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Cambridge 0580 IGCSE  ·  Topic C2.2

C2.2 Algebraic Manipulation: Simplifying, Expanding and Factorising

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01 — Explanation

How Do I Simplify by Collecting Like Terms?

Like terms contain exactly the same letter(s) raised to the same power — 3x3x and −5x-5x are like terms, but 3x3x and 3x23x^2 are not, no matter how similar they look. Add or subtract the coefficients of like terms only, leaving everything else untouched:

2a+3b+5a−9b=(2a+5a)+(3b−9b)=7a−6b2a + 3b + 5a - 9b = (2a+5a) + (3b-9b) = 7a - 6b

A useful habit is to underline or circle each family of like terms in a different way before combining them — it stops terms with different letters accidentally getting merged together, especially in longer expressions with three or four different letters mixed in.

02 — Explanation

How Do I Expand a Single Bracket?

Multiply the term outside the bracket by every term inside it:

a(b+c)=ab+ace.g. 3x(2x−4y)=6x2−12xya(b+c) = ab+ac \qquad \text{e.g. } 3x(2x-4y) = 6x^2-12xy

Exam Tip

When the term outside is negative, e.g. −3(x−5)-3(x-5), imagine the minus sign is "attached" to the 33: (−3)(x)+(−3)(−5)=−3x+15(-3)(x) + (-3)(-5) = -3x + 15. A negative times a negative gives a positive — this single step accounts for more lost marks than any other in this topic.

03 — Explanation

How Do I Expand Two Brackets Multiplied Together?

Multiply every term in the first bracket by every term in the second — four multiplications in total (FOIL: First, Outer, Inner, Last), then collect the two middle (like) terms:

(x+p)(x+q)=x2⏟First+qx⏟Outer+px⏟Inner+pq⏟Last(x+p)(x+q) = \underbrace{x^2}_{\text{First}} + \underbrace{qx}_{\text{Outer}} + \underbrace{px}_{\text{Inner}} + \underbrace{pq}_{\text{Last}}
(x+4)(x−7)=x2−7x+4x−28=x2−3x−28(x+4)(x-7) = x^2 -7x+4x-28 = x^2-3x-28

If both brackets contain a minus sign, work through each pair of terms carefully rather than trying to do it in your head — (x−3)(x−6)=x2−6x−3x+18=x2−9x+18(x-3)(x-6) = x^2-6x-3x+18 = x^2-9x+18, where two negatives multiplying together (giving +18+18) is easy to get wrong under time pressure.

04 — Explanation

How Do I Factorise an Expression Fully?

Factorising reverses expanding: pull the Highest Common Factor (HCF) of every term — numbers and letters — out to the front, leaving the rest in a bracket.

9x2+15xy=3x(3x+5y)(HCF of 9x2 and 15xy is 3x)9x^2 + 15xy = 3x(3x+5y) \qquad (\text{HCF of } 9x^2 \text{ and } 15xy \text{ is } 3x)

"Factorise fully" is the key instruction: check the bracket afterwards for anything — a number or a letter — that could still be taken out. This also works when every term is negative, or when the common factor itself is negative — factorising −4x2−8x-4x^2-8x as −4x(x+2)-4x(x+2) is often more useful in later working than leaving a leading negative x2x^2 term.

05 — Explanation

Exam-Style Question 1

Factorise fully 12a2b+18ab212a^2b + 18ab^2.

Solution

  1. The HCF of the numbers 1212 and 1818 is 66.
  2. Every term contains at least one aa and one bb, so the full HCF is 6ab6ab.
  3. Divide each term by 6ab6ab: 12a2b6ab=2a\dfrac{12a^2b}{6ab}=2a and 18ab26ab=3b\dfrac{18ab^2}{6ab}=3b.
  4. Write the factorised form:
6ab(2a+3b)6ab(2a+3b)

Check by expanding back: 6ab(2a)=12a2b6ab(2a) = 12a^2b and 6ab(3b)=18ab26ab(3b) = 18ab^2 ✓ — this quick check catches almost every factorising error before you move on.

Exam Tip

If you can still spot a common factor sitting inside your bracket (e.g. you stopped at 6(2a2b+3ab2)6(2a^2b+3ab^2), missing the shared aa), the answer isn't finished — "fully" means nothing can be taken out any further.

06 — Explanation

Exam-Style Question 2

Expand and simplify 4(2x−1)−3(x−5)4(2x-1) - 3(x-5).

Solution

  1. Expand each bracket separately, being careful with the negative sign in front of the second one:
4(2x−1)=8x−4−3(x−5)=−3x+154(2x-1) = 8x - 4 \qquad\qquad -3(x-5) = -3x+15
  1. Combine the two results:
8x−4−3x+158x - 4 - 3x + 15
  1. Collect like terms (xx-terms together, numbers together):
(8x−3x)+(−4+15)=5x+11(8x-3x) + (-4+15) = 5x + 11

Expanding both brackets before collecting like terms — rather than trying to subtract the brackets as whole blocks first — keeps each sign change explicit and easy to check.