Polynomial Inequalities

Solve polynomial inequalities using sign analysis, test points, and graphical methods.

Polynomial Inequalities

Introduction

A polynomial inequality involves a polynomial expression with an inequality sign ( $<$ , $>$ , $\leq$ , or $\geq$ ).

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:

Move all terms to one side (get 0 on the other side)
Factor the polynomial completely
Find critical values (zeros/roots of the polynomial)
Create a sign chart to test intervals
Determine solution intervals based on the inequality
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 $a < b < c$ :

| Interval | $(x-a)$ | $(x-b)$ | $(x-c)$ | $f(x)$ | |----------|---------|---------|---------|--------| | $x < a$ | $-$ | $-$ | $-$ | $-$ | | $a < x < b$ | $+$ | $-$ | $-$ | $+$ | | $b < x < c$ | $+$ | $+$ | $-$ | $-$ | | $x > c$ | $+$ | $+$ | $+$ | $+$ |

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 ( ))

📚 Practice Problems

1Problem 1easy

❓ Question:

Solve the inequality $x^2 - 5x + 6 \leq 0$ and express the solution in interval notation.

💡 Show Solution

Solution:

Given: $x^2 - 5x + 6 \leq 0$

Step 1: Factor the polynomial $x^2 - 5x + 6 = (x - 2)(x - 3)$

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

| Interval | Test Point | $(x-2)$ | $(x-3)$ | $(x-2)(x-3)$ | |----------|-----------|---------|---------|--------------| | $x < 2$ | $x = 0$ | $-$ | $-$ | $+$ | | $2 < x < 3$ | $x = 2.5$ | $+$ | $-$ | $-$ | | $x > 3$ | $x = 4$ | $+$ | $+$ | $+$ |

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 $x = 3$ : $(3)^2 - 5(3) + 6 = 9 - 15 + 6 = 0$ ✓

2Problem 2medium

❓ Question:

Solve $x^3 - 4x > 0$ and express the solution in interval notation.

💡 Show Solution

Solution:

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

| Interval | Test | $x$ | $(x-2)$ | $(x+2)$ | Product | |----------|------|-----|---------|---------|---------| | $x < -2$ | $-3$ | $-$ | $-$ | $-$ | $-$ | | $-2 < x < 0$ | $-1$ | $-$ | $-$ | $+$ | $+$ | | $0 < x < 2$ | $1$ | $+$ | $-$ | $+$ | $-$ | | $x > 2$ | $3$ | $+$ | $+$ | $+$ | $+$ |

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): $(3)^3 - 4(3) = 27 - 12 = 15 > 0$ ✓

3Problem 3hard

❓ Question:

Solve $(x - 1)^2(x + 3) \geq 0$ and express the solution in interval notation.

💡 Show Solution

Solution:

Given: $(x - 1)^2(x + 3) \geq 0$

Step 1: Identify factors (already factored)

The polynomial is already factored:

$(x - 1)^2$ 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 $(x + 3)$ :

| Interval | $(x-1)^2$ | $(x+3)$ | Product | |----------|-----------|---------|---------| | $x < -3$ | $+$ | $-$ | $-$ | | $-3 < x < 1$ | $+$ | $+$ | $+$ | | $x = 1$ | $0$ | $+$ | $0$ | | $x > 1$ | $+$ | $+$ | $+$ |

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 $x = 2$ : $(2-1)^2(2+3) = 1(5) = 5 > 0$ ✓

🎴

Practice with Flashcards

Review key concepts with our flashcard system

📖

Browse All Topics

Explore other calculus topics