IB Maths AA HL Topic 2 — Functions Paper 1 & 2 HL only ~7 min read

Polynomial Inequalities

When both sides are polynomials, you don’t need a GDC — the algebra is doable. Two methods work: the graph method (sketch the polynomial, read off where it sits above or below the axis) and the sign table method (split the number line at the roots, check the sign in each interval). Both rely on the same first step: rearrange so one side is zero, then factorise.

📘 What you need to know

Polynomial inequality: both sides involve only constants and positive integer powers of x. Things like √x, 1/x, or x⁻² don’t count.
Standard set-up: rearrange to P(x) ≤ 0 or P(x) ≥ 0 form.
Find the roots by factorising — use the factor theorem if needed for cubics or higher.
Method 1 — Graph: sketch the polynomial; read off intervals where the curve is above (≥ 0) or below (≤ 0) the x-axis.
Method 2 — Sign table: split the real line at the roots; in each interval, check the sign of the polynomial (or the sign of each factor).
Repeated roots: a root with even multiplicity touches the axis without changing sign; one with odd multiplicity crosses and changes sign.
Endpoints: ≤/≥ inequalities include the roots; </> exclude them.
Some inequalities have isolated solutions: (x − 3)² ≤ 0 is satisfied only at x = 3.

What counts as a polynomial inequality

Polynomial

x⁴ − x² > x + 6
2x³ ≤ 5x − 1

only constants and positive integer powers of x

Not polynomial

√x > 4
1/x ≥ 3x

fractional or negative powers — different methods needed

The general approach

Step 1 — get one side to zero f(x) ≤ g(x) ⟺ P(x) ≤ 0, where P(x) = f(x) − g(x)

Once you have P(x) on one side, the strategy is the same regardless of degree: find the roots, then determine the sign of P in each interval between roots.

Method 1 — Graph method

Sketch the polynomial using what you know about roots, multiplicities, and end behaviour (from Section 2.6). Then:

For P(x) ≤ 0

read off intervals where curve is below or on axis

“below” includes y = 0 for ≤; excludes it for <

For P(x) ≥ 0

read off intervals where curve is above or on axis

“above” includes y = 0 for ≥; excludes it for >

The graph method is fastest when you can sketch the polynomial confidently — usually after factorising. End behaviour and multiplicity tell you the shape; everything else falls into place.

Method 2 — Sign table

Useful when sketching feels risky or the polynomial has many factors. Split the real line at the roots, pick a test value in each interval, and record the sign of either the whole polynomial or of each factor.

Example sign table — P(x) = (x − 2)(x + 4)(2x − 1)

Interval	x < −4	−4 < x < 1/2	1/2 < x < 2	x > 2
x − 2	−	−	−	+
x + 4	−	+	+	+
2x − 1	−	−	+	+
P(x)	−	+	−	+

From the table, P(x) > 0 in the second and fourth intervals. The product of three factors changes sign each time you cross a (simple) root.

Quick alternative: pick one numerical value from each interval and substitute into P(x) directly. Same answer, fewer columns.

🧭 Recipe — solving a polynomial inequality

Rearrange to P(x) ≤ 0 or P(x) ≥ 0 — get one side to zero by adding and subtracting only.
Factorise P(x) fully. Use the factor theorem to find a first root if needed.
List the roots in order, noting any with multiplicity > 1.
Pick a method: sketch the polynomial (graph method) or build a sign table.
Read off the solution as one or more intervals.
Match endpoint type: ≤/≥ uses inclusive (closed); </> uses strict (open).

Worked examples

WE 1

Solve a quadratic inequality

Solve x² − 5x + 6 > 0.

Step 1: Already in standard form. Factorise (x − 2)(x − 3) > 0 Step 2: Roots are 2 and 3 Step 3: Sketch — parabola opens up curve is above x-axis to the left of 2 and to the right of 3 x < 2 or x > 3 opens-up parabola → positive outside the roots; opens-down → positive between roots

WE 2

Quadratic with a negative leading coefficient

Solve −x² + 4x − 3 ≥ 0.

Step 1: Find the roots — set −x² + 4x − 3 = 0 x² − 4x + 3 = 0 → (x − 1)(x − 3) = 0 roots: x = 1, x = 3 Step 2: Sketch — parabola opens down curve is above x-axis between the roots Step 3: ≥ includes the roots 1 ≤ x ≤ 3 opens-down parabola → positive between roots; sign of leading coefficient determines the shape

WE 3

Cubic inequality — three distinct roots

Solve x³ − 4x² − x + 4 ≤ 0.

Step 1: Find a root via factor theorem try x = 1: 1 − 4 − 1 + 4 = 0 ✓ → (x − 1) is a factor Step 2: Divide to find the quadratic factor x³ − 4x² − x + 4 = (x − 1)(x² − 3x − 4) = (x − 1)(x − 4)(x + 1) Step 3: Roots in order: −1, 1, 4. Sketch cubic positive leading, three simple roots → curve weaves through axis below x-axis: x < −1 and 1 < x < 4 Step 4: ≤ includes the roots x ≤ −1 or 1 ≤ x ≤ 4 simple roots → sign flips each time the curve crosses; the pattern is alternating across intervals

WE 4

Inequality with a repeated root

Solve (x − 3)²(x + 1) > 0.

Step 1: Roots are x = 3 (mult 2) and x = −1 (mult 1) Step 2: Note (x − 3)² ≥ 0 always — equals zero only at x = 3 so sign of P(x) is determined by sign of (x + 1) when (x − 3)² > 0 Step 3: Need P(x) strictly > 0 → both factors strictly nonzero x + 1 > 0 → x > −1 (x − 3)² > 0 → x ≠ 3 −1 < x < 3 or x > 3 at x = 3 the curve touches the axis without crossing — sign doesn’t flip there, just hits zero momentarily

WE 5

Use a sign table to solve a factorised inequality

Solve (x − 2)(x + 4)(2x − 1) > 0 using a sign table.

Step 1: Roots in order: −4, 1/2, 2 Step 2: Build sign table by testing each interval x < −4 (try −5): (−7)(−1)(−11) = −77 < 0 −4 < x < 1/2 (try 0): (−2)(4)(−1) = 8 > 0 ✓ 1/2 < x < 2 (try 1): (−1)(5)(1) = −5 < 0 x > 2 (try 3): (1)(7)(5) = 35 > 0 ✓ Step 3: Pick out the intervals where P > 0 −4 < x < 1/2 or x > 2 sign alternates across simple roots — once you’ve found the sign in one interval, the rest follow

WE 6

Cubic with double root — solve algebraically

Solve x³ + 3x² − 9x − 27 ≤ 0.

Step 1: Find a root via factor theorem try x = 3: 27 + 27 − 27 − 27 = 0 ✓ → (x − 3) is a factor Step 2: Divide to find quadratic factor x³ + 3x² − 9x − 27 = (x − 3)(x² + 6x + 9) = (x − 3)(x + 3)² Step 3: Roots are x = 3 (mult 1) and x = −3 (mult 2) Step 4: (x + 3)² ≥ 0 always — sign of P depends on (x − 3) P(x) ≤ 0 ⟺ (x − 3) ≤ 0 (with allowance for x = −3 where P = 0) x − 3 ≤ 0 → x ≤ 3 x ≤ 3 double root at x = −3 makes the curve touch the axis without crossing — no sign flip there, so no extra interval

💡 Top tips

Always rearrange to P(x) ≤ 0 or ≥ 0 first. Both methods need one side equal to zero.
Factorise fully. Multiplicity matters — and a fully factorised polynomial makes the sign analysis trivial.
Sign alternates across simple roots: once you’ve found one interval’s sign, the rest follow by flipping each time you cross a multiplicity-1 root.
Even-multiplicity roots don’t flip the sign: the curve just touches the axis. Skip them when alternating.
Use the graph method when you can sketch confidently — usually fastest for cubics with simple roots.
Use the sign table when you have many factors or are unsure of the shape — the columns make it mechanical.
Check the endpoint convention: ≤/≥ closes the boundary, </> leaves it open.

⚠ Common mistakes

Multiplying both sides by x or another variable. The sign of the variable changes the inequality direction — never multiply by something whose sign isn’t fixed.
Forgetting to flip when multiplying by −1. If you multiply −x² + 4x − 3 ≥ 0 by −1, the new inequality is x² − 4x + 3 ≤ 0.
Treating a multiplicity-2 root as a sign change. The graph touches the axis there but doesn’t cross.
Mixing inclusive and strict endpoints. The inequality type (≤ vs <) must match in the answer.
Stopping after finding one root when factorising. Cubics need three roots (or one root and an irreducible quadratic).
Using the wrong method: graph method on a polynomial you can’t easily sketch leads to errors. If unsure, use the sign table.
Forgetting that some inequalities have isolated solutions: (x − k)² ≤ 0 has just one solution, x = k.

And that closes Section 2.9 — Inequalities. With both methods at your disposal — graphical for messy mixed-function inequalities, algebraic with sign tables for polynomial ones — you can attack any inequality the IB throws at you. The next section, Modulus & Further Transformations, brings in the absolute value function and the more advanced graph manipulations like |f(x)|, f(|x|), and 1/f(x).

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