Mixed Factorisations
Mixed Factorisations brings every factorising method together, so the main skill is choosing the right one. In NSW Year 9 Advanced maths the checklist is a common factor first, then a difference of two squares, a perfect square, a trinomial or grouping in pairs (NESA outcome MA5‑ALG‑P‑02).
You will factorise fully expressions such as \(45 - 5x^2\), \(4x^2 - 4x - 48\) and \(-3x^2 - x + 14\), and group four-term expressions like \(6a^2 + 15ac + 8ab + 20bc\). Harder questions combine methods, as in \(x^2 - y^2 + x + y\) and \(3(m + n)^2 - 27(m - n)^2\), and \(2x^4 - 26x^2 + 72\) needs two steps to reach four factors. It is part of Year 9 Advanced Algebra: Expanding and Factorising.
Theory
Mixed factorisation means choosing the right method: take out the HCF first, then count the terms to pick difference of two squares, a trinomial or grouping, and keep going until nothing factorises. This Year 9 Advanced maths page covers content that leads into Advanced and Extension maths: multi-stage problems, four-term expressions and quadratics in \(x^2\).
An expression is factorised fully when it is a product and no factor can be factorised again. \(2(x^2 - 9)\) is not fully factorised; \(2(x - 3)(x + 3)\) is.
Most questions need two methods: an HCF first, then a pattern. The number of terms left in the bracket tells you which pattern to try.
A four-term expression is usually grouped in pairs. Sometimes the pairs are a difference of two squares and two other terms, as in \(m^2 - n^2 + 5m + 5n\).
An expression such as \(x^4 - 10x^2 + 9\) is a quadratic in \(x^2\): it factorises like \(u^2 - 10u + 9\) with \(u = x^2\).
The toolkit:
| Form | Factorisation |
|---|---|
| Common factor | \(ab + ac = a(b + c)\) |
| Difference of two squares | \(a^2 - b^2 = (a - b)(a + b)\) |
| Perfect squares | \(a^2 \pm 2ab + b^2 = (a \pm b)^2\) |
| Monic trinomial | \(x^2 + (p + q)x + pq = (x + p)(x + q)\) |
| Non-monic trinomial | split \(bx\) into two terms with product \(ac\), then group |
| Four terms | \(ax + ay + bx + by = (x + y)(a + b)\) |
How to factorise any expression
- HCF first. Take out the HCF of all the terms. If the leading term is negative, take out a negative.
- Two terms: look for a difference of two squares, including powers such as \(x^4 = (x^2)^2\).
- Three terms: test for a perfect square, then use sum and product (monic) or split the middle term (non-monic).
- Four terms: group in pairs. If the brackets do not match, reorder, or pair two squares together.
- Repeat on every bracket until none factorises. Then check by expanding.
A negative leading term
- Take out \(-1\): \(-2x^2 + 7x - 3\) \(= -(2x^2 - 7x + 3)\) \(= -(2x - 1)(x - 3)\).
- Absorb the minus into one bracket if you like: \(-(2x - 1)(x - 3) = (2x - 1)(3 - x)\). Both forms are correct.
The HCF is \(3x\). The bracket is a monic quadratic: product \(-12\), sum \(-4\), so \(-6\) and \(2\).
| \(3x^3 - 12x^2 - 36x\) | \(=\) | \(3x(x^2 - 4x - 12)\) |
| \(\) | \(=\) | \(\boldsymbol{3x(x - 6)(x + 2)}\) |
Check with \(x = 1\): \(3 - 12 - 36 = -45\) and \(3(-5)(3) = -45\).
No HCF for all four terms, so group in pairs.
| \(2x^3 - 6x^2 + 5x - 15\) | \(=\) | \(2x^2(x - 3) + 5(x - 3)\) |
| \(\) | \(=\) | \(\boldsymbol{(x - 3)(2x^2 + 5)}\) |
\(2x^2 + 5\) is a sum, not a difference, so it does not factorise further.
Group the squares together and the other two terms together. Each pair has a factor \((a - 3b)\).
| \(a^2 - 9b^2 + 2a - 6b\) | \(=\) | \((a - 3b)(a + 3b) + 2(a - 3b)\) |
| \(\) | \(=\) | \((a - 3b)\big[(a + 3b) + 2\big]\) |
| \(\) | \(=\) | \(\boldsymbol{(a - 3b)(a + 3b + 2)}\) |
Grouping \(a^2 + 2a\) and \(-9b^2 - 6b\) gives \(a(a + 2) - 3b(3b + 2)\), which does not share a bracket. Try another pairing when that happens.
Think of \(x^4\) as \((x^2)^2\). The expression is a quadratic in \(x^2\): product \(16\), sum \(-17\), so \(-1\) and \(-16\).
| \(x^4 - 17x^2 + 16\) | \(=\) | \((x^2 - 1)(x^2 - 16)\) |
| \(\) | \(=\) | \(\boldsymbol{(x - 1)(x + 1)(x - 4)(x + 4)}\) |
Both brackets were differences of two squares, so each factorised again. Four linear factors means it is fully factorised.
Common pitfalls
Frequently asked questions
How do I know which factorising method to use?
Take out the highest common factor first, then count the terms. Two terms suggests a difference of two squares, three terms a perfect square or quadratic trinomial, and four terms grouping in pairs.
What does factorise fully mean?
It means keep factorising until no bracket can be factorised any further. Check each bracket after every step.
How do you factorise x to the power 4 minus 81?
Treat it as a difference of two squares: (x squared minus 9)(x squared plus 9). Then x squared minus 9 factorises again to (x minus 3)(x plus 3). The answer is (x minus 3)(x plus 3)(x squared plus 9).
How do you factorise four terms?
Group them in pairs so each pair has a common factor, factorise each pair to get the same bracket twice, then take out that bracket. If it does not work, try a different pairing.
Is -(x - 2)(x + 5) the same as (2 - x)(x + 5)?
Yes. Multiplying \((x - 2)\) by \(-1\) gives \((2 - x)\), so both expressions are equal. Either form is a full factorisation.