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.
Definition
A polynomial inequality has the form
P(x)>0
(or <, ≤, ≥), where P(x) is a polynomial of degree n≥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=1 and n=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 n sets an upper bound: a polynomial of degree n has at most n real roots, and these roots create at most n+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)=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). Choose any convenient test point within the interval and evaluate the sign — either by substituting into P(x) directly or, if 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)>0, take the positive intervals. For P(x)<0, take the negative ones.
Step 5: Decide on endpoint inclusion. For non-strict inequalities (≤ or ≥), include the roots themselves because P(r)=0 satisfies ≤0 and ≥0. For strict inequalities (< or >), exclude them.
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) numerically. Instead, the sign of each individual factor is tracked, and the signs are multiplied together.
Consider (x+2)(x−1)(x−5)<0. The roots are −2, 1, and 5, 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): all three factors are negative. Three negatives multiply to a negative. The product is negative.
In (−2,1): the factor (x+2) is positive, while (x−1) and (x−5) are negative. One positive and two negatives: the product is positive.
In (1,5): the factors (x+2) and (x−1) are positive, (x−5) is negative. Two positives and one negative: the product is negative.
In (5,∞): all three factors are positive. The product is positive.
The inequality <0 selects the negative intervals: (−∞,−2)∪(1,5). No numerical evaluation of 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 x-axis, and the intervals on either side have opposite signs. A factor (x−r)1 is negative to the left of r 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 x-axis and turns back without crossing. A factor (x−r)2 is positive on both sides of r (and zero at r itself), so crossing it does not flip the sign of the product.
Consider (x+1)(x−3)2>0. The roots are x=−1 (multiplicity 1) and x=3 (multiplicity 2). In (−∞,−1): (x+1) is negative and (x−3)2 is positive — product is negative. In (−1,3): (x+1) is positive and (x−3)2 is positive — product is positive. In (3,∞): both are positive — product is positive. The sign changes at x=−1 (odd multiplicity) but not at x=3 (even multiplicity). The solution is (−1,3)∪(3,∞), which simplifies to (−1,∞)∖{3} — all x>−1 except x=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.
End Behavior
The sign of a polynomial for very large values of ∣x∣ is controlled entirely by the leading term anxn. 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>0: the polynomial is positive for both very large positive and very large negative x. Both outer intervals are positive.
For even degree with an<0: the polynomial is negative on both outer intervals.
For odd degree with an>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<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 an for odd degree, or positive for even degree with an>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.
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=0 has the single real root x=32≈1.26. For the inequality x3−2>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<32 and positive for x>32. The solution is (32,∞).
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 — 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≤0. The polynomial factors as (x−1)(x−2)(x−3). The roots 1, 2, 3 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,∞) is positive. Moving left: (2,3) is negative, (1,2) is positive, (−∞,1) is negative. The inequality ≤0 selects the negative intervals and includes the roots: (−∞,1]∪[2,3].
A quartic with a repeated root: solve (x+2)2(x−1)(x−4)>0. The roots are −2 (multiplicity 2), 1 (multiplicity 1), and 4 (multiplicity 1). The leading term is x4 (positive), so both outermost intervals are positive. Moving inward from the right: (4,∞) is positive. At x=4 (odd), the sign flips: (1,4) is negative. At x=1 (odd), the sign flips: (−2,1) is positive. At x=−2 (even), the sign stays: (−∞,−2) is positive. The solution to >0 is (−∞,−2)∪(−2,1)∪(4,∞), which simplifies to (−∞,1)∪(4,∞) with the single exclusion x=−2 already handled by the strict inequality — but since P(−2)=0, the strict inequality excludes −2 automatically. The solution is (−∞,−2)∪(−2,1)∪(4,∞).
A polynomial with an irrational root: solve x3−3x+1>0. The rational root theorem yields no rational roots. Numerical methods show three real roots near x≈−1.88, x≈0.35, and x≈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,∞).