Expand:
3(x+4).
We get:
3x+12.
Factorisation runs that relationship backwards.
Factorising rewrites an expression from an additive form into a product by exposing multiplicative structure that was already present.
So:
3x+12 = 3(x+4).
The value has not changed.
Only the representation has.
The quick answer: look for multiplicative structure
- Check for a common numerical factor.
- Check for a common variable factor.
- Factor out the greatest useful common factor.
- If no simple common factor exists, inspect special identities or quadratic structure.
- Expand the result to verify that the original expression returns.
Common-factor factorisation
Factorise:
6x+9.
Both terms contain factor 3.
So:
6x+9 = 3(2x+3).
Check by expansion:
3×2x=6x.
3×3=9.
Factor out variables as well as numbers
Factorise:
8x²+12x.
Numerical HCF of 8 and 12 is 4.
Both terms also contain x.
Greatest common factor:
4x.
So:
8x²+12x = 4x(2x+3).
A common factor must divide every term completely. Factorising is multiplicative division applied across the entire additive structure.
Negative common factors can improve structure
Factorise:
−6x+12.
One valid form is:
6(−x+2).
Another is:
−6(x−2).
The second form may be cleaner because the leading term inside the bracket is positive.
Worked example: several variables
Factorise:
15a²b−20ab².
Common numerical factor:
5.
Common variable factors:
a and b.
So:
5ab(3a−4b).
Difference of squares
The identity:
a²−b²=(a+b)(a−b)
comes from expansion:
(a+b)(a−b)=a²−ab+ab−b²=a²−b².
Example:
x²−25.
Recognise:
x²−5².
Factorise:
(x+5)(x−5).
Not every subtraction is a difference of squares
x²−5 is not a difference of two obvious perfect squares over integer coefficients.
The pattern requires both terms to be squares in the number system being used.
Perfect-square identities
From expansion:
(a+b)²=a²+2ab+b².
Therefore:
a²+2ab+b²=(a+b)².
Similarly:
a²−2ab+b²=(a−b)².
Worked example: recognise a perfect square
Factorise:
x²+10x+25.
x² is x².
25 is 5².
Middle term:
2×x×5 = 10x.
So:
(x+5)².
Simple monic quadratics
Factorise:
x²+7x+12.
We seek two numbers whose:
- product is 12;
- sum is 7.
3 and 4 work.
So:
(x+3)(x+4).
Why?
Expanding gives:
x²+4x+3x+12=x²+7x+12.
Quadratics with a negative constant
Factorise:
x²+x−12.
Need product −12 and sum 1.
4 and −3 work.
So:
(x+4)(x−3).
The sign pair is constrained by both sum and product.
Quadratics with both negative middle and positive constant
Factorise:
x²−7x+12.
Need product +12 and sum −7.
−3 and −4 work.
Result:
(x−3)(x−4).
Factor out a common factor before using a quadratic pattern
Factorise:
2x²+14x+24.
First factor out 2:
2(x²+7x+12).
Then:
x²+7x+12=(x+3)(x+4).
Final:
2(x+3)(x+4).
Before using a more complicated factorisation method, always ask whether a simple common factor can reduce the expression first.
Factorisation by grouping
Consider:
ax+ay+bx+by.
Group:
a(x+y)+b(x+y).
Now x+y is common:
(a+b)(x+y).
Grouping creates a shared factor that was not initially visible as one single term factor.
Worked grouping example
Factorise:
3x+6+2xy+4y.
Group:
3(x+2)+2y(x+2).
Factor common bracket:
(x+2)(3+2y).
Factorisation can solve equations
Suppose:
x²−5x+6=0.
Factorise:
(x−2)(x−3)=0.
Using the zero-product principle:
- x−2=0 → x=2;
- x−3=0 → x=3.
The product form exposes the roots.
Factorisation can expose forbidden values
Consider:
(x²−9)/(x−3).
Factor numerator:
(x−3)(x+3)/(x−3).
For x≠3, this simplifies to x+3.
But x=3 was forbidden in the original expression because the denominator became zero.
Factorisation can simplify the form without erasing the original domain restriction.
Expansion is the strongest routine check
If you claim:
x²−x−12=(x−4)(x+3),
expand:
x²+3x−4x−12=x²−x−12.
The original returns exactly.
This is a structural inverse check.
Substitution can catch sign errors quickly
Suppose a learner claims:
x²−7x+12=(x+3)(x+4).
Test x=0.
- original=12;
- proposed factors=12.
This does not expose the error yet.
Test x=1:
- original=6;
- proposed=20.
The sign error becomes obvious.
Common misconception 1: factorising means dividing every term by something and stopping
The factor must be written outside a bracket so expansion reproduces the original.
Common misconception 2: only numerical common factors count
Variables and powers can also be common factors.
Common misconception 3: x²−25=(x−5)²
Difference of squares gives:
(x+5)(x−5).
Common misconception 4: use sum/product pairs without checking signs
Both the middle coefficient and constant determine the sign pair.
Common misconception 5: simplification removes original domain restrictions
Cancelling a factor does not make a previously forbidden denominator value valid.
A factorisation diagnostic ladder
- Can the learner identify the greatest common numerical factor?
- Can the learner identify common variable factors?
- Can the learner factor a negative common factor deliberately?
- Can the learner recognise difference of squares?
- Can the learner recognise perfect-square identities?
- Can the learner factor simple monic quadratics?
- Can the learner factor out a common factor before a quadratic step?
- Can the learner use grouping?
- Can the learner connect factorisation to solving equations?
- Can the learner preserve domain restrictions in algebraic fractions?
- Can the learner verify by expansion?
How this fits Secondary Mathematics
Factorisation connects arithmetic factors, the distributive law, algebraic identities, quadratics, equations and algebraic fractions. Its depth grows as expressions become more structured, but the central move remains the same: expose a product hidden inside an additive form.
Exact G2 and G3 factorisation requirements should be checked against the relevant current SEAB syllabus.
The deeper lesson: factorisation changes what the structure reveals
x²−5x+6 hides its roots in expanded form.
(x−2)(x−3) exposes them.
Factorisation is powerful because it does more than shorten an expression. It changes the representation so common structure, roots, cancellation opportunities and identities become visible.