Sec Maths Tuition

Difference of Squares – Sec 2 G3 Question Explained

Source: Factorisation Sec 2 G3

Introduction

This difference of squares question is a key lower secondary algebra skill that students should recognise quickly. Many students try to expand or guess factors, but this type of expression follows a standard pattern. Once you spot the difference of squares form, factorisation becomes fast and systematic.

 

difference of squares Sec 2 G3 explained

 

The Question / Scenario Explanation

Source: Factorisation Sec 2 G3

Question (as shown): Factorise \(a^2 – 64b^2\).

 

Step-by-Step Solution / Explanation

Step 1: Recognise the difference of squares pattern

This expression is:

\(a^2 – 64b^2\)

We can rewrite \(64b^2\) as:

\((8b)^2\)

So the expression becomes:

\(a^2 – (8b)^2\)

This is now clearly in the difference of squares form:

\(x^2 – y^2\)

Step 2: Apply the formula

The standard formula is:

\(x^2 – y^2 = (x+y)(x-y)\)

Here, \(x = a\) and \(y = 8b\).

So:

\(a^2 – (8b)^2 = (a+8b)(a-8b)\)

✅ Final Answer: \((a+8b)(a-8b)\)

Step 3: Quick check

Expand the factors:

\((a+8b)(a-8b)\)

\(= a^2 – 8ab + 8ab – 64b^2\)

\(= a^2 – 64b^2\)

So the factorisation is correct.

 

Key Concepts Students Must Know

  • A difference of squares has the form \(x^2 – y^2\).
  • The factorisation formula is \(x^2 – y^2 = (x+y)(x-y)\).
  • Before factorising, check whether each term is a perfect square.
  • \(64b^2\) is a perfect square because \(64b^2 = (8b)^2\).
  • This pattern only works when there is subtraction, not addition.

 

Exam Tips / Common Mistakes

Exam Tips

  • Always check if both terms are perfect squares first.
  • In a difference of squares question, rewrite the second term clearly if needed, such as \(64b^2 = (8b)^2\).
  • Memorise the formula \(x^2 – y^2 = (x+y)(x-y)\).
  • Do a quick expansion check if you are unsure.

Common Mistakes

  • Writing \((a+64b)(a-64b)\), which is incorrect because \(64b^2 = (8b)^2\), not \((64b)^2\).
  • Trying to factorise it as \((a-8b)^2\), which would expand differently.
  • Forgetting that the pattern is for subtraction, not addition.
  • Not recognising \(64b^2\) as a perfect square term.

 

Parent Insight

This difference of squares question may look simple, but it is an important algebra foundation for secondary Maths. Students who recognise standard factorisation patterns quickly become more confident in algebra and make fewer careless mistakes. With regular practice, these patterns become automatic and save a lot of exam time.

 

Conclusion

To factorise this difference of squares expression, we first rewrote \(64b^2\) as \((8b)^2\). Then we used the formula \(x^2 – y^2 = (x+y)(x-y)\). So the correct factorisation of \(a^2 – 64b^2\) is \((a+8b)(a-8b)\).

 

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Frequently Asked Questions

Because \(64 = 8^2\) and \(b^2\) is already a square term. So \(64b^2\) can be written as \((8b)^2\), which helps us use the difference of squares formula.

 Because \((a-8b)^2\) expands to \(a^2 – 16ab + 64b^2\), which is different from \(a^2 – 64b^2\). The middle term should not appear in this question.

Use it when both terms are perfect squares and there is a subtraction sign between them, such as \(x^2 – y^2\).