Polynomial Inequalities
You can solve quadratic inequalities by sketching a parabola. But what about cubics? Quartics? The sign-table method scales to any degree — and once you understand the sign-change rule at odd vs even multiplicity roots, inequalities become almost mechanical. In this lesson you'll master the complete method and learn exactly when signs change.
How would you solve $(x - 1)(x + 2) > 0$? Without looking anything up — what role do the roots $x = 1$ and $x = -2$ play in deciding where the inequality holds? Write your prediction before reading on.
Everything in this lesson rests on two ideas. Lock them in before anything else.
Idea 1 — Roots divide the number line. Every root of a polynomial splits the real line into intervals. The sign of the polynomial is constant within each interval — it can only change at a root.
Idea 2 — Multiplicity controls sign change. At an odd-multiplicity root (simple, triple, …) the sign changes. At an even-multiplicity root (double, quadruple, …) the sign stays the same — the graph touches the axis and bounces back.
Key facts
- Sign changes occur only at odd-multiplicity roots
- Even-multiplicity roots leave the sign unchanged
- The sign table divides the number line at every root
Concepts
- Why the polynomial sign is constant within each interval
- How multiplicity of a root determines the graph's behaviour there
- The connection between inequality solutions and graph regions above/below the axis
Skills
- Construct a complete sign table for any factorised polynomial
- Solve inequalities of the form $P(x) > 0$, $P(x) \ge 0$, $P(x) < 0$, $P(x) \le 0$
- Handle repeated roots correctly in a sign table
Every polynomial inequality can be solved with the same five steps:
- Rearrange so all terms are on one side: $P(x) > 0$ (or $\ge 0$, $< 0$, $\le 0$).
- Factorise $P(x)$ completely, identifying each root and its multiplicity.
- Mark roots on a number line — these are the critical points that divide it into intervals.
- Determine the sign in each interval using one test point per interval.
- Select intervals that satisfy the inequality; include endpoints for $\le$ or $\ge$.
Five steps: rearrange → factorise → mark roots → test each interval → select solutions; Sign changes at odd-multiplicity roots; sign stays the same at even-multiplicity roots
Pause — copy the polynomial inequality method into your book: (1) rearrange to one side; (2) factorise; (3) mark all roots on a number line; (4) test one point per interval; (5) select intervals with the required sign; note sign changes at odd-multiplicity roots only.
Quick check: The polynomial $(x-2)(x+3)$ is negative for which values?
We just saw the five-step method: rearrange, factorise, mark roots, test intervals, select solutions. That raises a question: when there are three or more roots, testing a point in each interval by hand is tedious — is there a more systematic layout? This card answers it → a sign table lists each factor's sign across every interval, then multiplies signs to determine the product's sign, making the solution obvious at a glance.
A sign table organises the sign of each factor and the product across every interval. Consider $(x-1)(x+2)(x-3) \ge 0$.
Roots: $x = -2,\ 1,\ 3$ (all simple — odd multiplicity, so sign changes at each).
Intervals: $(-\infty,-2)$, $(-2,1)$, $(1,3)$, $(3,+\infty)$.
| Interval | $(x-1)$ | $(x+2)$ | $(x-3)$ | Product |
|---|---|---|---|---|
| $x < -2$ | $-$ | $-$ | $-$ | $-$ |
| $-2 < x < 1$ | $-$ | $+$ | $-$ | $+$ |
| $1 < x < 3$ | $+$ | $+$ | $-$ | $-$ |
| $x > 3$ | $+$ | $+$ | $+$ | $+$ |
We want $\ge 0$ (product positive or zero), so the solution is $-2 \le x \le 1$ or $x \ge 3$.
Draw a number line with each root marked; label the intervals between them; For each interval: substitute a test point, determine sign of each factor, multiply signs for product
Pause — copy the sign-table layout into your book: mark all roots on a number line; for each interval, determine the sign of each factor; multiply signs for the product; sign changes at odd-multiplicity roots, stays the same at even-multiplicity roots.
Did you get this? True or false: for a polynomial with all simple roots, the sign alternates between consecutive intervals.
Worked examples · 3 in a row, reveal as you go
Solve $(x - 2)(x + 1)(x - 4) > 0$.
Solve $(x - 1)^2(x + 3) < 0$.
Solve $x^3 - 4x^2 + 3x \le 0$.
Fill the gap: For $(x-2)^2(x+1) > 0$, the factor $(x-2)^2$ is always (or zero), so the sign of the product equals the sign of . The solution is $x > $ excluding $x = 2$.
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: the solution to $(x-3)^2(x+1) \ge 0$ includes $x = 3$.
Odd one out: Which of the following is the odd one out?
Activities · practice with the ideas
Solve $(x+4)(x-1) \le 0$ using a sign table. State the solution.
Solve $(x-2)(x+1)(x-5) > 0$. List the positive intervals.
Solve $x^3 - x \ge 0$. Fully factorise first, then build the sign table.
Solve $(x-3)^2(x+2) < 0$. Explain what happens at $x = 3$.
Without solving, state how many intervals you would need to test for a polynomial with roots at $x = -3,\ 0,\ 2,\ 5$ (all simple). Explain your reasoning.
Earlier you were asked: How would you solve $(x-1)(x+2) > 0$, and what role do the roots play?
The roots $x = -2$ and $x = 1$ are the critical points that split the number line into three intervals. Testing $x=-3$ gives $(-)(-)=+$; testing $x=0$ gives $(-)(+)=-$; testing $x=2$ gives $(+)(+)=+$. So the solution is $x < -2$ or $x > 1$. The roots don't satisfy the strict inequality — they make the product zero.
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. Solve $(x - 2)(x + 1)(x - 4) > 0$. Show the sign table and state the solution. (2 marks)
Q2. Solve $x^3 - 4x^2 + 3x \le 0$. Factorise fully and state the solution. (3 marks)
Q3. Solve $(x - 1)^2(x + 3) < 0$ and explain why the sign does not change at $x = 1$. (2 marks)
Comprehensive answers (click to reveal)
Activity 1: 1. Roots $-4, 1$; sign table gives $-$ in $(-\infty,-4)$, $+$ in $(-4,1)$, $-$ in $(1,\infty)$ for $(x+4)(x-1)$; solution $-4 \le x \le 1$. 2. Roots $-1,2,5$; positive in $(-1,2)$ and $(5,\infty)$. 3. $x(x-1)(x+1) \ge 0$; roots $-1,0,1$; positive/zero in $[-1,0]$ and $[1,\infty)$. 4. $(x-3)^2 \ge 0$ always, so sign equals sign of $(x+2)$; solution $x < -2$; at $x=3$ the factor $(x-3)^2=0$ so product is $0$, not negative. 5. Five intervals (4 roots create 5 regions).
Q1 (2 marks): Roots $-1,2,4$; sign table: $-,-,+,-,+$ cycling; positive intervals: $-1 < x < 2$ or $x > 4$. [2]
Q2 (3 marks): $x(x-1)(x-3)=0$ [1]; roots $0,1,3$; sign table: $-,+,-,+$; $P\le 0$ when $x \le 0$ or $1 \le x \le 3$ [2].
Q3 (2 marks): Roots $-3$ (simple), $1$ (double); test $x=-4$: $(25)(-1)=-$; test $x=0$: $(1)(3)=+$; test $x=2$: $(1)(5)=+$; solution $x < -3$ [1]. At $x=1$, the factor $(x-1)^2=0$ but is always non-negative (even multiplicity) so the sign of the product is determined by $(x+3)$, which is positive near $x=1$ — the sign stays positive and does not change [1].
Five timed questions. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arena- Factorise completely → mark roots → build sign table → select intervals.
- Sign changes at odd-multiplicity roots only; stays the same at even-multiplicity roots.
- Use test points — one per interval — to determine the sign.
- Include endpoints for $\le$ or $\ge$; exclude for $<$ or $>$.
- Next lesson: Applications of Polynomials — putting these techniques into real-world context.
Mark lesson as complete
Tick when you've finished the practice and review.