MathsICSE Class 9 · Chapter 4

Factorisation ICSE Class 9: Difference of Squares, Cubes and Splitting the Middle Term

Updated 15 October 2026 · Fact-checked

Factorisation writes an expression as a product of factors; it reverses expansion. ICSE Class 9 uses a² − b² = (a − b)(a + b), a³ ± b³ = (a ± b)(a² ∓ ab + b²), and splitting the middle term of ax² + bx + c into two parts whose product is ac, after taking out common factors.

What Factorisation is about

Factorising an expression means writing it as a product of factors, each of which cannot be broken down further. It is expansion run backwards: since $(x + 2) (x + 3) = x^{2} + 5 x + 6$ , the factors of $x^{2} + 5 x + 6$ are $(x + 2)$ and $(x + 3)$ .

The CISCE 2028 Class IX syllabus names three forms: the difference of two squares $a^{2} - b^{2}$ , the sum and difference of two cubes $a^{3} \pm b^{3}$ , and the quadratic trinomial $a x^{2} + b x + c$ by splitting the middle term. Taking out a common factor and grouping, from earlier classes, are the first steps in almost every question.

In the school exam you will see 2 to 3-mark 'factorise completely' questions, longer ones that combine two methods (grouping plus difference of squares, or a substitution followed by splitting), and quick arithmetic such as $100 1^{2} - 99 9^{2}$ . Factorisation is also the main tool for quadratic equations in Class X, so accuracy here pays later.

Weightage: CISCE publishes no chapter weightage and no question-paper layout for ICSE Class 9 Mathematics. Class IX is examined by your school: one 80-mark paper of 3 hours plus 20 marks of internal assessment. Factorisation is in the 3. Algebra (ii) part of the 2028 syllabus.Source: CISCE ICSE 2028 Mathematics syllabus, Class IX

Key concepts

Common factors and grouping

Always look first for a factor common to every term: $6 x^{2} y - 9 x y^{2} = 3 x y (2 x - 3 y)$ .

If there is no common factor to all terms, group terms in pairs so each pair has a common factor, and the brackets that remain are the same:

$a x + a y + b x + b y = a (x + y) + b (x + y) = (x + y) (a + b)$ .

If the brackets do not match, try a different grouping or change the sign you take out.

Worked example

Factorise $x^{2} - x z + x y - y z$ .

Try it first, then show the solutionSolution

Group: $(x^{2} - x z) + (x y - y z)$ .
$= x (x - z) + y (x - z)$ .
$= (x - z) (x + y)$ .

Answer: $(x - z) (x + y)$

Difference of two squares

$a^{2} - b^{2} = (a - b) (a + b)$ .

Recognise each term as a perfect square, with a minus sign between them. The squares can be expressions: $(x + y)^{2} - z^{2} = (x + y - z) (x + y + z)$ .

Keep going until no factor can be split further: $x^{4} - 16 = (x^{2} - 4) (x^{2} + 4) = (x - 2) (x + 2) (x^{2} + 4)$ . A sum of two squares such as $x^{2} + 4$ does not factorise over real numbers in this course.

It also speeds up arithmetic: $100 1^{2} - 99 9^{2} = (1001 - 999) (1001 + 999) = 2 \times 2000 = 4000$ .

Worked example

Factorise $9 (a + b)^{2} - 4 (a - b)^{2}$ .

Try it first, then show the solutionSolution

$= [3 (a + b)]^{2} - [2 (a - b)]^{2}$ .
$= [3 a + 3 b - 2 a + 2 b] [3 a + 3 b + 2 a - 2 b]$ .
$= (a + 5 b) (5 a + b)$ .

Answer: $(a + 5 b) (5 a + b)$

Perfect-square trinomials and differences of squares together

A trinomial of the form $a^{2} \pm 2 ab + b^{2}$ is a perfect square $(a \pm b)^{2}$ . When such a trinomial appears with another square subtracted, combine them:

$x^{2} + 2 x y + y^{2} - z^{2} = (x + y)^{2} - z^{2} = (x + y - z) (x + y + z)$ .

Look for three terms that form a perfect square and a fourth (or a group) that is a square being subtracted.

Worked example

Factorise $4 a^{2} - 4 ab + b^{2} - 9 c^{2}$ .

Try it first, then show the solutionSolution

$4 a^{2} - 4 ab + b^{2} = (2 a - b)^{2}$ .
Expression $= (2 a - b)^{2} - (3 c)^{2}$ .
$= (2 a - b - 3 c) (2 a - b + 3 c)$ .

Answer: $(2 a - b - 3 c) (2 a - b + 3 c)$

Sum and difference of two cubes

$a^{3} + b^{3} = (a + b) (a^{2} - ab + b^{2})$

$a^{3} - b^{3} = (a - b) (a^{2} + ab + b^{2})$

Sign pattern: the binomial factor has the same sign as the expression; the middle term of the trinomial has the opposite sign; the last term is always $+$ . (Remember 'SOAP': Same, Opposite, Always Positive.)

The trinomial factor $a^{2} \mp ab + b^{2}$ does not factorise further in this course.

Write each term as a cube first: $27 x^{3} = (3 x)^{3}$ , $64 = 4^{3}$ , $\frac{1}{8} = (\frac{1}{2})^{3}$ .

Worked example

Factorise $27 x^{3} - 64$ .

Try it first, then show the solutionSolution

$= (3 x)^{3} - 4^{3}$ .
$= (3 x - 4) [(3 x)^{2} + (3 x) (4) + 4^{2}]$ .
$= (3 x - 4) (9 x^{2} + 12 x + 16)$ .

Answer: $(3 x - 4) (9 x^{2} + 12 x + 16)$

Splitting the middle term of ax² + bx + c

To factorise $a x^{2} + b x + c$ :
1. Find the product $a c$ .
2. Find two numbers $p$ and $q$ with $p \times q = a c$ and $p + q = b$ .
3. Write $b x = p x + q x$ .
4. Group in pairs and take out common factors.

Sign guide: if $a c > 0$ , $p$ and $q$ have the same sign as $b$ ; if $a c < 0$ , they have opposite signs and the larger (in size) has the sign of $b$ .

Check by expanding the answer.

Worked example

Factorise $6 x^{2} + 7 x - 3$ .

Try it first, then show the solutionSolution

$a c = 6 \times (- 3) = - 18$ ; need $p + q = 7$ : $p = 9$ , $q = - 2$ .
$6 x^{2} + 9 x - 2 x - 3$ .
$= 3 x (2 x + 3) - 1 (2 x + 3)$ .
$= (2 x + 3) (3 x - 1)$ .

Answer: $(2 x + 3) (3 x - 1)$

Substitution for repeated expressions

When the same bracket appears more than once, replace it by a single letter, factorise, then put it back.

For example, in $(x^{2} - 3 x)^{2} - 8 (x^{2} - 3 x) - 20$ , let $y = x^{2} - 3 x$ . Then $y^{2} - 8 y - 20 = (y - 10) (y + 2)$ , and each bracket may factorise again.

This turns long-looking questions into ordinary splitting-the-middle-term questions.

Worked example

Factorise $(2 x + 3)^{2} - 5 (2 x + 3) - 14$ .

Try it first, then show the solutionSolution

Let $y = 2 x + 3$ : $y^{2} - 5 y - 14 = (y - 7) (y + 2)$ .
Replace $y$ : $(2 x + 3 - 7) (2 x + 3 + 2) = (2 x - 4) (2 x + 5)$ .
Take out 2: $2 (x - 2) (2 x + 5)$ .

Answer: $2 (x - 2) (2 x + 5)$

Maths in real life: Factorisation

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Difference of squares as an area

Suppose a square photo of side 43 cm is mounted at the centre of a square board of side 57 cm, leaving a border of board showing all round.

Find the area of the visible board border without squaring 57 and 43 separately.

Try it first, then show the solutionSolution

Border area $= 5 7^{2} - 4 3^{2} = (57 - 43) (57 + 43)$ .
$= 14 \times 100 = 1400$ cm².

Answer: 1,400 cm².

Memory hook: Big square minus small square = (difference) × (sum).

Splitting the middle term to find dimensions

Imagine a rectangular plot whose area is given by $6 x^{2} + 7 x - 3$ square metres, where the sides are linear expressions in $x$ with whole-number coefficients.

Find expressions for the sides, and the actual sides when $x = 5$ .

Try it first, then show the solutionSolution

$6 x^{2} + 7 x - 3 = (2 x + 3) (3 x - 1)$ .
At $x = 5$ : sides $13$ m and $14$ m (check: $13 \times 14 = 182 = 150 + 35 - 3$ ).

Answer: (2x + 3) m and (3x − 1) m; 13 m and 14 m.

Memory hook: Area expression factorises into length × breadth.

Difference of cubes as volume

Suppose a solid cube of edge 5 cm has a cube of edge 3 cm cut out of one corner.

Find the remaining volume using $a^{3} - b^{3}$ .

Try it first, then show the solutionSolution

$5^{3} - 3^{3} = (5 - 3) (25 + 15 + 9)$ .
$= 2 \times 49 = 98$ cm³.

Answer: 98 cm³.

Memory hook: Cube minus cube: (a − b) times the 'friendly trinomial'.

Topics in this chapter

Factorisation formula sheet

Difference of squares: $a^{2} - b^{2} = (a - b) (a + b)$; a² + b² does not factorise here.
Sum of cubes: $a^{3} + b^{3} = (a + b) (a^{2} - ab + b^{2})$; SOAP: same, opposite, always positive.
Difference of cubes: $a^{3} - b^{3} = (a - b) (a^{2} + ab + b^{2})$; Trinomial factor is final.
Perfect square trinomial: $a^{2} \pm 2 ab + b^{2} = (a \pm b)^{2}$; Check the middle term is exactly 2ab.
Splitting the middle term: $a x^{2} + b x + c : p + q = b, pq = a c$; Then group in pairs.

How the school exam asks Factorisation

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Difference of squares	Factorise 49x² − 81y², x⁴ − 16, or (a + b)² − 4c².	Write both terms as squares; factorise; keep going while a factor is still a difference of squares.
Sum or difference of cubes	Factorise 8a³ + 27b³, x³ − 1/x³ or a³ − b³ − a + b.	Write each term as a cube; apply SOAP; group if extra terms appear.
Splitting the middle term	Factorise 3x² − 10x − 8, 12x² − 7xy − 12y².	Find p, q with pq = ac and p + q = b; group; check by expanding.
Combined methods	Perfect square minus a square, grouping then cubes, or x⁶ − y⁶.	Look for a perfect-square group or a cube pattern; factorise in stages.
Substitution	Repeated brackets such as (x² − 3x)² − 8(x² − 3x) − 20.	Replace the bracket by y, factorise, substitute back and factorise again.
Arithmetic shortcuts	Evaluate 1001² − 999² or 57² − 43² using factorisation.	(a − b)(a + b) turns hard squares into easy products.

Tricks to remember Factorisation

Short memory aids. Say each one aloud once, then use it on a question the same day.

SOAP for cubes
Same sign, Opposite sign, Always Positive: a³ − b³ = (a − b)(a² + ab + b²).
Use it when: Signs in cube factorisation.
Product–sum hunt
For ax² + bx + c: multiply the ends (ac), add to the middle (b).
Use it when: Splitting the middle term.
Common factor first
CF first, then pattern: like taking off a lid before looking inside.
Use it when: Every factorisation question.
Squares don't add
Minus between squares: factorise. Plus between squares: stop.
Use it when: Knowing when a factor is final.

Factorisation: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Which is a factor? · 1 mark · MCQExam level

One factor of $x^{2} - x - 42$ is:
(a) $x - 6$
(b) $x + 7$
(c) $x + 6$
(d) $x - 42$

Try it first, then show the solutionSolution

Need $pq = - 42$ , $p + q = - 1$ : $- 7$ and $6$ .
$x^{2} - x - 42 = (x - 7) (x + 6)$ .

Answer: (c)

Example 2: Factorise completely · 2 marks · Short answerExam level

Factorise completely: $3 x^{3} - 48 x$ .

Try it first, then show the solutionSolution

Common factor: $3 x (x^{2} - 16)$ .
Difference of squares: $3 x (x - 4) (x + 4)$ .

Answer: $3 x (x - 4) (x + 4)$

Example 3: Two-variable trinomial · 3 marks · Short answerExam level

Factorise $12 x^{2} - 7 x y - 12 y^{2}$ .

Try it first, then show the solutionSolution

$a c = 12 \times (- 12) = - 144$ ; $p + q = - 7$ : $p = - 16$ , $q = 9$ .
$12 x^{2} - 16 x y + 9 x y - 12 y^{2} = 4 x (3 x - 4 y) + 3 y (3 x - 4 y)$ .
$= (3 x - 4 y) (4 x + 3 y)$ .

Answer: $(3 x - 4 y) (4 x + 3 y)$

Example 4: Cubes plus grouping · 3 marks · Short answerHarder exam level

Factorise $a^{3} - b^{3} - a + b$ .

Try it first, then show the solutionSolution

Group: $(a^{3} - b^{3}) - (a - b)$ .
$= (a - b) (a^{2} + ab + b^{2}) - (a - b)$ .
$= (a - b) (a^{2} + ab + b^{2} - 1)$ .

Answer: $(a - b) (a^{2} + ab + b^{2} - 1)$

Example 5: Sixth powers · 4 marks · Long answerHarder exam level

Factorise $x^{6} - y^{6}$ completely.

Try it first, then show the solutionSolution

Treat as a difference of squares: $(x^{3})^{2} - (y^{3})^{2} = (x^{3} - y^{3}) (x^{3} + y^{3})$ .
$x^{3} - y^{3} = (x - y) (x^{2} + x y + y^{2})$ .
$x^{3} + y^{3} = (x + y) (x^{2} - x y + y^{2})$ .
$x^{6} - y^{6} = (x - y) (x + y) (x^{2} + x y + y^{2}) (x^{2} - x y + y^{2})$ .

Answer: $(x - y) (x + y) (x^{2} + x y + y^{2}) (x^{2} - x y + y^{2})$

Example 6: Substitution twice · 4 marks · Long answerHarder exam level

Factorise $(x^{2} - 3 x)^{2} - 8 (x^{2} - 3 x) - 20$ .

Try it first, then show the solutionSolution

Let $y = x^{2} - 3 x$ : $y^{2} - 8 y - 20 = (y - 10) (y + 2)$ .
$y - 10 = x^{2} - 3 x - 10 = (x - 5) (x + 2)$ .
$y + 2 = x^{2} - 3 x + 2 = (x - 1) (x - 2)$ .
Answer: $(x - 5) (x + 2) (x - 1) (x - 2)$ .

Answer: $(x - 5) (x + 2) (x - 1) (x - 2)$

Example 7: Adding and subtracting a term · 4 marks · Long answerHOTS / competency

Factorise $a^{4} + a^{2} b^{2} + b^{4}$ .

Try it first, then show the solutionSolution

Add and subtract $a^{2} b^{2}$ : $a^{4} + 2 a^{2} b^{2} + b^{4} - a^{2} b^{2}$ .
$= (a^{2} + b^{2})^{2} - (ab)^{2}$ .
$= (a^{2} + b^{2} - ab) (a^{2} + b^{2} + ab)$ .

Answer: $(a^{2} - ab + b^{2}) (a^{2} + ab + b^{2})$

Case-study practice (Section B style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 10-mark question split into 3 + 3 + 4 marks.

Designing a storage box

A carpenter designs open boxes from square sheets. The volume of one design is $V = 4 x^{3} - 9 x$ cm³, and the area of the base of another is $2 x^{2} - 7 x + 6$ cm². For a display, a large cube of edge $(x + 2)$ cm is placed beside a small cube of edge $(x - 2)$ cm.

(i) · 3 marks
Factorise $4 x^{3} - 9 x$ completely.
Answer
Answer: $x (2 x - 3) (2 x + 3)$
$x (4 x^{2} - 9) = x (2 x - 3) (2 x + 3)$ .
(ii) · 3 marks
Factorise $2 x^{2} - 7 x + 6$ and find the base sides when $x = 7$ .
Answer
Answer: $(2 x - 3) (x - 2)$ ; 11 cm and 5 cm
$a c = 12$ , $p + q = - 7$ : $- 4, - 3$ . $2 x^{2} - 4 x - 3 x + 6 = 2 x (x - 2) - 3 (x - 2) = (x - 2) (2 x - 3)$ . At $x = 7$ : 5 and 11.
(iii) · 4 marks
Show that the difference of the volumes of the two cubes is $4 (3 x^{2} + 4)$ using $a^{3} - b^{3}$ .
Answer
Answer: $12 x^{2} + 16$
$a = x + 2$ , $b = x - 2$ ; $a - b = 4$ . $a^{2} + ab + b^{2} = (x^{2} + 4 x + 4) + (x^{2} - 4) + (x^{2} - 4 x + 4) = 3 x^{2} + 4$ . Difference $= 4 (3 x^{2} + 4)$ .

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Factorise $2 x^{2} - 7 x + 6$ . (3 marks)

Step	Marks
$a c = 12$ ; numbers $- 4$ and $- 3$ (sum $- 7$ )	1
$2 x^{2} - 4 x - 3 x + 6 = 2 x (x - 2) - 3 (x - 2)$	1
$(x - 2) (2 x - 3)$	1

Factorise $8 a^{3} - 27 b^{3} - 2 a + 3 b$ . (4 marks)

Step	Marks
Group: $(8 a^{3} - 27 b^{3}) - (2 a - 3 b)$	1
$8 a^{3} - 27 b^{3} = (2 a - 3 b) (4 a^{2} + 6 ab + 9 b^{2})$	1
Take out $(2 a - 3 b)$	1
$(2 a - 3 b) (4 a^{2} + 6 ab + 9 b^{2} - 1)$	1

Factorise completely: $x^{6} - 64$ . (5 marks)

Step	Marks
$(x^{3})^{2} - 8^{2} = (x^{3} - 8) (x^{3} + 8)$	1
$x^{3} - 8 = (x - 2) (x^{2} + 2 x + 4)$	1
$x^{3} + 8 = (x + 2) (x^{2} - 2 x + 4)$	1
Combine all four factors	1
$(x - 2) (x + 2) (x^{2} + 2 x + 4) (x^{2} - 2 x + 4)$	1

Where students lose marks in Factorisation

The slips examiners see most often, and what to write instead.

Mistake: Not factorising completely
Stopping at $(x^{2} - 4) (x^{2} + 4)$ leaves $x^{2} - 4$ , which still factorises.
Wrong working
$x^{4} - 16 = (x^{2} - 4) (x^{2} + 4)$
Right working
$x^{4} - 16 = (x - 2) (x + 2) (x^{2} + 4)$
Do this: After each step, test every factor again for a common factor or a pattern.
Usually costs: 1 mark.
Mistake: Missing the common factor
Starting with patterns before taking out the HCF gives messy or wrong factors.
Do this: Write 'HCF = …' as the first line of every factorisation.
Usually costs: 1 mark.
Mistake: Wrong signs in cube factors
Writing $a^{3} - b^{3} = (a - b) (a^{2} - ab + b^{2})$ .
Wrong working
$x^{3} - 8 = (x - 2) (x^{2} - 2 x + 4)$
Right working
$x^{3} - 8 = (x - 2) (x^{2} + 2 x + 4)$
Do this: SOAP: Same, Opposite, Always Positive. Check by expanding.
Usually costs: 1 to 2 marks.
Mistake: Splitting with numbers whose product is c, not ac
For $6 x^{2} + 7 x - 3$ , students look for numbers multiplying to $- 3$ ; this only works when $a = 1$ .
Do this: Always compute $a c$ first and write it down.
Usually costs: The whole question.
Mistake: Losing a sign while taking out a negative
In $2 x (x - 2) - 3 (x - 2)$ students write $- 3 (x + 2)$ , so the brackets do not match.
Do this: When you take out a negative factor, change the sign of every term inside the bracket, then check the brackets match.
Usually costs: 1 mark and the final answer.

Practice questions on Factorisation

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQExam level
The factors of $a^{3} + 8$ are:
(a) $(a + 2) (a^{2} + 2 a + 4)$
(b) $(a + 2) (a^{2} - 2 a + 4)$
(c) $(a - 2) (a^{2} + 2 a + 4)$
(d) $(a + 2)^{3}$
Hint
SOAP.
Answer
(b)
Q2 · 2 marks · Short answerExam level
Factorise $49 x^{2} - 81 y^{2}$ and $x^{2} + 5 x - 36$ .
Hint
Squares; pq = −36, p + q = 5.
Answer
$(7 x - 9 y) (7 x + 9 y)$ ; $(x + 9) (x - 4)$ .
Q3 · 2 marks · Short answerExam level
Using factorisation, find $100 1^{2} - 99 9^{2}$ .
Hint
(a − b)(a + b).
Answer
$2 \times 2000 = 4000$ .
Q4 · 1 mark · Assertion-reasonHarder exam level
Assertion (A): $x^{2} + 4$ cannot be written as a product of two real linear factors.
Reason (R): $x^{2} + 4 > 0$ for every real $x$ , so it is never zero.
(a) (A) is true and (R) is false.
(b) (A) is false and (R) is true.
(c) Both (A) and (R) are true, and (R) is the correct explanation for (A).
(d) Both (A) and (R) are true, but (R) is not the correct explanation for (A).
Hint
A real linear factor (x − k) would make the expression zero at x = k.
Answer
(c). If $x^{2} + 4 = (x - k) (x - m)$ with real $k$ , it would be 0 at $x = k$ ; but $x^{2} + 4 \geq 4$ always.
Q5 · 2 marks · Short answerExam level
Factorise $3 x^{2} - 10 x - 8$ .
Hint
ac = −24.
Answer
$3 x^{2} - 12 x + 2 x - 8 = (x - 4) (3 x + 2)$ .
Q6 · 3 marks · Short answerHarder exam level
Factorise $x^{2} - y^{2} - 2 y - 1$ .
Hint
Group the last three terms.
Answer
$x^{2} - (y + 1)^{2} = (x - y - 1) (x + y + 1)$ .
Q7 · 3 marks · Short answerHarder exam level
Factorise $a^{3} - \frac{1}{a ^{3}}$ and hence find its value when $a - \frac{1}{a} = 2$ .
Hint
(a − 1/a)(a² + 1 + 1/a²).
Answer
$(a - \frac{1}{a}) (a^{2} + 1 + \frac{1}{a ^{2}})$ . $a^{2} + \frac{1}{a ^{2}} = 4 + 2 = 6$ , so value $= 2 \times 7 = 14$ .
Q8 · 3 marks · Short answerHarder exam level
Factorise $3 x^{2} + 11 x + 63$ .
Hint
ac = √3 × 6√3 = 18.
Answer
$p = 9$ , $q = 2$ : $3 x^{2} + 9 x + 2 x + 63 = 3 x (x + 33) + 2 (x + 33) = (x + 33) (3 x + 2)$ .
Q9 · 4 marks · Long answerHarder exam level
Factorise $(a + b)^{3} - 8 (a - b)^{3}$ .
Hint
Let p = a + b, q = 2(a − b).
Answer
$(p - q) (p^{2} + pq + q^{2})$ with $p - q = - a + 3 b$ , $p^{2} = a^{2} + 2 ab + b^{2}$ , $pq = 2 a^{2} - 2 b^{2}$ , $q^{2} = 4 a^{2} - 8 ab + 4 b^{2}$ . Result $(3 b - a) (7 a^{2} - 6 ab + 3 b^{2})$ .
Q10 · 4 marks · Long answerHOTS / competency
Factorise $x^{4} + 4 y^{4}$ by adding and subtracting a suitable term.
Hint
Add and subtract 4x²y².
Answer
$(x^{2} + 2 y^{2})^{2} - (2 x y)^{2} = (x^{2} - 2 x y + 2 y^{2}) (x^{2} + 2 x y + 2 y^{2})$ .
Q11 · 3 marks · Short answerHarder exam level
Factorise $4 x^{2} - 9 y^{2} - 2 x - 3 y$ .
Hint
Difference of squares, then a common bracket.
Answer
$(2 x - 3 y) (2 x + 3 y) - (2 x + 3 y) = (2 x + 3 y) (2 x - 3 y - 1)$ .
Q12 · 4 marks · Long answerHOTS / competency
Without expanding, show that $(x + 1) (x + 2) (x + 3) (x + 4) + 1$ is a perfect square, and find its square root.
Hint
Pair (x + 1)(x + 4) and (x + 2)(x + 3).
Answer
$(x^{2} + 5 x + 4) (x^{2} + 5 x + 6) + 1$ . Let $y = x^{2} + 5 x$ : $(y + 4) (y + 6) + 1 = y^{2} + 10 y + 25 = (y + 5)^{2}$ . Square root $= x^{2} + 5 x + 5$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Factorise a² − b².Show answer
(a − b)(a + b).
2.Factorise a³ + b³.Show answer
(a + b)(a² − ab + b²).
3.Factorise a³ − b³.Show answer
(a − b)(a² + ab + b²).
4.For 3x² + 10x + 8, what are pq and p + q?Show answer
pq = 24, p + q = 10 (p = 6, q = 4).
5.Does x² + 9 factorise in this course?Show answer
No; a sum of two squares does not factorise over real numbers here.
6.What is the first step in any factorisation?Show answer
Take out the highest common factor.
7.Evaluate 51² − 49².Show answer
2 × 100 = 200.
8.Factorise x² + 2xy + y² − 9.Show answer
(x + y − 3)(x + y + 3).

Best order to study it

Revise common factors and grouping.
Practise the difference of two squares, including repeated use.
Learn a³ ± b³ with the SOAP sign rule.
Practise splitting the middle term with positive and negative ac.
Finish with combined methods and substitution questions.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Do you take out the HCF first?
Can you spot a difference of squares hidden in brackets?
Do you use SOAP for a³ ± b³?
Do you find ac before splitting the middle term?
Can you combine a perfect square with a difference of squares?
Do you substitute for repeated brackets?
Do you check every factor for further factorisation?

Factorisation: frequently asked questions

Is the factor theorem in the Class IX syllabus?

No. The Class IX syllabus covers a² − b², a³ ± b³ and splitting the middle term. The factor and remainder theorems come in Class X.

What if I cannot find p and q for splitting the middle term?

Recheck ac and its factor pairs. If no pair adds to b, the trinomial does not factorise with whole numbers, which is rare in exam questions.

Can a² − ab + b² be factorised further?

Not in this course. Leave it as the final trinomial factor.

How do I check my factorisation?

Expand the factors, or substitute a simple value like x = 1 into both the original expression and your answer.

Do I need to factorise expressions with surds?

Occasionally, as a harder question (for example √3x² + 11x + 6√3). Splitting the middle term works the same way.

Based on: CISCE ICSE 2028 Mathematics (51) syllabus (Classes IX and X); CISCE Regulations for ICSE Examination Year 2028