Rational inequalities: studying the sign

End of September 2026, Year 3 at a liceo classico (the Italian high school focused on Latin, Greek and the humanities): the factorisation revision is wrapping up and you are starting rational inequalities (inequalities with fractions), before moving on to quadratic equations and the parabola. The first test falls in five to seven weeks, so between the end of October and the beginning of November. With 2 hours of maths a week there is very little class time to practise: if you arrive at the test with a written method, not an improvised one, you make up the difference on your own.

This chapter is not an isolated episode. It is the first time you really do sign analysis, the technique that comes back in January, identical — same scheme, same logic — when you have to read the sign of a parabola to solve a quadratic inequality. With 2 hours a week the parabola takes 5-6 weeks and the test on that chapter falls right in January, just before the report card: sign analysis with the parabola is the topic most often failed in the last three years of the classical track, and the most common reason is not the parabola itself, but the fact that basic sign analysis — what you are doing now, with linear factors — was never solid.

Why you don’t “multiply by the denominator”

With a rational equation you can multiply everything by the denominator (after writing the existence conditions) because you are comparing two equal quantities, and multiplying both by the same number does not change the equality — provided that number is not zero. With an inequality you cannot do it, for a precise reason: if you multiply both sides by an expression that contains $x$ , you do not know whether that expression is positive or negative for the value of $x$ you are looking for, and the direction of the inequality depends on exactly that. Multiplying by a negative denominator without knowing it flips the inequality without you noticing.

The solution is never to multiply by the denominator: you bring everything to one side, reduce everything to a single fraction, and study the sign of that fraction without ever removing the denominator from the calculation.

The procedure in four steps

Bring everything to the left-hand side. Whatever the starting inequality, subtract everything on the right-hand side, so that you have a single fraction compared with zero: $\frac{N ( x )}{D ( x )} ≷ 0$ .
Reduce to a common denominator, but do not remove the denominator. If there are several fractions, find the lowest common multiple of the denominators and write everything as a single fraction. The denominator stays where it is: you never cancel it, unlike what you would do in an equation.
Factorise numerator and denominator. This is where the revision of the last few weeks pays off: common factor, difference of squares, square of a binomial, special trinomial. Ruffini’s rule stays a last resort, for the cases where none of the earlier techniques works — it is not the priority for this test.
Study the sign of each factor, then combine. You do not study the sign of the whole expression in one go: you study it factor by factor, put each sign in a table, and read the sign of the product (or quotient) row by row, as in the multiplication of signs you already know.

Existence conditions: always first, never after

Even before factorising to study the sign, write the existence conditions (E.C.): the denominator, in the whole expression reduced to one side, must be different from zero. You find them by factorising the denominator and requiring each factor to be different from zero — exactly as you already did for rational equations, with one important difference: in an inequality, the zero of the denominator must always be excluded from the final solution, whatever the direction of the inequality. There is no case in which a value that makes the denominator zero can appear in the solution set.

The sign table: how to build it

Once the expression is factorised into factors — say $\frac{f _{1} ( x ) \cdot f _{2} ( x )}{f _{3} ( x )}$ — you proceed like this:

Find the zero of each factor by solving $f_{1} (x) = 0$ , $f_{2} (x) = 0$ , $f_{3} (x) = 0$ separately. They are linear equations, so immediate.
Draw one row for each factor, with the zeros placed in increasing order on the same number line.
On each row mark the sign of the factor: negative to the left of the zero, positive to the right, if the factor has a positive coefficient of $x$ (for example $x - 3$ is negative for $x < 3$ and positive for $x > 3$ ); the opposite if the coefficient is negative.
Add a final row for the sign of the whole product/quotient: column by column, multiply the signs of the rows above with the usual rule (same signs → positive, different signs → negative).
Read the result on the last row, taking into account the direction of the starting inequality and the E.C.

How to read the result: filled dots and exclusions

On the real number line, the dot is filled when the value is part of the solution (inequality with $\geq$ or $\leq$ , and the value is not excluded by the E.C.), open when it is excluded (inequality with $>$ or $<$ , or a value that makes the denominator zero). One important point, specific to rational inequalities: a zero of the denominator always has an open dot, even if the inequality uses $\geq$ or $\leq$ . The $\geq$ includes the zeros of the numerator, never those of the denominator.

The final solution is written as intervals, with round brackets where the value is excluded and square brackets where it is included: for example $(- 2, 3]$ means “from $- 2$ excluded to $3$ included”.

Four worked examples

1. A simple rational inequality

\frac{2 x - 6}{x + 1} < 0

Numerator and denominator are already reduced to the minimum, but it helps to take out the 2 so the zero reads more clearly:

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

E.C.: $x \neq = - 1$ . Zeros: numerator $x = 3$ , denominator $x = - 1$ (excluded).

	$x < - 1$	$- 1 < x < 3$	$x > 3$
$x - 3$	$-$	$-$	$+$
$x + 1$	$-$	$+$	$+$
quotient	$+$	$-$	$+$

We look for where the quotient is negative (because the inequality is $< 0$ ): $- 1 < x < 3$ . Both endpoints are excluded, $- 1$ because it is an E.C. and $3$ because the inequality is strict:

- 1 < x < 3

2. With a squared factor

\frac{( x + 2 ) ^{2} ( 3 - x )}{x - 1} \geq 0

E.C.: $x \neq = 1$ . Zeros: $x = - 2$ (from the squared factor), $x = 3$ (numerator), $x = 1$ (denominator, excluded).

The factor $(x + 2)^{2}$ is a square: it never changes sign, it is always positive or zero, whatever $x$ is. In the table it is marked with $+$ everywhere, with an isolated zero at $x = - 2$ .

	$x < - 2$	$- 2 < x < 1$	$1 < x < 3$	$x > 3$
$(x + 2)^{2}$	$+$	$+$	$+$	$+$
$3 - x$	$+$	$+$	$+$	$-$
$x - 1$	$-$	$-$	$+$	$+$
quotient	$-$	$-$	$+$	$-$

We look for where the quotient is $\geq 0$ : the interval $1 < x < 3$ is fine in full, but you also have to add the points where the quotient is exactly zero, that is, the zeros of the numerator that are not also zeros of the denominator. $x = - 2$ makes the numerator zero (through the square) without making the denominator zero: it is acceptable, and has to be included on its own. $x = 3$ makes the numerator zero and is the right-hand endpoint of the positive interval: it must be included. $x = 1$ stays excluded, it is an E.C.

x = - 2 or 1 < x \leq 3

The isolated point $x = - 2$ is the easiest part to forget in this exercise: without the squared factor, that zero would have made the row change sign and would have ended up inside a continuous interval; with the even exponent, the sign before and after stays the same, and that zero remains an isolated solution.

3. Three factors, non-strict inequality

\frac{( x - 2 ) ( x + 1 )}{x ^{2} - 9} \leq 0

Factorise the denominator, a difference of squares: $x^{2} - 9 = (x - 3) (x + 3)$ . E.C.: $x \neq = 3$ and $x \neq = - 3$ .

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

Zeros in order: $- 3$ (excluded, E.C.), $- 1$ (numerator), $2$ (numerator), $3$ (excluded, E.C.).

	$x < - 3$	$- 3 < x < - 1$	$- 1 < x < 2$	$2 < x < 3$	$x > 3$
$x - 2$	$-$	$-$	$-$	$+$	$+$
$x + 1$	$-$	$-$	$+$	$+$	$+$
$x - 3$	$-$	$-$	$-$	$-$	$+$
$x + 3$	$-$	$+$	$+$	$+$	$+$
quotient	$+$	$-$	$+$	$-$	$+$

We look for where the quotient is $\leq 0$ , so the negative intervals plus the points where the numerator is zero (without making the denominator zero): $- 3 < x < - 1$ and $2 < x < 3$ , with $x = - 1$ and $x = 2$ included (zeros of the numerator) and $- 3$ , $3$ excluded (E.C.):

- 3 < x \leq - 1 \lor 2 \leq x < 3

4. With ≥ and a factor that appears in both numerator and denominator

\frac{x ^{2} - 4}{x ^{2} - 5 x + 6} \geq 0

Factorise both: $x^{2} - 4 = (x - 2) (x + 2)$ , $x^{2} - 5 x + 6 = (x - 2) (x - 3)$ (I look for two numbers with product $6$ and sum $- 5$ : they are $- 2$ and $- 3$ ).

\frac{( x - 2 ) ( x + 2 )}{( x - 2 ) ( x - 3 )} \geq 0

Here the factor $(x - 2)$ appears identically above and below. E.C.: $x \neq = 2$ and $x \neq = 3$ . The factor $(x - 2)$ must not be cancelled before the sign table: keep it and treat it like any other factor, otherwise you lose sight of the fact that $x = 2$ is still excluded from the original domain.

	$x < - 2$	$- 2 < x < 2$	$2 < x < 3$	$x > 3$
$x - 2$	$-$	$-$	$+$	$+$
$x + 2$	$-$	$+$	$+$	$+$
$x - 3$	$-$	$-$	$-$	$+$
quotient	$-$	$+$	$-$	$+$

We look for where the quotient is $\geq 0$ : $- 2 < x < 2$ and $x > 3$ , plus the point $x = - 2$ (zero of the numerator, not of the denominator, included). The point $x = 2$ must be excluded, and it is already naturally excluded because it is the endpoint of the open interval — but the reason it is open there is not “the sign changes”, it is that it is an E.C.:

- 2 \leq x < 2 \lor x > 3

If someone cancelled the factor $(x - 2)$ straight away, getting $\frac{x + 2}{x - 3} \geq 0$ , and forgot to carry the exclusion $x \neq = 2$ into the final answer, the numerical result would still coincide with the one above (because $x = 2$ already falls inside an excluded interval), but that is a lucky coincidence: change the exercise even slightly and that omission produces a wrong result. The rule is to keep the factor as it is until the table is complete.

Typical mistakes

Cancelling the $x$ on both sides. In $\frac{x ^{2} - 4}{x} \geq x$ you cannot “cancel the $x$ ”: first you bring everything to the left, then you reduce to a single fraction, then you study the sign. Cancelling an $x$ that appears in several terms of an inequality, as if it were a factor common to the whole expression, is the same mistake as cancelling a term in a fraction: you only do it with factors of a product, never with terms of a sum.
Forgetting the minus sign in front of a factor. $3 - x$ does not have the same sign as $x - 3$ : they are opposite. If you take out a $- 1$ to rewrite $3 - x = - (x - 3)$ , that minus sign has to be carried into the table, not left implicit.
Including the zero of the denominator. However the inequality is written, $\geq$ or $\leq$ , a zero of the denominator never enters the solution: it is always excluded, with an open dot and a round bracket.

How to tell whether you are ready

Try redoing exercise 4 with the solution covered, starting only from the question. If in under two minutes you have correctly written the E.C., factorised both polynomials and drawn the table with all three rows (without having cancelled the common factor too early), the method is solid. If you get stuck on factorising the denominator, it is worth going back to the revision of factorisation and algebraic fractions, which covers in full the decision tree (common factor, difference of squares, special trinomial) on which this whole chapter rests.

On this specific topic — Year 3 rational inequalities — we do not yet have a dedicated practice test in the catalogue. To practise in the meantime on the logic of inequalities and on reading intervals, there is the Year 2 test on quadratic inequalities: it is not the same topic — there the sign is read from the parabola, here from linear factors — but it is good follow-up training on the same skill of reading intervals, once this chapter is solid. The complete index of tests by topic is at /simulazioni/verifiche-liceo/.

Why this chapter comes back in January

The sign analysis you have just learned with linear factors is exactly the same procedure you will use in January with quadratic inequalities: only instead of reading the sign of each linear factor from a table, you will read the sign of a whole trinomial from the graph of a parabola. The logic — find the zeros, split the line into intervals, read the sign interval by interval — is identical. If you arrive in January with this chapter solid, you only have to learn “how to read the sign from a parabola”; if you arrive with sign analysis still shaky, you have to learn two things at once, and that is where the difficulty builds up that makes that chapter the most failed of the last three years. The next step in the syllabus, after quadratic equations, will be precisely the parabola: when you get there, the reading tools you use here — filled and open dots, intervals, sign table — do not change, only what you are reading does. In Year 4, the same tools for reading a graph come back for the domain of composite functions, as described in functions in Year 4 at the classical high school.

The full maths syllabus is on the maths page. If before the end-of-October test you want a hand putting your method in order, start with the free introductory call: half an hour to work out where to concentrate your revision. When an absolute value or a root appears in the inequality, the case-by-case reasoning is in absolute value and irrational equations in Year 3.

Rational inequalities in Year 3: studying the sign