Solving Polynomial Inequalities
Solve polynomial inequalities by finding zeros and using a sign chart to determine intervals where the polynomial is positive or negative. Express solutions in interval notation.
A lesson of Algebra 2, in Polynomial Functions, about 10 minutes.
By the end of this lesson, you will be able to solve polynomial inequalities of any degree using the sign-chart method, express solutions in interval notation, and interpret polynomial inequalities graphically.
The Big Idea
Solving a polynomial inequality extends the quadratic inequality approach to higher degrees. The key steps are unchanged: find zeros, divide the number line into intervals, test the sign in each interval. What changes is the number of intervals and the complexity of testing — but the logic is identical.
The Sign-Chart Method
- Move all terms to one side: $p(x) , \square , 0$
- Factor $p(x)$ completely
- Find all zeros (real roots)
- Mark zeros on a number line → creates test intervals
- Test a point in each interval (or use multiplicity rules)
- Determine where the inequality is satisfied
- Write in interval notation (include endpoints for $\leq$ or $\geq$)
Using Multiplicity to Avoid Testing
At a zero of odd multiplicity: the polynomial changes sign (crosses $x$-axis).
At a zero of even multiplicity: the polynomial does not change sign (bounces).
If you know the sign in one interval, you can propagate across zeros using this rule — no testing required.
Example 1 — Degree 3 Inequality
Solve: $x^3 - 4x^2 + x + 6 > 0$
Factor: Test $x = -1$: $-1 - 4 - 1 + 6 = 0$ ✓ → factor $(x+1)$.
Synthetic division gives: $(x+1)(x^2 - 5x + 6) = (x+1)(x-2)(x-3)$
Zeros: $x = -1, 2, 3$ (all mult. 1 — all cross)
Sign chart:
| Interval | Test point | Sign of $p$ |
|---|---|---|
| $(-\infty, -1)$ | $x = -2$ | $(-)(-)(-) = -$ |
| $(-1, 2)$ | $x = 0$ | $(+)(-)(-) = +$ |
| $(2, 3)$ | $x = 2.5$ | $(+)(+)(-) = -$ |
| $(3, \infty)$ | $x = 4$ | $(+)(+)(+) = +$ |
Solution: $(-1, 2) \cup (3, \infty)$
Example 2 — Even Multiplicity
Solve: $(x-1)^2(x+3) \leq 0$
Zeros: $x = 1$ (mult. 2 → bounce, sign unchanged); $x = -3$ (mult. 1 → cross)
Start right: $x = 10$: $(+)^2(+) > 0$
Cross from right: $x = 1$ → bounce (still $> 0$ to the left of $1$... wait, $x=0$: $(+1)(+3) = +3 > 0$, correct).
$x = -3$: cross → negative for $x < -3$.
Sign: