Interval Method

The interval method (also called the sign chart method or test-point method) generalizes the approach from quadratic inequalities to any polynomial or rational inequality.

The Method

⚡Theorem — Interval Method Steps

To solve $P(x) > 0$ (or $\geq$ , $<$ , $\leq$ ):

Move everything to one side so you have $P(x) > 0$ .
Factor completely into linear and irreducible quadratic factors.
Find all real roots and mark them on a number line.
Test one point in each interval to determine the sign of $P(x)$ there.
Account for equality: include endpoints for $\geq$ or $\leq$ ; exclude them for $>$ or $<$ .

ℹNote — Multiplicity Matters

The sign alternates at roots with odd multiplicity (simple roots, triple roots, etc.) but does not alternate at roots with even multiplicity (double roots, etc.). An even-multiplicity root is where the graph touches the axis but does not cross it.

Example 1: Simple Polynomial

✎Example — Three Simple Roots

Solve $(x - 1)(x + 2)(x - 3) > 0$ .

Solution. Roots: $x = -2, 1, 3$ (all simple). These divide the number line into four intervals.

Test $x = 0$ : $(-1)(2)(-3) = 6 > 0$ . So the interval $(-2, 1)$ is positive.

Since signs alternate at simple roots:

Interval	$(-\infty, -2)$	$(-2, 1)$	$(1, 3)$	$(3, +\infty)$
Sign	$-$	$+$	$-$	$+$

Solution: $(-2, 1) \cup (3, +\infty)$ .

Example 2: Even Multiplicity

✎Example — Double Root

Solve $x^2(x - 1)(x + 2) \geq 0$ .

Solution. Roots: $x = -2$ (simple), $x = 0$ (double), $x = 1$ (simple).

Test each interval:

$x = -3$ : $9 \cdot (-4) \cdot (-1) = 36 > 0$
$x = -1$ : $1 \cdot (-2) \cdot 1 = -2 < 0$
$x = 0.5$ : $0.25 \cdot (-0.5) \cdot 2.5 = -0.3125 < 0$
$x = 2$ : $4 \cdot 1 \cdot 4 = 16 > 0$

Interval	$(-\infty, -2)$	$(-2, 0)$	$(0, 1)$	$(1, +\infty)$
Sign	$+$	$-$	$-$	$+$

The sign does not change at $x = 0$ (even multiplicity). Include endpoints since $\geq 0$ .

Solution: $(-\infty, -2] \cup \{0\} \cup [1, +\infty)$ .

Rational Inequalities

The method extends to rational inequalities $\dfrac{P(x)}{Q(x)} > 0$ . Treat the roots of $Q(x) = 0$ as additional critical points, but always exclude them from the solution (the expression is undefined there).

✎Example — Rational Inequality

Solve $\dfrac{x + 1}{x - 2} \leq 0$ .

Solution. Critical points: $x = -1$ (numerator zero), $x = 2$ (denominator zero, excluded).

Test: $x = 0$ : $\dfrac{1}{-2} = -0.5 < 0$ . So the interval $(-1, 2)$ is negative.

Interval	$(-\infty, -1)$	$(-1, 2)$	$(2, +\infty)$
Sign	$+$	$-$	$+$

Include $x = -1$ (where expression equals $0$ ), exclude $x = 2$ (undefined).

Solution: $[-1, 2)$ .