Polynomial Inequalities (AA HL)
Once an inequality involves a quadratic, a cubic, or a fraction of two expressions, you can't just "do the same thing to both sides" the way you can for a straight line - the sign of what you're multiplying by matters. This page covers the sign-chart method that handles all of these safely, whether the graphs are drawn for you or you're finding everything algebraically. It's part of the broader Inequalities & Modulus topic.
39 questions on this sub-topic.
The sign-chart method
Covered under IB syllabus reference AHL2.15. Neither idea below is a formula you look up - both are prior-knowledge techniques you're expected to apply from memory, on paper or with a GDC graph as a check.
Quadratics and cubics
Critical values + sign chart
Factorise (or find the roots), mark the critical values on a number line, then test the sign of the expression in each interval between them.
Rational inequalities
\[\dfrac{p(x)}{q(x)}\ \bowtie\ 0\]
Never cross-multiply by an unknown-sign expression - move everything to one side and combine into a single fraction first, then apply the same sign-chart method to the numerator and denominator together.
Need the full syllabus wording and formula-booklet reference table, or the modulus versions of these inequalities? See Inequalities & Modulus. A GDC graph is a quick way to confirm which intervals satisfy the inequality - see the parent topic's GDC guidance.
Worked examples
Solve \(x^2 - 2x - 8 \le 0.\)
Worked solution
\((x-4)(x+2) \le 0.\) M1
The upward parabola is \(\le 0\) between the roots A1
\(-2 \le x \le 4.\) A1
Solve \(\dfrac{x - 1}{x + 2} \ge 0\).
Worked solution
numerator zero at \(x = 1\); denominator zero at \(x\) M1 \(= -2\) (excluded). A1
across \(x<-2,\ -2<x<1,\ x>1\): the quotient is \(+,\ -,\ +.\) M1 A1
(include \(x=1\), exclude \(x=-2\)): \(x < -2\) or \(x \ge 1.\) A1
Show that \(x^2-6x+11>0\) for all real \(x.\)
Worked solution
\(x^2 - 6x + 11\) M1 \(= (x-3)^2 + 2.\) A1
\((x-3)^2 \ge 0\) M1 so the expression \(\ge 2.\) A1 Hence \(> 0\) for all real \(x.\) R1 ∎
Common mistakes
- Cross-multiplying a rational inequality by an unknown-sign expression. Multiplying \(\dfrac{x-3}{x+1}<2\) straight through by \((x+1)\) silently flips the inequality whenever \(x+1\) is negative - always move everything to one side and combine into a single fraction instead.
- Including a value that makes the denominator zero. In \(\dfrac{x-1}{x+2}\ge0\), \(x=-2\) must always be excluded from the solution, even though \(\ge\) would normally include a boundary - a rational expression is simply undefined there.
- Forgetting the parabola's direction changes which region satisfies the inequality. An upward parabola is negative between its roots and positive outside them; a downward one is the other way round - check the sign of the leading coefficient before reading off the interval.
Ready to practise properly?
39 polynomial and rational inequality questions, marked instantly like the real exam.
Quick answers
How do you solve a quadratic or cubic inequality?
Move everything to one side, factorise to find the critical values (the roots), then use a sign chart or the shape of the graph to work out which intervals satisfy the inequality.
Why can't you cross-multiply in a rational inequality?
Multiplying both sides by an expression of unknown sign, like \(x+2\), can silently flip the inequality when that expression is negative. Instead move everything to one side, combine into a single fraction, and apply a sign chart to the result.