Rational Inequalities: Sign Charts and Excluded Zeros

Imagine a bridge with a resting platform and a missing section. You can stand on the platform. You cannot stand on the gap, however neatly it appears on your map.

Zeros in rational inequalities behave a little like that. A numerator zero can be an allowed value. A denominator zero is a gap in the expression’s domain. Treating both as ordinary endpoints is one of the easiest ways to lose an otherwise correct answer.

Let’s make the distinction clear, build sign charts that actually explain the answer, and tackle the cases where the usual shortcuts become unreliable.

The rule to remember first

A rational expression is a quotient of polynomials, such as (x - 3)/(x + 2). We work with real values of x throughout this guide.

For an inequality written as P(x)/Q(x) ≥ 0 or P(x)/Q(x) ≤ 0:

• A zero of P(x) is included if Q(x) is nonzero there.
• A zero of Q(x) is always excluded.

For P(x)/Q(x) > 0 or P(x)/Q(x) < 0, zeros of P(x) are excluded too, because the inequality is strict.

What happens at x = a?	Value of the quotient	Can a be a solution?
Numerator is zero, denominator is nonzero	Zero	Yes, if equality with zero is allowed
Denominator is zero, numerator is nonzero	Undefined	Never
Both numerator and denominator are zero	Undefined in the original expression	Never

The phrase written with zero on one side matters. If the question compares a fraction with 2, its numerator zeros alone do not tell you where equality holds. First subtract 2 and combine the result into one fraction.

Why cross-multiplication can give the wrong answer

Consider 1/(x - 2) > 0.

If you multiply by x - 2 and simply write 1 > 0, you might conclude that every value except 2 works. But x = 1 gives -1 > 0, which is false.

The problem is that x - 2 does not have a fixed sign:

• When x > 2, it is positive. Multiplication preserves the inequality sign.
• When x < 2, it is negative. Multiplication reverses the inequality sign.
• When x = 2, the original fraction is undefined.

The correct answer is x > 2.

You may multiply an inequality by a variable expression if you know its sign on the domain you are considering. Otherwise, split into cases or use a sign chart. In most classroom questions, the chart is the cleaner route.

A reliable sign-chart method

Use this sequence whenever the denominator contains x:

1. Record the original restrictions. Find every value that makes any original denominator zero.
2. Move everything to one side. Leave zero on the other side.
3. Combine and factorise. Write one fraction and factor its numerator and denominator.
4. Mark the boundary values. Include real numerator zeros, denominator zeros and any restrictions lost during simplification.
5. Test each open interval. Find whether the expression is positive or negative there.
6. Choose the intervals and check individual boundary points. Include allowed zeros only when the inequality permits equality.

This method is consistent with OpenStax’s lesson on solving rational inequalities. All the worked calculations below use original examples.

Why is one test point enough? Between consecutive marked values, a rational function is continuous and has no zeros or undefined points. Its sign cannot change within that interval. That reasoning depends on finding all the relevant real boundary values first.

If you are still getting comfortable with intervals, start with our guide to quadratic inequalities. The extra responsibility here is keeping denominator restrictions visible.

Example 1: One numerator zero and one denominator zero

Solve (x - 3)/(x + 2) ≥ 0.

The denominator gives x ≠ -2. The numerator is zero at x = 3. Mark -2 and 3, which create three open intervals.

Interval	Test x	Sign of x - 3	Sign of x + 2	Quotient
x < -2	-3	Negative	Negative	Positive
-2 < x < 3	0	Negative	Positive	Negative
x > 3	4	Positive	Positive	Positive

We want a non-negative quotient, so we choose the first and third intervals.

Now check the boundaries separately:

• At -2, the denominator is zero. Exclude it.
• At 3, the expression is 0/5 = 0. Include it.

Answer: x < -2 or x ≥ 3.

In interval notation: (-∞, -2) ∪ [3, ∞).

Notice that the square bracket belongs at 3, not at -2. The symbol ≥ does not give permission to include an undefined value.

Example 2: A nonzero right-hand side

Solve (x + 1)/(x - 2) ≤ 2.

First, x ≠ 2. Next, subtract 2:

(x + 1)/(x - 2) - 2 ≤ 0

Combining the terms gives:

[x + 1 - 2(x - 2)]/(x - 2) ≤ 0

Therefore:

(5 - x)/(x - 2) ≤ 0

The boundary values are 2 and 5.

Interval	Sign of 5 - x	Sign of x - 2	Quotient
x < 2	Positive	Negative	Negative
2 < x < 5	Positive	Positive	Positive
x > 5	Negative	Positive	Negative

We choose x < 2 and x > 5, then include 5 because equality is allowed. We still exclude 2.

Answer: (-∞, 2) ∪ [5, ∞).

A useful check is to substitute 5 into the original question: 6/3 = 2, so equality holds. The numerator zero of the original fraction was -1, but that was not the boundary at which the fraction equalled 2.

Example 3: Several factors and a strict inequality

Solve [(x + 3)(x - 2)]/[(x + 1)(x - 4)] < 0.

The restrictions are x ≠ -1, 4. The numerator zeros are -3 and 2. Arrange all four values in increasing order:

-3, -1, 2, 4

A test value in each interval gives the following result:

Interval	Test x	Numerator sign	Denominator sign	Quotient sign
x < -3	-4	Positive	Positive	Positive
-3 < x < -1	-2	Negative	Positive	Negative
-1 < x < 2	0	Negative	Negative	Positive
2 < x < 4	3	Positive	Negative	Negative
x > 4	5	Positive	Positive	Positive

We want the negative intervals.

Answer: (-3, -1) ∪ (2, 4).

All four endpoints are excluded, but for two different reasons. The numerator zeros -3 and 2 give zero, which fails the strict inequality. The denominator zeros -1 and 4 are undefined.

If the question had used ≤ instead, the answer would have been [-3, -1) ∪ [2, 4). Only the numerator endpoints would change.

Example 4: Cancelling a factor does not restore a missing value

Solve (x² - 9)/(x - 3) ≥ 0.

Before factorising, record x ≠ 3.

Now:

(x² - 9)/(x - 3) = [(x - 3)(x + 3)]/(x - 3) = x + 3, provided x ≠ 3.

The simplified inequality is x + 3 ≥ 0, which gives x ≥ -3. But the original restriction still applies.

Answer: [-3, 3) ∪ (3, ∞).

At x = -3, the original expression is 0/(-6), so inclusion is valid. At x = 3, it is 0/0, which is undefined.

Graphically, the expression follows the line y = x + 3 except for a hole at x = 3. The hole is not a vertical asymptote. OpenStax discusses this distinction in its rational functions chapter.

This also matters when several fractions are combined. A denominator restriction from either original fraction remains in force even if the combined numerator later cancels the troublesome factor.

Example 5: A repeated numerator zero can be an isolated answer

Solve (x - 2)²/(x + 1) ≤ 0.

The restriction is x ≠ -1. The numerator is positive everywhere except at x = 2, where it is zero.

For x < -1, the denominator is negative, so the quotient is negative. For x > -1, the denominator is positive, so the quotient is positive except at 2.

The negative interval is x < -1. However, we also need to include the zero at x = 2, because the question allows equality.

Answer: (-∞, -1) ∪ {2}.

The braces mean that 2 is a single included value. There is no small interval around 2 in the solution.

Try x = 1 and x = 3. Both give positive fractions and fail the inequality. At x = 2, the fraction is zero and succeeds.

This is why you should not merely shade some intervals and stop. Inspect the numerator zeros individually, even when both neighbouring intervals are rejected.

Example 6: A squared denominator does not change sign

Solve (x - 4)/(x + 2)² > 0.

The restriction is x ≠ -2. At every other real value, the denominator is positive. Therefore, the quotient’s sign matches the sign of x - 4.

We need x - 4 > 0, so x > 4.

Answer: (4, ∞).

The expression is negative on both sides of -2. Crossing a denominator zero does not automatically reverse the sign. Here the squared factor stays positive on both sides, although the expression is undefined at the point itself.

A common shortcut says to alternate signs at every marked value. That works for distinct simple factors, but it fails when a relevant factor has even multiplicity.

For a factor (x - a) raised to a power:

• An odd power changes sign across a.
• An even power keeps the same sign across a.

For a reduced rational expression, this rule applies to uncancelled numerator and denominator factors. Keep cancelled restrictions marked as excluded points, then analyse the remaining factors. Always determine at least one interval’s actual sign; do not assume the rightmost interval is positive if there is a negative constant factor.

Example 7: A denominator that never vanishes

Solve (x - 1)/(x² + 4) ≤ 0.

Because x² ≥ 0, we have x² + 4 > 0 for every real x. There are no real denominator restrictions and no denominator boundary values.

The fraction is non-positive exactly when x - 1 ≤ 0.

Answer: (-∞, 1].

Do not force every denominator into two real linear factors. Sometimes noticing that it is always positive makes the whole question shorter.

A commerce application: keeping the practical domain

Suppose a workshop’s total cost for q notebooks is C(q) = 1800 + 30q rupees. For this simplified model, q must be a positive integer.

The average cost per notebook is:

A(q) = (1800 + 30q)/q = 1800/q + 30

Find the production quantities for which the average cost is at most ₹45.

We need:

1800/q + 30 ≤ 45

Thus 1800/q ≤ 15. Here multiplying by q is safe because the practical domain explicitly tells us q > 0.

Therefore, 1800 ≤ 15q, giving q ≥ 120.

The answer is 120 or more notebooks, with q an integer. At 120 notebooks, total cost is ₹5,400 and average cost is ₹45. At 100 notebooks, average cost is ₹48, so the target is not met.

If you mechanically solved the corresponding rational inequality over all real numbers, negative quantities could appear among the algebraic solutions. The business meaning rules them out. A correct interval answer must still respect the question’s domain.

How to write the final answer clearly

Use round brackets for excluded finite endpoints and square brackets for included finite endpoints. Infinity always takes a round bracket because it is not a real endpoint you can include.

The union symbol ∪ means “or”: either interval is allowed. Braces such as {2} identify an individual value.

On a number line, use open circles for excluded values and filled circles for included ones. Shade only the intervals that satisfy the inequality. If one excluded value sits inside an otherwise accepted region, show the hole instead of drawing an unbroken shaded line through it.

A concise written solution should show the restrictions, the combined fraction, the ordered boundary values, the interval signs and the final set. This makes the reasoning visible without requiring pages of substitutions.

Try these before reading the answers

1. Solve (x + 4)/(x - 1) > 0.
2. Solve (x - 5)/(x + 2) ≤ 0.
3. Solve (x² - 1)/(x - 1) < 0.
4. Solve (x + 2)²/(x - 3) ≥ 0.
5. Solve (x - 1)/(x + 4)² ≤ 0.
6. Solve 2/(x - 1) ≥ 1.
7. Solve (x² + 1)/(x - 2)² < 0.
8. Solve (x - 2)/(x - 2) ≤ 1.

Answers with a quick reason

1. (-∞, -4) ∪ (1, ∞). The factors have the same sign there. Both endpoints are excluded.
2. (-2, 5]. The quotient is negative between -2 and 5, and zero at 5.
3. (-∞, -1). Simplify to x + 1 while keeping x ≠ 1. The strict inequality excludes -1.
4. {-2} ∪ (3, ∞). The quotient is positive for x > 3 and zero at -2. The denominator excludes 3.
5. (-∞, -4) ∪ (-4, 1]. The denominator is positive except at -4, so require x ≤ 1 and remove -4.
6. (1, 3]. Combining gives (3 - x)/(x - 1) ≥ 0. The equality point is 3, while 1 is forbidden.
7. No real solution. The expression is positive wherever defined. The excluded value 2 cannot help.
8. (-∞, 2) ∪ (2, ∞). The expression equals 1 everywhere in its original domain. It is still undefined at 2.

Before moving on, explain why Questions 4 and 8 cannot be answered by blindly alternating signs. If you can do that, you understand the method rather than just the pattern.

Further reading

For extra practice with domain restrictions and cancellation, see OpenStax’s lesson on simplifying rational expressions. For the related idea of using signs to describe a function’s behaviour, read our guide to increasing and strictly increasing functions.

Frequently asked questions

What is a rational inequality?

It is an inequality involving a ratio of polynomials. Solving it means finding every permitted value of the variable for which the comparison is true, usually as intervals or a union of intervals.

Why are denominator zeros always excluded?

Division by zero is undefined. An inequality can compare real values, but an undefined fraction does not supply a real value to compare. Equality being allowed does not change this restriction.

Are numerator zeros always included for ≥ or ≤?

Only after the inequality has been written as one fraction compared with zero, and only if the original expression is defined at that value. If the denominator is also zero, the value is excluded.

What happens when the same factor appears on top and bottom?

You may cancel it to simplify the calculation, but you must retain the restriction from the original denominator. The simplified expression agrees with the original only on the original domain.

Can I cross-multiply a rational inequality?

Yes, when the multiplier’s sign is known and nonzero. Preserve the inequality for a positive multiplier and reverse it for a negative one. If its sign depends on x, use cases or a sign chart.

Does the sign always change at a numerator or denominator zero?

No. An uncancelled factor with even multiplicity does not change sign across its zero. An odd-multiplicity factor does. If several factors share a zero, simplify and analyse their combined effect while retaining domain restrictions.

Can a solution contain just one isolated point?

Yes. For example, (x - 2)²/(x + 1) ≤ 0 includes x = 2 even though nearby values on both sides fail. This is why boundary points deserve their own check after the intervals.

Is every denominator zero a vertical asymptote?

No. A cancelled denominator factor can leave a hole instead. Both holes and vertical asymptotes correspond to excluded inputs in the original expression, but their graphs behave differently nearby.

Why do we put zero on one side before making a chart?

A sign chart tells us where an expression is positive, negative or zero. Moving all terms to one side turns the original comparison into exactly that question and identifies the correct equality points.

How can I check my answer quickly?

Substitute one value from each selected interval and one from each rejected interval into the original inequality. Then check every finite endpoint separately and compare the result with the original domain restrictions. These checks are most useful after a complete sign chart, not as a replacement for it.