Solving Polynomial and Rational Inequalities

Solving polynomial and rational inequalities involves finding where an expression is positive (greater than zero) or negative (less than zero). Instead of relying on algebra alone, the most reliable method is to use a sign chart (or number line) to test intervals between the zeros and undefined points of the expression.

Solving Polynomial Inequalities

To solve a polynomial inequality, follow these steps:

1. Move all terms to one side so the other side is zero.
2. Factor the polynomial completely to find its roots (zeros).
3. Plot these roots on a number line to divide it into intervals.
4. Pick a test value in each interval and plug it into the factored polynomial to check if the result is positive or negative.

Example: Solve $x^3 - 4x > 0$

Step 1 & 2: Factor the polynomial $x(x^2 - 4) > 0$ $x(x - 2)(x + 2) > 0$

The roots are $x = 0$, $x = 2$, and $x = -2$.

Step 3: Create intervals on a number line The roots divide the number line into four intervals: $(-\infty, -2)$, $(-2, 0)$, $(0, 2)$, and $(2, \infty)$.

Step 4: Test each interval

• For $(-\infty, -2)$, test $x = -3$: $(-3)(-5)(-1) = -15$ (Negative)
• For $(-2, 0)$, test $x = -1$: $(-1)(-3)(1) = 3$ (Positive)
• For $(0, 2)$, test $x = 1$: $(1)(-1)(3) = -3$ (Negative)
• For $(2, \infty)$, test $x = 3$: $(3)(1)(5) = 15$ (Positive)

Since the inequality is $> 0$, we want the positive intervals.

Solution: $x \in (-2, 0) \cup (2, \infty)$

Solving Rational Inequalities

Rational inequalities involve fractions with variables in the denominator. Important Rule: Never multiply both sides by a variable expression, because you don't know if it's positive or negative (which would flip the inequality sign).

Instead, use this method:

1. Move all terms to one side so the other side is zero.
2. Combine fractions over a common denominator.
3. Find the roots of the numerator (where the expression equals zero) and the roots of the denominator (where the expression is undefined).
4. Plot all these critical values on a number line and test the intervals.

Example: Solve $\frac{x - 1}{x + 3} \leq 2$

Step 1: Move everything to one side $\frac{x - 1}{x + 3} - 2 \leq 0$

Step 2: Find a common denominator $\frac{x - 1 - 2(x + 3)}{x + 3} \leq 0$ $\frac{x - 1 - 2x - 6}{x + 3} \leq 0$ $\frac{-x - 7}{x + 3} \leq 0$

To make it easier, you can multiply by $-1$ (remember to flip the inequality sign!): $\frac{x + 7}{x + 3} \geq 0$

Step 3: Find critical values

• Numerator root: $x + 7 = 0 \implies x = -7$
• Denominator root: $x + 3 = 0 \implies x = -3$ (Note: $x$ can never equal $-3$ because it makes the fraction undefined).

Step 4: Test intervals The intervals are $(-\infty, -7)$, $(-7, -3)$, and $(-3, \infty)$.

• Test $x = -8$: $\frac{-8 + 7}{-8 + 3} = \frac{-1}{-5} = \frac{1}{5}$ (Positive)
• Test $x = -4$: $\frac{-4 + 7}{-4 + 3} = \frac{3}{-1} = -3$ (Negative)
• Test $x = 0$: $\frac{0 + 7}{0 + 3} = \frac{7}{3}$ (Positive)

We want $\geq 0$ (positive or zero). The interval must include $x = -7$ (since it makes the numerator zero) but exclude $x = -3$ (since it makes the denominator undefined).

Solution: $x \in (-\infty, -7] \cup (-3, \infty)$