Factorising
Master factorising for Queensland Year 11 Mathematical Methods (QCAA). Factorising rewrites a quadratic as a product of brackets — the reverse of expanding.
You will learn to take out the highest common factor, recognise a difference of two squares, and factorise monic and non-monic quadratics by finding the right pair of numbers — the essential first step for solving equations and sketching parabolas.
Theory
In Year 11 Mathematical Methods (QCAA, Unit 1), factorising is the reverse of expanding: it writes a quadratic as a product of factors. This page covers taking out the highest common factor, the difference of two squares, factorising monic quadratics \(x^2+bx+c\) and non-monic quadratics \(ax^2+bx+c\) — all with real coefficients.
To factorise is to write an expression as a product. It undoes expanding, so a check is always to expand your factors and confirm you get back the original.
The highest common factor (HCF) is the largest numeric and algebraic factor shared by every term; always take it out first. A difference of two squares \(a^2-b^2\) factorises to \((a-b)(a+b)\). A monic quadratic has leading coefficient \(1\); a non-monic quadratic has \(a\ne 1\).
For a monic \(x^2+bx+c\), find two numbers that multiply to \(c\) and add to \(b\). For a non-monic \(ax^2+bx+c\), split the middle term using two numbers that multiply to \(ac\) and add to \(b\), then factorise by grouping.
The key factorising forms:
How to factorise a quadratic
- Take out the HCF of all terms first, if there is one greater than \(1\).
- Recognise the type: a difference of two squares \(a^2-b^2\), a monic trinomial \(x^2+bx+c\), or a non-monic trinomial \(ax^2+bx+c\).
- Apply the matching rule — sum-and-product for a monic trinomial, or split the middle term and group for a non-monic one — then check by expanding.
Find the HCF of \(12x^2\) and \(18x\): the numeric HCF of \(12\) and \(18\) is \(6\), and both terms have an \(x\), so the HCF is \(6x\).
| \(12x^2+18x\) | \(=\) | \(6x(2x)+6x(3)\) |
Write the HCF outside the bracket:
| \(=\) | \(6x(2x+3)\) |
\(12x^2+18x=6x(2x+3)\).
Write each term as a square: \(9x^2=(3x)^2\) and \(25=5^2\).
| \(9x^2-25\) | \(=\) | \((3x)^2-(5)^2\) |
Apply \(a^2-b^2=(a-b)(a+b)\) with \(a=3x,\ b=5\):
| \(=\) | \((3x-5)(3x+5)\) |
\(9x^2-25=(3x-5)(3x+5)\).
Find two numbers that multiply to \(c=12\) and add to \(b=-7\). Test the factor pairs of \(12\): \(-3\) and \(-4\) give \((-3)(-4)=12\) and \(-3+(-4)=-7\).
| \(-3\times -4\) | \(=\) | \(12\) |
| \(-3+(-4)\) | \(=\) | \(-7\) |
Write the brackets with those numbers:
| \(x^2-7x+12\) | \(=\) | \((x-3)(x-4)\) |
\(x^2-7x+12=(x-3)(x-4)\).
There is no common factor. Compute \(ac=6\times(-10)=-60\), and find two numbers that multiply to \(-60\) and add to \(b=11\): they are \(15\) and \(-4\).
| \(ac\) | \(=\) | \(6\times(-10)=-60\) |
| \(15\times(-4)\) | \(=\) | \(-60\) |
| \(15+(-4)\) | \(=\) | \(11\) |
Split the middle term \(11x=15x-4x\), then group in pairs:
| \(6x^2+11x-10\) | \(=\) | \(6x^2+15x-4x-10\) |
| \(=\) | \(3x(2x+5)-2(2x+5)\) |
Take out the common bracket \((2x+5)\):
| \(=\) | \((2x+5)(3x-2)\) |
\(6x^2+11x-10=(2x+5)(3x-2)\).
Common pitfalls
Frequently asked questions
What is the first step when factorising?
Always look for a highest common factor and take it out first. This simplifies the numbers and often exposes a standard form underneath.
How do I factorise a monic quadratic \(x^2+bx+c\)?
Find two numbers that multiply to \(c\) and add to \(b\), then write \((x+p)(x+q)\) with those numbers.
How is the difference of two squares factorised?
\(a^2-b^2=(a-b)(a+b)\). Write each term as a square, then use one bracket with a minus and one with a plus.
Does \(x^2+1\) factorise?
Not over the real numbers. A sum of squares has no real factors, so in Year 11 it is left as it is.
How do I factorise a non-monic quadratic \(ax^2+bx+c\)?
Find two numbers multiplying to \(ac\) and adding to \(b\), split the middle term with them, then factorise by grouping.
How can I check my factors are right?
Expand them. If the expansion returns the original quadratic, the factorisation is correct.