Factorisation Using Identities
Factorisation is the process of writing an algebraic expression as a product of its factors. One of the most powerful methods of factorisation uses algebraic identities.
You have already studied algebraic identities like (a + b)², (a − b)², and (a² − b²). In factorisation, you use these identities in reverse — instead of expanding, you recognise patterns and write the expression as a product.
This method is especially useful for factorising quadratic expressions and simplifying complex algebraic expressions. It is much faster than trial and error once you learn to recognise the patterns.
For example, if you see x² + 6x + 9, you can recognise that it matches a² + 2ab + b² with a = x and b = 3. So x² + 6x + 9 = (x + 3)². Similarly, 4x² − 25 matches a² − b² with a = 2x and b = 5, giving (2x + 5)(2x − 5).
In this topic, you will learn to recognise patterns that match the four main identities, factorise expressions systematically using these identities, handle cases where a common factor must be taken out first, apply factorisation to numerical calculations, and verify your answers by expanding the factors back.
What is Factorisation Using Identities?
Definition: Factorisation using identities means recognising that an expression matches the expanded form of a known identity and then rewriting it as the product form of that identity.
Key identities used for factorisation:
Identity 1:
a² + 2ab + b² = (a + b)²
Identity 2:
a² − 2ab + b² = (a − b)²
Identity 3:
a² − b² = (a + b)(a − b)
Identity 4:
x² + (a + b)x + ab = (x + a)(x + b)
Methods
Steps to factorise using identities:
- Look at the expression and identify its form.
- Check if it matches one of the standard identities.
- Identify a and b (or x, a, b) by comparing with the identity.
- Write the factored form using the identity.
- Verify by expanding the factors to confirm.
How to recognise each identity:
Pattern for (a + b)²:
- Expression has three terms.
- First and last terms are perfect squares.
- Middle term = 2 × √(first term) × √(last term).
- Middle term is positive.
Pattern for (a − b)²:
- Same as above, but middle term is negative.
Pattern for a² − b²:
- Expression has exactly two terms.
- Both terms are perfect squares.
- They are connected by a minus sign (difference).
Pattern for (x + a)(x + b):
- Expression is x² + px + q.
- Find two numbers a and b such that a + b = p and a × b = q.
Solved Examples
Example 1: Example 1: Using (a + b)²
Problem: Factorise x² + 6x + 9.
Solution:
Identify the pattern:
- x² = (x)² ✓
- 9 = (3)² ✓
- 6x = 2 × x × 3 = 2ab ✓
This matches a² + 2ab + b² = (a + b)²
- a = x, b = 3
Answer: x² + 6x + 9 = (x + 3)².
Example 2: Example 2: Using (a − b)²
Problem: Factorise 4x² − 20x + 25.
Solution:
Identify the pattern:
- 4x² = (2x)² ✓
- 25 = (5)² ✓
- 20x = 2 × 2x × 5 = 2ab ✓
- Middle term is negative → use (a − b)²
This matches a² − 2ab + b² = (a − b)²
- a = 2x, b = 5
Answer: 4x² − 20x + 25 = (2x − 5)².
Example 3: Example 3: Using a² − b²
Problem: Factorise 9x² − 16.
Solution:
Identify the pattern:
- 9x² = (3x)² ✓
- 16 = (4)² ✓
- Difference of two squares → use a² − b² = (a + b)(a − b)
With a = 3x, b = 4:
- 9x² − 16 = (3x + 4)(3x − 4)
Answer: 9x² − 16 = (3x + 4)(3x − 4).
Example 4: Example 4: Using (x + a)(x + b)
Problem: Factorise x² + 7x + 12.
Solution:
Find two numbers that:
- Add up to 7 (coefficient of x)
- Multiply to give 12 (constant term)
The numbers are 3 and 4:
- 3 + 4 = 7 ✓
- 3 × 4 = 12 ✓
Answer: x² + 7x + 12 = (x + 3)(x + 4).
Example 5: Example 5: Common factor first
Problem: Factorise 3x² − 27.
Solution:
Step 1: Take out common factor:
- 3x² − 27 = 3(x² − 9)
Step 2: Apply a² − b² to (x² − 9):
- x² − 9 = x² − 3² = (x + 3)(x − 3)
Final answer:
- 3x² − 27 = 3(x + 3)(x − 3)
Answer: 3x² − 27 = 3(x + 3)(x − 3).
Example 6: Example 6: With larger coefficients
Problem: Factorise 25a² + 30ab + 9b².
Solution:
Identify the pattern:
- 25a² = (5a)² ✓
- 9b² = (3b)² ✓
- 30ab = 2 × 5a × 3b = 2ab ✓
This matches a² + 2ab + b² = (a + b)²
- a = 5a, b = 3b
Answer: 25a² + 30ab + 9b² = (5a + 3b)².
Example 7: Example 7: Difference of squares with variables
Problem: Factorise 49p² − 64q².
Solution:
Identify:
- 49p² = (7p)² ✓
- 64q² = (8q)² ✓
- Difference of squares
Using a² − b² = (a + b)(a − b):
- a = 7p, b = 8q
Answer: 49p² − 64q² = (7p + 8q)(7p − 8q).
Example 8: Example 8: Negative middle term with (x + a)(x + b)
Problem: Factorise x² − 5x + 6.
Solution:
Find two numbers that:
- Add up to −5
- Multiply to give +6
The numbers are −2 and −3:
- (−2) + (−3) = −5 ✓
- (−2) × (−3) = +6 ✓
Answer: x² − 5x + 6 = (x − 2)(x − 3).
Example 9: Example 9: Numerical factorisation
Problem: Factorise 10201 using the identity (a + b)².
Solution:
Recognise:
- 10201 = 10000 + 200 + 1
- = 100² + 2 × 100 × 1 + 1²
- This matches a² + 2ab + b² with a = 100, b = 1
Therefore:
- 10201 = (100 + 1)² = 101²
Answer: 10201 = 101².
Example 10: Example 10: Numerical difference of squares
Problem: Evaluate 73² − 27² using factorisation.
Solution:
Using a² − b² = (a + b)(a − b):
- a = 73, b = 27
- 73² − 27² = (73 + 27)(73 − 27)
- = 100 × 46
- = 4,600
Answer: 73² − 27² = 4,600.
Real-World Applications
Real-world applications of factorisation using identities:
- Quick calculations: 997² = (1000 − 3)² = 1000000 − 6000 + 9 = 994009. Identities make mental maths faster.
- Simplifying expressions: In physics and engineering, complex expressions are simplified by factorising them using identities.
- Solving equations: Quadratic equations like x² − 9 = 0 are solved by factorising: (x+3)(x−3) = 0, giving x = ±3.
- Area problems: An area expression like x² + 10x + 25 can be factorised as (x+5)², showing it is a perfect square.
- Number theory: Factorisation helps prove divisibility and find factors of large numbers.
- Computer science: Algebraic simplification through factorisation reduces computational steps in algorithms.
- Financial calculations: Compound interest formulas use expansion and factorisation of binomial expressions.
Key Points to Remember
- a² + 2ab + b² = (a + b)² — perfect square (positive middle term).
- a² − 2ab + b² = (a − b)² — perfect square (negative middle term).
- a² − b² = (a + b)(a − b) — difference of two squares.
- x² + (a+b)x + ab = (x + a)(x + b) — sum and product factorisation.
- Always take out the common factor first before applying identities.
- a² + b² CANNOT be factorised using real-number identities.
- To verify: expand the factored form and check it matches the original expression.
- Check the middle term carefully: it must be exactly 2ab for perfect square identities.
- For x² + px + q: find two numbers with sum = p and product = q.
- Factorisation using identities is the reverse of expansion.
Practice Problems
- Factorise: x² + 10x + 25.
- Factorise: 16y² − 24y + 9.
- Factorise: 36a² − 49b².
- Factorise: x² + 9x + 20.
- Factorise: 2x² − 50.
- Factorise: x² − 11x + 30.
- Evaluate 105² − 95² using the difference of squares identity.
- Factorise: 4x² + 12xy + 9y².
Frequently Asked Questions
Q1. What is factorisation using identities?
It means recognising that an expression matches the expanded form of a standard algebraic identity, and then rewriting it as the product (factored) form of that identity.
Q2. Which identities are used for factorisation in Class 8?
Four main identities: (1) a² + 2ab + b² = (a+b)², (2) a² − 2ab + b² = (a−b)², (3) a² − b² = (a+b)(a−b), and (4) x² + (a+b)x + ab = (x+a)(x+b).
Q3. Can a² + b² be factorised?
No. The sum of two squares a² + b² cannot be factorised using real numbers. Only the difference of two squares a² − b² can be factorised.
Q4. How do you check if a trinomial is a perfect square?
Check if (1) the first and last terms are perfect squares, AND (2) the middle term equals ±2 × √(first term) × √(last term). If both conditions are met, it is a perfect square trinomial.
Q5. What should you do before applying identities?
Always check for a common factor first. Take out the common factor, then apply the identity to the remaining expression.
Q6. Is (a − b)² the same as (b − a)²?
Yes. (a − b)² = (b − a)² because squaring removes the sign. However, (a − b) ≠ (b − a); they differ by a negative sign.
Q7. How do you factorise x² + px + q?
Find two numbers a and b such that a + b = p and a × b = q. Then x² + px + q = (x + a)(x + b).
Q8. What if no two numbers satisfy the sum and product conditions?
Then the expression cannot be factorised into binomials with integer coefficients. You may need the quadratic formula or it may be a prime expression.
Q9. How is factorisation the reverse of expansion?
Expansion takes (x + 3)² and gives x² + 6x + 9. Factorisation takes x² + 6x + 9 and gives back (x + 3)². They are inverse operations.
Q10. Can identities be used for numerical calculations?
Yes. For example, 48² = (50 − 2)² = 2500 − 200 + 4 = 2304. And 83² − 17² = (83+17)(83−17) = 100 × 66 = 6600.
