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Module 2 · L13 of 15 ~35 min ⚡ +95 XP available

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.

Today's hook, Does the sign always change at a root? Most students say yes, and lose marks every time they hit a repeated root. By the end of this lesson you'll know exactly why $(x-2)^2(x+1) > 0$ behaves differently from $(x-2)(x-1)(x+1) > 0$, and you'll never confuse them again.

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Recall, your gut answer first

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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.

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Orientation and key terms

The two core ideas

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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.

Comparison of odd vs even root behaviour

Sign changes $\Leftrightarrow$ odd multiplicity root

Factorise first

You cannot draw a sign table until the polynomial is fully factorised. Always factorise before anything else.

One test point per interval

Choose a convenient value inside each interval. Substitute into the factorised form, you only need the sign, not the exact value.

Include endpoints for ≤ or ≥

For strict inequalities ($>$, $<$) exclude roots. For non-strict ($\ge$, $\le$) include them, they make $P(x) = 0$, which satisfies equality.

What you'll master

Know

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

Understand

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

Can do

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

Key terms

Sign tableA table showing the sign of each factor (and the polynomial) across intervals determined by the roots.

Test pointA value chosen inside an interval to determine the sign of the polynomial there.

Critical pointA root of the polynomial, the boundary between intervals where the sign may change.

Odd multiplicityA root that appears an odd number of times in the factorisation; the graph crosses the axis here and the sign changes.

Even multiplicityA root that appears an even number of times; the graph touches but does not cross the axis, so the sign does not change.

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The five-step method

core concept

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 $\ge$.

Why does this work? A polynomial is continuous, so it can only change sign by passing through zero. Between any two consecutive roots there are no zeros, meaning the sign is the same throughout that interval. One test point is enough to identify the sign for the whole interval.

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?

Reading a sign table

core concept

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 $x \ge 3$.

$$-2 \le x \le 1 \quad\text{or}\quad x \ge 3$$

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.

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Worked examples

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · CUBIC INEQUALITY

Solve $(x - 2)(x + 1)(x - 4) > 0$.

Roots: $x = -1,\ 2,\ 4$ (all simple). Intervals: $(-\infty,-1)$, $(-1,2)$, $(2,4)$, $(4,+\infty)$.

Identify critical points from the factorised form. Three simple roots create four intervals.

PROBLEM 2 · REPEATED ROOT

Solve $(x - 1)^2(x + 3) < 0$.

Roots: $x = 1$ (multiplicity 2, even) and $x = -3$ (multiplicity 1, odd). Intervals: $(-\infty,-3)$, $(-3,1)$, $(1,+\infty)$.

Note the multiplicities carefully, they determine whether the sign changes at each root.

PROBLEM 3 · FACTORISE FIRST

Solve $x^3 - 4x^2 + 3x \le 0$.

$x^3 - 4x^2 + 3x = x(x^2 - 4x + 3) = x(x-1)(x-3)$.

Factor out $x$ first, then factorise the quadratic. Always fully factorise before proceeding.

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 $x = 2$.

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Misconceptions to fix

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Sign always changes at a root

Students assume the sign always flips at a root. It only flips at odd-multiplicity roots. At even-multiplicity roots (e.g. $(x-2)^2$) the graph touches the axis and the sign stays the same.

Trap 02

Forgetting to include/exclude endpoints

For $P(x) > 0$ (strict), roots are excluded, they make $P(x)=0$. For $P(x) \ge 0$ (non-strict), roots are included. Getting this wrong costs a mark in every exam question involving this topic.

Trap 03

Forgetting to fully factorise first

Solving before fully factorising misses roots and creates incorrect intervals. Always reduce the inequality to a fully factorised form $P(x)$ on one side with 0 on the other before building the sign table.

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?

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Activities

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.

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Revisit your thinking

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.

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Multiple choice

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Pick your answer, then rate your confidence. That tells the system what to drill next. Each retry pulls a fresh mix from the bank.

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Short answer

ApplyBand 42 marks

Q1. Solve $(x - 2)(x + 1)(x - 4) > 0$. Show the sign table and state the solution. (2 marks)

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ApplyBand 43 marks

Q2. Solve $x^3 - 4x^2 + 3x \le 0$. Factorise fully and state the solution. (3 marks)

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AnalyseBand 52 marks

Q3. Solve $(x - 1)^2(x + 3) < 0$ and explain why the sign does not change at $x = 1$. (2 marks)

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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].

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Review and finish

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Key takeaways

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.

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