Solving Cubic Equations
Learn to solve cubic equations for Queensland Year 11 Mathematical Methods (QCAA). A cubic equation can have up to three solutions, found by rearranging to equal zero.
You will learn to take out a common factor, factorise by grouping, apply the null factor law to read each solution from the factorised form, use technology to solve, and handle a repeated solution where the curve touches the x-axis — key QCAA techniques.
Every question with a fully worked solution.
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Theory
In Year 11 Mathematical Methods (QCAA, Unit 1), a cubic equation is solved by first writing it as \(=0\) and factorising, then applying the null factor law: a product is zero only when a factor is zero. This page covers solving factorised cubics, taking out a common factor, grouping, repeated roots, and the factor theorem, with each root shown as an \(x\)-intercept.
A cubic equation has the form \(ax^3+bx^2+cx+d=0\) with \(a\neq0\); it has up to three real solutions, called roots. The key tool is the null factor law: if a product of factors equals zero, then at least one factor is zero.
So the plan is always the same: rearrange to \(=0\), factorise fully, then set each factor to zero and solve. The roots are exactly the \(x\)-intercepts of the graph \(y=ax^3+bx^2+cx+d\).
A repeated root (from a squared factor such as \((x-2)^2\)) gives one \(x\)-value where the curve touches the axis rather than crossing it. An irreducible quadratic factor (negative discriminant) contributes no further real roots.
The general cubic equation:
The null factor law (the engine of every solution):
Reading roots straight off factors:
How to solve a cubic equation
- Set to zero: move every term to one side so the equation reads \(\dots=0\).
- Factorise fully: take out a common factor, then group or use the factor theorem, and factorise any remaining quadratic.
- Solve each factor: set every factor equal to zero by the null factor law and solve. Report a repeated root once, and note any irreducible quadratic gives no further real solutions.
The product is already zero, so set each factor to zero (null factor law):
Solve each factor:
| \(x+1\) | \(=\) | \(0\ \Rightarrow\ x=-1\) |
| \(x-2\) | \(=\) | \(0\ \Rightarrow\ x=2\) |
| \(2x-1\) | \(=\) | \(0\ \Rightarrow\ x=\dfrac{1}{2}\) |
\(x=-1,\ \dfrac{1}{2},\ 2\).
Take out the common factor \(2x\):
| \(2x^3-2x^2-12x\) | \(=\) | \(2x(x^2-x-6)\) |
Factorise the quadratic:
| \(2x(x^2-x-6)\) | \(=\) | \(2x(x-3)(x+2)\) |
Set each factor to zero:
| \(2x\) | \(=\) | \(0\ \Rightarrow\ x=0\) |
| \(x-3\) | \(=\) | \(0\ \Rightarrow\ x=3\) |
| \(x+2\) | \(=\) | \(0\ \Rightarrow\ x=-2\) |
\(x=-2,\ 0,\ 3\).
Common factor \(x\), then factorise the quadratic:
| \(x^3-4x^2+4x\) | \(=\) | \(x(x^2-4x+4)\) |
| \(=\) | \(x(x-2)^2\) |
Set each factor to zero:
| \(x\) | \(=\) | \(0\ \Rightarrow\ x=0\) |
| \((x-2)^2\) | \(=\) | \(0\ \Rightarrow\ x=2\) |
The factor \((x-2)^2\) gives a repeated root \(x=2\), where the graph touches the \(x\)-axis instead of crossing it.
\(x=0\) and \(x=2\) (a repeated root).
There is no common factor or easy grouping, so test values. Try \(x=1\):
| \((1)^3-2(1)^2-5(1)+6\) | \(=\) | \(1-2-5+6\) |
| \(=\) | \(0\) |
So \((x-1)\) is a factor. Dividing gives the quotient \(x^2-x-6\).
Factorise the quotient and set each factor to zero:
| \(x^2-x-6\) | \(=\) | \((x-3)(x+2)\) |
| \(x-1\) | \(=\) | \(0\ \Rightarrow\ x=1\) |
| \(x-3\) | \(=\) | \(0\ \Rightarrow\ x=3\) |
| \(x+2\) | \(=\) | \(0\ \Rightarrow\ x=-2\) |
\(x=-2,\ 1,\ 3\).
Common pitfalls
Frequently asked questions
How do you solve a cubic equation?
Rearrange so it reads \(=0\), factorise fully, then use the null factor law: set each factor to zero and solve for \(x\).
What is the null factor law?
If a product of factors equals zero, then at least one of the factors is zero. So \((x+1)(x-2)=0\) means \(x=-1\) or \(x=2\).
Why should you not divide a cubic by \(x\)?
Dividing by \(x\) discards the solution \(x=0\). Factor \(x\) out instead, so the root \(x=0\) is kept.
What is a repeated root of a cubic?
A root that comes from a squared factor such as \((x-2)^2\). The curve touches the \(x\)-axis there rather than crossing it; you report the \(x\)-value once.
How does the factor theorem help solve a cubic?
Test values until \(P(a)=0\); then \((x-a)\) is a factor. Divide it out to get a quadratic, factorise that, and solve all the factors.
What if a quadratic factor will not factorise?
If its discriminant is negative it is irreducible over the reals, so it contributes no further real solutions. Report only the roots you have.