Polynomial and Rational Inequalities: Sign Charts

Q: What is a sign chart?

A number line marked with the critical numbers of an expression, showing whether the expression is positive or negative on each interval between them.

Q: What are the critical numbers of a rational expression?

The zeros of the numerator and the zeros of the denominator. The expression can change sign only at those numbers.

Q: Why can't I multiply both sides of a rational inequality by the denominator?

The denominator may be negative for some x, which would flip the inequality there. Move everything to one side and use a sign chart instead.

Q: Are the zeros of the denominator ever in the solution?

Never. The expression is undefined there, so they are always excluded, even for ≤ or ≥.

Q: Why doesn't a squared factor change the sign?

A square is never negative. (x − 1)² is positive on both sides of x = 1, so the expression keeps its sign as it passes through 1.

Quick answer

A sign chart solves any polynomial or rational inequality. Move everything to one side, factor, and mark the critical numbers: the zeros of every factor, and for a rational expression the zeros of the denominator too. The expression keeps one sign between critical numbers, so one test value per interval decides it. Zeros of the numerator are included for ≤ or ≥; zeros of the denominator never are, because the expression is undefined there. A squared factor touches zero without changing the sign.

What you'll learn

Build a sign chart from the factors of a polynomial
Solve rational inequalities without multiplying by the denominator
Decide which critical numbers belong in the solution
Recognize factors that do not change sign

From quadratics to anything that factors

Quadratic inequalities were solved by cutting the number line at the zeros and testing each piece. The same method works for any expression that factors, with one addition for fractions. Mark the critical numbers:

the zeros of every factor of the numerator, and
for a rational expression, the zeros of the denominator.

Between critical numbers the expression keeps one sign, so a sign chart, one test value per interval, solves the inequality.

y = x³ − 4x

The sign of x³ − 4x = x(x − 2)(x + 2)

Example 1 — A cubic

Solve $x^3 - 4x > 0$ .

Factor: $x(x - 2)(x + 2) > 0$ . The critical numbers $-2$ , $0$ and $2$ make four intervals, and the signs alternate, as the graph shows:

Interval	$x < -2$	$-2 < x < 0$	$0 < x < 2$	$x > 2$
Sign of $x(x - 2)(x + 2)$	$-$	$+$	$-$	$+$

The solution is $-2 < x < 0$ or $x > 2$ .

Why the sign changes only at critical numbers

Each factor $x - c$ is negative to the left of $c$ and positive to the right, so it switches sign at $c$ and nowhere else. The sign of a product or quotient depends only on how many of its factors are negative. That count changes only when some factor switches, which happens only at a critical number. The sign of a factored expression is a matter of counting negative factors, and the count can change only at a critical number.

The same count explains two special cases. A squared factor such as $(x - 1)^2$ is never negative, so the sign does not change at $1$ . And a factor in the denominator switches sign like any other, which is why its zero is a critical number. But the expression is undefined there, so that number is never a solution.

Worked examples

Common mistakes

Practice problems

Solve $x(x - 3)(x + 1) < 0$ .

Answer

$x < -1$ or $0 < x < 3$

Full solution

The critical numbers are $-1$ , $0$ and $3$ . Testing $-2$ , $-0.5$ , $1$ and $4$ gives $-$ , $+$ , $-$ and $+$ .
Solve $x^3 \ge x$ .

Answer

$-1 \le x \le 0$ or $x \ge 1$

Full solution

$x^3 - x = x(x - 1)(x + 1) \ge 0$ . The signs on the four intervals are $-$ , $+$ , $-$ , $+$ , and the critical numbers are included.
Solve $(x + 4)^2(x - 1) < 0$ .

Answer

$x < 1$ , except $x = -4$

Full solution

The squared factor is positive except at $-4$ , where the expression is $0$ and so not less than $0$ . Elsewhere the sign is the sign of $x - 1$ .
Solve $x^4 - 16 > 0$ .

Answer

$x < -2$ or $x > 2$

Full solution

$(x - 2)(x + 2)\left(x^2 + 4\right) > 0$ . The factor $x^2 + 4$ is always positive, so only $\pm 2$ are critical.
Solve $\displaystyle\frac{x - 4}{x + 1} < 0$ .

Answer

$-1 < x < 4$

Full solution

The critical numbers are $-1$ and $4$ ; the quotient is negative between them.
Solve $\displaystyle\frac{2x - 1}{x + 3} \ge 0$ .

Answer

$x < -3$ or $x \ge \tfrac{1}{2}$

Full solution

The critical numbers are $-3$ (excluded) and $\tfrac{1}{2}$ (included). The quotient is positive outside them.
Solve $\displaystyle\frac{x}{x - 2} > 3$ .

Answer

$2 < x < 3$

Full solution

$\tfrac{x}{x - 2} - 3 = \tfrac{x - 3(x - 2)}{x - 2} = \tfrac{6 - 2x}{x - 2} > 0$ . The critical numbers are $2$ and $3$ , and the quotient is positive between them.
Solve $\displaystyle\frac{x^2 - 9}{x} \le 0$ .

Answer

$x \le -3$ or $0 < x \le 3$

Full solution

$\tfrac{(x - 3)(x + 3)}{x}$ has critical numbers $-3$ , $0$ and $3$ . Signs: $-$ , $+$ , $-$ , $+$ . Include $\pm 3$ , exclude $0$ .
Solve $\displaystyle\frac{3}{x - 1} \ge 1$ .

Answer

$1 < x \le 4$

Full solution

$\tfrac{3}{x - 1} - 1 = \tfrac{4 - x}{x - 1} \ge 0$ . The critical numbers are $1$ (excluded) and $4$ (included), and the quotient is positive between them.
A student solves $\displaystyle\frac{x + 1}{x - 2} > 0$ by multiplying both sides by $x - 2$ , getting $x + 1 > 0$ , so $x > -1$ . What went wrong?

Hint

Test $x = 0$ in the original inequality.

Answer

Multiplying by $x - 2$ is wrong when $x - 2 < 0$ . The solution is $x < -1$ or $x > 2$ .

Full solution

At $x = 0$ the quotient is $\tfrac{1}{-2} < 0$ , so $0$ is not a solution, though the student’s answer includes it. The sign chart with critical numbers $-1$ and $2$ gives a positive quotient outside them.

Frequently asked questions

What is a sign chart?

A number line marked with the critical numbers of an expression, showing whether the expression is positive or negative on each interval between them.

What are the critical numbers of a rational expression?