Sign Charts for Products and Quotients, foundations for calculus lesson

To decide whether a product is positive you do not need its value, only the signs of its factors. Two negatives make a positive, one negative makes a negative. A sign chart organizes that idea on a number line, and it lets you solve inequalities like $\frac{x + 2}{x - 3} \ge 0$ without the risky step of multiplying by something whose sign you do not know. Calculus uses the same chart to decide where a derivative is positive or negative.

Key terms

Zero of a factor

Definition A zero of a factor is a real number at which that factor equals $0$ . A linear factor $ax + b$ with $a \ne 0$ has exactly one zero, $x = -\frac{b}{a}$ . It is negative on one side of that zero and positive on the other. A factor with no real zeros, such as $x^2 + 4$ , keeps a single sign for every real $x$ .

In plain words Each factor switches sign only where it equals zero, so its zero is the only place to watch.

Example The factor $2x - 3$ is zero at $x = \frac{3}{2}$ . It is negative for $x < \frac{3}{2}$ , as $2 \cdot 0 - 3 = -3$ shows, and positive for $x > \frac{3}{2}$ . The factor $x^2 + 4$ is at least $4$ everywhere, so it is always positive and never affects the sign of a product.

Critical input

Definition For an expression written as a product or quotient of factors, a critical input is a real number where some factor of the numerator is zero or some factor of the denominator is zero. Before simplifying anything, list every real zero of the numerator and every real zero of the denominator of the original expression. When every factor is a polynomial, these are the only inputs where the sign of the expression can change.

In plain words Critical inputs are the break points of the number line. Between them, nothing can switch the sign.

Example For

R(x) = \frac{x + 2}{x - 3}

the numerator is zero at $x = -2$ and the denominator is zero at $x = 3$ . The critical inputs are $-2$ and $3$ , and they split the line into $(-\infty, -2)$ , $(-2, 3)$ and $(3, \infty)$ .

Denominator exclusion

Definition An input where any denominator of the original expression is zero is not in the domain of the expression, so it can never belong to the solution of an inequality involving that expression. This stays true after cancellation. If a factor cancels, its zero is still excluded, since the original expression is undefined there.

In plain words A zero in a denominator means the expression has no value at that input, so the input can satisfy no inequality. Cancelling a factor does not bring it back.

Example Solve

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

For $x \ne 1$ the left side equals $x + 2$ , so the inequality becomes $x + 2 \ge 0$ , that is $x \ge -2$ . But $x = 1$ is excluded from the start. The solution set is $[-2, 1) \cup (1, \infty)$ , not $[-2, \infty)$ .

Sign chart

Definition A sign chart is a number line marked with all critical inputs, with one row for each factor and one row for the whole expression, showing the sign on each open interval into which the critical inputs cut the line, including the two unbounded intervals at the ends. It is valid because no factor is zero or undefined inside any of those open intervals. A polynomial factor with no zero in an interval keeps one sign on that interval, since its graph has no breaks and could only change sign by passing through $0$ . So each row has a constant sign on each interval, and the sign of the expression follows by multiplying and dividing the signs.

In plain words Mark the break points, write a plus or minus for each factor on each piece, and combine them. An even number of minus signs gives a plus.

Example For $R(x) = \frac{x + 2}{x - 3}$ the chart reads as follows.

Interval	$x + 2$	$x - 3$	$R(x)$
$x < -2$	$-$	$-$	$+$
$-2 < x < 3$	$+$	$-$	$-$
$x > 3$	$+$	$+$	$+$

One test value per interval confirms it. The values are $R(-3) = \frac{1}{6}$ , $R(0) = -\frac{2}{3}$ and $R(4) = 6$ .

Positive and negative intervals

Definition The positive intervals of an expression are the open intervals between critical inputs on which it is greater than $0$ , and the negative intervals are those on which it is less than $0$ . The critical inputs themselves belong to neither, since there the expression is either $0$ or undefined.

In plain words Once the chart is filled in, read off the pieces with a plus and the pieces with a minus.

Example From the chart above, $R$ is positive on $(-\infty, -2)$ and on $(3, \infty)$ and negative on $(-2, 3)$ . At $-2$ it equals $0$ , and at $3$ it is undefined. On a graph, this means the curve is above the $x$ -axis left of $-2$ and right of $3$ , and below it in between.

Product inequality

Definition A product inequality has the form $P(x) > 0$ , $P(x) \ge 0$ , $P(x) < 0$ or $P(x) \le 0$ , where $P$ is a product of polynomial factors whose real zeros you can find. Move every term to one side so the other side is $0$ , factor, build the sign chart, and select the intervals with the required sign. For $\ge$ or $\le$ , also include the zeros of $P$ .

In plain words Compare with zero, factor, and let the signs decide. Never compare a product with a nonzero number factor by factor.

Example Solve $(x - 1)(x + 3) \le 0$ . The critical inputs are $-3$ and $1$ . Left of $-3$ both factors are negative and the product is positive. Between $-3$ and $1$ the factor $x + 3$ is positive and $x - 1$ is negative, so the product is negative. Right of $1$ both are positive. The product is negative on $(-3, 1)$ and zero at $-3$ and $1$ , so the solution set is $[-3, 1]$ .

Rational inequality

Definition A rational inequality compares a quotient of polynomials with a number or with another expression. To solve it safely, subtract to get $0$ on one side, combine into a single fraction $\frac{N(x)}{D(x)}$ , factor $N$ and $D$ , and use a sign chart. Multiplying both sides by a denominator is not safe when its sign is unknown, because multiplying by a negative number reverses an inequality.

In plain words Do not clear the denominator. Bring everything to one side, make one fraction, and read the signs.

Example Solve

\frac{x}{x - 1} \ge 2.

Subtract $2$ and combine over the common denominator.

\begin{aligned} \frac{x}{x - 1} - 2 &= \frac{x - 2(x - 1)}{x - 1} \\[0.4em] &= \frac{2 - x}{x - 1} \ge 0. \end{aligned}

The critical inputs are $1$ and $2$ . The fraction is negative for $x < 1$ , positive for $1 < x < 2$ and negative for $x > 2$ . Including the numerator zero $2$ and excluding the denominator zero $1$ gives $(1, 2]$ . Multiplying by $x - 1$ at the start would have produced $x \le 2$ , which wrongly includes $x = 0$ , where the left side is $0$ and $0 \ge 2$ is false.

Endpoint test

Definition After choosing intervals from the sign chart, decide each critical input separately using the original inequality.

A zero of the numerator is included for $\ge$ or $\le$ when every denominator is nonzero there, since the expression then equals $0$ . It is excluded for $>$ or $<$ .
A zero of any denominator is always excluded, whatever the inequality sign.

In plain words The chart gives the open pieces. The endpoints need their own check, and division by zero always rules a point out.

Example For $\frac{x + 2}{x - 3} \ge 0$ the chart picks $(-\infty, -2)$ and $(3, \infty)$ . At $x = -2$ the expression is $\frac{0}{-5} = 0$ , and $0 \ge 0$ is true, so $-2$ is included. At $x = 3$ the denominator is zero, so $3$ is excluded. The solution set is $(-\infty, -2] \cup (3, \infty)$ .

Check yourself

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

Show the answer

Factoring gives $(x - 3)(x + 2) > 0$ . The product is positive when both factors have the same sign, which happens for $x < -2$ and for $x > 3$ . The inequality is strict, so the zeros are excluded and the solution set is $(-\infty, -2) \cup (3, \infty)$ .

Solve $\frac{x - 4}{(x + 1)^2} \le 0$ .

Show the answer

The denominator is positive for every $x \ne -1$ , so the sign of the fraction is the sign of $x - 4$ . That is negative for $x < 4$ and zero at $x = 4$ , which is allowed because the denominator is $25$ there. The input $-1$ is excluded. The solution set is $(-\infty, -1) \cup (-1, 4]$ .

Solve $\frac{3}{x - 2} < 1$ .

Show the answer

Subtracting $1$ and combining gives

\frac{3 - (x - 2)}{x - 2} = \frac{5 - x}{x - 2} < 0.

The critical inputs are $2$ and $5$ . The fraction is negative for $x < 2$ , positive between $2$ and $5$ , and negative for $x > 5$ . The inequality is strict and $2$ is excluded, so the solution set is $(-\infty, 2) \cup (5, \infty)$ .

Sign Charts for Products and Quotients

Key terms

Zero of a factor

Critical input

Denominator exclusion

Sign chart

Positive and negative intervals

Product inequality

Rational inequality

Endpoint test

Check yourself

How this lesson connects

Builds on

Used later in

Where this leads in Calculus