MathsICSE Class 8 · Factorisation

Factorisation: Common Factors, Perfect Squares and Difference of Squares (ICSE Class 8)

Updated 17 October 2026 · Fact-checked

Factorising writes an expression as a product. First take out the highest common factor; then look for $a^2 \pm 2ab + b^2 = (a \pm b)^2$ or $a^2 - b^2 = (a - b)(a + b)$. Keep factorising until no factor can be split further.

Take out the HCF first, then spot the identity, and keep splitting until no factor can break further; multiply back to check.

Illustration drawn for QuestionBuddy.

Understand Common Factors and Identity-Based Factorisation

The CISCE curriculum asks for factorisation in simple cases, of the types $a (x + y)$ , $(x \pm y)^{2}$ , $a^{2} - b^{2}$ and $(x + a) (x + b)$ . This page handles the first three types; the quadratic $x^{2} + p x + q$ type has its own page.

Step 1: highest common factor (HCF). Find the HCF of the numerical coefficients and the lowest power of each common letter: $6 x^{2} y - 9 x y^{2} = 3 x y (2 x - 3 y)$ .

Step 2: look for an identity.

Perfect square trinomial: $a^{2} + 2 ab + b^{2} = (a + b)^{2}$ and $a^{2} - 2 ab + b^{2} = (a - b)^{2}$ . Test: first and last terms are perfect squares, and the middle is twice the product of their roots. $4 x^{2} + 20 x + 25$ : roots $2 x$ and $5$ , twice product $= 20 x$ , so $(2 x + 5)^{2}$ .
Difference of two squares: $a^{2} - b^{2} = (a - b) (a + b)$ . A sum of squares $a^{2} + b^{2}$ does not factorise over the numbers used here.

Step 3: factorise completely. After removing the HCF, check each remaining factor again: $18 x^{3} - 50 x = 2 x (9 x^{2} - 25) = 2 x (3 x - 5) (3 x + 5)$ . And $x^{4} - 81 = (x^{2} - 9) (x^{2} + 9) = (x - 3) (x + 3) (x^{2} + 9)$ .

Using factors in arithmetic. $100 1^{2} - 99 9^{2} = (1001 - 999) (1001 + 999) = 2 \times 2000 = 4000$ .

Checking. Multiply the factors back; you must get the original expression.

Common Factors and Identity-Based Factorisation: formulas to remember

Perfect square trinomial: $a^{2} \pm 2 ab + b^{2} = (a \pm b)^{2}$; Middle term is twice the product of the roots.
Difference of squares: $a^{2} - b^{2} = (a - b) (a + b)$; Applies repeatedly, e.g. $x^{4} - 81$ .
Common factor: $ab + a c = a (b + c)$; Take the HCF out first.

How to solve Common Factors and Identity-Based Factorisation questions

Common factor first, identity second, and keep going until nothing splits.

1Find the HCF of all terms and take it out.
2Count the terms left: two terms suggest $a^{2} - b^{2}$ ; three suggest a perfect square.
3Check the square test (first, last, twice the product).
4Factorise each bracket again if it still splits.
5Multiply back to verify.

Common Factors and Identity-Based Factorisation in real life

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.

Quick differences of squares

Suppose a square tile floor of side 75 units is placed next to a smaller square of side 25 units, and you want the difference in their areas.

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 square units.

Memory hook: Difference of squares: subtract the roots, add the roots.

Tricks to remember Common Factors and Identity-Based Factorisation

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

HCF before identity
Pull out the common number first; the leftover often becomes an identity.
Use it when: Expressions like $18 x^{3} - 50 x$ .
Square test
Middle term = 2 x (root of first) x (root of last).
Use it when: Spotting perfect squares.

Common Factors and Identity-Based Factorisation: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Common factor · 1 mark · MCQBoard level

Factorise $6 x^{2} y - 9 x y^{2}$ .
(a) $3 x y (2 x + 3 y)$ (b) $3 x y (2 x - 3 y)$ (c) $x y (6 x + 9 y)$ (d) $3 x (2 x y - 3 y)$

Try it first, then show the solutionSolution

HCF $= 3 x y$ . Dividing: $6 x^{2} y \div 3 x y = 2 x$ and $9 x y^{2} \div 3 x y = 3 y$ , with the minus sign kept.

Answer: (b)

Example 2: Perfect square · 3 marks · Short answerBoard level

Factorise $4 x^{2} + 20 x + 25$ .

Try it first, then show the solutionSolution

$4 x^{2} = (2 x)^{2}$ and $25 = 5^{2}$ .
Middle term: $2 \times 2 x \times 5 = 20 x$ , matches.
So $4 x^{2} + 20 x + 25 = (2 x + 5)^{2}$ .

Answer: $(2 x + 5)^{2}$

Example 3: HCF then difference of squares · 3 marks · Short answerHarder board level

Factorise completely $18 x^{3} - 50 x$ .

Try it first, then show the solutionSolution

HCF $= 2 x$ : $2 x (9 x^{2} - 25)$ .
$9 x^{2} - 25 = (3 x)^{2} - 5^{2} = (3 x - 5) (3 x + 5)$ .
$18 x^{3} - 50 x = 2 x (3 x - 5) (3 x + 5)$ .

Answer: $2 x (3 x - 5) (3 x + 5)$

Example 4: Repeated difference of squares · 4 marks · Short answerHarder board level

Factorise completely $x^{4} - 81$ .

Try it first, then show the solutionSolution

$x^{4} - 81 = (x^{2})^{2} - 9^{2} = (x^{2} - 9) (x^{2} + 9)$ .
$x^{2} - 9 = (x - 3) (x + 3)$ ; $x^{2} + 9$ does not factorise further.
$x^{4} - 81 = (x - 3) (x + 3) (x^{2} + 9)$ .
Check at $x = 2$ : LHS $= 16 - 81 = - 65$ ; RHS $= (- 1) (5) (13) = - 65$ .

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

Example 5: A divisibility result · 4 marks · Long answerHOTS / competency

Prove that the difference of the squares of two consecutive odd numbers is a multiple of 8, and check with 11 and 9.

Try it first, then show the solutionSolution

Let the odd numbers be $2 k - 1$ and $2 k + 1$ .
$(2 k + 1)^{2} - (2 k - 1)^{2} = [(2 k + 1) - (2 k - 1)] [(2 k + 1) + (2 k - 1)] = 2 \times 4 k = 8 k$ .
For $11$ and $9$ : $2 k + 1 = 11$ gives $k = 5$ ; $121 - 81 = 40 = 8 \times 5$ .

Answer: $8 k$ , a multiple of 8; $1 1^{2} - 9^{2} = 40 = 8 \times 5$ .

Where students lose marks in Common Factors and Identity-Based Factorisation

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

Mistake: Stopping before factorising completely
A factor like $(x^{2} - 9)$ still splits.
Wrong working
$x^{4} - 81 = (x^{2} - 9) (x^{2} + 9)$ .
Right working
$(x - 3) (x + 3) (x^{2} + 9)$ .
Do this: Inspect every bracket again.
Usually costs: The last mark.
Mistake: Treating a sum of squares as a difference
$a^{2} + b^{2}$ has no such factors.
Wrong working
$x^{2} + 9 = (x + 3) (x + 3)$ .
Right working
$(x + 3)^{2} = x^{2} + 6 x + 9$ ; $x^{2} + 9$ does not factorise.
Do this: Multiply back to test.
Usually costs: Answer mark.

Exam tips

Always take the HCF first, even a numerical one.
Verify by multiplying your factors.
Write the final answer as a product, never as a sum.

Practice questions on Common Factors and Identity-Based 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 · MCQBoard level
$x^{2} - 49$ factorises as:
(a) $(x - 7) (x + 7)$ (b) $(x - 7)^{2}$ (c) $(x + 7)^{2}$ (d) $x (x - 49)$
Hint
Difference of squares.
Answer
(a)
Q2 · 2 marks · Short answerBoard level
Factorise $15 a^{2} b - 25 a b^{2}$ .
Hint
HCF is 5ab.
Answer
$5 ab (3 a - 5 b)$ .
Q3 · 3 marks · Short answerHarder board level
Factorise $9 x^{2} - 30 x + 25$ .
Hint
Check the middle term.
Answer
$(3 x)^{2} - 2 (3 x) (5) + 5^{2} = (3 x - 5)^{2}$ .
Q4 · 3 marks · Short answerHarder board level
Factorise completely $3 a^{4} - 48$ .
Hint
Take out 3, then a^4 - 16.
Answer
$3 (a^{4} - 16) = 3 (a^{2} - 4) (a^{2} + 4) = 3 (a - 2) (a + 2) (a^{2} + 4)$ .
Q5 · 4 marks · Long answerHOTS / competency
If $x^{2} - y^{2} = 45$ and $x - y = 5$ , find $x + y$ , then $x$ , $y$ and $x^{2} + y^{2}$ .
Hint
Factorise the left side.
Answer
$(x - y) (x + y) = 45$ , so $x + y = 9$ . Adding: $2 x = 14$ , $x = 7$ ; $y = 2$ . $x^{2} + y^{2} = 49 + 4 = 53$ .

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 a^2 - b^2.Show answer
(a - b)(a + b).
2.Factorise x^2 + 6x + 9.Show answer
(x + 3)^2.
3.Does x^2 + 9 factorise?Show answer
No.
4.What is the HCF of 12x^2y and 18xy^2?Show answer
6xy.

Common Factors and Identity-Based Factorisation: frequently asked questions

Why take the HCF first?

It simplifies the remaining expression and often reveals an identity.

What does factorise completely mean?

No factor can be split further.