Expansion and factorisation are reverse ways of representing the same algebraic relationship. Expansion exposes individual terms. Factorisation exposes multiplicative structure. Neither form is automatically better; the useful form depends on what the problem asks you to see next.
This thirty-second Secondary 4 Mathematics Learning Guide develops expansion, factorisation, identities and equivalent forms as one algebraic-control system. It belongs to the Secondary Mathematics Hub and S1–S4 Capability Map.
The guide focuses on collecting like terms, expanding brackets, extracting common factors, factorising quadratic expressions, recognising identities, preserving restrictions and choosing the representation that makes the next operation easiest.
Equivalent expressions look different but represent the same value
The expressions 3(x+4) and 3x+12 are equivalent for every x. The first highlights a common factor 3; the second displays the separate terms.
Algebraic manipulation should change form without changing value, except where domain restrictions or equation-solving steps introduce additional conditions.
Change the form. Preserve the relationship.
Collect like terms before doing unnecessary work
Like terms have the same variable structure. 5x and −2x are like terms. 3x² and 7x² are like terms. But x and x² are not.
Worked Example 1 | Simplify structurally
Simplify 4x²+3x−7+5x²−8x+2.
Collect x² terms, x terms and constants:
9x²−5x−5.
Single-bracket expansion is distribution
For a(b+c), multiply a by every term inside the bracket:
a(b+c)=ab+ac.
If the outside factor is negative, the sign distributes too.
Worked Example 2 | Negative factor
Expand −3(2x−5).
−3(2x−5)=−6x+15.
The second term is positive because negative 3 multiplied by negative 5 gives positive 15.
Double brackets create four products
For (a+b)(c+d), each term in the first bracket multiplies each term in the second.
Worked Example 3 | Expand two binomials
Expand (x+4)(x−7).
x·x=x², x·(−7)=−7x, 4·x=4x, 4·(−7)=−28.
x²−3x−28.
The middle terms combine only after all products are written correctly.
Special products are identities, not magic patterns
- (a+b)²=a²+2ab+b²;
- (a−b)²=a²−2ab+b²;
- (a+b)(a−b)=a²−b².
These can always be verified by ordinary expansion. Recognising them saves time, but understanding the underlying products prevents sign errors.
Worked Example 4 | Difference of two squares
Expand (3x+5)(3x−5).
(3x)²−5²=9x²−25.
The cross terms cancel because they are equal in magnitude and opposite in sign.
Factorisation reverses expansion
If expansion turns 5(x+3) into 5x+15, factorisation turns 5x+15 back into 5(x+3).
The first factorisation check should always be: is there a common factor shared by every term?
Worked Example 5 | Extract the highest common factor
Factorise 12x²y−18xy².
The numerical HCF is 6. Both terms contain x and y.
12x²y−18xy²=6xy(2x−3y).
Expand to verify: 6xy·2x=12x²y and 6xy·(−3y)=−18xy².
Factorising x²+bx+c
For a monic quadratic x²+bx+c, seek two numbers whose product is c and whose sum is b.
Worked Example 6 | Monic quadratic
Factorise x²+7x+12.
We need two numbers with product 12 and sum 7: 3 and 4.
(x+3)(x+4).
Signs can be predicted before trial
If c is positive, the two factor constants have the same sign. If b is positive, both are positive; if b is negative, both are negative. If c is negative, the two signs differ.
Worked Example 7 | Negative constant
Factorise x²−5x−24.
We need product −24 and sum −5: −8 and 3.
(x−8)(x+3).
Factorising ax²+bx+c
When the coefficient of x² is not 1, one route is to seek two binomial factors directly and verify by expansion. Another route is to split the middle term using numbers whose product is ac and sum is b, then factor by grouping.
Worked Example 8 | Non-monic quadratic
Factorise 6x²+11x+3.
ac=18. We need two numbers with product 18 and sum 11: 9 and 2.
6x²+9x+2x+3
=3x(2x+3)+1(2x+3).
(3x+1)(2x+3).
Difference of squares factorisation
If an expression is A²−B², then:
A²−B²=(A+B)(A−B).
Worked Example 9 | Recognise structure before expanding
Factorise 49x²−16.
(7x+4)(7x−4).
No middle term appears because the cross products cancel under expansion.
Perfect-square trinomials
An expression such as x²+10x+25 matches (x+5)². The middle coefficient 10 is twice the product x·5 in symbolic form.
Worked Example 10 | Perfect square
Factorise 9x²−24x+16.
9x²=(3x)² and 16=4². The middle term −24x is −2(3x)(4).
(3x−4)².
Why factorisation helps solve equations
If a product equals zero, at least one factor must be zero. This zero-product principle turns a quadratic equation into simpler linear equations.
Worked Example 11 | From factorisation to roots
Solve x²−5x−24=0.
Factorise:
(x−8)(x+3)=0.
Therefore:
x=8 or x=−3.
The separate Quadratic Equations and Algebraic Fractions guide develops equation solving further.
Equivalent forms reveal different features
| Form | What it often reveals |
|---|---|
| x²−5x−24 | Expanded coefficients |
| (x−8)(x+3) | Roots 8 and −3 |
| 3(x+4) | Common factor 3 |
| 3x+12 | Separate additive terms |
Good algebra often means choosing the form that exposes the next useful feature.
Substitution is a strong equivalence check
If you think two expressions are equivalent, substitute a few convenient values. This does not prove an identity by itself, but it can quickly detect an error.
Worked Example 12 | Catch a false expansion
A learner claims (x+2)²=x²+4. Test x=1.
Left side=(1+2)²=9.
Claimed right side=1²+4=5.
The claim is false. Correct expansion:
(x+2)²=x²+4x+4.
Factor before cancelling algebraic fractions
Cancellation applies to common factors, not to terms separated by addition or subtraction.
For example:
(x²−9)/(x−3) = (x−3)(x+3)/(x−3)=x+3, for x≠3.
The restriction x≠3 remains because the original denominator was zero there.
Worked Example 13 | Simplify by factorisation
Simplify (x²+5x+6)/(x+2).
Factor numerator:
x²+5x+6=(x+2)(x+3).
Expression=x+3, with original restriction x≠−2.
Common failure modes
| Error | Cause | Repair |
|---|---|---|
| −3(2x−5)=−6x−15 | Negative sign not distributed correctly | Multiply outside factor by every term |
| (x+4)(x−7)=x²−28 | Middle products omitted | Write all four products |
| x²+7x+12=(x+2)(x+6) | Product checked but sum not checked | Verify both product and middle coefficient |
| 49x²−16=(7x−4)² | Difference of squares confused with perfect square | Expand proposed factorisation to check |
| Cancels x from (x+3)/x | Terms confused with factors | Cancel only multiplicative common factors |
| Drops domain restriction after cancellation | Simplified form treated as original domain | Record excluded denominator values before simplification |
Independent practice
- Simplify 7x²−4x+3+2x²+9x−8.
- Expand −4(3x−2).
- Expand (x−5)(x+6).
- Factorise 15x²−20x.
- Factorise x²+9x+20.
- Factorise 2x²+7x+3.
- Factorise 25x²−9.
- Simplify (x²−4)/(x−2), stating the restriction.
Explained answers
1. 9x²+5x−5.
2. −12x+8.
3. x²+x−30.
4. 5x(3x−4).
5. Product 20, sum 9: (x+4)(x+5).
6. 2x²+6x+x+3=(2x+1)(x+3).
7. (5x+3)(5x−3).
8. (x−2)(x+2)/(x−2)=x+2, with x≠2.
Teaching sequence: expand to see terms, factorise to see structure
Begin with distribution and like terms. Then move into double brackets and special identities, always verifying by ordinary multiplication.
Next reverse the process through common factors and quadratic factorisation. Finish by asking learners to choose between expanded and factorised forms depending on whether the next task is substitution, solving, simplification or structural comparison.
Connect this guide to Algebraic Formulae, nth-Term Patterns and Change of Subject, Quadratic Equations and Algebraic Fractions, and Algebra, Functions and Graphs Under Mixed-Topic Conditions.
Final thought
Algebraic fluency is not the ability to expand everything. It is the ability to move between equivalent forms without losing meaning and to choose the form that reveals the structure needed next.
Expand when you need the terms. Factorise when you need the structure. Verify that both forms still describe the same mathematics.
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