MathsICSE Class 8 · Chapter 12

Factorisation ICSE Class 8: Common Factors, Identities and x² + px + q

Updated 15 October 2026 · Fact-checked

Factorisation writes an algebraic expression as a product of its factors, the reverse of expanding. ICSE Class 8 covers: taking out a common factor, a(x + y); perfect squares, (x ± y)²; the difference of two squares, a² − b² = (a − b)(a + b); and trinomials x² + px + q = (x + a)(x + b).

What Factorisation is about

Factorising means writing an expression as a product. It is expansion done backwards:

expanding: $(x + 3) (x + 4) \to x^{2} + 7 x + 12$ ,
factorising: $x^{2} + 7 x + 12 \to (x + 3) (x + 4)$ .

The CISCE Class VIII curriculum asks for simple cases only, of these types:
1. Common factor: $a x + a y = a (x + y)$ .
2. Perfect square: $x^{2} \pm 2 x y + y^{2} = (x \pm y)^{2}$ .
3. Difference of two squares: $a^{2} - b^{2} = (a - b) (a + b)$ .
4. Trinomial: $x^{2} + (a + b) x + ab = (x + a) (x + b)$ .

A good habit is to ask the same questions in the same order every time: Is there a common factor? Is it a difference of squares? Is it a perfect square? Is it a trinomial $x^{2} + p x + q$ ? Keep going until no bracket can be factorised further (factorise completely), and check by expanding.

Weightage: CISCE publishes no question paper, marking scheme or chapter weightage for Class VIII Mathematics: your school sets and marks the exam. Factorisation sits in the Algebra theme of the CISCE Curriculum for Upper Primary Classes (VI-VIII).Source: CISCE Curriculum for Upper Primary Classes (VI-VIII), RDCD 2016

Key concepts

Taking out a common factor

Find the HCF of the coefficients and the lowest power of each common letter. Write it outside the bracket, and divide each term by it to get what goes inside.

The common factor can also be a bracket: $a (x + y) + b (x + y) = (x + y) (a + b)$ .

Always look for a common factor first; it makes everything else easier.

Worked example

Factorise $6 x^{2} y - 9 x y^{2}$ .

Try it first, then show the solutionSolution

HCF of 6 and 9 is 3; common letters: $x$ (lowest power 1) and $y$ (lowest power 1).
Common factor $= 3 x y$ .
$6 x^{2} y \div 3 x y = 2 x$ and $9 x y^{2} \div 3 x y = 3 y$ , so $6 x^{2} y - 9 x y^{2} = 3 x y (2 x - 3 y)$ .

Answer: $3 x y (2 x - 3 y)$

Perfect square trinomials

Recognise the pattern $a^{2} \pm 2 ab + b^{2}$ :

the first and last terms are perfect squares ( $a^{2}$ and $b^{2}$ ),
the middle term is twice the product of their square roots.

Then $a^{2} + 2 ab + b^{2} = (a + b)^{2}$ and $a^{2} - 2 ab + b^{2} = (a - b)^{2}$ . The sign in the bracket matches the sign of the middle term.

Test $9 a^{2} - 12 ab + 4 b^{2}$ : $9 a^{2} = (3 a)^{2}$ , $4 b^{2} = (2 b)^{2}$ , and $2 (3 a) (2 b) = 12 ab$ ✓.

Worked example

Factorise $x^{2} + 10 x + 25$ and $9 a^{2} - 12 ab + 4 b^{2}$ .

Try it first, then show the solutionSolution

$x^{2} + 10 x + 25$ : $x^{2}$ and $5^{2}$ , middle $2 (x) (5) = 10 x$ ✓, so $(x + 5)^{2}$ .
$9 a^{2} - 12 ab + 4 b^{2}$ : $(3 a)^{2}$ and $(2 b)^{2}$ , middle $2 (3 a) (2 b) = 12 ab$ with a minus sign, so $(3 a - 2 b)^{2}$ .

Answer: $(x + 5)^{2}$ ; $(3 a - 2 b)^{2}$

Difference of two squares

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

To use it you need:

exactly two terms,
both perfect squares,
a minus sign between them.

A sum of two squares such as $x^{2} + 9$ cannot be factorised in this way.

Take out any common factor first: $3 x^{2} - 27 = 3 (x^{2} - 9) = 3 (x - 3) (x + 3)$ .

Worked example

Factorise $49 x^{2} - 16 y^{2}$ .

Try it first, then show the solutionSolution

$49 x^{2} = (7 x)^{2}$ and $16 y^{2} = (4 y)^{2}$ .
So $49 x^{2} - 16 y^{2} = (7 x - 4 y) (7 x + 4 y)$ .

Answer: $(7 x - 4 y) (7 x + 4 y)$

Trinomials x² + px + q

Since $(x + a) (x + b) = x^{2} + (a + b) x + ab$ , to factorise $x^{2} + p x + q$ find two numbers $a$ and $b$ with:

product $ab = q$ (the constant term),
sum $a + b = p$ (the coefficient of $x$ ).

Sign guide:

$q$ positive: both numbers have the sign of $p$ .
$q$ negative: the numbers have opposite signs; the larger one (in size) takes the sign of $p$ .

List the factor pairs of $q$ and pick the pair that gives the right sum.

Worked example

Factorise $x^{2} + 7 x + 12$ and $x^{2} - 2 x - 15$ .

Try it first, then show the solutionSolution

$x^{2} + 7 x + 12$ : product 12, sum 7: $3$ and $4$ . So $(x + 3) (x + 4)$ .
$x^{2} - 2 x - 15$ : product $- 15$ , sum $- 2$ : $- 5$ and $3$ . So $(x - 5) (x + 3)$ .
Check: $(x - 5) (x + 3) = x^{2} - 2 x - 15$ ✓.

Answer: $(x + 3) (x + 4)$ ; $(x - 5) (x + 3)$

Factorising completely

An expression is completely factorised when no factor can be factorised again. Often you need two steps:
1. Take out the common factor.
2. Use an identity on what is left.

Sometimes a factor can be broken again: $x^{4} - 81 = (x^{2} - 9) (x^{2} + 9) = (x - 3) (x + 3) (x^{2} + 9)$ . The factor $x^{2} + 9$ is a sum of squares and stays as it is.

Worked example

Factorise completely $2 x^{3} - 8 x$ .

Try it first, then show the solutionSolution

Common factor $2 x$ : $2 x (x^{2} - 4)$ .
$x^{2} - 4$ is a difference of squares: $(x - 2) (x + 2)$ .
So $2 x^{3} - 8 x = 2 x (x - 2) (x + 2)$ .

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

Using factorisation

Factorisation helps to:

divide expressions: $\frac{x ^{2} - 9}{x + 3} = \frac{( x - 3 ) ( x + 3 )}{x + 3} = x - 3$ (for $x \neq = - 3$ );
simplify fractions by cancelling common factors (factors only, never single terms);
calculate quickly: $7 5^{2} - 2 5^{2} = (75 - 25) (75 + 25) = 50 \times 100 = 5000$ .

Worked example

Find $7 5^{2} - 2 5^{2}$ by factorising.

Try it first, then show the solutionSolution

$7 5^{2} - 2 5^{2} = (75 - 25) (75 + 25)$ .
$= 50 \times 100 = 5000$ .

Answer: 5000

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.

Perfect square

Suppose a square photo has an area of $(x^{2} + 6 x + 9)$ cm².

Find its side, and its side when $x = 7$ .

Try it first, then show the solutionSolution

$x^{2} + 6 x + 9 = (x + 3)^{2}$ , so the side is $(x + 3)$ cm.
When $x = 7$ : side $= 10$ cm (area $49 + 42 + 9 = 100$ ✓).

Answer: $(x + 3)$ cm; 10 cm

Memory hook: A square area that is a perfect square trinomial gives the side directly.

Difference of squares

Imagine you cut a 3 cm square out of the corner of a 13 cm square card.

What area of card is left?

Try it first, then show the solutionSolution

Left $= 1 3^{2} - 3^{2} = (13 - 3) (13 + 3)$ .
$= 10 \times 16 = 160$ cm².

Answer: 160 cm²

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

Trinomial factors

Suppose $x^{2} + 7 x + 12$ tiles are laid in a rectangle with $(x + 3)$ rows.

How many tiles are in each row? Check with $x = 5$ .

Try it first, then show the solutionSolution

$x^{2} + 7 x + 12 = (x + 3) (x + 4)$ , so each row has $(x + 4)$ tiles.
$x = 5$ : 8 rows of 9 tiles $= 72$ , and $25 + 35 + 12 = 72$ ✓.

Answer: $(x + 4)$ tiles per row

Memory hook: Factors are the length and breadth of a rectangle of area equal to the expression.

Topics in this chapter

Factorisation formula sheet

Common factor: $a x + a y = a (x + y)$; Always try this first.
Perfect square (plus): $x^{2} + 2 x y + y^{2} = (x + y)^{2}$; Middle term = twice the product.
Perfect square (minus): $x^{2} - 2 x y + y^{2} = (x - y)^{2}$; Last term positive.
Difference of squares: $a^{2} - b^{2} = (a - b) (a + b)$; Two terms, minus sign.
Trinomial: $x^{2} + (a + b) x + ab = (x + a) (x + b)$; Product q, sum p.

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
Common factor	Factorise 12a²b − 18ab² + 6ab, or p(q − r) + s(q − r).	HCF of coefficients and lowest powers of common letters; a common bracket counts too.
Difference of squares	Factorise 25x² − 36y², or 50a² − 2b².	Common factor first, then (a − b)(a + b).
Perfect squares	Factorise 16x² − 24xy + 9y².	Check the first and last are squares and the middle is twice the product.
x² + px + q	Factorise x² + 4x − 21 or x² − 11x + 28.	Two numbers: product q, sum p; mind the signs.
Apply factorisation	Divide or simplify using factors; evaluate 87² − 13².	Factorise top and bottom, cancel factors, or use a² − b².

Tricks to remember Factorisation

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

C-D-P-T order
Common factor, Difference of squares, Perfect square, Trinomial: check in this order every time.
Use it when: Choosing the method.
Product q, sum p
For x² + px + q: 'multiply to q, add to p'.
Use it when: Trinomials.
Minus between squares
Only a minus between two squares splits into (a − b)(a + b). A plus does not split.
Use it when: Avoiding x² + 9 = (x + 3)(x + 3).
Expand to check
Multiply your factors back out; you must get the original expression.
Use it when: Every answer.

Factorisation: exam-level solved questions

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

Example 1: Pick the factors · 1 mark · MCQExam level

$x^{2} - 5 x + 6$ factorises as:
(a) $(x + 2) (x + 3)$
(b) $(x - 2) (x - 3)$
(c) $(x - 1) (x - 6)$
(d) $(x + 1) (x - 6)$

Try it first, then show the solutionSolution

Need product $+ 6$ and sum $- 5$ .
$- 2$ and $- 3$ work.

Answer: (b)

Example 2: Common factor of three terms · 3 marks · Short answerExam level

Factorise $12 a^{2} b - 18 a b^{2} + 6 ab$ .

Try it first, then show the solutionSolution

HCF of 12, 18, 6 is 6; common letters $a$ and $b$ (power 1 each). Common factor $6 ab$ .
Divide each term: $2 a$ , $- 3 b$ , $1$ .
Answer: $6 ab (2 a - 3 b + 1)$ .

Answer: $6 ab (2 a - 3 b + 1)$

Example 3: Difference of squares · 3 marks · Short answerExam level

Factorise $25 x^{2} - 36 y^{2}$ .

Try it first, then show the solutionSolution

$25 x^{2} = (5 x)^{2}$ and $36 y^{2} = (6 y)^{2}$ .
Two squares with a minus sign between them.
$(5 x - 6 y) (5 x + 6 y)$ .

Answer: $(5 x - 6 y) (5 x + 6 y)$

Example 4: Trinomial with a negative constant · 3 marks · Short answerHarder exam level

Factorise $x^{2} + 4 x - 21$ .

Try it first, then show the solutionSolution

Product $- 21$ , sum $+ 4$ : opposite signs, larger positive.
Pairs of 21: $1 \times 21$ , $3 \times 7$ . $7 - 3 = 4$ ✓, so $+ 7$ and $- 3$ .
$(x + 7) (x - 3)$ .

Answer: $(x + 7) (x - 3)$

Example 5: Perfect square and a value · 4 marks · Long answerHarder exam level

Factorise $16 x^{2} - 24 x y + 9 y^{2}$ and hence find its value when $x = 2$ , $y = 1$ .

Try it first, then show the solutionSolution

$16 x^{2} = (4 x)^{2}$ , $9 y^{2} = (3 y)^{2}$ , $2 (4 x) (3 y) = 24 x y$ ✓.
Middle sign minus, so $(4 x - 3 y)^{2}$ .
At $x = 2$ , $y = 1$ : $(8 - 3)^{2}$ .
$= 25$ .

Answer: $(4 x - 3 y)^{2}$ ; value 25

Example 6: Factorise completely · 4 marks · Long answerHarder exam level

Factorise completely (i) $3 a^{3} - 48 a$ (ii) $x^{4} - 16$ .

Try it first, then show the solutionSolution

(i) Common factor: $3 a (a^{2} - 16)$ .
$a^{2} - 16 = (a - 4) (a + 4)$ , so $3 a (a - 4) (a + 4)$ .
(ii) $x^{4} - 16 = (x^{2})^{2} - 4^{2} = (x^{2} - 4) (x^{2} + 4)$ .
$x^{2} - 4 = (x - 2) (x + 2)$ , so $(x - 2) (x + 2) (x^{2} + 4)$ .

Answer: (i) $3 a (a - 4) (a + 4)$ (ii) $(x - 2) (x + 2) (x^{2} + 4)$

Example 7: Simplify by factorising · 5 marks · Long answerHOTS / competency

Simplify $\frac{x ^{2} + 5 x + 6}{x ^{2} - 9} \times \frac{x - 3}{x + 2}$ , where $x \neq = 3, - 3, - 2$ .

Try it first, then show the solutionSolution

$x^{2} + 5 x + 6 = (x + 2) (x + 3)$ .
$x^{2} - 9 = (x - 3) (x + 3)$ .
Expression $= \frac{( x + 2 ) ( x + 3 )}{( x - 3 ) ( x + 3 )} \times \frac{x - 3}{x + 2}$ .
Cancel the common factors $(x + 2)$ , $(x + 3)$ and $(x - 3)$ .
Value $= 1$ (check with $x = 5$ : $\frac{56}{16} \times \frac{2}{7} = 1$ ✓).

Answer: 1

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.

A park with a pond

Suppose a square park has side $a$ metres. A square pond of side $b$ metres is dug inside it, and the rest of the park is to be turfed.

(i) · 3 marks
Write the turfed area as an expression and factorise it.
Answer
Answer: $a^{2} - b^{2} = (a - b) (a + b)$ m²
Turfed area $=$ park $-$ pond $= a^{2} - b^{2} = (a - b) (a + b)$ .
(ii) · 3 marks
Find the turfed area when $a = 45$ and $b = 15$ , using the factorised form.
Answer
Answer: 1800 m²
$(45 - 15) (45 + 15) = 30 \times 60 = 1800$ m².
(iii) · 4 marks
Turf costs ₹12 per m². Find the cost of turfing, and explain why the factorised form made the calculation easier.
Answer
Answer: ₹21,600
Cost $= 1800 \times 12 = ₹21600$ . The factorised form replaced squaring 45 and 15 ( $2025 - 225$ ) with the simple product $30 \times 60$ .

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 $9 x^{2} - 30 x + 25$ . (3 marks)

Step	Marks
$9 x^{2} = (3 x)^{2}$ and $25 = 5^{2}$	1
Middle term check: $2 (3 x) (5) = 30 x$ , sign minus	1
$(3 x - 5)^{2}$	1

Factorise completely $2 x^{3} - 50 x$ . (4 marks)

Step	Marks
Common factor $2 x$	1
$2 x (x^{2} - 25)$	1
$x^{2} - 25 = (x - 5) (x + 5)$	1
Answer: $2 x (x - 5) (x + 5)$	1

Factorise $x^{2} + 2 x - 24$ . Hence divide it by $(x + 6)$ , and check your answer by putting $x = 2$ . (5 marks)

Step	Marks
Product $- 24$ , sum $+ 2$ : numbers $+ 6$ and $- 4$	1
$x^{2} + 2 x - 24 = (x + 6) (x - 4)$	1
Quotient $= x - 4$	1
At $x = 2$ : dividend $= 4 + 4 - 24 = - 16$ ; divisor $= 8$	1
$- 16 \div 8 = - 2 = 2 - 4$ ✓	1

Where students lose marks in Factorisation

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

Mistake: Not factorising completely
A factor that can still be broken earns only part marks.
Wrong working
$2 x^{3} - 8 x = 2 x (x^{2} - 4)$
Right working
$2 x (x - 2) (x + 2)$
Do this: After each step, test every bracket again.
Usually costs: 1 mark.
Mistake: Factorising a sum of squares
$x^{2} + 9$ has no such factors.
Wrong working
$x^{2} + 9 = (x + 3) (x + 3)$
Right working
$(x + 3) (x + 3) = x^{2} + 6 x + 9$ , so $x^{2} + 9$ stays as it is
Do this: Only a difference of squares splits.
Usually costs: The question.
Mistake: Wrong signs in the trinomial
With a negative constant the two numbers must have opposite signs.
Wrong working
$x^{2} - 2 x - 15 = (x - 5) (x - 3)$
Right working
$(x - 5) (x + 3)$
Do this: Check product and sum, including signs.
Usually costs: 1 to 2 marks.
Mistake: Missing the 1 when a whole term is the common factor
$6 ab \div 6 ab = 1$ , not 0.
Wrong working
$12 a^{2} b - 18 a b^{2} + 6 ab = 6 ab (2 a - 3 b)$
Right working
$6 ab (2 a - 3 b + 1)$
Do this: The bracket must have as many terms as the original.
Usually costs: 1 mark.
Mistake: Cancelling terms instead of factors
In $\frac{x ^{2} + 5 x + 6}{x + 2}$ you cannot cancel the $x$ or the 6 separately.
Do this: Factorise first; cancel only whole factors.
Usually costs: The simplification.

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
$4 x^{2} - 1$ factorises as:
(a) $(2 x - 1)^{2}$
(b) $(4 x - 1) (x + 1)$
(c) $(2 x - 1) (2 x + 1)$
(d) $(2 x + 1)^{2}$
Hint
Difference of squares.
Answer
(c)
Q2 · 1 mark · Very short answerExam level
Factorise $7 ab + 21 b$ .
Hint
HCF 7b.
Answer
$7 b (a + 3)$ .
Q3 · 2 marks · Short answerExam level
Factorise $p (q - r) + s (q - r)$ .
Hint
The common factor is a bracket.
Answer
$(q - r) (p + s)$ .
Q4 · 2 marks · Short answerHarder exam level
Factorise $x^{2} - 11 x + 28$ .
Hint
Product 28, sum −11.
Answer
$(x - 4) (x - 7)$ .
Q5 · 1 mark · Assertion-reasonHarder exam level
Assertion (A): $x^{2} + 6 x + 9 = (x + 3)^{2}$ .
Reason (R): It has the form $a^{2} + 2 ab + b^{2}$ with $a = x$ and $b = 3$ .
(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
Check 2ab.
Answer
(c)
Q6 · 3 marks · Short answerHarder exam level
Factorise completely $50 a^{2} - 2 b^{2}$ .
Hint
Common factor 2 first.
Answer
$2 (25 a^{2} - b^{2}) = 2 (5 a - b) (5 a + b)$ .
Q7 · 2 marks · Short answerHarder exam level
Using factorisation, find $8 7^{2} - 1 3^{2}$ .
Hint
$(87 - 13) (87 + 13)$ .
Answer
$74 \times 100 = 7400$ .
Q8 · 2 marks · Short answerHarder exam level
Factorise $x^{2} - 3 x - 40$ .
Hint
Product −40, sum −3.
Answer
$(x - 8) (x + 5)$ .
Q9 · 3 marks · Short answerHarder exam level
Factorise $(a + b)^{2} - (a - b)^{2}$ and simplify the result.
Hint
Treat $(a + b)$ and $(a - b)$ as single terms.
Answer
$[(a + b) - (a - b)] [(a + b) + (a - b)] = (2 b) (2 a) = 4 ab$ .
Q10 · 2 marks · Short answerHarder exam level
Factorise $4 x^{2} + 20 x y + 25 y^{2}$ .
Hint
$(2 x)^{2}$ and $(5 y)^{2}$ .
Answer
$(2 x + 5 y)^{2}$ .
Q11 · 4 marks · Long answerHOTS / competency
Factorise completely $x^{4} - 13 x^{2} + 36$ .
Hint
Treat $x^{2}$ as a single letter first.
Answer
Product 36, sum −13: $- 4$ and $- 9$ . $(x^{2} - 4) (x^{2} - 9) = (x - 2) (x + 2) (x - 3) (x + 3)$ .
Q12 · 4 marks · Long answerHOTS / competency
The area of a rectangle is $(x^{2} + 9 x + 20)$ cm², and one side is $(x + 4)$ cm. Find the other side and the perimeter. Check both when $x = 3$ .
Hint
Factorise the area.
Answer
$x^{2} + 9 x + 20 = (x + 4) (x + 5)$ , so the other side is $(x + 5)$ cm. Perimeter $= 2 (2 x + 9) = (4 x + 18)$ cm. At $x = 3$ : sides 7 and 8, area 56 ( $9 + 27 + 20$ ✓), perimeter 30 ( $12 + 18$ ✓).

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 x² + 2x + 1.Show answer
(x + 1)².
3.Factorise 5x + 15.Show answer
5(x + 3).
4.Factorise x² + 5x + 6.Show answer
(x + 2)(x + 3).
5.Can x² + 4 be factorised in Class 8?Show answer
No: it is a sum of squares.
6.Find 51² − 49².Show answer
200.
7.What two numbers have product −12 and sum 1?Show answer
4 and −3.
8.Factorise 9 − y².Show answer
(3 − y)(3 + y).

Best order to study it

Revise the identities from the previous chapter.
Practise taking out common factors, including common brackets.
Learn the difference of squares and perfect squares.
Practise x² + px + q with all four sign patterns.
Finish with complete factorisation and using factors to simplify and divide.

Quick revision checklist

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

Did you take out the common factor first?
Are both terms perfect squares with a minus between them?
Is the middle term exactly twice the product?
Do your two numbers multiply to q and add to p?
Can any bracket be factorised again?
Did you expand to check?

Factorisation: frequently asked questions

Is factorisation the opposite of expansion?

Yes. Expanding turns a product into a sum of terms; factorising turns the sum back into a product.

Which methods are in the Class 8 syllabus?

Simple cases of four types: a(x + y), (x ± y)², a² − b² and (x + a)(x + b).

How do I know I have factorised completely?

When none of your factors can be factorised again using these four types. For example x² + 4 cannot be broken further.

What if I cannot find two numbers for x² + px + q?

Recheck the factor pairs of q and the signs. Questions at this level are set so that whole-number pairs exist.

Why do we write x ≠ −3 when simplifying fractions?

Because the factor we cancel, (x + 3), would be 0 at x = −3, and dividing by 0 is not allowed.

Based on: CISCE Curriculum for Upper Primary Classes (VI-VIII), Mathematics (Class VIII); CISCE RDCD: Curriculum Pre-School to Class VIII