Recipe

Solve a polynomial inequality

Use when: a polynomial is compared to 0 (or can be rearranged to be) and you need every x making it positive, negative, or zero

Goal: the complete solution set in interval notation, with every boundary correctly included or excluded

Taught in Polynomial inequalities · a later section extends this recipe with more cases

1. 1

   Rearrange the inequality so one side reads 0 and the other side is a single polynomial.

   Example

   x^2 > 5x - 6 becomes x^2 - 5x + 6 > 0 after subtracting 5x - 6 from both sides.

   Why: every later step reads the polynomial's sign directly off it. A nonzero value sitting on the other side says nothing about that sign.

2. 2

   Factor the polynomial completely; each factor's own zero is a critical value, a point where the sign could change.

   Example

   x^2 - 5x + 6 = (x-2)(x-3): critical values x=2 and x=3.

   • factors over the integers — read the critical values straight off the factors.
   • doesn't factor over the integers — find the critical values with the quadratic formula (or another root-finding tool) instead; the rest of the procedure is unchanged.
   • Not yet: the inequality involves a ratio of two polynomials, not a single one, and each critical value needs identifying as an x-intercept, a vertical asymptote, or a hole — covered in Asymptotes, holes, and graphs of rational functions

   Why: a polynomial is defined and continuous for every real number, so it can only flip sign by passing through zero. There's no "undefined" checkpoint the way a rational expression has one.

3. 3

   Plot the critical values on a number line; they split it into intervals. Test one point from each interval to find the polynomial's sign there.

   Example

   Critical values 2 and 3 give intervals (-\infty,2), (2,3), (3,\infty); testing x=0, 2.5, 4 gives signs +, -, +.

   Why: the sign is constant across a whole interval once a polynomial can't flip without crossing zero, so one test point speaks for the entire stretch.

4. 4

   Check the power on each factor before trusting that every checkpoint flips the sign.

   Example

   For (x+2)^2(x-1) \ge 0, the checkpoint x=-2 (squared) doesn't flip the sign: negative on both sides, while x=1 (power 1) does flip.

   • factor's power is odd — an ordinary sign flip, same as Step 3.
   • factor's power is even — the sign is the same on both sides of that checkpoint; it can still earn a spot in the answer by itself, on a non-strict inequality.

   Why: an even power is never negative, so it can't be the thing that carries the whole product from positive to negative or back.

5. 5

   Collect every interval whose sign matches the inequality, then decide each critical value's fate: included when the inequality is non-strict (\le, \ge), excluded when it's strict (<, >).

   Example

   For x^2 - 5x + 6 > 0 (strict), the answer is (-\infty, 2) \cup (3, \infty), with both critical values excluded.

   Why: a strict inequality never keeps its own boundary. The boundary itself makes the expression exactly zero, which is neither positive nor negative.