Secondary Mathematics · Worked Repair Guide 42
Expansion and factorisation are inverse ways of viewing the same algebraic structure. Expansion turns products into sums. Factorisation turns sums into products. Identities sit between them: statements that remain true for every allowed value of the variables.
This guide develops one central habit: expand to verify and factorise to expose structure. A proposed factorisation is not trustworthy because the brackets look familiar. Expanding the answer should reproduce every term of the original expression, including signs and coefficients.
All examples below are original teaching constructions. This page concentrates on algebraic identity and factorisation transfer, complementing Signed Numbers, Brackets and Algebraic Structure and Quadratic Equations, Factorisation and Roots.
1. An identity is true for every allowed input
The statement (x+3)²=x²+6x+9 is an identity. It is true for every real x.
By contrast, x²=9 is an equation that is true only for x=3 or x=−3.
Entry check: substituting x=2 into the identity gives25=25; substituting x=−5 gives4=4.
One numerical check can catch an error, but an identity is established algebraically, not by testing a few values.
2. The distributive law is the engine underneath expansion
a(b+c)=ab+ac.
For 3x(2x−5), distribute 3x to both terms: 6x²−15x.
For −2y(3y+4), the result is −6y²−8y.
A frequent error is to multiply the first term but not the second, or to lose the negative sign on distribution.
3. Double brackets require every term to meet every term
(x+4)(x+7)=x²+7x+4x+28=x²+11x+28.
(2x−3)(x+5)=2x²+10x−3x−15=2x²+7x−15.
The middle term is a sum of two cross-products.
A grid or area model is useful when sign errors recur because it forces all four products to be recorded.
4. Square identities contain a doubled cross-term
(a+b)²=a²+2ab+b².
This follows from (a+b)(a+b): the cross-products ab and ba add to2ab.
Therefore (x+5)²=x²+10x+25, not x²+25.
Likewise (2x+3)²=4x²+12x+9.
5. The negative square identity still has a positive final square
(a−b)²=a²−2ab+b².
The final b² is positive because (−b)(−b)=b².
Example: (x−6)²=x²−12x+36.
Example: (3x−2)²=9x²−12x+4.
6. Difference of squares removes the middle terms
(a+b)(a−b)=a²−b².
The cross-products −ab and +ab cancel.
Therefore x²−49=(x−7)(x+7).
And 9x²−16=(3x−4)(3x+4).
7. Sum of squares is not the same factorisation over the reals
x²+49 is not a difference of squares.
Over the real numbers, it does not factorise into two real linear factors.
The visual similarity between plus and minus signs is not enough to apply the identity.
Always check the operation between the squares.
8. Common factors should usually be extracted first
6x²−18x=6x(x−3).
12a³b−8a²b²=4a²b(3a−2b).
Extracting the greatest useful common factor simplifies what remains inside the brackets.
Check by multiplying the factor back through the bracket.
9. A negative common factor can make the bracket easier to read
−3x²+12x=−3x(x−4).
One could also write3x(−x+4), but the first form leaves a positive leading x term inside the bracket.
Factorisation is not unique in appearance, but equivalent factored forms should expand to the same expression.
The sign of the extracted factor controls every sign inside.
10. Factorising x²+bx+c reverses the expansion pattern
To factorise x²+9x+20, find two numbers with sum9 and product20:4 and5.
Thus x²+9x+20=(x+4)(x+5).
For x²−x−20, seek sum−1 and product−20:4 and−5.
Thus x²−x−20=(x+4)(x−5).
11. Leading coefficients change the factor search
For 2x²+7x+3, one factorisation is(2x+1)(x+3).
Expand to check:2x²+6x+x+3=2x²+7x+3.
For 3x²−10x−8, (3x+2)(x−4)=3x²−12x+2x−8=3x²−10x−8.
The cross-products must combine to the required middle coefficient.
12. Perfect-square trinomials have a specific middle term
x²+14x+49=(x+7)² because49=7² and14x=2·x·7.
4x²−20x+25=(2x−5)² because the middle term is−2(2x)(5)=−20x.
Matching only the first and last squares is insufficient.
The middle coefficient confirms whether the square identity really applies.
13. Grouping can reveal a repeated bracket
Factorise ax+ay+bx+by.
Group: a(x+y)+b(x+y).
Then factor the repeated bracket: (a+b)(x+y).
Grouping is useful when no single common factor spans every term but pairs share structure.
14. Rearrangement can expose a hidden identity
Consider x²+6x+9−y².
The first three terms form(x+3)², giving(x+3)²−y².
Now use difference of squares:
(x+3−y)(x+3+y).
Multi-stage factorisation often means recognising one structure inside another.
15. Identities can simplify numerical calculation
99²=(100−1)²=10000−200+1=9801.
103×97=(100+3)(100−3)=10000−9=9991.
The identities are algebraic, but they can compress arithmetic.
The same structure works with variables because numbers and symbols obey the same distributive laws.
16. Factorisation can prove divisibility statements
For any integer n, n²−n=n(n−1).
The product of two consecutive integers is even, so n²−n is always even.
Likewise n²−1=(n−1)(n+1), the product of two integers two apart.
Factorisation exposes arithmetic structure that is difficult to see in expanded form.
17. Factorisation can solve equations only after one side is zero
x²−9=0 becomes(x−3)(x+3)=0, so x=3 or−3.
But(x−3)(x+3)=8 does not allow setting each factor equal to8.
The zero-product property is the bridge from factorisation to roots.
This is why rearranging an equation to zero is often the first solving step.
18. Capstone: factor a layered expression completely
Factorise 4x²−12x+9−25y².
The first three terms form(2x−3)².
Thus the expression is(2x−3)²−(5y)².
Difference of squares gives:
(2x−3−5y)(2x−3+5y).
Expanding these factors reproduces the original expression, completing the verification.
19. Independent practice
- Expand3x(2x−5).
- Expand−2y(3y+4).
- Expand(x+4)(x+7).
- Expand(2x−3)(x+5).
- Expand(x+5)².
- Expand(3x−2)².
- Factorise x²−49.
- Factorise9x²−16.
- Factorise6x²−18x.
- Factorise12a³b−8a²b².
- Factorise x²+9x+20.
- Factorise x²−x−20.
- Factorise2x²+7x+3.
- Factorise3x²−10x−8.
- Factorise x²+14x+49.
- Factorise4x²−20x+25.
- Factorise ax+ay+bx+by.
- Factorise x²+6x+9−y² completely.
- Use an identity to calculate103×97.
- Factorise4x²−12x+9−25y² completely.
20. Worked answers
1. 6x²−15x.
2. −6y²−8y.
3. x²+11x+28.
4. 2x²+7x−15.
5. x²+10x+25.
6. 9x²−12x+4.
7. (x−7)(x+7).
8. (3x−4)(3x+4).
9. 6x(x−3).
10. 4a²b(3a−2b).
11. (x+4)(x+5).
12. (x+4)(x−5).
13. (2x+1)(x+3).
14. (3x+2)(x−4).
15. (x+7)².
16. (2x−5)².
17. (a+b)(x+y).
18. (x+3−y)(x+3+y).
19. 9991. (100+3)(100−3)=10000−9.
20. (2x−3−5y)(2x−3+5y).
21. Diagnose factorisation errors by expanding the answer
Common failures include dropping a cross-term in a square, treating a sum of squares as a difference, forgetting a common factor, choosing factor pairs whose products work but whose cross-products do not, or stopping before a nested difference of squares has been fully factorised.
A useful correction note says “expand to verify”, “square has two cross-products”, “common factor first”, or “check the middle coefficient”.
Then present the same algebra in expanded and factored form and ask which representation makes the next task easier. Factorisation is learned more deeply when the learner knows why the form is useful.
22. Continue through the BTT learning routes
Return to the BTT Mathematics Hub or BTT Mathematical Lab. Use Signed Numbers, Brackets and Algebraic Structure, Quadratic Equations, Factorisation and Roots and Algebraic Fractions, Formulae and Substitution for neighbouring algebra.
Within Batch11, continue to Congruence Tests, Triangle Correspondence and Geometric Proof, Fractional Equations, Denominator Restrictions and Quadratic Reduction, or Pyramids, Prisms, Nets, Slant Heights and Mensuration.
23. Sources and scope
The identities, examples and practice questions are original teaching material. The algebra follows the distributive law and standard real polynomial factorisation.
For current Singapore Secondary curriculum and assessment scope, consult the relevant MOE/SEAB syllabus for the learner’s subject level and year.