History

Long before modern symbols, people solved problems that required multiplying lengths and collecting areas — the same ideas behind expanding and factorising. Babylonian scribes (around 1800 BCE) recorded rectangle problems on clay tablets: given area and side relationships, find unknown lengths. Their recipes are early cousins of “expand, then simplify.”

Illustrated timeline: Babylonian area on clay tablets, al-Khwarizmi's algebra, Viète's difference of squares, Enlightenment science, and modern expand–factorise steps
From geometric area puzzles to symbolic algebra: how expansion and factorisation became standard tools in school mathematics.

In the 9th century, al-Khwarizmi organised methods for equations and products in Arabic texts that later spread to Europe. Renaissance mathematicians such as Viète introduced letters for unknowns and constants, making it easier to see patterns like \(a^2 - b^2\). By the 17th–18th centuries, polynomial algebra was central to calculus and science; factorisation became a routine step before solving equations or simplifying formulas.

Derivation

Area proof of \((a+b)(c+d)\)

Draw a rectangle with width \(a+b\) and height \(c+d\). The total area is \((a+b)(c+d)\). Split the rectangle into four smaller rectangles with areas \(ac\), \(ad\), \(bc\), and \(bd\). The sum of partial areas equals the whole: \[ (a+b)(c+d) = ac + ad + bc + bd. \] This is the distributive law in two dimensions — the same picture as in the area-model widget above.

Deriving standard identities

Set \(c = a\) and \(d = b\) in the area formula for a square of side \(a+b\): \[ (a+b)^2 = a^2 + ab + ab + b^2 = a^2 + 2ab + b^2. \] For a square of side \(a-b\) (with \(a > b\)), remove a strip of width \(b\) from a square of side \(a\); the remaining L-shape rearranges to show \[ (a-b)^2 = a^2 - 2ab + b^2. \] Difference of squares: a large square of side \(a\) minus a small square of side \(b\) leaves two congruent rectangles, which rearrange to a rectangle of dimensions \((a+b)\) by \((a-b)\): \[ a^2 - b^2 = (a+b)(a-b). \]

Grouping walkthrough

Factorise \(3x^2 + 12x - 2x - 8\).

  1. Group: \((3x^2 + 12x) + (-2x - 8)\).
  2. Factor each pair: \(3x(x+4) - 2(x+4)\).
  3. Common bracket \((x+4)\): \((3x-2)(x+4)\).

Checkpoints

Pause and reason before continuing. Discuss with a partner or write your work in a notebook.

Checkpoint

In \(4x^3 - 2x^2 + 5\), what is the degree? Which terms are like terms with \(4x^3\)?

Checkpoint

Expand \((2x-1)(x+4)\) without skipping steps. How many terms do you get before collecting?

Checkpoint

How would you factorise \(x^2 - 16\)? Which identity applies?

Checkpoint

Expand \(x(x+1) - (x+1)\). What common factor appears? What is the factorised form?

Applications

Geometry: area and perimeter

A rectangle has sides \((x+3)\) cm and \((2x-1)\) cm. Its area is \[ (x+3)(2x-1) = 2x^2 + 5x - 3 \quad\text{cm}^2. \] The perimeter is \(2\bigl((x+3)+(2x-1)\bigr) = 2(3x+2) = 6x+4\) cm. Expansion links algebraic descriptions of sides to measurable quantities.

Revenue and pricing

If a shop sells \(n\) items at price \((50 - 0.5n)\) dollars each, revenue is \(n(50 - 0.5n) = 50n - 0.5n^2\). Expanding a product of two linear factors often appears before finding maximum revenue (a quadratic in \(n\)).

Number patterns

Notice that \(11^2 = 121\), \(111^2 = 12321\). In general, expanding \((10^k + 10^{k-1} + \cdots + 1)^2\) explains palindromic squares — a playful link between identities and arithmetic patterns.

Question bank

Work each question, then open the solution to check your reasoning. Difficulty increases through the sets.

Easy

Expand \(3(x+2)\).

Answer: \(3x + 6\).

Steps: \(3 \cdot x + 3 \cdot 2 = 3x + 6\).

Simplify \(2x + 5x - 1\).

Answer: \(7x - 1\).

Explanation: Combine like terms \(2x + 5x = 7x\).

Factorise \(6x + 9\).

Answer: \(3(2x + 3)\).

Steps: HCF is \(3\); \(6x \div 3 = 2x\), \(9 \div 3 = 3\).

Intermediate

Expand \((x+4)(x-2)\).

Answer: \(x^2 + 2x - 8\).

Steps: \(x^2 - 2x + 4x - 8 = x^2 + 2x - 8\).

Factorise \(x^2 - 9\).

Answer: \((x+3)(x-3)\).

Explanation: Difference of squares: \(x^2 - 3^2\).

Factorise \(x^2 + 6x + 9\).

Answer: \((x+3)^2\).

Explanation: Perfect square: \(3 \times 3 = 9\), \(3 + 3 = 6\).

Difficult

Factorise \(2x^2 + 6x - x - 3\) by grouping.

Answer: \((2x-1)(x+3)\).

Steps: \(2x(x+3) - 1(x+3) = (2x-1)(x+3)\).

Factorise \(2x^2 + 7x + 3\).

Answer: \((2x+1)(x+3)\).

Steps: \(ac = 6\); split \(7x = 6x + x\); \(2x(x+3) + 1(x+3) = (2x+1)(x+3)\).

Hardcore

Expand and then factorise \(x(x+2) - (x+2)\).

Answer: \((x+2)(x-1)\) after factorising (expansion gives \(x^2 + x - 2\)).

Steps: Factor common \((x+2)\): \((x+2)(x-1)\). Check: \((x+2)(x-1) = x^2 + x - 2\).

A rectangle has sides \((x+3)\) cm and \((2x-1)\) cm. Find its area in expanded form.

Answer: \((2x^2 + 5x - 3)\) cm².

Calculation: \((x+3)(2x-1) = 2x^2 - x + 6x - 3 = 2x^2 + 5x - 3\).

References