2.14 Polynomial and Rational Inequalities (HL)
1. Polynomial Inequalities
To systematically evaluate polynomial inequalities (e.g., $f(x) > 0$), factorizing $f(x)$ fundamentally allows a sign table or rough graphical sketch to quickly map positive and negative domains.
When factorizing a polynomial, expressions consist of:
- Linear factors indicating roots of the form $(x-a)$.
- Irreducible quadratic factors indicating imaginary roots of the form $(x^2+bx+c)$ where $\Delta < 0$. These inherently maintain a consistent positive or negative sign and never trigger an axis crossing.
Root Multiplicity Impact:
- A root repeated an odd number of times (e.g., $(x-a)^1, (x-a)^3$) behaves identically to a single root, forcing a strict structural sign change across the axis.
- A root repeated an even number of times (e.g., $(x-a)^2, (x-a)^4$) behaves precisely like a vertex boundary, maintaining the function's structural sign without crossing the axis.
EXAMPLE 1
Solve the targeted inequality $2x^3 - 7x^2 - 17x + 10 > 0$.
Solution: The cubic function factorizes cleanly into three isolated single roots: $-2, 0.5$, and $5$. The structured inequality reads $2(x+2)(x-0.5)(x-5) > 0$.
| $x$ | $-\infty$ | $-2$ | $0.5$ | $5$ | $+\infty$ |
|---|---|---|---|---|---|
| $f(x)$ | $-$ | $+$ | $-$ | $+$ |
The domain interval satisfying positive outputs calculates as $x \in ]-2, 0.5[ \cup ]5, +\infty[$.
EXAMPLE 2
Solve the following structural variations evaluating root multiplicity:
The double root at $x=1$ prevents a sign flip. The output switches strictly to positive past $x=5$. Result: $\mathbf{x > 5}$.
Identical domain parameters apply, but zero-values are now tolerated. The root $x=1$ evaluates to zero cleanly. Result: $\mathbf{x=1 \text{ or } x \ge 5}$.
The quadratic component $(x^2+1)$ lacks real roots and maintains a constant positive value globally. It triggers zero influence on the function's final sign table layout. Result replicates identically: $\mathbf{x=1 \text{ or } x \ge 5}$.
2. Rational Inequalities
An elementary algebraic property establishes that the fraction $\frac{f(x)}{g(x)}$ shares identical sign properties with the polynomial product $f(x)g(x)$.
To safely evaluate rational limits like $\frac{f(x)}{g(x)} \ge 0$, build a continuous sign table using the aggregated roots of both numerator and denominator. Critically, any valid solutions evaluating $g(x) = 0$ must be rigorously rejected to prevent mathematical undefined states.
EXAMPLE 3
Solve $\frac{(x+2)}{(x-1)(x-3)} \le 0$.
Solution: Extract the cumulative critical boundary roots: $-2$ (numerator), $1, 3$ (denominator). Assess the combined product factor $(x+2)(x-1)(x-3)$.
| $x$ | $-\infty$ | $-2$ | $1$ | $3$ | $+\infty$ |
|---|---|---|---|---|---|
| Sign | $-$ | $+$ | $-$ | $+$ |
Extracting intervals mapping to $\le 0$ isolates $x \le -2$ and $1 \le x \le 3$. Because $x=1$ and $x=3$ crash the denominator, they are strictly removed. Final established bounds: $\mathbf{x \in ]-\infty, -2] \cup ]1, 3[}$.
EXAMPLE 4 (Equations vs Inequalities)
Compare the algebraic mechanisms required for equations against identical inequalities.
Cross-multiplication is mathematically valid.
$x+1 = (x-2)(x-3) \Rightarrow x+1 = x^2 - 5x + 6 \Rightarrow x^2 - 6x + 5 = 0$. Roots correctly identify as $\mathbf{x=1 \text{ or } x=5}$.
Cross-multiplication is strictly invalid due to unknown denominator polarity altering the logical operator. The terms must be manually relocated to standard zero bounds:
$\frac{x+1}{x-2} - (x-3) \ge 0$
$\frac{x+1 - (x-2)(x-3)}{x-2} \ge 0 \Rightarrow \frac{x+1 - x^2 + 5x - 6}{x-2} \ge 0 \Rightarrow \frac{-x^2 + 6x - 5}{x-2} \ge 0$
$\frac{-(x-1)(x-5)}{x-2} \ge 0$.
Evaluating cumulative roots $1, 2, 5$ on a sign table produces the resulting valid subset $x \in ]-\infty, 1] \cup ]2, 5]$. Note $x=2$ remains strictly open to avoid mathematical failure.