Math Core

When is a profit positive? When does a projectile stay above a certain height? Questions like these ask where a function is above or below zero, and for polynomial and rational functions there is one reliable method: find the few places where the sign can change, then test the pieces in between.

Why a sign chart works

A polynomial's graph is unbroken, so by the intermediate value theorem it can only change from positive to negative by passing through a zero. A rational function can also change sign by jumping across a vertical asymptote. Everywhere else, the sign stays the same.

Definition

Critical numbers

The critical numbers of an inequality $f(x) > 0$ (or $\ge$ , $<$ , $\le$ ) are the real zeros of $f$ and, for a rational function, the zeros of its denominator. They are the only places where $f$ can change sign.

The critical numbers cut the number line into intervals. On each interval $f$ has a single sign, so testing one value per interval tells you the sign on the whole interval.

Solving an inequality with a sign chart

Move everything to one side so the other side is $0$ . For a rational inequality, combine into a single fraction.
Factor the numerator and denominator completely.
Find the critical numbers and mark them on a number line.
Test one value in each interval (or use multiplicities) to find the sign of $f$ there.
Choose the intervals that match the inequality. Include zeros of $f$ for $\le$ or $\ge$ ; never include zeros of a denominator.

Polynomial inequalities

Worked example: A cubic inequality

Solve $x^3 - x^2 - 6x > 0$ .

Factor: $x^3 - x^2 - 6x = x(x^2 - x - 6) = x(x - 3)(x + 2)$ . The critical numbers are $-2$ , $0$ and $3$ .

interval	$x$ below $-2$	$-2$ to $0$	$0$ to $3$	$x$ above $3$
test $x$	$-3$	$-1$	$1$	$4$
value	$-18$	$4$	$-6$	$24$
sign	$-$	$+$	$-$	$+$

The inequality is strict, so the zeros themselves are excluded. The solution is

-2 < x < 0 \quad \text{or} \quad x > 3.

On the graph, these are exactly the $x$ -values where the curve is above the $x$ -axis.

y = x³ − x² − 6x is above the x-axis for −2 < x < 0 and for x > 3.Open in grapher →

−2 < x < 0 or x > 3

A shortcut with multiplicities

You don't always need test values. Start at the far right, where the sign matches the leading coefficient (for large $x$ , the leading term wins). Then move left across each critical number: the sign flips at a zero of odd multiplicity and stays the same at a zero of even multiplicity.

Worked example: A repeated factor

Solve $(x - 1)^2(x + 3)(x - 4) \le 0$ .

The critical numbers are $-3$ , $1$ and $4$ . The leading term is $x^2 \cdot x \cdot x = x^4$ , so the sign is $+$ to the right of $4$ .

Crossing $4$ (multiplicity $1$ ): flip to $-$ on the interval from $1$ to $4$ .
Crossing $1$ (multiplicity $2$ ): stay $-$ on the interval from $-3$ to $1$ .
Crossing $-3$ (multiplicity $1$ ): flip to $+$ for $x$ below $-3$ .

The expression is negative on $-3 < x < 1$ and $1 < x < 4$ , and it equals $0$ at $-3$ , $1$ and $4$ . Because of the "or equal to," all three zeros are included, and the pieces join up:

-3 \le x \le 4.

If the inequality had been strict, $(x - 1)^2(x + 3)(x - 4) < 0$ , you would have to remove $x = 1$ , where the expression is $0$ : the answer would be $-3 < x < 1$ or $1 < x < 4$ .

Rational inequalities

Rational inequalities use the same chart, with two extra rules. First, the zeros of the denominator are critical numbers too, and they are always excluded, even for $\le$ and $\ge$ , because $f$ is undefined there. Second, you must get $0$ on one side before you do anything else.

Common mistake

Never multiply both sides of an inequality by an expression like $x - 3$ . Its sign depends on $x$ : when it's negative the inequality should flip, and when it's positive it shouldn't, and you can't know which in advance. Subtract to get $0$ on one side and combine into a single fraction instead.

Worked example: Getting zero on one side

Solve $\dfrac{x + 1}{x - 3} \ge 2$ .

Subtract $2$ and use the common denominator $x - 3$ :

\frac{x + 1}{x - 3} - \frac{2(x - 3)}{x - 3} \ge 0 \quad\Longrightarrow\quad \frac{x + 1 - 2x + 6}{x - 3} \ge 0 \quad\Longrightarrow\quad \frac{7 - x}{x - 3} \ge 0.

The critical numbers are $7$ (a zero of the numerator, which is allowed) and $3$ (a zero of the denominator, which is never allowed).

interval	$x$ below $3$	$3$ to $7$	$x$ above $7$
test $x$	$0$	$5$	$8$
value of $\dfrac{7 - x}{x - 3}$	$-\dfrac{7}{3}$	$1$	$-\dfrac{1}{5}$
sign	$-$	$+$	$-$

The solution is $3 < x \le 7$ . Check an endpoint in the original: at $x = 7$ , $\dfrac{8}{4} = 2$ , and $2 \ge 2$ is true. ✓

3 < x ≤ 7

If you had multiplied both sides by $x - 3$ , you would have gotten $x + 1 \ge 2x - 6$ , or $x \le 7$ , which wrongly includes values like $x = 0$ (where the left side is $-\dfrac{1}{3}$ ).

Worked example: Four critical numbers

Solve $\dfrac{x^2 - 4}{x^2 - 2x - 3} < 0$ .

Factor:

\frac{(x - 2)(x + 2)}{(x - 3)(x + 1)} < 0.

The critical numbers are $-2$ , $-1$ , $2$ and $3$ , each with multiplicity $1$ . Far to the right the fraction behaves like $\dfrac{x^2}{x^2} = 1$ , which is positive. Every critical number has odd multiplicity, so the signs alternate moving left:

interval	$x$ below $-2$	$-2$ to $-1$	$-1$ to $2$	$2$ to $3$	$x$ above $3$
sign	$+$	$-$	$+$	$-$	$+$

The solution is $-2 < x < -1$ or $2 < x < 3$ . All endpoints are excluded: $-2$ and $2$ because the inequality is strict, and $-1$ and $3$ because they make the denominator $0$ .

Tip

Spot-check one interval with an actual number even when you use the multiplicity shortcut. If $x = 0$ lies in an interval you think is positive, $f(0)$ is quick to compute and catches most sign slips. Here $f(0) = \dfrac{-4}{-3} = \dfrac{4}{3} > 0$ , matching the chart.