Examples:

• $x^2 - 5x + 6 > 0$
• $x^3 - 4x \leq 0$
• $2x^4 - 8x^2 + 6 \geq 0$

Solution Strategy

The key steps for solving polynomial inequalities:

1. Move all terms to one side (get 0 on the other side)
2. Factor the polynomial completely
3. Find critical values (zeros/roots of the polynomial)
4. Create a sign chart to test intervals
5. Determine solution intervals based on the inequality
6. Write the solution in interval notation

Sign Analysis Method

Step-by-Step Process

Step 1: Set up the inequality

Rearrange so that the polynomial is on one side and 0 is on the other.

Step 2: Factor

Factor the polynomial completely to identify all zeros.

Step 3: Find critical values

Set each factor equal to zero and solve. These are the critical values that divide the number line into intervals.

Step 4: Create intervals

The critical values divide the number line into regions. Test a point from each region.

Step 5: Make a sign chart

Create a table showing:

• The intervals
• The sign of each factor in each interval
• The overall sign of the product

Step 6: Identify solution intervals

• For $> 0$ or $\geq 0$: Choose intervals where the product is positive
• For $< 0$ or $\leq 0$: Choose intervals where the product is negative
• Use $\leq$ or $\geq$ to include critical values where the expression equals 0
• Use $<$ or $>$ to exclude critical values

Sign Chart Template

For $f(x) = (x - a)(x - b)(x - c)$ where $a < b < c$:

Key Principles

Sign of Linear Factors

For $(x - a)$:

• Negative when $x < a$
• Zero when $x = a$
• Positive when $x > a$

Multiplicity Effects

• Odd multiplicity: Sign changes across the zero
• Even multiplicity: Sign stays the same across the zero

Example: $(x - 2)^2(x - 5)$

• At $x = 2$ (multiplicity 2): sign doesn't change
• At $x = 5$ (multiplicity 1): sign changes

Graphical Interpretation

The solution to a polynomial inequality corresponds to where the graph satisfies the condition:

• $f(x) > 0$: Graph is above the x-axis
• $f(x) < 0$: Graph is below the x-axis
• $f(x) \geq 0$: Graph is above or on the x-axis
• $f(x) \leq 0$: Graph is below or on the x-axis

Interval Notation

Symbols:

• $(a, b)$: Open interval (doesn't include endpoints)
• $[a, b]$: Closed interval (includes both endpoints)
• $[a, b)$: Half-open interval (includes $a$ but not $b$)
• $(a, \infty)$: Unbounded interval extending to positive infinity

Union: Use $\cup$ to combine disjoint intervals

Example: $(-\infty, -2] \cup [3, \infty)$

Special Cases

Always Positive or Always Negative

Some polynomials never change sign:

• $x^2 + 1 > 0$ for all real $x$ (always positive, never zero)
• $-(x^2 + 4) < 0$ for all real $x$ (always negative)

Includes Equality

Remember:

• $f(x) \geq 0$: Include values where $f(x) = 0$ (use brackets [ ])
• $f(x) > 0$: Exclude values where $f(x) = 0$ (use parentheses ( ))

Step 2: Find critical values

Set each factor equal to zero:

• $x - 2 = 0 \Rightarrow x = 2$
• $x - 3 = 0 \Rightarrow x = 3$

Critical values: $x = 2, 3$

Step 3: Create a sign chart

Test intervals: $(-\infty, 2)$, $(2, 3)$, $(3, \infty)$

Step 4: Identify where $f(x) \leq 0$

We need intervals where the product is negative or zero.

From the chart:

• Negative in interval $(2, 3)$
• Zero at $x = 2$ and $x = 3$

Since we have $\leq$ (includes equality), we include the critical values.

Answer: $[2, 3]$

Verification:

• At $x = 2$: $(2)^2 - 5(2) + 6 = 4 - 10 + 6 = 0$ ✓
• At $x = 2.5$: $(2.5)^2 - 5(2.5) + 6 = 6.25 - 12.5 + 6 = -0.25 < 0$ ✓
• At $x = 3$: $(3)^2 - 5(3) + 6 = 9 - 15 + 6 = 0$ ✓

Given: $x^3 - 4x > 0$

Step 1: Factor the polynomial $x^3 - 4x = x(x^2 - 4) = x(x - 2)(x + 2)$

Step 2: Find critical values

Set each factor equal to zero:

• $x = 0$
• $x - 2 = 0 \Rightarrow x = 2$
• $x + 2 = 0 \Rightarrow x = -2$

Critical values: $x = -2, 0, 2$ (in order)

Step 3: Create a sign chart

Test intervals: $(-\infty, -2)$, $(-2, 0)$, $(0, 2)$, $(2, \infty)$

Step 4: Identify where the product is positive

We need $> 0$ (strictly positive, exclude zeros).

From the chart, the product is positive in:

• $(-2, 0)$
• $(2, \infty)$

Since we have $>$ (strict inequality), we exclude the critical values.

Answer: $(-2, 0) \cup (2, \infty)$

Verification:

• At $x = -1$ (in first interval): $(-1)^3 - 4(-1) = -1 + 4 = 3 > 0$ ✓
• At $x = 1$ (in middle): $(1)^3 - 4(1) = 1 - 4 = -3 < 0$ ✓
• At $x = 3$ (in last interval): $(3)^3 - 4(3) = 27 - 12 = 15 > 0$ ✓