In short

A polynomial inequality like (x - 1)(x - 3)(x + 2) > 0 asks: for which values of x is the polynomial positive (or negative)? The sign scheme (also called the wavy curve or method of intervals) answers this by marking the roots on the number line and tracking where the polynomial changes sign. For linear inequalities, the answer is a single interval. For higher-degree polynomials, it is a union of intervals.

A shopkeeper stocks boxes of 12 mangoes each. She rejects a box if more than 2 mangoes are damaged. One morning she finds that the number of damaged mangoes in a box satisfies 3x - 7 > 5, where x is a count per box. Which boxes get rejected? You need to solve an inequality — find the set of values where a condition holds, not just a single solution.

Equations ask "where does this expression equal zero?" Inequalities ask "where is it positive?" or "where is it negative?" The answer is not a single number but a range — an interval or a union of intervals on the number line. The technique for finding these intervals systematically is the sign scheme, and it works for any polynomial.

Solving linear inequalities

A linear inequality has the form ax + b > 0 (or \geq, <, \leq). The rules for solving it are almost identical to solving a linear equation, with one critical difference.

Rule 1. You can add or subtract any number from both sides without changing the inequality direction.

Rule 2. You can multiply or divide both sides by a positive number without changing the direction.

Rule 3. Multiplying or dividing by a negative number reverses the inequality.

Take 3x - 7 > 5. Add 7: 3x > 12. Divide by 3 (positive): x > 4. The solution set is the interval (4, \infty).

Take -2x + 6 \leq 10. Subtract 6: -2x \leq 4. Divide by -2 (negative — flip the sign): x \geq -2. The solution set is [-2, \infty).

Two linear inequalities solved on the number lineTwo number lines. The first shows x greater than 4 with an open circle at 4 and a ray pointing right. The second shows x greater than or equal to negative 2 with a filled circle at negative 2 and a ray pointing right. Solutions of linear inequalities on the number line −∞ +∞ 4 x > 4 3x − 7 > 5: −∞ +∞ −2 x ≥ −2 −2x + 6 ≤ 10:
Linear inequality solutions on the number line. An open circle ($\circ$) means the endpoint is excluded ($>$ or $<$); a filled circle ($\bullet$) means it is included ($\geq$ or $\leq$).

The critical point to remember: dividing by a negative number flips the inequality. Forgetting this is the single most common error in inequality problems.

From quadratic to polynomial: the sign scheme

You already know from quadratic inequalities that a quadratic like x^2 - 5x + 6 > 0 can be factored as (x-2)(x-3) > 0, and the solution comes from checking the sign in each interval created by the roots x = 2 and x = 3.

The same idea scales to any polynomial. Consider:

(x - 1)(x - 3)(x + 2) > 0

The roots are x = -2, 1, 3. These three points divide the number line into four intervals: (-\infty, -2), (-2, 1), (1, 3), (3, \infty). In each interval, the polynomial has a constant sign (positive or negative) because it is continuous and has no roots inside the interval.

The sign scheme (also called the wavy curve method or method of intervals) determines the sign in each interval without plugging in test points for every single one. Here is how it works.

The wavy curve method — step by step

Step 1. Factor the polynomial completely. Write the inequality in the form p(x) > 0 (or < 0, \geq 0, \leq 0), where p(x) is a product of linear factors.

Step 2. Mark the roots on the number line. Plot all real roots in increasing order.

Step 3. Determine the sign in the rightmost interval. For large positive x, the sign of p(x) is determined by the leading coefficient. If the leading coefficient is positive, p(x) > 0 in the rightmost interval.

Step 4. Alternate signs as you cross each root. Each time you pass through a root of odd multiplicity (simple root), the sign flips. Each time you pass through a root of even multiplicity (repeated root), the sign stays the same.

Step 5. Read off the solution intervals where the sign matches the inequality.

Sign scheme (wavy curve) for (x-1)(x-3)(x+2) greater than 0A number line with roots at x = -2, 1, 3 marked. Above the line, a wavy curve alternates between positive and negative regions. From right to left: positive for x greater than 3, negative between 1 and 3, positive between -2 and 1, negative for x less than -2. The positive regions are highlighted. (x − 1)(x − 3)(x + 2) > 0: the wavy curve −∞ +∞ −2 1 3 − + − + Solution: x ∈ (−2, 1) ∪ (3, ∞)
The wavy curve for $(x-1)(x-3)(x+2)$. Starting from the right (where the polynomial is positive, since the leading coefficient of $x^3$ is positive), the sign alternates at each root. The solution to $p(x) > 0$ is the union of the positive intervals: $(-2, 1) \cup (3, \infty)$.

Why does the sign alternate? At a simple root x = a, the factor (x - a) changes from negative to positive (or vice versa) as x crosses a. All other factors are nonzero at x = a and do not change sign in a small neighbourhood of a. So the product flips sign exactly once.

Repeated roots — when the sign does not flip

If a factor appears with even multiplicity — say (x - 2)^2 — then the corresponding factor is always non-negative near x = 2. The polynomial touches the x-axis at x = 2 but does not cross it. The sign does not change.

Consider (x - 1)(x - 2)^2 > 0. The roots are x = 1 (simple) and x = 2 (double).

Sign scheme for (x-1)(x-2)-squared greater than 0, with a repeated rootA number line with roots at x=1 (open circle) and x=2 (filled circle). The wavy curve is negative for x less than 1, positive for x between 1 and 2, and positive for x greater than 2. At x=2 the curve touches the axis but does not cross it, so the sign does not change. (x − 1)(x − 2)² > 0: repeated root at x = 2 1 (simple) 2 (double) − + + touches, does not cross Solution: x ∈ (1, 2) ∪ (2, ∞) = (1, ∞) \ {2} for ≥ 0: include the roots → x ∈ [1, ∞)
At the double root $x = 2$, the polynomial touches the axis but does not change sign. The sign rule: flip at odd-multiplicity roots, stay at even-multiplicity roots. For strict inequality ($> 0$), exclude the roots; for $\geq 0$, include them.

The rule to remember: odd multiplicity = sign change; even multiplicity = no sign change. This is the same behaviour you see in the graph of a polynomial — the curve crosses the axis at simple roots and bounces off at double roots.

Graphical interpretation

The sign scheme is really a shortcut for reading the graph. When p(x) > 0, the graph of y = p(x) lies above the x-axis. When p(x) < 0, the graph lies below.

Graph of y = (x+2)(x-1)(x-3) with positive and negative regions shadedA coordinate plane with the graph of the cubic polynomial (x+2)(x-1)(x-3). The curve passes through x=-2, x=1, and x=3. The regions where the curve is above the x-axis (between -2 and 1, and to the right of 3) are highlighted. The regions below the axis are not highlighted. x y −2 1 3 p(x) > 0 p(x) > 0 p(x) < 0 p(x) < 0
The graph of $y = (x+2)(x-1)(x-3)$. The shaded regions above the $x$-axis are where $p(x) > 0$: the intervals $(-2, 1)$ and $(3, \infty)$. The sign scheme is a compressed version of this picture — it captures the same information on a single number line.

The graph makes clear why the wavy curve works: a polynomial is continuous, so it can only change sign by passing through zero. Between consecutive roots, the sign is constant. The sign scheme extracts exactly this structural information without needing to plot the full graph.

Interactive: build your own wavy curve

Drag the three root points along the number line to set the roots of a cubic (x - a)(x - b)(x - c). The figure shows the sign in each interval and highlights the positive regions. Watch how the positive and negative regions shift as you move the roots.

Interactive wavy curve for a cubic with three movable rootsA number line with three draggable red points representing roots. As the roots move, the sign indicators in each interval update. The positive regions are highlighted. x y ↔ drag the red points
Move the three roots to see how the sign pattern changes. The cubic always has four intervals, and the signs always alternate (since all roots here are simple). The solution to $p(x) > 0$ is always the union of the positive intervals.

Two worked examples

Example 1: Solve $(x + 3)(x - 1)(x - 4) \leq 0$

Step 1. Identify the roots. The roots are x = -3, 1, 4.

Why: each linear factor (x - a) is zero at x = a. These are the only points where the polynomial can change sign.

Step 2. Set up the sign scheme. Mark -3, 1, 4 on the number line. The leading coefficient is positive (x^3 term), so the polynomial is positive in the rightmost interval (4, \infty).

Why: for very large x, all three factors are positive, so the product is positive.

Step 3. Alternate signs moving left: positive in (4, \infty), negative in (1, 4), positive in (-3, 1), negative in (-\infty, -3).

Why: each root has multiplicity 1 (odd), so the sign flips at every root.

Step 4. Read the solution. The inequality is \leq 0, so include the negative intervals and the roots themselves.

x \in (-\infty, -3] \cup [1, 4]

Step 5. Verify with test points. At x = -4: (-1)(-5)(-8) = -40 < 0. At x = 0: (3)(-1)(-4) = 12 > 0. At x = 2: (5)(1)(-2) = -10 < 0. At x = 5: (8)(4)(1) = 32 > 0. Consistent with the sign scheme.

Result: x \in (-\infty, -3] \cup [1, 4].

Number line solution of (x+3)(x-1)(x-4) less than or equal to 0A number line with roots at -3, 1, 4. The solution intervals are highlighted: from negative infinity to -3 (inclusive) and from 1 to 4 (inclusive). Signs in each interval are marked: negative, positive, negative, positive from left to right. (x + 3)(x − 1)(x − 4) ≤ 0 −∞ +∞ −3 1 4 − + − + x ∈ (−∞, −3] ∪ [1, 4] filled circles: roots included (≤ allows equality)
The sign scheme for $(x+3)(x-1)(x-4)$. The negative intervals (where $p(x) \leq 0$) are $(-\infty, -3]$ and $[1, 4]$, shown as thick red segments. The roots are included because the inequality is $\leq$ (not strict $<$).

The test points confirm the sign scheme. In practice, once you trust the method, you only need the leading coefficient and the multiplicity of each root — no test points required.

Example 2: Solve $x^2(x - 2)^3(x + 1) \geq 0$

Step 1. Identify the roots and their multiplicities. x = 0 (multiplicity 2, even), x = 2 (multiplicity 3, odd), x = -1 (multiplicity 1, odd).

Why: multiplicities matter because even-multiplicity roots do not cause a sign change, while odd-multiplicity roots do.

Step 2. Set up the sign scheme. Mark -1, 0, 2 on the number line. The leading term is x^6 (degree 2 + 3 + 1 = 6, positive coefficient), so p(x) > 0 in the rightmost interval (2, \infty).

Why: for large x, x^2 > 0, (x-2)^3 > 0, (x+1) > 0, so the product is positive.

Step 3. Apply the sign-change rule moving left from (2, \infty):

  • At x = 2 (odd multiplicity 3): sign flips → negative in (0, 2).
  • At x = 0 (even multiplicity 2): sign stays → negative in (-1, 0).
  • At x = -1 (odd multiplicity 1): sign flips → positive in (-\infty, -1).

Why: the double root at x = 0 means x^2 is always non-negative near 0 — the polynomial does not cross the axis there.

Step 4. Read the solution. The inequality is \geq 0, so include all intervals where p(x) > 0 and all roots (where p(x) = 0).

Positive intervals: (-\infty, -1) and (2, \infty). Roots: x = -1, 0, 2.

x \in (-\infty, -1] \cup \{0\} \cup [2, \infty)

Result: x \in (-\infty, -1] \cup \{0\} \cup [2, \infty).

Sign scheme for x-squared times (x-2)-cubed times (x+1) greater than or equal to 0A number line with roots at -1, 0, 2. Above the line, signs are marked: positive for x less than -1, negative between -1 and 0, negative between 0 and 2, positive for x greater than 2. At x=0, the sign does not change because the root has even multiplicity. The solution includes negative infinity to -1, the isolated point 0, and 2 to positive infinity. x²(x − 2)³(x + 1) ≥ 0 −∞ +∞ −1 (odd) 0 (even) 2 (odd) + − − + no flip x ∈ (−∞, −1] ∪ {0} ∪ [2, ∞) the isolated point {0} is included because x² = 0 there, satisfying ≥ 0
The double root at $x = 0$ does not cause a sign change — the polynomial is negative on both sides of $0$, but equals $0$ at $x = 0$ itself. The solution to $\geq 0$ includes the isolated point $\{0\}$ alongside the two positive rays.

The isolated point \{0\} in the solution is the hallmark of an even-multiplicity root under a non-strict inequality. The polynomial is negative on both sides of 0, but exactly zero at 0 — so 0 satisfies \geq 0 as a lone included point.

Common confusions

Going deeper

If you can factor a polynomial, apply the sign scheme with correct multiplicity rules, and write the solution as a union of intervals, you have the complete method for polynomial inequalities. What follows extends the technique.

Rational inequalities

The wavy curve method extends naturally to inequalities of the form \frac{p(x)}{q(x)} > 0. The key insight: the sign of a fraction depends on the signs of the numerator and denominator. The steps are:

  1. Factor both p(x) and q(x) completely.
  2. Mark all roots of p(x) and q(x) on the number line.
  3. Apply the sign scheme as before — each root of p or q contributes a potential sign change based on its multiplicity.
  4. Exclude the roots of q(x) from the solution (they make the expression undefined).

For example, \frac{(x-1)(x+3)}{(x-2)} > 0 has critical points at x = -3, 1, 2. The sign alternates: -, +, -, + from left to right (leading coefficient positive, degree of numerator minus denominator is 1, and the rightmost interval has positive sign). The solution is (-3, 1) \cup (2, \infty), with x = 2 excluded.

Why "wavy curve" is the right mental image

The name "wavy curve" comes from the shape of the polynomial graph between its roots. A polynomial of degree n with n distinct real roots crosses the x-axis n times, creating n + 1 regions that alternate in sign. The "wave" is the graph itself, rising and falling as it passes through each root. The sign scheme is a simplified one-dimensional projection of this two-dimensional picture — it keeps only the essential information (the sign) and discards the shape.

For polynomials with repeated roots, the graph "bounces" at even-multiplicity roots (like a ball bouncing off a wall) and "passes through" at odd-multiplicity roots. This physical intuition makes the multiplicity rule easy to remember.

Where this leads next