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Polynomial Inequalities



Key Terms
Definition
The Sign Chart Method
Factored Form and Sign Analysis
Root Multiplicity and Sign Changes
End Behavior
Polynomials That Do Not Factor Over the Rationals
Worked Examples
Summary: Filling the Sign Chart Without Test Points



Sign Charts, Root Multiplicity, and the Shape of the Solution

When a polynomial inequality reaches degree three or higher, the parabola-based reasoning of the quadratic case no longer applies — the graph can cross the axis multiple times, flatten at certain roots, and reverse direction repeatedly. The sign chart becomes the primary tool: find every real root, partition the number line, and determine the sign in each interval. The new ingredient is multiplicity. A root where the polynomial merely touches zero without crossing behaves differently from one where it passes through, and the sign chart must reflect this distinction to produce the correct solution.

Key Terms

Core Concepts

Polynomial Inequality— P(x)>0P(x) > 0 for degree ≥3\geq 3; sign chart with multiplicity awareness
Sign Analysis— the primary method: find roots, partition, test each interval

From Polynomials Category

Multiplicity— odd multiplicity causes sign change, even does not
End Behavior— determines the sign in the outermost intervals
Factoring— required to locate all real roots

See All Algebra Definitions →


Definition

A polynomial inequality has the form

P(x)>0P(x) > 0


(or <<, ≤\leq, ≥\geq), where P(x)P(x) is a polynomial of degree n≥3n \geq 3. The inequality asks where the polynomial takes positive or negative values — not where it equals zero, but which stretches of the number line lie above or below the horizontal axis.

Linear and quadratic inequalities are special cases corresponding to n=1n = 1 and n=2n = 2. At those low degrees, the structure is simple enough that dedicated methods suffice. From degree three onward, the number of roots and the variety of sign patterns increase, and a general method is needed. The sign chart provides that method.

As with polynomial equations, the degree nn sets an upper bound: a polynomial of degree nn has at most nn real roots, and these roots create at most n+1n + 1 intervals on the number line. Within each interval, the polynomial maintains a constant sign. The task is to determine that sign in every interval and select those that satisfy the inequality.

The Sign Chart Method

The sign chart is a systematic procedure for solving any polynomial inequality once the real roots are known.

Step 1: Solve P(x)=0P(x) = 0 to find all real roots. Use factoring, the rational root theorem, or the quadratic formula after degree reduction — whatever combination of techniques produces the roots.

Step 2: Order the roots on the number line from smallest to largest. They divide the line into intervals: one to the left of the smallest root, one between each consecutive pair, and one to the right of the largest.

Step 3: In each interval, determine the sign of P(x)P(x). Choose any convenient test point within the interval and evaluate the sign — either by substituting into P(x)P(x) directly or, if P(x)P(x) is in factored form, by tracking the sign of each factor separately and multiplying.

Step 4: Select the intervals where the sign matches the inequality. For P(x)>0P(x) > 0, take the positive intervals. For P(x)<0P(x) < 0, take the negative ones.

Step 5: Decide on endpoint inclusion. For non-strict inequalities (≤\leq or ≥\geq), include the roots themselves because P(r)=0P(r) = 0 satisfies ≤0\leq 0 and ≥0\geq 0. For strict inequalities (<< or >>), exclude them.
Step Action Tool / detail
1 solve P(x) = 0 to find all real roots factoring; rational root theorem; quadratic formula after degree reduction; numerical methods if needed
2 order the roots on the number line smallest to largest; intervals form between consecutive roots
3 determine the sign of P(x) in each interval test points, factor signs from factored form, or end behavior + multiplicity
4 select intervals matching the inequality positive intervals for P(x) > 0; negative intervals for P(x) < 0
5 apply endpoint rules include roots for ≤ ⁄ ≥; exclude for < ⁄ >

Factored Form and Sign Analysis

When the polynomial is expressed as a product of linear factors, the sign in each interval can be determined without evaluating P(x)P(x) numerically. Instead, the sign of each individual factor is tracked, and the signs are multiplied together.

Consider (x+2)(x−1)(x−5)<0(x + 2)(x - 1)(x - 5) < 0. The roots are −2-2, 11, and 55, creating four intervals. In each interval, every factor is either positive or negative depending on whether the test point lies to the left or right of that factor's root.

In (−∞,−2)(-\infty, -2): all three factors are negative. Three negatives multiply to a negative. The product is negative.

In (−2,1)(-2, 1): the factor (x+2)(x + 2) is positive, while (x−1)(x - 1) and (x−5)(x - 5) are negative. One positive and two negatives: the product is positive.

In (1,5)(1, 5): the factors (x+2)(x + 2) and (x−1)(x - 1) are positive, (x−5)(x - 5) is negative. Two positives and one negative: the product is negative.

In (5,∞)(5, \infty): all three factors are positive. The product is positive.

The inequality <0< 0 selects the negative intervals: (−∞,−2)∪(1,5)(-\infty, -2) \cup (1, 5). No numerical evaluation of P(x)P(x) was needed — only sign tracking.

This approach is faster than substituting test points into the expanded polynomial, especially for higher-degree expressions with many terms. It requires the polynomial to be in factored form, which is why factoring is typically the first step.

Root Multiplicity and Sign Changes

Not every root causes the sign of the polynomial to change. Whether the sign flips at a root depends on the root's multiplicity.

At a root of odd multiplicity (simple root, triple root, etc.), the polynomial changes sign. The graph crosses the xx-axis, and the intervals on either side have opposite signs. A factor (x−r)1(x - r)^1 is negative to the left of rr and positive to the right, so crossing it flips the sign of the product.

At a root of even multiplicity (double root, quadruple root, etc.), the polynomial keeps its sign. The graph touches the xx-axis and turns back without crossing. A factor (x−r)2(x - r)^2 is positive on both sides of rr (and zero at rr itself), so crossing it does not flip the sign of the product.

Consider (x+1)(x−3)2>0(x + 1)(x - 3)^2 > 0. The roots are x=−1x = -1 (multiplicity 11) and x=3x = 3 (multiplicity 22). In (−∞,−1)(-\infty, -1): (x+1)(x + 1) is negative and (x−3)2(x - 3)^2 is positive — product is negative. In (−1,3)(-1, 3): (x+1)(x + 1) is positive and (x−3)2(x - 3)^2 is positive — product is positive. In (3,∞)(3, \infty): both are positive — product is positive. The sign changes at x=−1x = -1 (odd multiplicity) but not at x=3x = 3 (even multiplicity). The solution is (−1,3)∪(3,∞)(-1, 3) \cup (3, \infty), which simplifies to (−1,∞)∖{3}(-1, \infty) \setminus \{3\} — all x>−1x > -1 except x=3x = 3, where the expression equals zero.

This multiplicity rule eliminates the need to test points in every interval. Once the sign is known in any one interval, the sign in each adjacent interval is determined by whether the separating root has odd or even multiplicity.
Multiplicity at root r Example factor Sign behavior crossing r Graph behavior at r
Odd  (1, 3, 5, …) (x − r),  (x − r)³ flips — adjacent intervals have opposite signs graph crosses the x-axis
Even  (2, 4, 6, …) (x − r)²,  (x − r)⁴ preserved — adjacent intervals have the same sign graph touches the x-axis but does not cross (tangent)

End Behavior

The sign of a polynomial for very large values of ∣x∣|x| is controlled entirely by the leading term anxna_nx^n. This provides a guaranteed starting point for the sign chart: the sign in the outermost intervals is known before any test points are evaluated.

For even degree with an>0a_n > 0: the polynomial is positive for both very large positive and very large negative xx. Both outer intervals are positive.

For even degree with an<0a_n < 0: the polynomial is negative on both outer intervals.

For odd degree with an>0a_n > 0: the polynomial is negative far to the left and positive far to the right. The leftmost interval is negative and the rightmost is positive.

For odd degree with an<0a_n < 0: the polynomial is positive far to the left and negative far to the right. The signs are reversed from the previous case.

Knowing the outermost signs and applying the multiplicity rule inward determines the sign in every interval without any test-point computation. Start from the rightmost interval (whose sign is the sign of ana_n for odd degree, or positive for even degree with an>0a_n > 0), then flip or preserve the sign at each root moving leftward according to its multiplicity. This technique is faster and less error-prone than substituting numerical test points.
Case Leftmost interval (x → −∞) Rightmost interval (x → +∞)
Even degree, aₙ > 0 + +
Even degree, aₙ < 0 − −
Odd degree, aₙ > 0 − +
Odd degree, aₙ < 0 + −

Polynomials That Do Not Factor Over the Rationals

The sign chart method requires knowing the real roots, but it does not require those roots to be rational. When the polynomial has no rational roots — when the rational root theorem produces no valid candidates — irrational or approximate roots still serve as valid boundary points.

The polynomial x3−2=0x^3 - 2 = 0 has the single real root x=23≈1.26x = \sqrt[3]{2} \approx 1.26. For the inequality x3−2>0x^3 - 2 > 0, this root divides the number line into two intervals. Since the leading coefficient is positive and the degree is odd, the polynomial is negative for x<23x < \sqrt[3]{2} and positive for x>23x > \sqrt[3]{2}. The solution is (23,∞)(\sqrt[3]{2}, \infty).

When multiple irrational roots exist and exact values are unavailable, numerical approximations from graphing or iterative methods (bisection, Newton's method) identify the boundary points to whatever precision is needed. The sign chart is then constructed using these approximate values. The logical structure — roots create intervals, signs are constant within intervals — holds regardless of whether the roots are rational, irrational, or only known approximately.

For polynomials that factor partially — say, one rational root found and a remaining irreducible quadratic with Δ<0\Delta < 0 — the irreducible factor contributes no real roots and maintains constant sign. It affects the overall sign in every interval uniformly but creates no new boundary points.

Worked Examples

A cubic with three distinct real roots: solve x3−6x2+11x−6≤0x^3 - 6x^2 + 11x - 6 \leq 0. The polynomial factors as (x−1)(x−2)(x−3)(x - 1)(x - 2)(x - 3). The roots 11, 22, 33 create four intervals. All three roots have odd multiplicity, so the sign alternates. The leading coefficient is positive and the degree is odd, so the rightmost interval (3,∞)(3, \infty) is positive. Moving left: (2,3)(2, 3) is negative, (1,2)(1, 2) is positive, (−∞,1)(-\infty, 1) is negative. The inequality ≤0\leq 0 selects the negative intervals and includes the roots: (−∞,1]∪[2,3](-\infty, 1] \cup [2, 3].

A quartic with a repeated root: solve (x+2)2(x−1)(x−4)>0(x + 2)^2(x - 1)(x - 4) > 0. The roots are −2-2 (multiplicity 22), 11 (multiplicity 11), and 44 (multiplicity 11). The leading term is x4x^4 (positive), so both outermost intervals are positive. Moving inward from the right: (4,∞)(4, \infty) is positive. At x=4x = 4 (odd), the sign flips: (1,4)(1, 4) is negative. At x=1x = 1 (odd), the sign flips: (−2,1)(-2, 1) is positive. At x=−2x = -2 (even), the sign stays: (−∞,−2)(-\infty, -2) is positive. The solution to >0> 0 is (−∞,−2)∪(−2,1)∪(4,∞)(-\infty, -2) \cup (-2, 1) \cup (4, \infty), which simplifies to (−∞,1)∪(4,∞)(-\infty, 1) \cup (4, \infty) with the single exclusion x≠−2x \neq -2 already handled by the strict inequality — but since P(−2)=0P(-2) = 0, the strict inequality excludes −2-2 automatically. The solution is (−∞,−2)∪(−2,1)∪(4,∞)(-\infty, -2) \cup (-2, 1) \cup (4, \infty).

A polynomial with an irrational root: solve x3−3x+1>0x^3 - 3x + 1 > 0. The rational root theorem yields no rational roots. Numerical methods show three real roots near x≈−1.88x \approx -1.88, x≈0.35x \approx 0.35, and x≈1.53x \approx 1.53. All have odd multiplicity, so the sign alternates. The leading coefficient is positive, so the pattern from right to left is +,−,+,−+, -, +, -. The solution is approximately (−1.88,0.35)∪(1.53,∞)(-1.88, 0.35) \cup (1.53, \infty).

Summary: Filling the Sign Chart Without Test Points

The sign chart for a polynomial inequality of degree three or higher can be filled in entirely without test points by combining the root multiplicity rule (Section 4) with the end behavior of the polynomial (Section 5). The table below collects the resulting workflow: each step lists what is determined at that stage and the tool that supplies it, building the sign chart from the rightmost interval inward.
Step What you determine How
1 all real roots, ordered on the number line factor; rational root theorem; quadratic formula; numerical methods if needed
2 multiplicity at each root read directly from the factored form (exponent of each factor)
3 sign in the rightmost interval end behavior — sign of aₙ for odd degree; sign of aₙ for even degree (both ends share it)
4 sign in every other interval, working leftward flip the sign at each root of odd multiplicity; keep it the same at each root of even multiplicity
5 solution set select intervals matching the inequality; include roots for ≤ ⁄ ≥, exclude for < ⁄ >