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 upwhat 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 confidencethat 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.