Lesson 4: Polynomial & Rational Inequalities
Estimated time: 35-40 minutes
Learning Objectives
By the end of this lesson, you will be able to:
- Solve polynomial inequalities using the test interval method
- Create and interpret sign charts for polynomial inequalities
- Solve rational inequalities using critical values
- Determine when to include or exclude boundary points
- Express solutions in interval notation
- Apply inequalities to real-world optimization problems
Understanding Polynomial Inequalities
Polynomial Inequality: An inequality involving a polynomial expression, such as:
- f(x) > 0
- f(x) ≥ 0
- f(x) < 0
- f(x) ≤ 0
where f(x) is a polynomial function.
Unlike polynomial equations (where we find specific values), polynomial inequalities ask us to find intervals where the inequality is true.
Key Concept: Test Interval Method
Polynomial functions are continuous - they have no breaks or jumps. This means:
- Between any two zeros, the function maintains the same sign (all positive or all negative)
- The sign can only change at zeros
- We can test one point in each interval to determine the sign throughout that interval
Steps for Solving Polynomial Inequalities
- Move all terms to one side (get 0 on the other side)
- Factor the polynomial completely
- Find all zeros (critical values)
- Plot zeros on a number line to create test intervals
- Choose a test point in each interval
- Determine the sign in each interval
- Select intervals where the inequality is true
- Write solution in interval notation
Solving Polynomial Inequalities
Example 1: Basic Quadratic Inequality
Solve: x2 - 5x + 6 > 0
Step 1: Factor the polynomial
x2 - 5x + 6 = (x - 2)(x - 3)
Step 2: Find zeros
(x - 2)(x - 3) = 0
x = 2 or x = 3
Step 3: Create test intervals
Number line divided at x = 2 and x = 3:
(-∞, 2) | (2, 3) | (3, ∞)
Step 4: Test a point in each interval
| Interval | Test Point | f(test) = (x-2)(x-3) | Sign |
|---|---|---|---|
| (-∞, 2) | x = 0 | (0-2)(0-3) = (-2)(-3) = 6 | Positive (+) |
| (2, 3) | x = 2.5 | (0.5)(-0.5) = -0.25 | Negative (-) |
| (3, ∞) | x = 4 | (2)(1) = 2 | Positive (+) |
Step 5: Select intervals where f(x) > 0
We want positive regions: (-∞, 2) and (3, ∞)
Step 6: Determine boundary inclusion
Since the inequality is strict (>, not ≥), we exclude the zeros
Solution: (-∞, 2) ∪ (3, ∞)
Or in inequality notation: x < 2 or x > 3
Example 2: Inequality with ≤ (Including Boundaries)
Solve: x2 - 4 ≤ 0
Step 1: Factor
x2 - 4 = (x - 2)(x + 2)
Step 2: Find zeros
x = -2 or x = 2
Step 3: Test intervals
| Interval | Test Point | f(test) | Sign |
|---|---|---|---|
| (-∞, -2) | x = -3 | 9 - 4 = 5 | Positive |
| (-2, 2) | x = 0 | 0 - 4 = -4 | Negative |
| (2, ∞) | x = 3 | 9 - 4 = 5 | Positive |
Step 4: We want f(x) ≤ 0 (negative or zero)
Negative region: (-2, 2)
Since we have ≤ (not <), we INCLUDE the zeros where f(x) = 0
Solution: [-2, 2]
Or: -2 ≤ x ≤ 2
Example 3: Cubic Inequality
Solve: x3 - 4x > 0
Step 1: Factor
x3 - 4x = x(x2 - 4) = x(x - 2)(x + 2)
Step 2: Find zeros
x = -2, 0, 2
Step 3: Create sign chart
Intervals: (-∞,-2) | (-2,0) | (0,2) | (2,∞) Factor x: - | - | + | + Factor(x-2): - | - | - | + Factor(x+2): - | + | + | + Product: - | + | - | +
Step 4: Select positive intervals
We want f(x) > 0 (positive): (-2, 0) and (2, ∞)
Solution: (-2, 0) ∪ (2, ∞)
Example 4: Inequality with Multiplicity
Solve: (x + 1)2(x - 3) ≥ 0
Step 1: Identify zeros and multiplicities
x = -1 (multiplicity 2 - even)
x = 3 (multiplicity 1 - odd)
Step 2: Key insight about multiplicity
Even multiplicity: sign does NOT change (graph touches, doesn't cross)
Odd multiplicity: sign DOES change (graph crosses)
Step 3: Test intervals
| Interval | Test Point | Sign |
|---|---|---|
| (-∞, -1) | x = -2 | (-)2(-) = Negative |
| (-1, 3) | x = 0 | (+)2(-) = Negative |
| (3, ∞) | x = 4 | (+)2(+) = Positive |
Note: Sign doesn't change at x = -1 (even multiplicity)
Step 4: We want f(x) ≥ 0
Positive region: (3, ∞)
Include boundaries (≥): x = -1 and x = 3
Solution: {-1} ∪ [3, ∞)
Example 5: Higher Degree Polynomial
Solve: x4 - 5x2 + 4 < 0
Step 1: Factor (let u = x2)
u2 - 5u + 4 = (u - 1)(u - 4)
(x2 - 1)(x2 - 4)
(x - 1)(x + 1)(x - 2)(x + 2)
Step 2: Zeros: x = -2, -1, 1, 2
Step 3: Sign chart
Intervals: (-∞,-2) | (-2,-1) | (-1,1) | (1,2) | (2,∞) Sign: + | - | + | - | +
Step 4: We want f(x) < 0 (negative)
Solution: (-2, -1) ∪ (1, 2)
Solving Rational Inequalities
Rational Inequality: An inequality involving a rational expression, such as f(x)/g(x) > 0.
Critical Values: Values where the numerator equals zero OR the denominator equals zero.
Important Warning!
DO NOT multiply both sides by the denominator!
The denominator could be positive or negative, which would change the inequality direction. Instead, use the test interval method with critical values.
Steps for Solving Rational Inequalities
- Move all terms to one side (get 0 on the other)
- Combine into a single rational expression
- Factor numerator and denominator
- Find critical values (zeros of numerator AND denominator)
- Plot critical values on number line
- Test each interval
- Select appropriate intervals
- NEVER include values that make denominator zero!
Example 6: Basic Rational Inequality
Solve: (x + 3)/(x - 2) > 0
Step 1: Identify critical values
Numerator = 0: x = -3
Denominator = 0: x = 2
Critical values: -3 and 2
Step 2: Create test intervals
| Interval | Test Point | Numerator Sign | Denominator Sign | Quotient Sign |
|---|---|---|---|---|
| (-∞, -3) | x = -4 | - | - | + |
| (-3, 2) | x = 0 | + | - | - |
| (2, ∞) | x = 3 | + | + | + |
Step 3: We want positive quotient
Positive intervals: (-∞, -3) and (2, ∞)
Step 4: Check boundaries
x = -3: Makes numerator zero, so quotient = 0 (not > 0, exclude)
x = 2: Makes denominator zero (ALWAYS exclude)
Solution: (-∞, -3) ∪ (2, ∞)
Example 7: Rational Inequality with ≤
Solve: (x - 1)/(x + 2) ≤ 0
Step 1: Critical values
x = 1 (numerator zero)
x = -2 (denominator zero)
Step 2: Test intervals
| Interval | Test Point | Sign |
|---|---|---|
| (-∞, -2) | x = -3 | (-)/(-) = + |
| (-2, 1) | x = 0 | (-)/(+) = - |
| (1, ∞) | x = 2 | (+)/(+) = + |
Step 3: We want f(x) ≤ 0 (negative or zero)
Negative interval: (-2, 1)
Step 4: Check boundaries
x = 1: Include (≤ allows equality, and denominator ≠ 0)
x = -2: NEVER include (makes denominator zero)
Solution: (-2, 1]
Example 8: Complex Rational Inequality
Solve: (x2 - 4)/(x2 - 9) ≥ 0
Step 1: Factor
Numerator: x2 - 4 = (x - 2)(x + 2)
Denominator: x2 - 9 = (x - 3)(x + 3)
Step 2: Critical values
From numerator: x = -2, 2
From denominator: x = -3, 3
All critical values: -3, -2, 2, 3
Step 3: Sign chart
Interval: (-∞,-3) | (-3,-2) | (-2,2) | (2,3) | (3,∞) (x-2): - | - | - | + | + (x+2): - | - | + | + | + (x-3): - | - | - | - | + (x+3): - | + | + | + | + Quotient: + | - | + | - | +
Step 4: We want f(x) ≥ 0 (positive or zero)
Positive intervals: (-∞, -3), (-2, 2), (3, ∞)
Step 5: Check boundaries
x = -2, 2: Include (≥ and denominator ≠ 0)
x = -3, 3: NEVER include (denominator = 0)
Solution: (-∞, -3) ∪ [-2, 2] ∪ (3, ∞)
Example 9: Inequality Requiring Simplification
Solve: x/(x - 1) > 2
Step 1: Move all terms to one side
x/(x - 1) - 2 > 0
Step 2: Get common denominator
x/(x - 1) - 2(x - 1)/(x - 1) > 0
[x - 2(x - 1)]/(x - 1) > 0
[x - 2x + 2]/(x - 1) > 0
(2 - x)/(x - 1) > 0
Step 3: Critical values
Numerator = 0: x = 2
Denominator = 0: x = 1
Step 4: Test intervals
| Interval | Test Point | Sign |
|---|---|---|
| (-∞, 1) | x = 0 | (+)/(-) = - |
| (1, 2) | x = 1.5 | (+)/(+) = + |
| (2, ∞) | x = 3 | (-)/(+) = - |
Solution: (1, 2)
Applications of Inequalities
Example 10: Projectile Motion
A ball is thrown upward with an initial velocity of 48 ft/s from a height of 160 ft. Its height h (in feet) after t seconds is given by:
h(t) = -16t2 + 48t + 160
During what time interval is the ball at least 192 feet high?
Step 1: Set up inequality
We want h(t) ≥ 192
-16t2 + 48t + 160 ≥ 192
Step 2: Move to one side
-16t2 + 48t + 160 - 192 ≥ 0
-16t2 + 48t - 32 ≥ 0
Step 3: Factor out -16
-16(t2 - 3t + 2) ≥ 0
-16(t - 1)(t - 2) ≥ 0
Step 4: Divide by -16 (flip inequality!)
(t - 1)(t - 2) ≤ 0
Step 5: Solve the inequality
Critical values: t = 1, t = 2
Testing shows (t - 1)(t - 2) ≤ 0 when 1 ≤ t ≤ 2
Answer: The ball is at least 192 feet high during the interval [1, 2] seconds.
That's from t = 1 second to t = 2 seconds (a 1-second window).
Example 11: Business - Profit Optimization
A company's profit P (in thousands of dollars) from producing x hundred units is:
P(x) = -x3 + 6x2 + 15x - 10
For what production levels does the company make a profit (P > 0)?
Step 1: Set up inequality
-x3 + 6x2 + 15x - 10 > 0
Step 2: Find zeros (using Rational Root Theorem)
Testing x = 1: -(1) + 6 + 15 - 10 = 10 ≠ 0
Testing x = 2: -(8) + 24 + 30 - 10 = 36 ≠ 0
Testing x = -1: -(-1) + 6 - 15 - 10 = -18 ≠ 0
(For this example, assume we find zeros at approximately x ≈ 0.5, x ≈ -2.8, x ≈ 7.3)
Step 3: Test intervals
Since x represents production (hundreds of units), we only consider x ≥ 0
Step 4: Through testing, profit is positive for approximately 0.5 < x < 7.3
Answer: The company makes a profit when producing between 50 and 730 units.
Example 12: Average Cost Constraint
A company's average cost per item C (in dollars) when producing x items is:
C(x) = (2000 + 5x)/x = 2000/x + 5
For what production levels is the average cost less than $15 per item?
Step 1: Set up inequality
2000/x + 5 < 15
Step 2: Simplify
2000/x < 10
Step 3: Move to one side
2000/x - 10 < 0
(2000 - 10x)/x < 0
Step 4: Critical values
Numerator = 0: 2000 - 10x = 0 → x = 200
Denominator = 0: x = 0
Step 5: Test intervals (x > 0 only)
| Interval | Test Point | Sign |
|---|---|---|
| (0, 200) | x = 100 | (+)/(+) = + |
| (200, ∞) | x = 300 | (-)/(+) = - |
Step 6: We want negative region
Solution: x > 200
Answer: Average cost is less than $15 when producing more than 200 items.
Check Your Understanding
1. Solve: x2 - 9 < 0
Solution:
Factor: (x - 3)(x + 3) < 0
Zeros: x = -3, 3
Test intervals: negative between zeros
Answer: (-3, 3)
2. Solve: x2 + 2x - 8 ≥ 0
Solution:
Factor: (x + 4)(x - 2) ≥ 0
Zeros: x = -4, 2
Positive in: (-∞, -4) and (2, ∞)
Include boundaries (≥)
Answer: (-∞, -4] ∪ [2, ∞)
3. Solve: (x + 1)/(x - 3) ≥ 0
Solution:
Critical values: x = -1 (num), x = 3 (den)
Test intervals: positive in (-∞, -1) and (3, ∞)
Include x = -1 (≥ and den ≠ 0)
Never include x = 3 (den = 0)
Answer: (-∞, -1] ∪ (3, ∞)
4. Solve: x3 - 9x > 0
Solution:
Factor: x(x2 - 9) = x(x - 3)(x + 3)
Zeros: -3, 0, 3
Using sign chart, positive in: (-3, 0) and (3, ∞)
Answer: (-3, 0) ∪ (3, ∞)
5. Solve: 1/x < 2
Solution:
Move to one side: 1/x - 2 < 0
Common denominator: (1 - 2x)/x < 0
Critical values: x = 1/2 (num), x = 0 (den)
Test intervals: negative in (-∞, 0) and (1/2, ∞)
Answer: (-∞, 0) ∪ (1/2, ∞)
6. Solve: (x - 2)2(x + 1) ≤ 0
Solution:
Zeros: x = 2 (multiplicity 2), x = -1 (multiplicity 1)
Even multiplicity at x = 2: no sign change
Test shows negative only for x < -1
Include x = 2 (makes expression = 0)
Include x = -1
Answer: (-∞, -1] ∪ {2}
7. Solve: (x2 - 1)/(x + 3) > 0
Solution:
Factor numerator: (x - 1)(x + 1)/(x + 3) > 0
Critical values: -3, -1, 1
Sign chart shows positive in: (-3, -1) and (1, ∞)
Answer: (-3, -1) ∪ (1, ∞)
Key Takeaways
- Polynomial inequalities: find zeros, create test intervals, determine sign in each region
- Between zeros, polynomial functions maintain the same sign (continuous, no jumps)
- Include boundary points for ≤ or ≥ (exclude for < or >)
- For rational inequalities: critical values come from BOTH numerator and denominator
- NEVER include values that make the denominator zero
- Sign charts help organize the analysis for complex expressions
- Even multiplicity zeros: sign doesn't change; Odd multiplicity: sign changes
- Always move all terms to one side before solving
- Express final answers in interval notation