Polynomial and Rational Inequalities

You already know how to crack a quadratic inequality: factor it, find where it crosses zero, and decide which side you want. But what about a cubic like (x-1)(x+2)(x-4) > 0, with three factors instead of two? Or a fraction like \dfrac{x-1}{x+3} \ge 0, where the bottom can blow up to infinity?

Here is the beautiful part: one method handles them all. It doesn't care whether the degree is 2, 3, or 17, and it copes with fractions too. You draw one number line, mark a handful of key points on it, work out a sign in each gap, and the answer simply falls out. It's called a sign chart (or sign table), and once you've seen it move you'll never dread a high-degree inequality again.

The idea: critical values chop the line into sign-stable pieces

Think about a product of factors like (x-1)(x+2)(x-4). Each factor is a straight line that is zero at exactly one place and changes sign as x sweeps past that place: x-1 is negative for x < 1 and positive for x > 1. So the only places where the whole product can switch between positive and negative are those root points — the critical values.

Between two neighbouring critical values, nothing flips, so the sign is constant. That is the whole trick. The recipe:

  1. Get one side equal to zero (never leave stuff on both sides).
  2. Factor fully. For a fraction, factor the top and the bottom separately.
  3. Mark every critical value on a number line — the zeros of the numerator and the zeros of the denominator.
  4. Test one point inside each interval to find the sign there (a factor flips sign each time you cross its own root).
  5. Read off the intervals whose sign matches what you want (>0, \le 0, …).

Worked example 1 — a cubic

Solve (x-1)(x+2)(x-4) > 0.

It's already one side = 0 and fully factored, so straight to the critical values — the roots of each factor: x = 1, x = -2, x = 4. Put them in order on the line: -2 < 1 < 4. They split the number line into four intervals. Now pick an easy test point in each:

Notice the signs just alternate -,+,-,+ — because at each critical value exactly one factor flips. We asked for > 0, so we keep the positive intervals (and, being strict >, the endpoints are excluded):

-2 < x < 1 \quad\text{or}\quad x > 4.

Step through the chart being built below:

Worked example 2 — a rational inequality

Solve \frac{x-1}{x+3} \ge 0.

A fraction is zero when its top is zero and undefined when its bottom is zero. Both kinds of place are critical values, because the whole expression can only change sign at one of them:

Now test a point in each of the three intervals:

We want \ge 0, so keep the positive intervals:

x < -3 \quad\text{or}\quad x \ge 1.

Look carefully at the two endpoints: -3 is open (kicked out, undefined) while 1 is filled (the fraction is genuinely 0 there). Same inequality sign, opposite treatment — the reason is why the point is critical, not the \ge itself.

Solving = for a high-degree polynomial can be brutal, but an inequality only asks a much softer question: where is this thing positive, and where negative? And a product changes sign only where a factor is zero — nowhere else. So no matter how many factors you stack up, the number line breaks into a finite string of pieces, each with a single fixed sign. You never have to compute the actual value anywhere; a cheap test point per interval tells you everything.

The signs even alternate for simple roots (each crossing flips exactly one factor), so once you know the sign of the far-right interval — usually + if the leading coefficient is positive — you can often fill in the rest by just flipping +,-,+,- down the line. (The one wrinkle: a repeated factor like (x-2)^2 touches zero without crossing, so the sign does not flip there — always double-check with a test point.)

The single biggest mistake with fractions is to clear the denominator by multiplying both sides by it. With \dfrac{x-1}{x+3} \ge 0 it is very tempting to multiply both sides by (x+3) and write x - 1 \ge 0. This is wrong.

Why? Because (x+3) is sometimes negative (whenever x < -3), and multiplying an inequality by a negative number flips the sign. Since you don't know the sign of x+3 in advance, you don't know whether to flip — so the move is simply not allowed. It silently throws away the entire x < -3 branch of the answer.

Two things to hold onto:

 To solve a polynomial or rational inequality: