Factorisation

Every algebraic expression you have built so far — by adding terms, multiplying binomials, expanding squares — can also be taken apart. Factorisation is that reverse journey: writing an expression as a product of simpler pieces (its factors) instead of a sum of terms. You already do this with plain numbers without thinking twice — you know $30 = 2 \times 3 \times 5$ and that 2, 3, 5 are its prime factors. This chapter extends the same idea to algebraic expressions such as $2x + 4$ , $x^2 + 5x + 6$ , or $49p^2 - 36$ , teaching you to see the hidden multiplication inside a sum. Why bother unwriting a sum as a product? Because a factor form is often more useful. A product like $2(x+2)$ tells you instantly that 2 and $(x+2)$ are exact building blocks — nothing is left over. A factorised expression is easier to simplify, easier to check for common terms with another expression, and, crucially, is exactly what you need to divide one algebraic expression by another cleanly, the way $56 \div 8 = 7$ works because $56 = 8 \times 7$ . This chapter builds four factorisation tools in sequence and one new operation. First, the method of common factors, where every term shares an irreducible piece. Second, regrouping, for when no single factor is common to all terms but pairs of terms share one after rearrangement. Third, factorisation using identities — the perfect-square and difference-of-squares patterns you met in Class VII, now read backwards. Fourth, the special pattern $x^2 + (a+b)x + ab = (x+a)(x+b)$ , useful for trinomials that are not perfect squares. Finally, the chapter turns to division of algebraic expressions — monomial by monomial, polynomial by monomial, and polynomial by polynomial — showing that factorising both dividend and divisor, then cancelling, is the master method throughout. By the end you should look at any algebraic expression and ask, almost automatically: does it have a common factor, does regrouping help, does it match a known identity, or is it of the $(x+a)(x+b)$ type? That question, asked in that order, is the working method of this entire chapter.

Key takeaways

Factorisation writes an expression as a product of factors — numbers, variables, or expressions — that multiply back to give the original.
An irreducible factor cannot itself be broken into a further product of factors; e.g. $5 \times x \times y$ is the irreducible form of $5xy$ .
Method of common factors: write every term in irreducible factor form, pull out what is common to all terms, using the distributive law $ab + ac = a(b+c)$ .
Regrouping: when no factor is common to all terms, group terms in pairs (or other groups) that share a common factor, so a new common factor — a bracket — emerges across groups.
Identities for factorisation (read the familiar expansions backwards): $a^2 + 2ab + b^2 = (a+b)^2$ , $a^2 - 2ab + b^2 = (a-b)^2$ , $a^2 - b^2 = (a+b)(a-b)$ .
Trinomial pattern: $x^2 + (a+b)x + ab = (x+a)(x+b)$ — find two numbers whose product is the constant term and whose sum is the coefficient of $x$ .
Division of a monomial by a monomial: write both in irreducible factor form and cancel common factors, exactly as with $77 \div 7$ .
Division of a polynomial by a monomial: either divide every term separately by the monomial, or factor out the monomial first and cancel.
Division of a polynomial by a polynomial: factorise both completely; only common irreducible factors may be cancelled — never split the division term by term across the denominator.
Key idea: factorisation writes an expression as a product using common factors, regrouping, or the identities a squared plus/minus 2ab plus b squared and a squared minus b squared, or the x plus a times x plus b pattern, and this same factor form is exactly what lets you divide one expression by another by cancelling shared factors.

The core insight

Factorising an algebraic expression — by pulling out common factors, regrouping, matching an identity, or splitting a trinomial as $(x+a)(x+b)$ — turns a sum into a product, and that product form is precisely what allows clean division of one expression by another through cancellation of shared irreducible factors.

What this chapter covers

The chapter has two connected halves. Sections 12.1–12.2 build up factorisation: first recalling factors of numbers, then defining factors of algebraic expressions and irreducible form, then four methods — common factors, regrouping, identities, and the $(x+a)(x+b)$ pattern for trinomials. Sections 12.3–12.4 build division of algebraic expressions, starting from monomial-by-monomial division, moving to polynomial-by-monomial division, and finishing with polynomial-by-polynomial division, where the factor forms from the first half become essential tools. Throughout, keep NCERT's own worked example numbers (Examples 1–16) in view — this article reuses and extends them.

The core ideas

Factors of algebraic expressions and irreducible form

Just as $30 = 2 \times 3 \times 5$ writes 30 in prime factor form, an algebraic term such as $5xy$ can be written as a product of factors that cannot be broken down any further:

5xy = 5 \times x \times y.

where: the factors $5$ , $x$ , $y$ are each irreducible — none of them can be expressed as a further product. In algebra we say "irreducible" where, for numbers, we would say "prime." Note that $5 \times (xy)$ is not an irreducible factorisation of $5xy$ , because $xy$ can still be split into $x \times y$ . As with numbers, $1$ is technically a factor of every term, but we do not write it separately unless it is specially needed (for example, during regrouping). An expression such as $3x(x+2)$ is already in factor form; its irreducible factors are $3$ , $x$ , and $(x+2)$ . But an expression such as $2x + 4$ is a sum of two terms, and it is not obvious from that sum what its factors are. Turning such a sum into a product is exactly what factorisation does.

Method of common factors

The most basic route to factorisation rests on the distributive law used in reverse: since $a \times b + a \times c = a(b+c)$ , if every term of an expression shares a common irreducible factor, that factor can be pulled outside a bracket.

2x + 4 = (2 \times x) + (2 \times 2) = 2(x + 2).

where: the common factor here is $2$ ; after removing it, the leftover pieces $x$ and $2$ are combined by addition inside the bracket. The three-step recipe is always the same: write each term in irreducible factor form, identify the factor(s) common to every term, and combine the remaining pieces using the distributive law. Example: $12a^2b + 15ab^2$ has $12a^2b = 2\times2\times3\times a\times a\times b$ and $15ab^2 = 3\times5\times a\times b\times b$ ; the common factors are $3, a, b$ , giving $12a^2b + 15ab^2 = 3ab(4a+5b)$ .

Factorisation by regrouping

Sometimes no single factor is common to all terms, but the terms can be split into groups, each of which has its own common factor, and those groups in turn share a common bracket. Consider $2xy + 2y + 3x + 3$ : the first pair has common factor $2y$ , and the second pair has common factor $3$ .

2xy + 2y + 3x + 3 = 2y(x+1) + 3(x+1) = (x+1)(2y+3).

where: the bracket $(x+1)$ is the new common factor that appears after grouping — this is why the method is called regrouping. The order in which terms are arranged matters: the same expression written as $2xy + 3 + 2y + 3x$ hides the grouping, while rearranging it as $2xy + 2y + 3x + 3$ (or even $2xy + 3x + 2y + 3$ ) reveals it. There may be more than one valid way to regroup, but a correct regrouping always yields the same final factors, just possibly in a different order.

Factorisation using identities

Three identities from Class VII, read from right to left, turn perfect-square and difference-of-squares expressions directly into products:

a^2 + 2ab + b^2 = (a+b)^2 \qquad \text{(I)}

a^2 - 2ab + b^2 = (a-b)^2 \qquad \text{(II)}

a^2 - b^2 = (a+b)(a-b) \qquad \text{(III)}

where: to use these, check whether the given expression's terms match the right-hand pattern — two squared terms and (for I, II) a middle term equal to $\pm 2ab$ , or (for III) exactly two terms that are both perfect squares with a minus sign between them. Recognising $a$ and $b$ correctly is the entire skill; the factorisation then follows immediately.

Factors of the type $(x+a)(x+b)$

Trinomials such as $x^2 + 5x + 6$ are not perfect squares (the constant term $6$ is not a perfect square) and do not fit the difference-of-squares pattern either. They instead match a fourth identity from Class VII:

(x+a)(x+b) = x^2 + (a+b)x + ab \qquad \text{(IV)}

where: to factorise $x^2 + px + q$ using this pattern, you must find two numbers $a$ and $b$ such that their product $ab$ equals the constant term $q$ and their sum $a + b$ equals the coefficient $p$ of $x$ . Once found, the middle term $px$ is split as $ax + bx$ , and the expression is factorised by grouping: $x^2 + ax + bx + ab = x(x+a) + b(x+a) = (x+a)(x+b)$ .

Division of algebraic expressions

Division undoes multiplication: since $2x \times 3x^2 = 6x^3$ , it follows that $6x^3 \div 2x = 3x^2$ . To divide a monomial by a monomial, write both in irreducible factor form and cancel the factors common to numerator and denominator, exactly as $77 \div 7 = \frac{7 \times 11}{7} = 11$ . To divide a polynomial by a monomial, either divide each term of the polynomial separately by the monomial, or first write the polynomial in factor form (pulling out the monomial as a common factor) and cancel. To divide a polynomial by a polynomial, term-by-term division does not work — instead, factorise both the dividend and the divisor completely, then cancel their common irreducible factors, leaving the quotient.

Figure

Two rectangular tile diagrams side by side, the left showing 2x+4 split into two unequal blocks of width x and 2 each of height 2 illustrating the common factor 2, and the right showing an x by x square plus two x by 4 strips plus a 4 by 4 square combining into one x+4 by x+4 square illustrating x squared plus 8x plus 16 equals x+4 whole squared

Figure

A number line diagram showing that the difference of squares 7p squared minus 36 equals the product of two brackets 7p minus 6 and 7p plus 6, drawn as two labelled segments multiplying to a rectangle area

Derivation ADeriving the common-factor method from the distributive law

The goal of this derivation is to show, from first principles, exactly why pulling out a common factor is a valid and complete method of factorisation, and to state precisely the three steps that make it work every time rather than by trial. We build it directly from the distributive law of multiplication over addition, the same law used to expand brackets, applied here in the reverse direction.

Step 1. Recall the distributive law in its expanding direction: for any three quantities $m$ , $p$ , $q$ , we have $m(p+q) = mp + mq$ . This is how you multiply out a bracket — the factor $m$ is distributed to each term inside.

Step 2. Read the same law from right to left: $mp + mq = m(p+q)$ . This says that if two terms $mp$ and $mq$ share a common factor $m$ , their sum can be rewritten as $m$ times a bracket. Nothing new is being claimed here — it is the identical law, only applied starting from the sum instead of the product.

Step 3. To use Step 2 on a given expression, first express every term in its irreducible factor form, since a common factor might be hidden inside a term. For example, in $10x^2 - 18x^3 + 14x^4$ , write $10x^2 = 2\times5\times x\times x$ , $18x^3 = 2\times3\times3\times x\times x\times x$ , and $14x^4 = 2\times7\times x\times x\times x\times x$ .

Step 4. Identify the factors that appear in every one of these irreducible expansions. Here, $2$ and $x \times x$ appear in all three terms, so the common factor is $2x^2$ .

Step 5. Apply the distributive law of Step 2 to separate the common factor: $10x^2 - 18x^3 + 14x^4 = 2x^2(5) - 2x^2(9x) + 2x^2(7x^2) = 2x^2(5 - 9x + 7x^2)$ .

Step 6. Check by expanding: distributing $2x^2$ back over $(7x^2 - 9x + 5)$ must reproduce every original term exactly. Each step above is a reversible application of a law already trusted, so this method always gives a correct factor form, and the leftover bracket cannot be simplified further once no factor remains common to all its terms.

Go deeper · Derivation B · 1 minDeriving the (x+a)(x+b) rule for trinomials

Derivation BDeriving the (x+a)(x+b) rule for trinomials

The goal of this derivation is to establish, rigorously and not just by pattern-matching, why searching for two numbers whose product is the constant term and whose sum is the coefficient of $x$ correctly factorises a trinomial of the form $x^2 + px + q$ , and to justify why this particular pair of conditions — and no other — is the right one to search for.

Step 1. Start from the known expansion of a product of two linear binomials: $(x+a)(x+b) = x\cdot x + x\cdot b + a\cdot x + a\cdot b = x^2 + (a+b)x + ab$ . This is Identity IV, obtained purely by multiplying out brackets term by term, so it is already established and trustworthy.

Step 2. Compare this expanded form, $x^2 + (a+b)x + ab$ , with a general trinomial $x^2 + px + q$ that we wish to factorise. For the trinomial to match the expanded product, we need the coefficient of $x$ to match, giving $a + b = p$ , and the constant term to match, giving $ab = q$ .

Step 3. This comparison converts the problem of factorising into the problem of finding two numbers $a$ and $b$ satisfying both conditions simultaneously: their product must equal $q$ and their sum must equal $p$ . Because $q$ has only finitely many factor pairs, this is a finite (not infinite) search — list the factor pairs of $q$ and test which pair sums to $p$ .

Step 4. Once such $a$ and $b$ are found, rewrite the middle term $px$ as $ax + bx$ (valid since $a + b = p$ ), turning the trinomial into four terms: $x^2 + ax + bx + ab$ .

Step 5. Factorise this four-term expression by regrouping: the first two terms give $x(x+a)$ , and the last two terms give $b(x+a)$ , so $x^2 + ax + bx + ab = x(x+a) + b(x+a) = (x+a)(x+b)$ .

Step 6. Worked check with $x^2 + 5x + 6$ : here $q = 6$ and $p = 5$ . Testing factor pairs of 6 — $(6,1)$ gives sum $7$ , not $5$ ; $(2,3)$ gives sum $5$ , exactly matching. So $a=2$ , $b=3$ , and $x^2+5x+6 = (x+2)(x+3)$ , which expands back correctly by Step 1, confirming the derivation is consistent in both directions.

Worked examples

Board level - Factorise 6xy minus 4y plus 6 minus 9x (NCERT Example 3)

Step 1. Check whether a single factor is common to all four terms $6xy$ , $-4y$ , $6$ , $-9x$ . There is none, so the plain common-factor method alone will not work.

Step 2. Look for a grouping. The terms $6xy$ and $-4y$ share the common factor $2y$ : $6xy - 4y = 2y(3x-2)$ .

Step 3. For the remaining terms, rearrange $6 - 9x$ as $-9x + 6$ and factor: $-9x + 6 = -3(3x) + 3(2) = -3(3x-2)$ .

Step 4. Combine: $6xy - 4y - 9x + 6 = 2y(3x-2) - 3(3x-2) = (3x-2)(2y-3)$ .

The factors are $(3x-2)$ and $(2y-3)$ .

Board level - Factorise a squared minus 2ab plus b squared minus c squared (NCERT Example 7)

Step 1. Observe that the expression has four terms; the first three, $a^2 - 2ab + b^2$ , match Identity II exactly, with the fourth term $c^2$ left over as a perfect square with a minus sign.

Step 2. Apply Identity II to the first three terms: $a^2 - 2ab + b^2 = (a-b)^2$ , so the expression becomes $(a-b)^2 - c^2$ .

Step 3. This is now a difference of two squares, with $A = (a-b)$ and $B = c$ , matching Identity III: $A^2 - B^2 = (A+B)(A-B)$ .

Step 4. Substitute back: $(a-b)^2 - c^2 = \big[(a-b)+c\big]\big[(a-b)-c\big] = (a-b+c)(a-b-c)$ .

The factors are $(a-b+c)$ and $(a-b-c)$ .

Higher-order thinking (HOTS / olympiad-style) - Divide 44(x⁴ − 5x³ − 24x²) by 11x(x − 8) (NCERT Example 15)

Step 1. Factorise the dividend completely. Take out the common factor $x^2$ first: $44(x^4 - 5x^3 - 24x^2) = 44x^2(x^2 - 5x - 24)$ .

Step 2. Factorise the trinomial $x^2 - 5x - 24$ using the $(x+a)(x+b)$ method: we need $ab = -24$ and $a+b = -5$ . Testing pairs, $a = 3$ , $b = -8$ gives $a+b = -5$ and $ab = -24$ , both correct. So $x^2 - 5x - 24 = x^2 - 8x + 3x - 24 = x(x-8) + 3(x-8) = (x+3)(x-8)$ .

Step 3. Also break $44 = 2 \times 2 \times 11$ , so the fully factorised dividend is $2 \times 2 \times 11 \times x \times x \times (x+3) \times (x-8)$ .

Step 4. Write the divisor in factor form too: $11x(x-8) = 11 \times x \times (x-8)$ , already irreducible.

Step 5. Cancel the factors common to numerator and denominator — one factor of $11$ , one factor of $x$ , and the factor $(x-8)$ : $\dfrac{2\times2\times11\times x\times x\times(x+3)\times(x-8)}{11\times x\times(x-8)} = 2\times2\times x\times(x+3)$ .

The quotient is $4x(x+3)$ . Timing note: aim to complete this in 3 to 4 minutes in an exam — factorising the trinomial correctly is the only genuinely slow step; the cancellation itself should be almost immediate once both sides are in irreducible factor form.

Common traps

A common error is leaving a term only partly expanded before hunting for common factors — writing $5xy$ as $5 \times (xy)$ instead of $5 \times x \times y$ can hide a factor that is actually present. In regrouping, students often rearrange terms without checking that both resulting groups genuinely share a common factor; not every rearrangement works, and correct grouping sometimes needs trial and error. With the identities, a frequent slip is misreading the sign of the middle term — treating $4y^2-12y+9$ as identity I instead of identity II gives $(2y+3)^2$ instead of the correct $(2y-3)^2$ . In the $(x+a)(x+b)$ pattern, students sometimes pick a factor pair of $q$ that multiplies correctly but does not sum to $p$ — both conditions must hold together. Most dangerously, in division, students try to split a polynomial divided by a polynomial term by term across the denominator; this only works when the denominator is a single monomial.

You should be able to…

Write any given term or expression in its irreducible (fully factored) form, distinguishing this from a partly factored form.
Apply the method of common factors correctly, using the distributive law to pull a shared factor outside a bracket.
Recognise when regrouping is needed and successfully rearrange and group terms so a common bracket emerges.
State Identities I, II, III (and IV) from memory and identify which one, if any, an algebraic expression matches.
Factorise perfect-square trinomials and differences of squares using the identities, including multi-step cases like $a^2-2ab+b^2-c^2$ .
Factorise a trinomial of the form $x^2+px+q$ by finding a suitable pair of numbers with product $q$ and sum $p$ .
Divide a monomial by a monomial, and a polynomial by a monomial, using both the term-by-term method and the factor-and-cancel method.
Divide a polynomial by a polynomial by fully factorising both sides and cancelling common irreducible factors.
Verify any factorisation or division result by expanding or multiplying back to confirm it reproduces the original expression.

Ask the tutor: Want to work through a few more mixed factorisation and division problems together, step by step, until the pattern-spotting becomes automatic?

Why do we write 1 as a factor sometimes but not always

By convention, we do not write 1 as a separate factor of a term unless it is specially needed, exactly the way we do not usually say "30 is 1 times 2 times 3 times 5." However, during regrouping you sometimes must show the 1 explicitly. For instance, in $3x + 3 = 3(x) + 3(1) = 3(x+1)$ , the 1 has to appear so that both terms inside the later combined bracket have a matching partner term. Once the factorisation is complete, the 1 is simply part of the bracket $(x+1)$ and is not written as a separate outside factor.

How do I know when to use regrouping instead of the common factor method

Try the common factor method first: check if a single factor is common to every term of the expression. If you find one, factor it out and you are done. If no factor is common to all terms, look at whether the terms can be split into two (or more) groups where each group has its own common factor, and — this is the key test — the two groups, once factored, leave behind the same bracket. If that bracket matches, regrouping works; if it does not match after one attempt, try a different order of grouping before concluding the expression needs another method entirely.

Why does the difference of squares identity not apply to a sum of squares like x squared plus 9

Identity III, $a^2 - b^2 = (a+b)(a-b)$ , only applies when there is a minus sign between two perfect squares. A sum like $x^2+9$ cannot be factorised using the identities of this chapter — it also fails to match Identity I or II, since those always produce a middle term $\pm 2ab$ , which is missing here. So $x^2+9$ is simply left as it is at Class 8 level; only expressions matching one of the four identities, or the common-factor or regrouping patterns, are meant to be factorised in this chapter.

What is the difference between dividing a polynomial by a monomial and by another polynomial

When dividing a polynomial by a single monomial (like $(4y^3+5y^2+6y) \div 2y$ ), you can safely divide each term of the polynomial by the monomial separately, since each term shares that one monomial as a factor. When dividing a polynomial by another polynomial (like $(7x^2+14x)\div(x+2)$ ), that term-by-term shortcut breaks down completely, because the denominator itself has more than one term. The reliable method in that case is always to factorise both the dividend and the divisor fully into irreducible factors first, and then cancel whatever factors are common to both — the quotient is whatever factors remain in the numerator after cancellation.

Factorisation

What this chapter covers

The core ideas

Factors of algebraic expressions and irreducible form

Method of common factors

Factorisation by regrouping

Factorisation using identities

Factors of the type $(x+a)(x+b)$

Division of algebraic expressions

Worked examples

Board level - Factorise 6xy minus 4y plus 6 minus 9x (NCERT Example 3)

Board level - Factorise a squared minus 2ab plus b squared minus c squared (NCERT Example 7)

Higher-order thinking (HOTS / olympiad-style) - Divide 44(x⁴ − 5x³ − 24x²) by 11x(x − 8) (NCERT Example 15)

Common traps

You should be able to…

Before this

Where this leads

Factorisation

What this chapter covers

The core ideas

Factors of algebraic expressions and irreducible form

Method of common factors

Factorisation by regrouping

Factorisation using identities

Factors of the type (x+a)(x+b)(x+a)(x+b)(x+a)(x+b)

Division of algebraic expressions

Worked examples

Board level - Factorise 6xy minus 4y plus 6 minus 9x (NCERT Example 3)

Board level - Factorise a squared minus 2ab plus b squared minus c squared (NCERT Example 7)

Higher-order thinking (HOTS / olympiad-style) - Divide 44(x⁴ − 5x³ − 24x²) by 11x(x − 8) (NCERT Example 15)

Common traps

You should be able to…

Before this

Where this leads

Factors of the type $(x+a)(x+b)$