Detailed notes on Algebra for Cambridge Lower Secondary Mathematics, covering key concepts, explanations, examples, and exam-focused revision points.
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Expansion and Factorising — frequently asked questions
The things students keep getting wrong in this sub-topic, answered.
Expansion and Factorising — Cambridge Lower Secondary Maths, Checkpoint
Revise how to expand single and double brackets, and how to factorise expressions by spotting common factors and simple quadratic patterns — building confidence with both directions so your algebra stays tidy heading into your Checkpoint.
At a glance
Expanding removes brackets; factorising puts them back.
To expand a single bracket, multiply every inside term by the outside factor.
A negative outside factor flips the sign of every inside term.
To expand two brackets, use FOIL: First, Outside, Inside, Last.
After expanding, collect like terms — usually the two middle ones.
To factorise by a common factor, take out the largest number and lowest power of each shared letter.
Simple quadratics x2+bx+c factorise to (x+p)(x+q) where p+q=b and pq=c.
Always check by expanding your factorised answer back out.
What you’ll learn
Mapped to the Cambridge Lower Secondary Mathematics curriculum framework.
Expand single brackets and double brackets reliably, including with negatives.
Collect like terms after expanding to give a tidy answer.
Factorise expressions by taking out a common factor.
Factorise simple quadratic expressions of the form x2+bx+c.
Expanding a single bracket
Multiply every term inside the bracket by the factor outside — none gets left behind.
Expanding a bracket means rewriting it without the brackets, by multiplying every term inside by the factor outside. The rule that powers this is the distributive law: a(b+c)=ab+ac.
A few quick examples set the pattern:
Expanding double brackets (FOIL)
Multiply every term in the first bracket by every term in the second — four products in total.
When two brackets sit side by side, like (x+3)(x+5), you have to multiply every term in the first by every term in the second. The friendly memory trick is FOIL:
First terms: x×
Factorising by a common factor
Take out the biggest factor that divides every term — number and letters.
Factorising is the reverse of expanding. Instead of removing brackets, you put them back by spotting what each term has in common.
For an expression like 6x+9, both terms share a factor of 3. Take it out: 6x+9=. The inside of the bracket is what is left after dividing each original term by .
Factorising simple quadratics
x2+bx+c factorises to (x+p)(x, where and .
Checking and tidying your answer
Expand your factorised answer back out — it should match what you started with.
Whether you've expanded or factorised, the single most useful check is to do the opposite operation and see whether you land back where you began.
If you expanded (x+3)(x+5) and got x, factorise it back: two numbers that multiply to and add to — that's and , giving . The loop closes.
Where you'll use this next
Expanding and factorising are the engines of all later algebra.
Confident expansion and factorising open the door to:
Solving quadratic equations — most start with a factorised form, then use the rule "if AB=0 then A=0 or B=0".
Algebraic fractions — common factors cancel from top and bottom only if both are factorised first.
Quick recap
Expanding removes brackets; factorising puts them back.
Multiply every inside term by the outside factor, including its sign.
Use FOIL for double brackets and collect the middle terms.
Factorise by common factor: largest number, lowest power of each letter.
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Where you'll use this next
4(x+
3)=
4x+
12
−3(2y−5)=−6y+15 (the −3 multiplies both inside terms)
x(x+7)=x2+7x (a letter factor uses the index law)
Every inside term is multiplied — none stays untouched.
After expanding, collect like terms if the bigger expression has any. For example 4(x+3)+2(x−5)=4x+12+2x−10=6x+2. Two warnings worth remembering: a minus outside a bracket flips every inside sign, and a missing operation between the factor and the bracket is always multiplication.
Multiply every inside term by the outside factor.
A negative outside factor flips every inside sign.
A letter factor uses the index law: x×x=x2.
Expand first, then collect like terms.
x
=
x2
Outside terms: x×5=5x
Inside terms: 3×x=3x
Last terms: 3×5=15
Add the four products together: x2+5x+3x+15=x2+8x+15. The two middle terms always collect into one.
Four products from the four pairings — then add the like terms.
Watch the signs carefully. For (x−4)(x+6), the four products are x2, +6x, −4x and −24, giving x2+2x−24. The "Last" pair is the constant term and inherits the signs of both bracketed numbers.
Use FOIL — First, Outside, Inside, Last.
Get four products from every double-bracket expansion.
Add the Outside and Inside products into one middle term.
Keep every sign attached to its number throughout.
3(2x+
3)
3
When letters are involved too, take the largest number that divides every coefficient and the lowest power of each letter that appears in every term.
4x2+6x=2x(2x+3) — common factor 2x.
10y3−15y2=5y2(2y−3) — common factor 5y2.
Take out the biggest factor — number and letters — that divides every term.
After factorising, check by expanding your bracket back out. If it returns to the original expression, you've factorised completely. If you can still see another common factor inside the bracket, you have not gone far enough.
Factorising is the reverse of expanding.
Take the largest number that divides every coefficient.
Take the lowest power of each letter that appears in every term.
Check by expanding your answer back out.
+
q)
p+q=b
pq=c
A simple quadratic is an expression of the form x2+bx+c — three terms, with x2 as the highest power and no number in front of the x2. These factorise into a pair of brackets (x+p)(x+q), where two clever numbers p and q must satisfy:
p×q=c (their product is the constant), and
p+q=b (their sum is the coefficient of x).
So to factorise x2+7x+12, look for two numbers that multiply to 12 and add to 7. Try the pairs: 1 and (no, adds to ), and (no, adds to ), and (yes, adds to ). So .
List pairs, work out their sums, and pick the matching row.
Signs matter. If c is positive, both p and q have the same sign as b. If c is negative, p and have , and the bigger one carries the sign of .
x2−5x+6=(x−2)(x−3) — both negative because positive, negative.
As always, check by expanding your factorised answer.
x2+bx+c factorises to (x+p)(x+q).
Find p and q: their product is c and their sum is b.
If c>0, both signs match the sign of b.
If c<0, the signs are opposite; the bigger number takes the sign of b.
2
+
8x+
15
15
8
3
5
(x+3)(x+5)
If you factorised 4x2+6x and got 2x(2x+3), expand: 2x×2x=4x2, 2x×3=6x. Returns to the original — the answer is good.
One last habit. After factorising, scan the inside of the bracket for any further common factor. If you can spot one, factorise again. The aim is fully factorised, not just partly so.
Always check by doing the inverse operation.
If expanding, factorise back to check.
If factorising, expand back to check.
Look for any further common factor in the inside bracket.
Identities and proofs — proving two expressions are equal often comes down to expanding both sides.
Simplifying formulae — many physics, chemistry and finance formulae need factorising before rearranging.
If a later topic feels slow, it is often a missed sign in an expansion or an incomplete factorisation underneath. Come back to this guide whenever you need a steady reset — accuracy beats speed every time.
Quadratic equations start with a factorised form.
Algebraic fractions need factors on top and bottom.
Identities are proved by expanding both sides.
Many real-world formulae factorise before they rearrange.
p
)
(
x
+
q)
p+q=b
pq=c
Watch the signs of b and c to decide the signs of p and q.
Check every answer by doing the inverse operation.
6
9
3
Step 2
Take 3 outside a bracket and divide each original term by 3 to fill the bracket.
6x÷3=2x,9÷3=3
Step 3
Write the factorised form and check by expanding.
3(2x+3)
Answer
3(2x+3)
Step 1
First terms.
x×x=x2
Step 2
Outside and Inside terms.
x×5=5x,3×x=3x
Step 3
Last terms.
3×5=15
Step 4
Collect the two middle terms.
5x+3x=8x
Step 5
Write the tidy answer.
x2+8x+15
Answer
x2+8x+15
Step 1
Find the largest number that divides both coefficients: 2.
Step 2
Both terms contain at least one x, so the lowest power of x in common is x.
Step 3
The common factor is 2x. Divide each term by 2x.
4x2÷2x=2x,6x÷2x=3
Step 4
Write the factorised form.
4x2+6x=2x(2x+3)
Answer
2x(2x+3)
15
Step-by-step solution
Step 1
Look for two numbers whose product is −15 and whose sum is +2.
Step 2
Try pairs that multiply to −15: 1×−15, −1×15, 3×−5, −3×5.
Step 3
Check which pair sums to +2.
−3+5=2✓
Step 4
Write the factorised form using those two numbers.
x2+2x−15=(x+5)(x−
Step 5
Check by expanding: x2−3x+5x−15=x2 ✓
Answer
(x+5)(x−3)
+
12
15
▼
Why it happens
Only the First and Last products of FOIL are remembered; Outside and Inside go missing.
How to avoid it
Always write all four products before tidying: x2+5x+3x+15=x2+8x+15.
The negative is only applied to the first inside term.
How to avoid it
A negative outside flips every inside sign: −2(y−5)=−2y+10.
2
+
3x)
▼
Why it happens
The number factor is taken out but the shared x is missed.
How to avoid it
Check the inside bracket for further common factors. Here x still divides both terms, so the full factorisation is 2x(2x+3).
5)(x+
3)
▼
Why it happens
The two numbers are chosen but their signs are swapped.
How to avoid it
If the constant is negative, the two numbers have opposite signs; the larger one matches the sign of the middle term. Here +5 and −3 give a sum of +2, so the answer is (x+5)(x−3).
Expansion and Factorising — Cambridge Lower Secondary Mathematics — Checkpoint Revision Notes & Practice | Tutopiya
12
13
2
6
8
3
4
7
x2+7x+12=(x+3)(x+4)
q
opposite signs
b
c
b
x2+2x−15=(x+5)(x−3) — opposite signs because c negative; +5 wins because b is positive.