Recipe

Solve a rational inequality

Use when: you're given an inequality like $p(x)/q(x) > 0$ (or $<, \ge, \le$ ), a fraction of two polynomials

Goal: the exact solution set in interval notation, with every hole and vertical asymptote correctly excluded

Taught in Asymptotes, holes, and graphs of rational functions · extends Solve a polynomial inequality

Quick recipe

Get zero on one side and factor completely
Cancel shared factors, but flag what each one becomes
Build the sign chart from the reduced expression
Assemble the answer, then punch out every hole

1
Move every term to one side so the other side is $0$ , then combine into a single fraction and factor the numerator and denominator completely.

Example

$\dfrac{x^2-4}{x^2-x-2} \ge 0$ is already in this form: $\dfrac{(x-2)(x+2)}{(x-2)(x+1)} \ge 0$ .

Why: Factored form is what turns a fraction into a list of critical values in the next step. This is the exact same first move a polynomial inequality uses, just applied to a fraction instead of one polynomial.
2
Cancel any factor shared by the numerator and denominator. A cancelled value is a hole. A denominator zero that doesn't cancel is a vertical asymptote. Both are excluded from the domain; only the vertical asymptote is a place the sign can flip.

Example

$(x-2)$ cancels: $x=2$ is a hole. $(x+1)$ survives: $x=-1$ is a vertical asymptote. The reduced expression is $\dfrac{x+2}{x+1}$ .
- The factor cancels completely — mark that $x$ -value a hole, excluded from the answer, with no sign flip there.
- The factor survives in the denominator — mark that $x$ -value a vertical asymptote, excluded from the answer, where the sign can flip.
Why: This is the one step a purely polynomial inequality never needs, since a polynomial has no denominator and therefore no holes or asymptotes at all.
3
Mark the reduced numerator's zeros and the surviving denominator zero(s) on a number line. Test one point per region, or track each factor's multiplicity, exactly as with a polynomial inequality.

Example

Reduced $\dfrac{x+2}{x+1}$ has a zero at $x=-2$ and a vertical asymptote at $x=-1$ . Testing $x=-3,-1.5,0$ gives signs $+,-,+$ .

Why: Once the fraction is reduced, the sign-chart logic is identical to the polynomial case, one factor at a time, one region at a time.
4
Include a numerator zero exactly when the inequality is non-strict, same boundary rule as a polynomial inequality. Never include a vertical asymptote. Then, as a final rational-only step, remove every hole from the answer too, even if it sits inside a region that otherwise qualifies.

Example

The reduced inequality is non-negative on $(-\infty,-2]\cup(-1,\infty)$ . The hole at $x=2$ sits inside $(-1,\infty)$ and must be carved out, giving the final answer $(-\infty,-2] \cup (-1,2) \cup (2,\infty)$ .

Why: The original expression is undefined at a hole no matter what the reduced expression says there, so no inequality, strict or not, can ever be satisfied at a point with no output at all.

Common mistakes

Treating a hole exactly like a vertical asymptote in the sign chart. A hole never flips the sign, but it's still excluded from the answer.
Including a denominator zero because the inequality is non-strict. That boundary rule only ever applies to zeros of the numerator.
Forgetting to punch the hole back out of the final answer after reducing, since the reduced expression alone can't tell you it's missing.
Working the sign chart on the un-reduced expression's factors but reading the domain restrictions off the reduced one. Always check both against the original, un-cancelled denominator.