Solving Polynomial and Rational Inequalities (Precalculus 3.5 Study Notes)

Solving Polynomial and Rational Inequalities

Steps to Solving Polynomial and Rational Inequalities Algebraically

Solving inequalities involving polynomials and rational functions is a fundamental skill in precalculus. The following steps outline a systematic approach to solving these inequalities algebraically:

• Step 1: Get the function on the left and zero on the right. - For rational inequalities, write as a single fraction.

• Step 2: Determine the zeros and any numbers that make the function undefined. - These values are critical points that divide the number line into intervals.

• Step 3: Use the numbers found in Step 2 to separate a number line. - Mark these points and consider the intervals between them.

• Step 4: Evaluate a value between the intervals to see if the function is positive or negative. - The solution set will depend on the inequality sign (e.g., >, <, ≥, ≤).

Example 1: Solving a Polynomial Inequality

Problem: Solve the inequality and graph the solution set:

$x^2 - 5x + 12 > 0$

• Step 1: Set the inequality to zero: $x^2 - 5x + 12 > 0$

• Step 2: Find the zeros: Solve $x^2 - 5x + 12 = 0$ (Quadratic formula or factoring)

• Step 3: Plot the zeros on a number line (if real solutions exist).

• Step 4: Test intervals between zeros to determine where the inequality holds.

• Graph: The graph of $y = x^2 - 5x + 12$ is a parabola opening upwards. The solution set corresponds to the intervals where the graph is above the x-axis.

Example: If $x^2 - 5x + 12$ has no real zeros (discriminant $< 0$), then the expression is always positive, so the solution is all real numbers.

Example 2: Solving a Rational Inequality

Problem: Solve the inequality and graph the solution set:

$\frac{x+1}{x-2} \geq 3$

• Step 1: Bring all terms to one side: $\frac{x+1}{x-2} - 3 \geq 0$

• Step 2: Combine into a single fraction: $\frac{x+1 - 3(x-2)}{x-2} \geq 0$

• Step 3: Simplify numerator: $x+1 - 3x + 6 = -2x + 7$

• Step 4: The inequality becomes $\frac{-2x+7}{x-2} \geq 0$

• Step 5: Find zeros and undefined points:

  • Numerator zero: $-2x+7=0 \Rightarrow x=3.5$

  • Denominator zero: $x-2=0 \Rightarrow x=2$ (undefined)

• Step 6: Test intervals: $(-\infty, 2)$, $(2, 3.5)$, $(3.5, \infty)$

• Step 7: Determine where the fraction is positive or zero.

• Graph: The graph of $y = \frac{x+1}{x-2}$ shows a vertical asymptote at $x=2$ and crosses $y=3$ at $x=3.5$.

Solution Set: $x \leq 2$ (not including $x=2$) and $x \geq 3.5$

Difference of Cubes Formula

The difference of cubes is a useful factoring identity:

$a^3 - b^3 = (a - b)(a^2 + ab + b^2)$

• Application: Use this formula to factor cubic expressions when solving inequalities or equations.

• Example: Factor $x^3 - 8 = (x - 2)(x^2 + 2x + 4)$

Graphical Representation of Solution Sets

Graphs are useful for visualizing where a function is positive or negative, and for identifying solution intervals for inequalities.

• Polynomial Graphs: The sign of the function changes at its zeros.

• Rational Graphs: The function is undefined at points where the denominator is zero (vertical asymptotes).

• Solution Set: Mark intervals on the x-axis where the inequality holds.

Homework Practice

Practice problems reinforce the concepts of solving polynomial and rational inequalities. Refer to textbook page 217 and complete the following problems: 19, 23, 25, 28, 39, 48, 49.

Table: Pencil Sales Requirements (Application Example)

This table provides a real-world application of inequalities, relating the number of pencils sold to the requirements for a fundraising event.

Number of Pencils	Profit ($)
50	62.50
60	75.00
70	87.50
80	100.00
90	112.50
100	125.00

Application: Use inequalities to determine the minimum number of pencils needed to reach a profit goal.

Additional info: The table and context illustrate how inequalities are used in practical scenarios, such as fundraising or budgeting.