Polynomial Inequality Solver

Math Algebra • Inequalities

Written by STEM Calculators Team Published December 12, 2025 Updated February 24, 2026

Enter a polynomial and press Calculate. The solver will:

Form the Sturm sequence (with exact variation counts at \(\pm\infty\)).
Isolate and refine all real roots by bisection.
Test the sign of \(P(x)\) on each interval between roots.
Return the solution set as a union of intervals, and plot it.

Graph of \(y = P(x)\) (blue) with \(y = 0\) (orange dashed). Green segments on the \(x\)-axis show where the inequality holds.

Number line: thick green segments mark the solution intervals. Closed/open dots reflect \(\le,\ge\) vs <,>.

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6. Polynomial Inequality Solver — Theory

Method to solve polynomial inequalities \(P(x)\,\square\,0\) (\(\square\in\{<,\le,>,\ge\}\)) for polynomials up to degree 4.

1) Strategy overview

Find all real roots of \(P(x)\) (without multiplicity) and order them \(r_1 < r_2 < \dots < r_k\).
Interval partition: \((-\infty,r_1), (r_1,r_2), \dots, (r_k,\infty)\).
Test-point method: on each open interval \(P(x)\) has constant sign; evaluate once to decide.
Endpoints: include real roots for \(\le,\ge\); exclude for <,>.

2) Sturm sequence with exact counts at \(\pm\infty\)

Build the Sturm sequence \(P_0=P,\; P_1=P'\), and \(P_{k+1}=-\mathrm{rem}(P_{k-1},P_k)\) until constant. The number of real roots in \((a,b]\) is \(V(a)-V(b)\), where \(V(x)\) counts sign variations of \((P_0(x),\dots,P_m(x))\) ignoring zeros.

Using signs at \(\pm\infty\) of each \(P_i\), \[ \#\{\text{real roots}\}=V(-\infty)-V(+\infty), \] which avoids numerical fragility at large finite endpoints.

3) Bounding the search (Cauchy bound)

If \(P(x)=a_nx^n+\cdots+a_0\) with \(a_n\ne0\), then every real root satisfies \[ |x|\le 1+\max_{0\le i<n}\frac{|a_i|}{|a_n|}. \] The calculator splits \([-(B+1),B+1]\) to isolate each root, then refines by bisection.

4) Constructing the solution

With isolated roots (verified by \(|P(r)|\approx 0\)), order them and apply the test-point method on every open interval. Non-strict inequalities add the roots themselves. The final answer is the union of the kept intervals, shown also on a number line.

5) Special cases

Identically zero: \(P(x)\equiv 0\). Then \(P(x)\le0\) and \(\ge0\) hold for all \(x\); the strict ones never hold.
Constant nonzero: sign is constant on \(\mathbb{R}\); the solution is \(\mathbb{R}\) or \(\varnothing\).
Degree limit: this tool supports degree \(\le 4\).

6) Example

Solve \(x^4 - x^3 - 4x^2 + 4 \;>\; 0\).
Sturm counting gives the number of real roots, intervals are isolated and refined, then signs are tested to produce the union of solution intervals; the graph highlights them.

Frequently Asked Questions

What degrees of polynomials can the polynomial inequality solver handle?

It supports real polynomials up to degree 4 (quartic), using coefficients a4 through a0. Lower-degree polynomials work by leaving higher-degree coefficients as 0.

How does the calculator find the real roots of a polynomial inequality?

It builds a Sturm sequence to count and isolate real roots, then refines each root using bisection. The solution intervals are formed by testing the sign of P(x) between adjacent roots.

How are endpoints treated for < versus <= in polynomial inequalities?

Strict inequalities (< or >) exclude roots, so endpoints are open. Non-strict inequalities (<= or >=) include roots where P(x) = 0, so endpoints are closed.

Why is checking one test point per interval enough to solve P(x) compared to 0?

Between consecutive real roots, a polynomial is continuous and does not cross zero, so its sign stays constant on that interval. A single test point determines whether the whole interval satisfies the inequality.

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