Solving Cubic Inequalities
Master cubic inequalities for Victorian Year 11 Mathematical Methods (VCAA) — statements asking where a cubic curve lies above or below the horizontal axis.
You will learn to factorise the cubic, find its critical values, use a sign diagram to test each interval, and write the solution set in interval notation, reading the answer from the shape of the graph.
Theory
In Year 11 Mathematical Methods (Unit 1), a cubic inequality such as \((x-1)(x-2)(x-3)>0\) asks for every \(x\) that makes a degree-three polynomial positive or negative. You factorise, find the critical values (the zeros), then build a sign diagram to read off the solution as a union of intervals. This page shows the method, how repeated factors behave, and how strict versus non-strict signs change the endpoints.
A cubic inequality compares a cubic \(f(x)=ax^3+bx^2+cx+d\) with \(0\), using one of \(>\), \(<\), \(\ge\) or \(\le\). Solving it means finding all the \(x\)-values that make the statement true — usually a union of intervals, not a single number.
The critical values are the \(x\)-values where \(f(x)=0\): the zeros of the factors. Between two consecutive zeros the graph stays on one side of the \(x\)-axis, so the sign of \(f(x)\) is constant across each interval.
A sign diagram is a number line marked with the critical values and the sign (\(+\) or \(-\)) of \(f(x)\) in each interval. You test one convenient point per interval to fix each sign.
Write the cubic in fully factored form (leading coefficient \(a\)):
The sign of a product is the product of the signs, so \(f(x)\) changes sign at each single zero. At a repeated (squared) factor the sign does not change — the graph only touches the axis there:
How to solve a cubic inequality
- One side zero: rearrange so the inequality reads \(f(x)>0\) (or \(<,\ge,\le\)) with \(0\) on the right.
- Factorise fully: take out a common factor, group, or use the factor theorem (try \(x=\pm1,\pm2,\dots\)) to write \(f(x)\) as a product of linear factors.
- Critical values: set each factor to \(0\) to get the zeros, and mark them on a number line.
- Sign diagram: test one point in each interval to record the sign of \(f(x)\) there.
- Read off: select the intervals matching the inequality; include the endpoints for \(\ge/\le\), exclude them for \(>/<\), and write the answer as a union of intervals.
Critical values — set each factor to zero:
| \(x-1\) | \(\Rightarrow\) | \(x=1\) |
| \(x-2\) | \(\Rightarrow\) | \(x=2\) |
| \(x-3\) | \(\Rightarrow\) | \(x=3\) |
These split the number line into four intervals.
Test one point in each interval:
| \(f(0)\) | \(=\) | \((-1)(-2)(-3)=-6\;(-)\) |
| \(f(\tfrac32)\) | \(=\) | \((\tfrac12)(-\tfrac12)(-\tfrac32)=\tfrac{3}{8}\;(+)\) |
| \(f(\tfrac52)\) | \(=\) | \((\tfrac32)(\tfrac12)(-\tfrac12)=-\tfrac{3}{8}\;(-)\) |
| \(f(4)\) | \(=\) | \((3)(2)(1)=6\;(+)\) |
We want \(f(x)>0\), so take the \(+\) intervals. The inequality is strict, so the endpoints are excluded.
Solution: \((1,\,2)\cup(3,\,\infty)\).
Critical values — set each factor to zero:
| \(x\) | \(\Rightarrow\) | \(x=0\) |
| \(x-2\) | \(\Rightarrow\) | \(x=2\) |
| \(x-5\) | \(\Rightarrow\) | \(x=5\) |
Test one point in each interval:
| \(f(-1)\) | \(=\) | \((-1)(-3)(-6)=-18\;(-)\) |
| \(f(1)\) | \(=\) | \((1)(-1)(-4)=4\;(+)\) |
| \(f(3)\) | \(=\) | \((3)(1)(-2)=-6\;(-)\) |
| \(f(6)\) | \(=\) | \((6)(4)(1)=24\;(+)\) |
We want \(f(x)\le 0\), so take the \(-\) intervals. The sign is not strict, so the endpoints (where \(f(x)=0\)) are included.
Solution: \((-\infty,\,0]\cup[2,\,5]\).
Factorise — the factor theorem gives \(f(1)=f(2)=f(3)=0\):
| \(f(x)\) | \(=\) | \((x-1)(x-2)(x-3)\) |
Critical values:
| \(=\) | \(x=1,\quad x=2,\quad x=3\) |
Test one point in each interval:
| \(f(0)\) | \(=\) | \((-1)(-2)(-3)=-6\;(-)\) |
| \(f(\tfrac32)\) | \(=\) | \(\tfrac{3}{8}\;(+)\) |
| \(f(\tfrac52)\) | \(=\) | \(-\tfrac{3}{8}\;(-)\) |
| \(f(4)\) | \(=\) | \((3)(2)(1)=6\;(+)\) |
We want \(f(x)\le 0\): take the \(-\) intervals with the endpoints included.
Solution: \((-\infty,\,1]\cup[2,\,3]\).
Critical values — set each factor to zero:
| \((x-2)^2\) | \(\Rightarrow\) | \(x=2\) |
| \(x+1\) | \(\Rightarrow\) | \(x=-1\) |
The factor \((x-2)^2\) is squared, so the graph only touches the axis at \(x=2\) — the sign does not change there.
Test one point in each interval:
| \(f(-2)\) | \(=\) | \((-4)^2(-1)=-16\;(-)\) |
| \(f(0)\) | \(=\) | \((-2)^2(1)=4\;(+)\) |
| \(f(3)\) | \(=\) | \((1)^2(4)=4\;(+)\) |
We want \(f(x)<0\): only the first interval is negative. \(x=2\) is not included (it gives \(0\), and the sign stays \(+\) either side of it).
Solution: \((-\infty,\,-1)\).
Common pitfalls
Frequently asked questions
What is a cubic inequality?
It compares a cubic \(f(x)=ax^3+bx^2+cx+d\) with \(0\) using \(>\), \(<\), \(\ge\) or \(\le\); solving it finds every \(x\) that makes the statement true.
How do you solve a cubic inequality?
Move everything to one side, factorise fully, mark the critical values (zeros) on a number line, build a sign diagram by testing a point in each interval, then read off the intervals that match the inequality.
What is a sign diagram?
A number line marked with the zeros of \(f(x)\) and the sign of \(f(x)\) in each interval between them, found by testing one convenient point per interval.
Do you include the endpoints in the answer?
For \(\ge\) or \(\le\) yes — the zeros make \(f(x)=0\), which satisfies the inequality, so use closed dots and square brackets. For \(>\) or \(<\) the endpoints are excluded (open dots, round brackets).
What happens at a repeated (double) root?
The graph touches the axis but does not cross it, so \(f(x)\) keeps the same sign on both sides of a squared factor — the sign only flips at single factors.
Can I just divide both sides by \(x\)?
No. \(x\) can be negative (flipping the inequality) or zero (losing a solution). Keep every factor and use a sign diagram instead.