2.14 Polynomial and Rational Inequalities (HL)

1. Polynomial Inequalities

To solve a polynomial inequality such as $f(x)\gt 0$, first factorize the polynomial. The roots divide the real line into intervals. On each interval, the polynomial has a fixed sign, so a sign table can be used to decide where the inequality is true.

When factorizing a polynomial, the factors may include:

  • Linear factors, such as $(x-a)$, which give real roots.
  • Irreducible quadratic factors, such as $x^2+bx+c$ with $\Delta\lt 0$. These have no real roots and do not create boundary points in the sign table.

Root multiplicity matters:

  • If a root has odd multiplicity, for example $(x-a)^1$ or $(x-a)^3$, the sign of the polynomial changes across $x=a$.
  • If a root has even multiplicity, for example $(x-a)^2$ or $(x-a)^4$, the sign of the polynomial does not change across $x=a$.

METHOD: SOLVING A POLYNOMIAL INEQUALITY

  1. Move all terms to one side so the inequality is compared with $0$.
  2. Factorize the polynomial completely over the real numbers.
  3. Place all real roots on a sign table.
  4. Use multiplicity to decide whether signs change or stay the same.
  5. Select the intervals that satisfy the required sign.
  6. Include roots only for $\leq$ or $\geq$ inequalities, not for $\lt$ or $\gt$ inequalities.

EXAMPLE 1

Solve $$2x^3-7x^2-17x+10\gt 0.$$

Solution:

Factorize the polynomial: $$\begin{aligned} 2x^3-7x^2-17x+10 &=2(x+2)\left(x-\dfrac{1}{2}\right)(x-5). \end{aligned}$$
The critical values are $$x=-2,\qquad x=\dfrac{1}{2},\qquad x=5.$$ Each root has odd multiplicity, so the sign changes at each root.
Interval $(-\infty,-2)$ $-2$ $\left(-2,\dfrac{1}{2}\right)$ $\dfrac{1}{2}$ $\left(\dfrac{1}{2},5\right)$ $5$ $(5,\infty)$
Sign of $f(x)$ $-$ $0$ $+$ $0$ $-$ $0$ $+$
Since the inequality is strict, we do not include the roots. Therefore, $$x\in\left(-2,\dfrac{1}{2}\right)\cup(5,\infty).$$

EXAMPLE 2

Solve the following inequalities involving repeated roots.

(a) Solve $$3(x-1)^2(x-5)\gt 0.$$

The factor $(x-1)^2$ is always non-negative and has even multiplicity, so the sign does not change at $x=1$. The sign only changes at the odd-multiplicity root $x=5$.

Interval $(-\infty,1)$ $1$ $(1,5)$ $5$ $(5,\infty)$
Sign $-$ $0$ $-$ $0$ $+$

Hence, $$x\in(5,\infty).$$

(b) Solve $$3(x-1)^2(x-5)\geq 0.$$

This time zero values are allowed. Therefore $x=1$ and $x=5$ may be included. The positive interval is still to the right of $5$.

Hence, $$x=1\qquad\text{or}\qquad x\geq 5.$$ Equivalently, $$x\in\{1\}\cup[5,\infty).$$

(c) Solve $$3(x-1)^2(x-5)(x^2+1)\geq 0.$$

Since $x^2+1\gt 0$ for all real $x$, the factor $x^2+1$ does not change the sign of the expression. Therefore the solution is the same as in part (b):

$$x=1\qquad\text{or}\qquad x\geq 5.$$ Equivalently, $$x\in\{1\}\cup[5,\infty).$$

2. Rational Inequalities

A rational inequality involves a quotient such as $$\dfrac{f(x)}{g(x)}\gt 0,\qquad \dfrac{f(x)}{g(x)}\leq 0,\qquad \text{or similar forms.}$$ The signs of the numerator and denominator determine the sign of the fraction.

To solve $$\dfrac{f(x)}{g(x)}\geq 0,$$ build a sign table using all critical values from both the numerator and the denominator. However, values satisfying $g(x)=0$ must always be excluded because the expression is undefined.

The signs of $\dfrac{f(x)}{g(x)}$ and $f(x)g(x)$ are the same whenever $g(x)\neq 0$. This is useful for sign analysis, but denominator zeros must still be removed from the answer.

IMPORTANT WARNING

Do not cross-multiply an inequality by an expression whose sign is unknown. Multiplying by a negative quantity reverses the inequality sign, so cross-multiplication can produce incorrect solutions.

EXAMPLE 3

Solve $$\dfrac{x+2}{(x-1)(x-3)}\leq 0.$$

Solution:

The critical values are: $$x=-2\qquad\text{from the numerator,}$$ and $$x=1,\qquad x=3\qquad\text{from the denominator.}$$
The expression is zero at $x=-2$, but it is undefined at $x=1$ and $x=3$.
Interval $(-\infty,-2)$ $-2$ $(-2,1)$ $1$ $(1,3)$ $3$ $(3,\infty)$
Sign $-$ $0$ $+$ undefined $-$ undefined $+$
We need the expression to be less than or equal to zero. Therefore we take the negative intervals and include the numerator zero $x=-2$, but exclude denominator zeros $x=1$ and $x=3$.
Hence, $$x\in(-\infty,-2]\cup(1,3).$$

EXAMPLE 4 (Equations vs Inequalities)

Compare the algebraic method for solving a rational equation with the method for solving a rational inequality.

(a) Equation: $$\dfrac{x+1}{x-2}=x-3.$$

Since this is an equation, cross-multiplication is allowed, provided we later check that $x\neq 2$.

$$\begin{aligned} \dfrac{x+1}{x-2}&=x-3,\\ x+1&=(x-2)(x-3),\\ x+1&=x^2-5x+6,\\ 0&=x^2-6x+5,\\ 0&=(x-1)(x-5). \end{aligned}$$

Therefore, $$x=1\qquad\text{or}\qquad x=5.$$ Both values are valid because neither makes the denominator zero.

(b) Inequality: $$\dfrac{x+1}{x-2}\geq x-3.$$

For an inequality, do not multiply by $x-2$ directly because the sign of $x-2$ depends on $x$. Instead, move everything to one side and combine into one fraction.

$$\begin{aligned} \dfrac{x+1}{x-2}-(x-3)&\geq 0,\\ \dfrac{x+1-(x-2)(x-3)}{x-2}&\geq 0,\\ \dfrac{x+1-(x^2-5x+6)}{x-2}&\geq 0,\\ \dfrac{-x^2+6x-5}{x-2}&\geq 0,\\ \dfrac{-(x-1)(x-5)}{x-2}&\geq 0. \end{aligned}$$

The critical values are $x=1$, $x=2$, and $x=5$. The value $x=2$ must be excluded because the denominator is zero.

Interval $(-\infty,1)$ $1$ $(1,2)$ $2$ $(2,5)$ $5$ $(5,\infty)$
Sign $+$ $0$ $-$ undefined $+$ $0$ $-$

Therefore, $$x\in(-\infty,1]\cup(2,5].$$