Factorisation | ICSE Class 9 Maths Notes

This note covers the meaning of factorisation, common factors, grouping, difference of squares, repeated use of identities, sum and difference of cubes, splitting the middle term of quadratic expressions, choosing a suitable method and checking factors by multiplication.

What does factorising an algebraic expression mean?

How are terms different from factors?

An algebraic expression combines numbers and letters using mathematical operations. A variable is a letter representing a number whose value may vary. Here, x and y represent numbers. In 5xy + 3x, writing letters together means multiplication: xy means x × y.

The multiplication sign × indicates a product; + means addition, − means subtraction and = means equality. Brackets group an expression into one unit. Thus, 3x(x + 2) means that 3, x and the whole expression x + 2 are multiplied.

A term is a part of an expression separated from other parts by addition or subtraction, with its sign retained. A factor is a number or expression multiplied by another to form a product. In 5xy, the factors include 5, x and y.

Definition: Factorisation means writing an algebraic expression as a product of factors. These factors may be numbers, variables or algebraic expressions.

A coefficient is the numerical multiplier of a variable term. In 3x, the coefficient of x is 3. A constant term has no variable: in x + 2, it is 2. An exponent records repeated multiplication: x² means x × x and x³ means x × x × x.

What is the relationship with expansion?

Expansion multiplies out brackets; factorisation reverses that process. The expressions 2x + 4 and 2(x + 2) have equal values for every value of x. The first displays a sum of terms, while the second displays a product.

An irreducible factor cannot be broken down further into the kinds of factors being considered. The final aim is to expose the factors, rather than merely rearrange the original terms. Multiplying the proposed factors provides a direct check of the result.

How do common factors provide the first step?

Property: the distributive law in reverse

A common factor occurs in every term under consideration. The distributive law says that multiplying a bracket means multiplying each term inside it. In reverse, it allows a shared multiplier to be taken outside a bracket without changing the expression.

Worked example 1. Factorise 2x + 4.

Answer: 2x + 4 = 2 × x + 2 × 2 = 2(x + 2). The common factor is 2, and the remaining factors form x + 2.

The bracket contains what remains from each term after removing the common factor. Nothing is discarded. Expanding 2(x + 2) returns 2x + 4 because the outside 2 multiplies both x and 2.

Worked example 2. Factorise 5xy + 10x.

Answer: 5xy + 10x = 5x × y + 5x × 2 = 5x(y + 2). Both terms contain the numerical factor 5 and the variable factor x.

How should the method be organised?

1. Write each term as a product of its numerical and variable factors.
2. Identify the factors present in every term, including repeated variable factors.
3. Place their product outside a bracket and collect the remaining parts inside it.
4. Multiply back to check every coefficient, variable power and sign.

Let a and b now denote variables representing numbers. The same method works when both variables appear in each term, even if their exponents differ. Removing a repeated factor requires it to be present that many times in every term.

Worked example 3. Factorise 12a²b + 15ab².

Answer: 12a²b + 15ab² = 3ab × 4a + 3ab × 5b = 3ab(4a + 5b). The common factors are 3, a and b.

Taking out a common factor can expose another factorisation inside the bracket. It is therefore a useful opening check even when the expression later needs an identity or a split middle term.

How does grouping turn a sum into a product?

How is a common bracket created?

Grouping collects terms so that each group can be factorised separately. The useful outcome is the same bracket in both groups. That bracket then acts as a common factor, just as a number or a single variable did earlier.

Worked example 4. Factorise 2xy + 2y + 3x + 3.

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

There is no numerical or variable factor other than 1 shared by all four original terms. However, the first pair contains 2y and the second pair contains 3. Once these are removed, the common bracket x + 1 becomes visible.

The remaining 1 in x + 1 matters: 2y × 1 produces 2y, and 3 × 1 produces 3. Removing that 1 would change the expression. Factorisation preserves the complete value, including terms without a visible variable multiplier.

What happens when terms are rearranged?

Regrouping may require changing the order of terms. Each term must move with its sign. Different successful groupings can produce the same factors in a different order. However, any regrouping may not lead to factorisation; the desired grouping may need trial and error.

Worked example 5. Factorise 6xy − 4y + 6 − 9x.

Answer: 6xy − 4y + 6 − 9x = 2y(3x − 2) − 3(3x − 2) = (3x − 2)(2y − 3).

The last pair is written as −9x + 6. Taking out −3 leaves 3x − 2, since −3 × (−2) = 6. Choosing the negative outside factor makes both brackets agree and allows the second use of the distributive law.

How do you factorise a difference of two squares?

Identity: difference of squares

An identity is an equality true for every permitted value of its variables. A square is a quantity multiplied by itself. The difference-of-squares identity rewrites a subtraction of two squares as a product of their sum and difference.

Here a and b may represent numbers or complete algebraic expressions. The identity is a² − b² = (a − b)(a + b). Read each squared quantity as a whole before choosing its corresponding unsquared expression.

Expanding the product gives a² + ab − ab − b². The middle terms, ab and −ab, cancel because their sum is zero. What remains is a² − b². This cancellation explains why the two brackets have opposite signs.

Worked example 6. Factorise 49p² − 36, where p is a variable.

Answer: 49p² − 36 = (7p)² − 6² = (7p − 6)(7p + 6). The quantities being squared are 7p and 6.

Do not use the original coefficients as the unsquared quantities: the square of 7p is 49p². After identifying both squares, keep the same two quantities in both brackets. Change only the sign between them.

When can the identity be used again?

Let m denote a variable. The notation m⁴ means four factors of m multiplied together, so it can also be read as (m²)². A factor that remains after one application may itself be a difference of squares.

Worked example 7. Factorise m⁴ − 256.

Answer: m⁴ − 256 = (m² − 16)(m² + 16) = (m − 4)(m + 4)(m² + 16).

The first step uses 256 = 16². The next uses 16 = 4² in m² − 16. The factor m² + 16 is a sum of squares, so this difference-of-squares identity does not split it further.

Note: Check the sign between the squares. The identity for a² − b² cannot be applied unchanged to a² + b².

How can square identities reveal a hidden difference of squares?

Identity: square of a sum or difference

A perfect-square expression is the square of an algebraic expression. The identities a² + 2ab + b² = (a + b)² and a² − 2ab + b² = (a − b)² help recognise such expressions. The middle term is twice the product of a and b.

Checking the first and last terms is not enough. The middle term must also match in size and sign. The plus version contains +2ab, while the minus version contains −2ab. Both have a positive final term b².

Worked example 8. Factorise 4y² − 12y + 9.

Answer: 4y² − 12y + 9 = (2y)² − 2 × (2y) × 3 + 3² = (2y − 3)².

Here, 2y and 3 are the two quantities in the square identity. Their doubled product is 12y, exactly the numerical size of the middle term apart from its sign. The negative middle sign selects the square of a difference.

How can two identities work in sequence?

Let c denote another variable. In a² − 2ab + b² − c², the first three terms form one square, even though the whole expression contains four terms. Replacing that group by (a − b)² reveals a difference of squares.

Worked example 9. Factorise a² − 2ab + b² − c².

Answer: a² − 2ab + b² − c² = (a − b)² − c² = (a − b − c)(a − b + c).

The first squared quantity is the entire bracket a − b. Apply the difference-of-squares identity to that bracket and c. Retaining the bracket as one unit prevents the internal minus sign from being lost during the second factorisation.

How is a difference of cubes factorised?

Identity: difference of cubes

A cube is the product of three equal factors. Thus, a³ means a × a × a. A base is the quantity raised to an exponent; in a³, the base is a. For a and b representing quantities, the difference-of-cubes identity is a³ − b³ = (a − b)(a² + ab + b²).

A binomial has two terms, and a trinomial has three. In this factorisation, a − b is the binomial factor. The trinomial contains the square of the first quantity, the product of the two quantities and the square of the second quantity.

Why are both signs in the trinomial positive?

1. Multiply a by the trinomial to obtain a³ + a²b + ab².
2. Multiply −b by the trinomial to obtain −a²b − ab² − b³.
3. Add the results: the terms a²b and −a²b cancel, as do ab² and −ab².
4. The remaining terms are a³ − b³, proving the stated factorisation.

The cross terms are the terms involving both quantities. Their cancellation is the reason for the sign pattern. Giving the middle term of the trinomial a negative sign would prevent this cancellation for the difference-of-cubes formula.

The same identity can be written with x and y: x³ − y³ = (x − y)(x² + xy + y²). Changing the letters does not change the rule. The factors depend on the two quantities being cubed, rather than on a particular choice of letter.

Note: The difference a³ − b³ is not the same expression as (a − b)³. The latter expands to a³ − 3a²b + 3ab² − b³ and includes additional terms.

Recognise the cubes before writing either bracket. Then form the difference of their bases and the three-term factor in the prescribed order. Multiplication provides a check without assigning a particular numerical value to either variable.

How does the sum-of-cubes identity differ?

Identity: sum of cubes

The companion identity is a³ + b³ = (a + b)(a² − ab + b²). The sign in the first bracket matches the sign between the cubes. In the second bracket the product term is negative, while the two squared terms are positive.

To verify it, multiply a by the second bracket to obtain a³ − a²b + ab². Multiplication by b gives a²b − ab² + b³. Adding the two results cancels both pairs of mixed terms, leaving a³ + b³.

How do fractional coefficients fit the identity?

Let z be a variable. The slash / denotes division, so z/5 means z divided by 5. The numerator is the part above a fraction bar and the denominator is the part below it. A fractional coefficient is handled by cubing the complete fraction: (z/5)³ = z³/125. Brackets clarify that both numerator and denominator are involved.

Worked example 10. Factorise 64y³ + z³/125.

Answer: 64y³ + z³/125 = (4y)³ + (z/5)³ = (4y + z/5)(16y² − 4yz/5 + z²/25).

In this example, the two bases are 4y and z/5. Their squares give 16y² and z²/25. Their product gives 4yz/5, which enters the second bracket with a minus sign. The denominator 125 belongs to the original cube, not to every resulting term.

Identifying complete bases, including their numerical multipliers, is more reliable than copying the coefficients from the original expression. The same care is needed for the difference of cubes.

How do you split the middle term when the coefficient of the square is one?

How does the product-and-sum test arise?

A quadratic expression in x has highest exponent 2. A monic quadratic has coefficient 1 for x². Here p and q denote fixed numerical coefficients in x² + px + q; p multiplies x and q is the constant term.

For this method, let a and b denote numbers to be found. The identity (x + a)(x + b) = x² + (a + b)x + ab shows two requirements: their sum must equal p and their product must equal q.

Splitting the middle term replaces px by ax + bx using a + b = p. The four resulting terms can then be grouped. Satisfying the product condition as well makes the two groups share a bracket.

Worked example 11. Factorise x² + 5x + 6.

Answer: 2 × 3 = 6 and 2 + 3 = 5. Hence x² + 5x + 6 = x² + 2x + 3x + 6 = x(x + 2) + 3(x + 2) = (x + 2)(x + 3).

The pair 6 and 1 gives the correct product but the wrong sum, 7. A pair is suitable only when both conditions hold. The factors record the same two numbers used to split the coefficient of x.

How are negative signs handled?

Worked example 12. Factorise y² − 7y + 12.

Answer: (−3)(−4) = 12 and −3 − 4 = −7. Thus y² − 7y + 12 = y² − 3y − 4y + 12 = y(y − 3) − 4(y − 3) = (y − 3)(y − 4).

A positive product with a negative sum requires two negative numbers. With a negative product, the two numbers have opposite signs. Their signed sum still has to match the middle coefficient, including its sign.

Worked example 13. Factorise z² − 4z − 12.

Answer: (−6) × 2 = −12 and −6 + 2 = −4. Therefore z² − 4z − 12 = z(z − 6) + 2(z − 6) = (z − 6)(z + 2).

How do you split the middle term in a general quadratic?

What changes when the leading coefficient is not one?

In ax² + bx + c, x is the variable, while a, b and c are fixed coefficients. The leading coefficient a multiplies x² and must be non-zero for the expression to be quadratic. Here b multiplies x and c is the constant term.

The symbol ≠ means “is not equal to”, so the condition is a ≠ 0. Choose two numbers whose product is ac and whose sum is b. Use those numbers as the coefficients of the two terms replacing bx.

The product to inspect is ac, rather than c alone. When a is 1, this reduces to the monic method. After splitting, group the four terms and take out common factors until the same bracket appears in both groups.

Worked example 14. Factorise 6x² + 7x + 2.

Answer: ac = 6 × 2 = 12. Choose 3 and 4 because their product is 12 and their sum is 7. Then 6x² + 7x + 2 = 6x² + 3x + 4x + 2 = 3x(2x + 1) + 2(2x + 1) = (2x + 1)(3x + 2).

How does the method handle a negative constant?

Worked example 15. Factorise 10x² − 11x − 6.

Answer: ac = 10 × (−6) = −60. Choose −15 and 4, with sum −11. Thus 10x² − 11x − 6 = 10x² − 15x + 4x − 6 = 5x(2x − 3) + 2(2x − 3) = (2x − 3)(5x + 2).

The unequal coefficients of x in the final brackets matter. Their product produces 10x², and the two cross products produce −15x + 4x. Simply placing the splitting numbers into brackets of the form x plus a number would lose the leading coefficient.

If all terms share a numerical factor, remove it first. For the variable m, 3m² + 9m + 6 = 3(m² + 3m + 2) = 3(m + 1)(m + 2). The outside 3 remains part of the answer throughout.

How should you choose a method and check the result?

Which feature of the expression should you inspect?

Start by looking for a common factor. Next, inspect the remaining expression for an identity or a useful grouping. The method depends on the expression's form, including its signs and coefficients, rather than on its number of terms alone.

Some expressions need more than one step. A common factor can expose a quadratic; a square trinomial can expose a difference of squares; and one difference of squares can produce another. Inspect the factors obtained before deciding that the work is finished.

How does multiplication verify a factorisation?

Like terms have the same variables raised to the same powers. In a check, expand every bracket and combine like terms. For example, (2x − 3)(5x + 2) gives 10x² + 4x − 15x − 6 = 10x² − 11x − 6.

Compare that expansion with the original expression term by term. The squared term, middle term and constant must all agree. A correct first and last term cannot compensate for an incorrect middle coefficient.

Factorisation asks for an equivalent product. It does not require a numerical value of the variable. Keeping the full expression visible through the working makes it easier to preserve signs, retain outside factors and confirm that the final brackets reconstruct the starting expression.

Glossary

• Factorisation — Writing an algebraic expression as a product of numerical or algebraic factors.
• Factor — A number or expression multiplied by another to form a product.
• Term — A signed part of an expression separated by addition or subtraction.
• Coefficient — The numerical multiplier attached to a variable or a variable power.
• Constant term — A term containing no variable, whose value is fixed in the expression.
• Common factor — A factor present in every term or group currently being considered.
• Identity — An equality that holds for every permitted value of its variables.
• Grouping — Collecting terms so that extracting factors can reveal a shared bracket.
• Quadratic expression — An expression in one variable whose highest variable exponent is two.
• Monic quadratic — A quadratic expression with coefficient one for its squared variable term.
• Leading coefficient — The coefficient multiplying the highest power of the variable in an expression.
• Perfect-square expression — An expression that can be written as an algebraic expression multiplied by itself.
• Binomial — An algebraic expression consisting of two terms joined by addition or subtraction.
• Trinomial — An algebraic expression consisting of three terms with their associated signs.
• Irreducible factor — A factor that cannot be split further within the kinds of factors considered.

Common errors and misconceptions

• Misconception: Terms and factors are interchangeable names. Correct: Terms are added or subtracted; factors are multiplied. The terms 2x and 4 become the factors 2 and x + 2.
• Misconception: A factor can be removed from just some terms when placing it outside the whole expression. Correct: It must occur in every term of that expression.
• Misconception: Any grouping must produce factors. Correct: A chosen grouping may not work. Look for groups that produce a shared bracket, preserving the sign of every term.
• Misconception: A sum of squares uses the difference-of-squares formula unchanged. Correct: The identity requires subtraction; multiplying its brackets gives a² − b².
• Misconception: The difference of cubes is the cube of a difference. Correct: a³ − b³ has two terms; expanding (a − b)³ also produces mixed terms.
• Misconception: Both cube identities have the same trinomial sign pattern. Correct: a³ − b³ uses a² + ab + b², while a³ + b³ uses a² − ab + b².
• Misconception: A correct product alone identifies the numbers for splitting. Correct: Their sum must match the middle coefficient as well; for ax² + bx + c their product is ac.
• Misconception: One application of an identity finishes every factorisation. Correct: Inspect the resulting factors; in m⁴ − 256, the factor m² − 16 splits again.

Exam-style questions with model answers

Q1. Factorise 49p² − 36 using a suitable identity, where p is a variable. [2 marks]

1. Write both terms as squares: 49p² − 36 = (7p)² − 6².
2. Apply the difference-of-squares identity to obtain the product (7p − 6)(7p + 6).

Q2. Factorise 6xy − 4y + 6 − 9x by grouping, where x and y are variables. [3 marks]

1. Reorder the last two terms while preserving their signs: 6xy − 4y − 9x + 6. Group the first two and the last two terms.
2. Take out 2y from the first pair and −3 from the second: 2y(3x − 2) − 3(3x − 2).
3. The shared bracket is 3x − 2, so the factorised expression is (3x − 2)(2y − 3).

Q3. Factorise y² − 7y + 12 by splitting the middle term, where y is a variable. Show how the splitting numbers are chosen. [3 marks]

1. Find two numbers whose product is 12 and sum is −7. The numbers −3 and −4 satisfy both conditions, since (−3)(−4) = 12 and −3 − 4 = −7.
2. Split and group: y² − 7y + 12 = y² − 3y − 4y + 12 = y(y − 3) − 4(y − 3).
3. Take out the common bracket y − 3. The required factorisation is (y − 3)(y − 4).

Q4. Factorise 6x² + 7x + 2 by splitting the middle term, and verify your answer by multiplication. Here x is a variable. [5 marks]

1. The leading coefficient is 6 and the constant is 2, giving product 6 × 2 = 12. We need two numbers with this product and sum 7.
2. Choose 3 and 4: their product is 12 and their sum is 7. Therefore replace 7x by 3x + 4x.
3. Regroup the expression as (6x² + 3x) + (4x + 2). Taking out common factors gives 3x(2x + 1) + 2(2x + 1).
4. Both terms contain the bracket 2x + 1. Taking it outside gives the factorised expression (2x + 1)(3x + 2).
5. Multiply back: (2x + 1)(3x + 2) = 6x² + 4x + 3x + 2 = 6x² + 7x + 2, as required.

Q5. Factorise 64y³ + z³/125 using the sum-of-cubes identity and verify the result. Here y and z are variables. [5 marks]

1. Identify the two cubes: 64y³ = (4y)³ and z³/125 = (z/5)³. Thus the complete bases to use are 4y and z/5.
2. Use a³ + b³ = (a + b)(a² − ab + b²), where a and b stand for those two bases.
3. The binomial factor is 4y + z/5. The trinomial is 16y² − 4yz/5 + z²/25, using the two squares and their negative product.
4. Hence the required product is (4y + z/5)(16y² − 4yz/5 + z²/25). Its first and last products give 64y³ and z³/125.
5. On expansion, the remaining terms are −16y²z/5 + 4yz²/25 + 16y²z/5 − 4yz²/25. They cancel in pairs, verifying the original sum of cubes.

Q6. Factorise m⁴ − 256 completely using the difference-of-squares identity, where m is a variable. Explain why two applications are needed. [4 marks]

1. Recognise the original expression as (m²)² − 16². The first pair of squared quantities is therefore m² and 16.
2. Apply the identity to obtain (m² − 16)(m² + 16). This is an intermediate product because its first bracket still contains a difference of squares.
3. Write m² − 16 as m² − 4², giving the further factorisation (m − 4)(m + 4).
4. The final product is (m − 4)(m + 4)(m² + 16). The remaining sum m² + 16 does not use this difference-of-squares identity.

Q7. Factorise 10x² − 11x − 6 by splitting the middle term, where x is a variable. [4 marks]

1. Multiply the leading coefficient by the constant: 10 × (−6) = −60. The required splitting numbers must also have sum −11.
2. Choose −15 and 4 because (−15) × 4 = −60 and −15 + 4 = −11. Split −11x as −15x + 4x.
3. Group to obtain (10x² − 15x) + (4x − 6) = 5x(2x − 3) + 2(2x − 3).
4. Take out the shared bracket 2x − 3. The final product is (2x − 3)(5x + 2).

Q8. For variables a and b, establish the identity a³ − b³ = (a − b)(a² + ab + b²) by multiplication. [3 marks]

1. Multiply a by each term of the second bracket to obtain a³ + a²b + ab². Multiply −b by that bracket to obtain −a²b − ab² − b³.
2. Add the two expressions. The pair a²b and −a²b has sum zero, and the pair ab² and −ab² also has sum zero.
3. The uncancelled terms give a³ − b³. Therefore the proposed product equals the difference of cubes, establishing the identity.

Key takeaways

• Factorisation rewrites an expression as a product; expansion checks it by rebuilding the original terms.
• Look for a common factor before trying identities or splitting a quadratic's middle term.
• Grouping succeeds when the separate groups expose a shared bracket that can be taken outside.
• A difference of squares produces two brackets containing the same quantities with opposite signs.
• The difference-of-cubes trinomial has a positive product term; the sum-of-cubes trinomial has a negative product term.
• For ax² + bx + c, choose splitting numbers with product ac and sum b, retaining their signs.
• More than one factorisation step may be needed, so inspect each resulting factor before stopping.
• Check every coefficient and sign by multiplying the final factors and combining like terms.

Test yourself

What are the factors obtained by factorising 5xy + 10x?

The expression becomes 5x(y + 2), with common factor 5x and remaining bracket y + 2.

Why does 2xy + 2y become 2y(x + 1)?

Both terms contain 2y. Their remaining factors are x and 1, so both belong inside the bracket.

What are the factors of 49p² − 36?

They are 7p − 6 and 7p + 6, obtained from the difference of the squares (7p)² and 6².

What is the factorisation of a³ − b³?

It is (a − b)(a² + ab + b²); the two signs inside the trinomial are positive.

What changes in the factors when a³ − b³ becomes a³ + b³?

The first factor becomes a + b, while the trinomial becomes a² − ab + b².

Why are 6 and 1 unsuitable for splitting the middle term of x² + 5x + 6?

Their product is correctly 6, but their sum is 7 rather than the required middle coefficient 5.

Which two numbers split the middle term of z² − 4z − 12?

Use −6 and 2: their product is −12 and their sum is −4, matching both requirements.

What product and sum are needed for the splitting numbers in 6x² + 7x + 2?

The required product is 6 × 2 = 12 and the required sum is 7, so choose 3 and 4.