Solve Polynomial Inequalities - Tutorial

This tutorial explains how to solve polynomial inequalities using the sign table method. You'll learn to identify critical points, divide the number line into intervals, and determine the sign of polynomial expressions in each interval. Each step is illustrated with detailed examples to build a clear understanding of the process.

Review

A polynomial can change sign only at its real zeros. When ordered, the real zeros of a polynomial divide the real number line into intervals in which the polynomial does not change sign.

Here is an example: Below is shown the graph of the polynomial \[ y = 0.2 (x^2 - 3x - 4) (-x^2 + 2x - 1) (x - 3) (x + 3) \]

Note that the graph of \( y \) has 5 \( x \)-intercepts at the points: \[ A(-3,0) \quad B(-1,0) \quad C(1,0) \quad D(3,0) \quad E(4,0) \] which divide the number line into 6 intervals: \[ (-\infty, -3), (-3, -1), (-1, 1), (1, 3), (3, 4), (4, +\infty) \]

and the sign of the coordinate \( y \) within each interval is either negative or positive but is the same over the whole interval.

graph of a polynomial and its x intercepts

Examples & Step-by-Step Solutions

Example 1: Quadratic Inequality \( x^2 < -x + 6 \)

Problem: Solve the polynomial inequality

\[ x^2 < -x + 6 \]

Solution:

  1. Step 1: Move all terms to one side \[ x^2 + x - 6 < 0 \]
  2. Step 2: Factor the quadratic expression on the left side \[ x^2 + x - 6 = (x + 3)(x - 2) \] Now the inequality becomes: \[ (x + 3)(x - 2) < 0 \]
  3. Step 3: Identify critical points such that \( (x + 3)(x - 2) = 0 \): \[ x + 3 = 0 \implies x = -3 \quad \text{and} \quad x - 2 = 0 \implies x = 2 \]
  4. Step 4: Test the intervals made up of the critical points \( x = -3 \) and \( x = 2 \) using a sign chart: \[ \begin{array}{|c|c|c|c|c|} \hline \text{Interval} & \text{Test Value} & x + 3 & x - 2 & \text{Sign of \( (x + 3)(x - 2) \) } \\ \hline (-\infty, -3) & x = -4 & - & - & + \\ (-3, 2) & x = 0 & + & - & - \\ (2, \infty) & x = 3 & + & + & + \\ \hline \end{array} \]
  5. Step 5: Determine the solution set
    Because we are solving the inequality \( (x + 3)(x - 2) < 0 \), choose the interval where the product is negative. The inequality is satisfied when: \[ (x + 3)(x - 2) < 0 \implies x \in (-3, 2) \]

    Final Answer: \[ \boxed{(-3, 2)} \]

Example 2: Higher-Degree Polynomial Inequality \( (x^2 + 2)(x + 1)(x + 6) > 0 \)

Problem: Solve the polynomial inequality

\[ (x^2 + 2)(x + 1)(x + 6) > 0 \]

Solution:

  1. Step 1: Identify and analyze each factor:
    • \((x^2 + 2)\) is always positive for all real numbers.
    • \((x + 1)\) is zero at \(x = -1\), positive if \(x > -1\), and negative if \(x < -1\).
    • \((x + 6)\) is zero at \(x = -6\), positive if \(x > -6\), and negative if \(x < -6\).
  2. Step 2: Create a sign chart using test points in each interval determined by the critical points \(x = -6\) and \(x = -1\), including the sign of \(x^2 + 2\): \[ \begin{array}{|c|c|c|c|c|c|} \hline \textbf{Interval} & \textbf{Test Point} & x^2 + 2 & x + 6 & x + 1 & \text{Sign of \( (x^2 + 2)(x + 1)(x + 6) \)} \\ \hline (-\infty, -6) & x = -7 & + & - & - & + \\ \hline (-6, -1) & x = -3 & + & + & - & - \\ \hline (-1, \infty) & x = 0 & + & + & + & + \\ \hline \end{array} \]
  3. Step 3: Exclude points where the expression equals zero: \(x = -6\) and \(x = -1\).
    Final Answer: The \(x\)-values that make the expression positive are given by the interval: \[ (-\infty, -6) \cup (-1, \infty) \]

More References and Links

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