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\( \definecolor{colordef}{RGB}{249,49,84} \definecolor{colorprop}{RGB}{18,102,241} \)
International Baccalaureate · MYP 5

Expansion and Factorisation

⌚ ~90 min ▢ 9 blocks ✓ 41 exercises ➣ Prerequisites : Algebraic Expansion, Factorisation
I Expanding the Cube of a Sum and of a Difference
Proposition — Perfect Squares Expansion
The square of a sum and the square of a difference can be written as: $$(a+b)^2=a^2+2ab+b^2 \quad\text{and}\quad (a-b)^2=a^2-2ab+b^2.$$
Figure
Proposition — Cube of a Sum and of a Difference
For all numbers \(a\) and \(b\), the cube of a sum is $$(a+b)^3=a^3+3a^2b+3ab^2+b^3.$$ Replacing \(b\) by \(-b\) gives the cube of a difference: $$(a-b)^3=a^3-3a^2b+3ab^2-b^3.$$
Figure

$$\begin{aligned} (a+b)^3 &= (a+b)(a+b)(a+b) &&\text {(cube definition)} \\ &= (a^2+2ab+b^2)(a+b) &&\text {(using the square expansion)} \\ &= (a^2+2ab+b^2)a + (a^2+2ab+b^2)b &&\text {(expanding)} \\ &= a^3+2a^2b+ab^2 + a^2b+2ab^2+b^3 &&\text {(distributive law)} \\ &= a^3+3a^2b+3ab^2+b^3 &&\text {(combining)}\\ \end{aligned}$$

Example
Expand and simplify \((x+2)^3\)

In the identity for the cube of a sum, we substitute \(a=x\) and \(b=2\):$$\begin{aligned}(\textcolor{colordef}{x}+\textcolor{colorprop}{2})^{3}&= \textcolor{colordef}{x}^{3} + 3 \times \textcolor{colordef}{x}^2 \times \textcolor{colorprop}{2} + 3 \times \textcolor{colordef}{x} \times \textcolor{colorprop}{2}^2 + \textcolor{colorprop}{2}^{3} \\ &= x^{3} + 6x^2 + 12x + 8\end{aligned}$$

Example
Expand and simplify \((x-3)^3\).

In the identity for the cube of a difference, we substitute \(a=x\) and \(b=3\):$$\begin{aligned}(\textcolor{colordef}{x}-\textcolor{colorprop}{3})^{3}&= \textcolor{colordef}{x}^{3} - 3 \times \textcolor{colordef}{x}^2 \times \textcolor{colorprop}{3} + 3 \times \textcolor{colordef}{x} \times \textcolor{colorprop}{3}^2 - \textcolor{colorprop}{3}^{3} \\ &= x^{3} - 9x^2 + 27x - 27.\end{aligned}$$ So \((x-3)^3 = x^3 - 9x^2 + 27x - 27\).

Skills to practice
  • Expanding the Cube of a Sum
  • Expanding the Cube of a Difference
II Expanding Three Terms and Three Factors
Proposition — Square of a Trinomial
For all numbers \(a\), \(b\) and \(c\):$$(a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2ac + 2bc$$ — the three squares, plus twice each of the three possible products of two different terms.

Group the first two terms and apply the perfect-square identity twice: $$\begin{aligned}[t] (a+b+c)^2 &= \big((a+b) + c\big)^2 \\ &= (a+b)^2 + 2(a+b)c + c^2 && {\scriptstyle\text{(square of a sum, with } a+b \text{ as one block)}} \\ &= a^2 + 2ab + b^2 + 2ac + 2bc + c^2 && {\scriptstyle\text{(expand the block and distribute } 2c)} \\ &= a^2 + b^2 + c^2 + 2ab + 2ac + 2bc. \end{aligned}$$

Example
Expand \((x + y + 2)^2\). $$\begin{aligned}[t] (x+y+2)^2 &= x^2 + y^2 + 2^2 + 2xy + 2 \cdot x \cdot 2 + 2 \cdot y \cdot 2 && {\scriptstyle\text{(the trinomial identity)}} \\ &= x^2 + y^2 + 4 + 2xy + 4x + 4y. \end{aligned}$$
Method — Expanding a Product of Three Factors
To expand \((A)(B)(C)\):
  1. Expand the product of the first two factors.
  2. Multiply the result by the third factor, distributing every term.
  3. Collect like terms.
The order does not matter (multiplication is associative) — choose the pair that expands most easily.
Example
Expand \((x+1)(x+2)(x+3)\). $$\begin{aligned}[t] (x+1)(x+2)(x+3) &= \big(x^2 + 3x + 2\big)(x+3) && {\scriptstyle\text{(first two factors)}} \\ &= x^3 + 3x^2 + 3x^2 + 9x + 2x + 6 && {\scriptstyle\text{(distribute each term)}} \\ &= x^3 + 6x^2 + 11x + 6. && {\scriptstyle\text{(collect)}} \end{aligned}$$ Quick check at \(x = 1\): left side \(2 \cdot 3 \cdot 4 = 24\); right side \(1 + 6 + 11 + 6 = 24\).
Skills to practice
  • Expanding Further
  • Justifying an Expansion
III Choosing the Most Suitable Form
Proposition — Difference of Squares
\(\textcolor{colordef}{a}^2-\textcolor{colorprop}{b}^2 = (\textcolor{colordef}{a}-\textcolor{colorprop}{b})(\textcolor{colordef}{a}+\textcolor{colorprop}{b})\)
Method — Choosing the Most Suitable Form
The same expression can be written in several forms. Before computing, ask what the question needs:
  • the factorised form reveals the zeros: it serves to solve \(E(x) = 0\) (null factor law) and to find its sign;
  • the expanded and simplified form serves to compute \(E(0)\), to compare two expressions, or to combine them.
Changing form does not change the expression: both writings give the same value for every \(x\).
Example
Let \(E(x) = (x+3)^2 - 25\).
  1. Expand and simplify \(E(x)\).
  2. Factorise \(E(x)\).
  3. Choose the suitable form to: solve \(E(x) = 0\); compute \(E(0)\).

  1. $$\begin{aligned}[t] E(x) &= (x+3)^2 - 25 \\ &= x^2 + 6x + 9 - 25 && \text{(square of a sum)}\\ &= x^2 + 6x - 16. \end{aligned}$$
  2. $$\begin{aligned}[t] E(x) &= (x+3)^2 - 5^2 \\ &= (x+3-5)(x+3+5) && \text{(difference of two squares)}\\ &= (x-2)(x+8). \end{aligned}$$
  3. To solve \(E(x) = 0\): the factorised form gives \((x-2)(x+8) = 0\), so \(x = 2\) or \(x = -8\) (null factor law). To compute \(E(0)\): the expanded form gives \(E(0) = -16\) at a glance.

Skills to practice
  • Choosing the Right Form
IV Going further
Skills to practice
  • Reasoning and Problem-Solving
  • Solving Problems with Identities
  • Investigating Patterns
  • Modelling with Expansions
  • Quiz