page follows the chapter section by section.

NCERT Class 8 Maths Chapter 12 Factorisation PDF

This is the official file NCERT publishes for this chapter — the same PDF the printed textbook is set from, so the layout, figures and numbering all match your book.

Open the NCERT Class 8 Maths Chapter 12 Factorisation PDF to read the chapter exactly as NCERT printed it, or save the file for offline revision.

Chapter 12 Factorisation at a Glance

This table gives the chapter’s size at a glance — its sections, figures, exercise questions and the volume of algebra it works through.

The chapter moves in one direction: factorising methods first, from simple common factors up to the \((x + a)(x + b)\) pattern, then division in sections 12.3–12.4 that cancels the very factors you have just learned to find.

What the Chapter Covers: Four Factorising Methods, Then Division

Factorisation rewrites an expression as a product of factors — the reverse of expanding brackets. Division then cancels the common factors that factorisation exposes. That is why the chapter is deliberately built in this order: factorise first, divide second.

• Section 12.1 recalls factors of natural numbers — the prime factor form of 30 is \( 2 \times 3 \times 5 \) — and carries the same idea into algebra with irreducible factors.
• Section 12.2 is the working core: the four factorising methods — common factors, regrouping, the identities, and the \((x + a)(x + b)\) form.
• Sections 12.3 and 12.4 turn factorisation into a tool for division, from monomial divided by monomial up to polynomial divided by polynomial.

Each later block assumes the earlier ones, so if you get lost in the division section, the cause is almost always a factorisation you have not done.

Key Concepts in Factorisation Class 8, Method by Method

Every method below does the same job — turning one expression into a product — but each one is the fastest route in its own situation. Start with the word the whole chapter leans on.

What “irreducible” means

Just as 30 is written in prime factor form as \( 2 \times 3 \times 5 \), an algebraic term can be written as a product of factors that cannot split further. Those factors are its irreducible factors (NCERT, section 12.1.2).

Take \( 5xy = 5 \times x \times y \): the factors 5, \(x\) and \(y\) cannot break down further, so \( 5 \times x \times y \) is the irreducible form of \(5xy\). Writing \( 5 \times (xy) \) is not good enough, because \(xy\) can still split into \(x \times y\).

One special case: 1 is a factor of every term — \( 5xy = 1 \times 5 \times x \times y \) — but it is not written unless it is actually needed (NCERT, section 12.1.2).

Method of common factors

Use this when every term of the expression shares a factor. The chapter’s closing recap reduces it to three steps.

1. Write each term as a product of irreducible factors.
2. Separate the factors common to all the terms.
3. Combine what remains in each term using the distributive law.

Example: factorise \( 18x^3 – 27x^2 + 36x \).

\[ \begin{aligned} 18x^3 – 27x^2 + 36x &= (2 \times 3 \times 3 \times x \times x \times x) – (3 \times 3 \times 3 \times x \times x) + (2 \times 2 \times 3 \times 3 \times x) \\ &= 9x(2x^2) – 9x(3x) + 9x(4) \\ &= 9x(2x^2 – 3x + 4) \end{aligned} \]

Check by multiplying: \( 9x(2x^2 – 3x + 4) = 18x^3 – 27x^2 + 36x \), so the factorisation is correct. The chapter’s TRY THESE items — \( 12x + 36 \), \( 22y – 33z \), \( 14pq + 35pqr \) — all work the same way and give \( 12(x + 3) \), \( 11(2y – 3z) \) and \( 7pq(2 + 5r) \).

Regrouping

When no single factor is common to all terms, rearrange and group the terms so that each group has its own common factor. If a common bracket then appears in every group, take it out and the factorisation is complete (NCERT, section 12.2.2).

Example: factorise \( 2ab – 6b + 5a – 15 \). Group the first two terms and the last two:

\[ 2ab – 6b + 5a – 15 = 2b(a – 3) + 5(a – 3) = (a – 3)(2b + 5) \]

Two traps live here. First, the hidden 1: \( 3x + 3 \) must be written as \( 3(x + 1) \), because 1 carries the second term \( 3 \times 1 \) and keeps the common bracket \( (x + 1) \) alive — the book itself asks why 1 must be shown as a factor (NCERT, section 12.2.2).

Second, signs: if a group starts with a minus, reorder it so the shared factor appears. In \( 6xy – 4y + 6 – 9x \), write the last two terms as \( -9x + 6 = -3(3x – 2) \) so they match the \( 2y(3x – 2) \) from the first pair (NCERT, section 12.2.2).

Identities I to III

Three of the four identities turn a familiar shape straight into factors. Match the shape first, then write the factors (NCERT, section 12.2.3).

• Three terms, first and last perfect squares, positive middle term → Identity I: \( (a + b)^2 \).
• Three terms, first and last perfect squares, negative middle term → Identity II: \( (a – b)^2 \).
• Two square terms joined by a minus → Identity III: \( (a + b)(a – b) \).

Example: factorise \( 25x^2 – 20x + 4 \). Here \( 25x^2 = (5x)^2 \), \( 4 = 2^2 \) and \( 20x = 2 \times (5x) \times 2 \), so the middle term is exactly \( 2ab \). The expression fits Identity II:

\[ 25x^2 – 20x + 4 = (5x – 2)^2 \]

Example: factorise \( 64a^2 – 9b^2 \). Two squares, minus between → Identity III:

\[ 64a^2 – 9b^2 = (8a)^2 – (3b)^2 = (8a – 3b)(8a + 3b) \]

Watch for expressions that need an identity twice. The chapter’s example \( m^4 – 256 \) becomes \( (m^2 – 16)(m^2 + 16) \) first, then \( m^2 – 16 \) factorises again as \( (m – 4)(m + 4) \), so the full form is \( (m – 4)(m + 4)(m^2 + 16) \) (NCERT, section 12.2.3).

The (x + a)(x + b) form

For three-term expressions in one variable whose constant term is not a square — like \( x^2 + 5x + 6 \) — use Identity IV. Find two numbers \(a\) and \(b\) whose product is the constant term and whose sum is the coefficient of \(x\) (NCERT, section 12.2.4):

\[ x^2 + (a + b)x + ab = (x + a)(x + b) \]

Example: factorise \( y^2 + 2y – 24 \). Here \( ab = -24 \) and \( a + b = 2 \). The constant is negative, so one of \(a, b\) is negative. The pair 6 and \(-4\) works: \( 6 \times (-4) = -24 \) and \( 6 + (-4) = 2 \).

\[ y^2 + 2y – 24 = (y + 6)(y – 4) \]

When the coefficient of \(x^2\) is not 1, take out the common factor first. For \( 2n^2 + 10n + 12 \):

\[ 2n^2 + 10n + 12 = 2(n^2 + 5n + 6) = 2(n + 2)(n + 3) \]

The book models the same idea with \( 3m^2 + 9m + 6 \): take out 3, then find the pair 1 and 2 (NCERT, section 12.2.4).

Division of algebraic expressions

Division is the inverse of multiplication: since \( 7 \times 8 = 56 \), you get \( 56 \div 8 = 7 \). The same idea drives every division in this chapter (NCERT, section 12.3). There are three cases to recognise.

• Monomial ÷ monomial: write both in irreducible factor form and cancel the common factors. Example: \( 15a^3b^2 \div 5ab = 3a^2b \).
• Polynomial ÷ monomial: two allowed routes — divide each term of the polynomial by the monomial, or take out the common factor and cancel. The book works \( (4y^3 + 5y^2 + 6y) \div 2y \) both ways and gets \( 2y^2 + \frac{5}{2}y + 3 \) (NCERT, section 12.3.2).
• Polynomial ÷ polynomial: never divide term by term. Factorise both the dividend and the divisor, then cancel their common factors (NCERT, section 12.4).

Example: divide \( 6x^2 – 9x \) by \( 2x – 3 \). Factorise the dividend first:

\[ 6x^2 – 9x = 3x(2x – 3) \]

\[ (6x^2 – 9x) \div (2x – 3) = \frac{3x(2x – 3)}{2x – 3} = 3x \]

The answer checks by multiplication: \( 3x(2x – 3) = 6x^2 – 9x \).

The Chapter’s Closing Figure: Division with Remainder Zero

The chapter ends with an image that marks a restriction, and that restriction is why every exercise behaves the way it does.

Every division in this chapter is exact: you cancel common factors completely, so nothing is left over. That is why the relation used throughout is \( \text{Dividend} = \text{Divisor} \times \text{Quotient} \), with no remainder term.

The chapter does name the general form — \( \text{Dividend} = \text{Divisor} \times \text{Quotient} + \text{Remainder} \) — and then states, as the closing figure marks, that only the zero-remainder case is studied here (NCERT, section 12.4). This is why no exercise in the chapter expects long division: division always waits for factorisation first.

Key Terms and the Four Identities, Explained

These are the words the chapter uses on every page. Know them cold and the methods stop being confusing.

• Factor — a number or expression that multiplies with others to give a product. The factors of 30 are 1, 2, 3, 5, 6, 10, 15 and 30.
• Prime factor form — a number written as a product of its prime factors; \( 30 = 2 \times 3 \times 5 \).
• Irreducible factor — a factor that cannot be expressed further as a product of factors; the algebraic version of “prime”.
• Factorisation — writing an expression as a product of factors, which may be numbers, variables or expressions.
• Common factor method — the three-step process: irreducible factors, separate common factors, combine the rest by the distributive law.
• Regrouping — rearranging terms into groups so each group has its own common factor, and a common bracket emerges across the groups.
• Monomial — an expression with a single term, like \( 2x \) or \( 3x^2y \).
• Dividend, divisor, quotient, remainder — the four parts of a division: the quantity being divided, the quantity dividing it, the result, and what is left over.

The four identities are not new in this chapter — the book’s earlier algebra work introduced them, and this chapter assumes you know them. Factorisation uses them backwards: you start from the expression on the right and recover the factors on the left (NCERT, sections 12.2.3–12.2.4).

Common Mistakes in Factorisation Class 8, with Corrections

These are the points NCERT flags in its closing recap — read each one before you make it.

How to Practise: Exercise Map, Self-Checking, and Exam Sense

Each exercise practises one stage of the skill. Work them in order and you are following the book’s own build-up.

Verify every answer without an answer key. Factorisation is the reverse of multiplication, so expand your answer with the distributive law — if you get the original expression, the factorisation is right. For division, check \( \text{Dividend} = \text{Divisor} \times \text{Quotient} \); the check comes out exact in this chapter because the remainder is always zero.

Exercise 12.3 Q5 — “factorise the expressions and divide them as directed” — is the point where the chapter’s two halves come together: you must factorise first, then divide. It is the hardest part of the chapter, and a slip in either half loses the answer.

One syllabus note: textbook contents and the examinable syllabus are not always identical — check the current official CBSE syllabus to see what is examinable this session.

What Have We Discussed: The Chapter’s Recap in Plain Words

The chapter closes with its own eleven-point recap. Rebuilt as a checklist you can run on yourself, it reads like this — can you do each line without opening the book?

1. Factorisation writes an expression as a product of factors — numbers, variables or expressions.
2. An irreducible factor cannot be split into smaller factors.
3. Common factor method: irreducible form per term → separate common factors → combine the rest by the distributive law.
4. When no factor is common to all terms, regroup so each group has its own common factor and a shared bracket emerges.
5. Not every regrouping works — expect trial and error; the wrong grouping hides the bracket instead of exposing it.
6. Four identities are enough for most factorisations: I and II for the perfect-square shapes, III for the difference of squares, IV for \((x + a)(x + b)\).
7. For \((x + a)(x + b)\): the constant gives \(ab\); choose \(a\), \(b\) so that their sum, signs included, is the coefficient of \(x\).
8. Division is the inverse of multiplication, for algebra exactly as for numbers.
9. Polynomial ÷ monomial has two allowed routes: divide each term, or use the common factor method.
10. Polynomial ÷ polynomial: factorise both and cancel common factors — never divide term by term.
11. All divisions in this chapter leave remainder zero, so Dividend = Divisor × Quotient, with no remainder term.

Related Chapters and Resources

This listing is maintained for the 2026-27 academic session using the NCERT textbook information available to us. NCERT remains the authority for confirming the latest edition.

Factorisation is one of the last algebra chapters in the Class 8 book, and the pages around it share its vocabulary. The four identities this chapter assumes come from the book’s earlier work on algebraic expressions — if they feel unfamiliar, return to that chapter before practising here.

• Steps and notes for every chapter of this book — Class 8 Mathematics notes
• Everything for Class 8 across subjects — Class 8 study hub
• The chapter before this one — Direct and Inverse Proportions (Chapter 11)
• The chapter after this one — Introduction to Graphs (Chapter 13)
• All classes and subjects — CBSE notes index

For the official full book, browse the complete NCERT Class 8 Mathematics textbook on ncert.nic.in to confirm the edition and open any chapter’s PDF. For what is examinable this session, see the current CBSE syllabus.

Sources and Data Verification

The contents described on this page come from the current NCERT Class 8 Mathematics textbook, rationalised edition, published on ncert.nic.in. This page covers that NCERT textbook only — it is not a statement of the full CBSE subject scheme. This listing is maintained for the current academic session using the NCERT information available to us.

NCERT settles textbooks, editions and PDFs; CBSE settles curriculum, syllabus and examinations. The section references above point into the NCERT chapter itself, so you can follow every claim with the PDF open.

Reference: NCERT Class 8 Mathematics textbook, chapter 12, official edition on ncert.nic.in.

Common Questions on Factorisation Class 8

What is the difference between a factor and an irreducible factor in factorisation?

A factor is any number or expression that multiplies with others to give a product. An irreducible factor is one that cannot be split further — the algebraic version of a prime number.

In \( 5xy = 5 \times x \times y \), the factors 5, \(x\) and \(y\) are irreducible; writing \( 5 \times (xy) \) is not irreducible form because \(xy\) can still be split into \(x \times y\) (NCERT, section 12.1.2).

Why does 3x + 3 have to be written as 3(x + 1) when regrouping?

Because 3 is a factor of both terms: \( 3x + 3 = 3 \times x + 3 \times 1 \). The 1 is the hidden factor of the second term, and without it the common bracket disappears and regrouping cannot finish. The book itself stops to ask why 1 must be shown as a factor (NCERT, section 12.2.2).

How do I pick a and b when the constant term is negative, like z² − 4z − 12?

Start with \( ab = -12 \): a negative constant means one of \(a, b\) is negative. Then use \( a + b = -4 \) to fix the signs — the factor with the larger numerical value takes the sign of the middle coefficient, so \( a = -6 \), \( b = 2 \) is right and \( a = -4 \), \( b = 3 \) is wrong, because it gives \( a + b = -1 \). The factorised form is \( (z – 6)(z + 2) \) (NCERT, section 12.2.4).

How do I decide which identity to use for factorising in Class 8?

Look at the shape of the expression. Three terms with the first and last perfect squares → Identity I if the middle term is positive, Identity II if it is negative. Two square terms with a minus between → Identity III.

Three terms in one variable with a constant that is not a square → Identity IV, and you search for the pair \(a, b\). Apply the same identity twice when a factor still fits the shape, as with \( m^4 – 256 \) (NCERT, section 12.2.3).

Why can’t I divide a polynomial by a polynomial term by term?

Because term-by-term division does not account for the divisor as a whole — the recap states it directly: for polynomial ÷ polynomial, factorise both and cancel their common factors, never divide each term of the dividend by the divisor (NCERT, chapter recap, point 10).

For example, in \( (6x^2 – 9x) \div (2x – 3) \), factorise the dividend to \( 3x(2x – 3) \), cancel \( (2x – 3) \), and the quotient is \(3x\).

Why do all the divisions in this chapter have a zero remainder?

Because the chapter deliberately restricts itself to exact divisions. Its closing points state that in the divisions studied here, Dividend = Divisor × Quotient, and then name the general form Dividend = Divisor × Quotient + Remainder — adding that only the zero-remainder case is considered in this chapter (NCERT, chapter recap, point 11).

That is why every division is done by factorising and cancelling rather than by long division.