Polynomial Inequalities

A polynomial inequality asks: for which values of $x$ is a polynomial expression positive, negative, or zero? Unlike equations (which have finitely many solutions), inequalities have solutions that span entire intervals. In this lesson you will learn to solve polynomial inequalities using the zero-finding and sign analysis method, write solutions in interval notation, and handle higher-degree polynomials with confidence.

The Strategy: Zeros and Test Intervals

The key insight is that a polynomial can only change sign at its zeros (roots). Between consecutive zeros, the polynomial is either entirely positive or entirely negative. This leads to a systematic process:

1. Move everything to one side so you have $p(x) > 0$ (or $\ge$, $<$, $\le$)
2. Factor the polynomial to find its zeros
3. Plot the zeros on a number line — they divide it into intervals
4. Test one value in each interval to determine the sign
5. Write the solution in interval notation, including or excluding endpoints based on $\ge$ vs. $>$

Worked Examples

Example 1: Quadratic Inequality

Solve $x^2 - 5x + 6 > 0$.

Step 1 — Factor:

$x^2 - 5x + 6 = (x - 2)(x - 3) > 0$

Step 2 — Find zeros: $x = 2$ and $x = 3$.

Step 3 — Test intervals:

Step 4 — Write solution: We want $> 0$ (positive), so:

$x \in (-\infty, 2) \cup (3, \infty)$

Endpoints are excluded because the inequality is strict ($>$, not $\ge$).

Example 2: Quadratic with $\le$

Solve $x^2 - 4x \le 0$.

Factor: $x(x - 4) \le 0$. Zeros: $x = 0$ and $x = 4$.

Test intervals:

We want $\le 0$ (negative or zero):

$x \in [0, 4]$

Endpoints are included because $\le$ allows equality.

Example 3: Higher-Degree Polynomial

Solve $(x + 1)(x - 2)(x - 5) \ge 0$.

Zeros: $x = -1$, $x = 2$, $x = 5$. Four intervals.

We want $\ge 0$ (positive or zero):

$x \in [-1, 2] \cup [5, \infty)$

Example 4: Repeated Zeros

Solve $x^2(x - 3) > 0$.

Zeros: $x = 0$ (multiplicity 2) and $x = 3$ (multiplicity 1).

Notice: $x^2$ is always non-negative, so it does not change sign at $x = 0$. The polynomial is negative on both sides of $x = 0$ (and zero at $x = 0$).

We want $> 0$: $x \in (3, \infty)$.

Rule of thumb: A factor with even multiplicity touches the $x$-axis but does not cross it. A factor with odd multiplicity crosses the $x$-axis.

Example 5: Not in Factored Form

Solve $x^3 - 4x^2 - x + 4 > 0$.

Step 1 — Factor by grouping:

$x^2(x - 4) - 1(x - 4) = (x^2 - 1)(x - 4) = (x-1)(x+1)(x-4)$

Zeros: $x = -1$, $x = 1$, $x = 4$.

Test intervals:

Solution: $x \in (-1, 1) \cup (4, \infty)$.

Sign Analysis Without a Table

Once you are comfortable, you can determine signs quickly:

1. Count the number of negative factors in each interval
2. Even number of negatives → positive product. Odd number → negative product.

Alternatively, use the leading coefficient test: for a polynomial of odd degree with positive leading coefficient, the far left is negative and the far right is positive, alternating at each simple zero.

Writing Solutions in Interval Notation

Note: Always use parentheses (not brackets) with $\infty$ and $-\infty$, because infinity is not a number.

Real-World Application: Engineering — Beam Load Capacity

A simply supported beam deflects downward when the deflection function is negative. For a certain beam, the deflection is modeled by:

$d(x) = -x(x - 2)(x - 8)$

where $x$ is the distance (in meters) from the left support. The beam spans from $x = 0$ to $x = 8$.

Where does the beam deflect downward ($d(x) < 0$)?

We need $-x(x-2)(x-8) < 0$, which is equivalent to $x(x-2)(x-8) > 0$.

Zeros: $x = 0$, $x = 2$, $x = 8$.

On the physical beam domain $[0, 8]$: $x(x-2)(x-8) > 0$ on $(0, 2)$, so $d(x) < 0$ on $(0, 2)$.

The beam deflects downward between $x = 0$ and $x = 2$, then deflects upward from $x = 2$ to $x = 8$. (This unusual behavior might indicate the beam is also supported at $x = 2$.)

Common Mistakes

1. Forgetting to set one side to zero. Always rearrange to $p(x) > 0$ (or $\ge$, $<$, $\le$) before factoring.
2. Dividing both sides by a variable expression. Never divide by $x$ (or any expression containing $x$) — it might be negative, which flips the inequality, or zero, which is undefined. Factor instead.
3. Ignoring repeated roots. A factor like $(x-1)^2$ is always non-negative and does not change sign.
4. Wrong interval notation. Use $[$ $]$ for $\le$ / $\ge$ (endpoints included) and $( )$ for $<$ / $>$ (endpoints excluded).

Practice Problems

Key Takeaways

• A polynomial can only change sign at its zeros — between zeros, it is entirely positive or negative
• The process: move to one side, factor, find zeros, test intervals, write in interval notation
• Even-multiplicity zeros touch the axis (no sign change); odd-multiplicity zeros cross the axis
• Use $[$ $]$ for $\le$/$\ge$ (include endpoints) and $( )$ for $<$/$>$ (exclude endpoints)
• Never divide both sides of an inequality by a variable expression
• This method extends directly to rational inequalities, with an additional consideration for undefined points

Return to College Algebra for more topics in this section.