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 ( ))

• $(x - 1)^2$ with multiplicity 2
• $(x + 3)$ with multiplicity 1

Step 2: Find critical values

• $x - 1 = 0 \Rightarrow x = 1$ (multiplicity 2)
• $x + 3 = 0 \Rightarrow x = -3$ (multiplicity 1)

Critical values: $x = -3, 1$

Step 3: Analyze signs

Important: $(x - 1)^2$ is always non-negative and equals 0 only at $x = 1$.

Since $(x - 1)^2 \geq 0$ always, the sign of the product depends on $(x + 3)$:

Step 4: Identify where product $\geq 0$

We need non-negative values (positive or zero).

From analysis:

• Negative when $x < -3$
• Zero at $x = -3$ (included with $\geq$)
• Positive when $-3 < x < 1$
• Zero at $x = 1$ (included with $\geq$)
• Positive when $x > 1$

Answer: $[-3, \infty)$

Key insight: The even multiplicity at $x = 1$ means the sign doesn't change there. The factor $(x-1)^2$ touches the x-axis but doesn't cross it.

Verification:

• At $x = -4$: $(-4-1)^2(-4+3) = 25(-1) = -25 < 0$ ✓
• At $x = -3$: $(-3-1)^2(-3+3) = 16(0) = 0$ ✓
• At $x = 0$: $(0-1)^2(0+3) = 1(3) = 3 > 0$ ✓
• At $x = 1$: $(1-1)^2(1+3) = 0(4) = 0$ ✓
• At $x = 2$: $(2-1)^2(2+3) = 1(5) = 5 > 0$ ✓