Expanding and Factorising in IGCSE Maths
Expanding single and double brackets, then factorising in reverse: common factors, quadratics and the difference of two squares. Clear worked examples, area-model diagrams and an exam-style question, set out the way the mark scheme rewards.
0580 · 2.2Expanding a single bracket
Expanding a bracket means multiplying every term inside by the term outside. This is the distributive law:
\[a(b + c) = ab + ac\]The area model shows why: a rectangle of height \(3\) and width \(x + 4\) splits into a piece of area \(3x\) and a piece of area \(12\), so \(3(x + 4) = 3x + 12\).
Expand \(5(2x - 3)\).
A negative outside the bracket flips the sign of every term inside. For example, \(-2(x - 4) = -2x + 8\).
Expanding two brackets
To expand two brackets, multiply each term in the first bracket by each term in the second. A useful memory aid is FOIL: First, Outer, Inner, Last. The grid (area) method keeps every product in its own box:
Adding the four boxes gives \(x^2 + 3x + 2x + 6\), which simplifies to \(x^2 + 5x + 6\).
Expand and simplify \((x + 3)(x + 2)\).
Two patterns are worth memorising:
\[(x + a)^2 = x^2 + 2ax + a^2\] \[(x + a)(x - a) = x^2 - a^2\]The second is the difference of two squares, which we use again when factorising.
Factorising with common factors
Factorising means writing an expression as a product. The first step is always to take out the highest common factor (HCF) of all the terms.
Factorise \(6x^2 + 9x\).
The highest common factor of \(6x^2\) and \(9x\) is \(3x\):
\[6x^2 + 9x = 3x(2x + 3)\]Check by expanding: \(3x(2x + 3) = 6x^2 + 9x\).
With four terms, factorise in two pairs, then take out the common bracket:
\[xy + 2x + 3y + 6 = x(y + 2) + 3(y + 2) = (x + 3)(y + 2)\]Factorising quadratics
ExtendedTo factorise \(x^2 + bx + c\), find two numbers that multiply to \(c\) and add to \(b\). Those two numbers go into the brackets. This is the area-model grid worked in reverse.
Factorise \(x^2 + 7x + 12\).
Two numbers that multiply to \(12\) and add to \(7\) are \(3\) and \(4\):
\[x^2 + 7x + 12 = (x + 3)(x + 4)\]Factorise \(x^2 - 2x - 15\).
Two numbers that multiply to \(-15\) and add to \(-2\) are \(-5\) and \(+3\):
\[x^2 - 2x - 15 = (x - 5)(x + 3)\]Factorising ax² + bx + c
ExtendedWhen \(a \neq 1\), find two numbers that multiply to \(a \times c\) and add to \(b\). Use them to split the middle term, then factorise by grouping.
Factorise \(2x^2 + 7x + 3\).
Here \(a \times c = 2 \times 3 = 6\). Two numbers that multiply to \(6\) and add to \(7\) are \(6\) and \(1\). Split the middle term:
\[2x^2 + 6x + 1x + 3\] \[= 2x(x + 3) + 1(x + 3)\] \[= (2x + 1)(x + 3)\]Difference of two squares
ExtendedWhen one square is subtracted from another, it factorises into a sum times a difference:
\[a^2 - b^2 = (a + b)(a - b)\]Factorise \(x^2 - 49\).
Write \(49\) as \(7^2\), so \(x^2 - 49 = x^2 - 7^2\):
\[x^2 - 49 = (x + 7)(x - 7)\]Factorise \(9x^2 - 25\).
Write it as \((3x)^2 - 5^2\):
\[9x^2 - 25 = (3x + 5)(3x - 5)\]Exam-style question
Extended(a) Expand and simplify \((2x - 3)(x + 5)\). [2]
(b) Factorise fully \(12x^2 - 8x\). [2]
(c) Factorise \(x^2 - 5x - 14\). [2]
Show solution (a)
Show solution (b)
The highest common factor of \(12x^2\) and \(8x\) is \(4x\):
\[12x^2 - 8x = 4x(3x - 2)\]Show solution (c)
Two numbers that multiply to \(-14\) and add to \(-5\) are \(-7\) and \(+2\):
\[x^2 - 5x - 14 = (x - 7)(x + 2)\]Expanding and factorising underpins almost all of algebra, from solving quadratics to simplifying fractions. The word "fully" in part (b) is a signal to take out the highest common factor first. Always check a factorised answer by expanding it back.
Common mistakes
When expanding, multiply the outside term by every term inside, and with two brackets work out all four products. Missing one is the most common slip.
A negative in front of a bracket changes the sign of everything inside: \(-(x - 3) = -x + 3\), not \(-x - 3\).
Always take out the highest common factor. \(2x^2 + 4x = 2x(x + 2)\), not \(x(2x + 4)\), which is only partly factorised.
For a quadratic the two numbers must multiply to \(c\) and add to \(b\), with the correct signs. Check both conditions before writing the brackets.
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