Polynomial and Rational Inequalities

A polynomial inequality compares a polynomial with zero, or can be rearranged so that it does, for example ${x^3-x\gt 0}$ or ${3x^4+10x}\le{11x^3+4.}$ A rational inequality has the unknown in a denominator, for example ${\frac{x-3}{x+4}}\ge{\frac{x+2}{x-5}.}$ Both kinds are solved with the same idea: find the numbers where the expression is zero or undefined, cut the number line at those numbers, and find the sign of the expression on each piece. This is the sign chart method, also called the method of intervals.

On this page: Polynomial inequalities · Alternating signs · Repeated roots · Rational inequalities · Common mistakes · Practice problems

Polynomial inequalities

After moving all terms to one side, a polynomial inequality has one of the forms ${P(x)\gt 0,}$ ${P(x)\ge 0,}$ ${P(x)\lt 0}$ or ${P(x)\le 0,}$ where ${P(x)}$ is a polynomial. A quadratic inequality such as ${3x^2-2x-5\gt 0}$ is the case of degree 2. Other examples: ${x^3-x\gt 0,}$   ${-2x^4+x^2-3}\lt{7,}$   ${4x^3-2x^2}\gt{{5x+7}.}$

The sign chart method

1. Move all terms to one side, so that the other side is ${0.}$
2. Solve the related equation ${P(x)=0}$: factor the polynomial and find its roots.
3. Mark the roots on the number line. They cut it into intervals.
4. Find the sign of ${P(x)}$ on each interval: substitute one test value from the interval, or use the rule of alternating signs below.
5. Choose the intervals with the sign you need. With ${\ge}$ or ${\le}$ the roots belong to the answer (square brackets); with ${\gt}$ or ${\lt}$ they do not (round brackets).

Example 1

Solve ${x^3-x\gt 0.}$

The right side is already ${0.}$ Factor the left side:

${x^3-x}={x(x^2-1)}={x(x+1)(x-1)}$

The roots are ${-1,}$ ${0}$ and ${1.}$ They divide the number line into four intervals. Take one test value in each:

Interval	${x}$	${x^3-x}$
${(-\infty,\,-1)}$	${-2}$	${-6}$
${(-1,\,0)}$	${-0.5}$	${0.375}$
${(0,\,1)}$	${0.5}$	${-0.375}$
${(1,\,\infty)}$	${2}$	${6}$

The sign chart of ${x^3-x.}$ The empty circles show that the roots are not solutions.

We need ${x^3-x\gt 0,}$ so we take the two intervals marked “+”. At the roots the expression equals ${0,}$ and ${0\gt 0}$ is false, so the roots are left out.

Answer: $x\in(-1,\,0)\cup(1,\,\infty),$ that is, ${-1\lt x\lt 0}$ or ${x\gt 1.}$

The graph of ${y=x^3-x}$ shows the same thing: the curve is above the ${x}$-axis exactly over these two intervals.

The graph of ${y=x^3-x.}$ The answer is marked in orange on the ${x}$-axis.

The rule of alternating signs

Test values always work, but once the polynomial is factored there is a faster way. Write it as

${P(x)}={a(x-x_1)(x-x_2)\cdots(x-x_n),}$   where $x_1\lt x_2\lt\dots\lt x_n.$

To the right of the largest root every bracket is positive, so there ${P(x)}$ has the sign of the leading coefficient ${a.}$ Each time we move left past a root, exactly one bracket changes sign, so the sign of the product changes too.

In Example 1, ${x^3-x}={x(x+1)(x-1)}$ has ${a=1\gt 0,}$ so from right to left the signs are ${+,}$ ${-,}$ ${+,}$ ${-,}$ exactly what the test values showed.

Example 2

Solve ${3x^4+10x}\le{11x^3+4.}$

Subtract ${11x^3+4}$ from both sides:

${3x^4-11x^3+10x-4}\le 0$

Algebraic solution

We solve the related equation ${3x^4-11x^3+10x-4}=0.$ Try small numbers first: ${P(-1)}={3+11-10-4}={{0},}$ so ${x=-1}$ is a root and the polynomial is divisible by ${x+1.}$ Dividing (for example with Horner's method) gives

${3x^4-11x^3+10x-4}={(x+1)(3x^3-14x^2+14x-4)}$

A rational root ${\frac pq}$ of a polynomial with integer coefficients has ${p}$ dividing the constant term and ${q}$ dividing the leading coefficient. For the cubic, the candidates are ${\pm1,}$ ${\pm2,}$ ${\pm4,}$ ${\pm\frac13,}$ ${\pm\frac23,}$ ${\pm\frac43,}$ and ${x=\frac23}$ works:

${3\cdot\frac{8}{27}-14\cdot\frac49+14\cdot\frac23-4}={\frac{8-56+84-36}{9}}={0}$

Dividing the cubic by ${3x-2}$ leaves ${x^2-4x+2,}$ whose roots are ${x}={\frac{4\pm\sqrt{16-8}}{2}}={{2\pm\sqrt2}.}$ So

${3x^4-11x^3+10x-4}={(x+1)(3x-2)(x^2-4x+2)}$

and the four roots, from the smallest, are

${-1}\lt{2-\sqrt2}\lt{\frac23}\lt{2+\sqrt2,}$   or approximately   ${-1,}$ ${0.586,}$ ${0.667,}$ ${3.414.}$

The leading coefficient is ${3\gt 0}$ and all four roots are different, so from right to left the signs are ${+,}$ ${-,}$ ${+,}$ ${-,}$ ${+}$:

Test values confirm the chart (the value ${0.6}$ lies between ${0.586}$ and ${0.667}$):

Interval	${x}$	${P(x)}$
${(-\infty,\,-1)}$	${-2}$	${112}$
${(-1,\,2-\sqrt2)}$	${0}$	${-4}$
${(2-\sqrt2,\,\frac23)}$	${0.6}$	${0.0128}$
${(\frac23,\,2+\sqrt2)}$	${1}$	${-2}$
${(2+\sqrt2,\,\infty)}$	${4}$	${100}$

We need ${P(x)\le 0}$: the two intervals marked “−”, together with their endpoints, because the inequality is not strict.

Answer: $x\in\left[-1,\,2-\sqrt2\,\right]\cup\left[\frac23,\,2+\sqrt2\,\right].$

Graphical solution

The graph of $y={3x^4-11x^3+10x-4}$ crosses the ${x}$-axis at ${x=-1}$ and near ${x=3.414.}$ Between ${x=0}$ and ${x=1}$ the curve only seems to touch the axis.

Enlarging the circled spot shows that the curve really crosses the axis twice there, at ${2-\sqrt2}\approx 0.586$ and at ${\frac23}\approx 0.667.$ Between these roots there is a tiny hump above the axis; its highest point is only about ${0.022.}$

The graph is on or below the ${x}$-axis over ${[-1,\ 0.586]}$ and ${[0.667,\ 3.414],}$ which agrees with the algebraic answer. A graph is a good check, but it gives only approximate values and can hide roots that lie close together. The algebra gives the exact critical values; the graph shows at a glance which intervals satisfy the inequality.

Example 3

Solve ${x^4-5x^2+4}\lt 0.$

The left side is biquadratic, a quadratic in ${x^2.}$ Factor it in two steps:

${x^4-5x^2+4}={(x^2-1)(x^2-4)}={(x+2)(x+1)(x-1)(x-2)}$

The roots ${-2,}$ ${-1,}$ ${1,}$ ${2}$ are all different and the leading coefficient is ${1,}$ so the signs from right to left are ${+,}$ ${-,}$ ${+,}$ ${-,}$ ${+.}$

Answer: $x\in(-2,\,-1)\cup(1,\,2).$

Repeated roots

If a bracket appears twice, as in ${(x-2)^2,}$ then ${2}$ is a double root. A square is never negative, so the sign does not change there.

Example 4. Solve ${x^3-4x^2+4x}\le 0.$

${x^3-4x^2+4x}={x(x^2-4x+4)}={x(x-2)^2}$

The root ${0}$ is simple and the root ${2}$ is double. To the right of ${2}$ the sign is ${+.}$ Passing ${2}$ it stays ${+;}$ passing ${0}$ it changes to ${-.}$

The expression is negative for ${x\lt 0}$ and equals ${0}$ at ${x=0}$ and at ${x=2.}$ All of these satisfy ${\le 0.}$

Answer: $x\in(-\infty,\,0]\cup\{2\}.$

Do not forget the single point ${x=2.}$ For the strict inequality ${x(x-2)^2\lt 0}$ the answer would be just ${x\lt 0.}$ For ${x(x-2)^2\gt 0}$ it would be $x\in(0,\,2)\cup(2,\,\infty);$ there ${2}$ must be left out.

The graph of ${y=x(x-2)^2}$ crosses the ${x}$-axis at the simple root ${0}$ and only touches it at the double root ${2.}$

Factors that are never zero

A quadratic factor with a negative discriminant, such as ${x^2+x+1}$ (${D=1-4=-3\lt 0}$), has no real roots. It keeps the sign of its leading coefficient for every ${x,}$ so it never affects the sign chart. A positive factor can simply be dropped, because dividing by a positive number does not change an inequality.

Example 5. Solve ${(x^2+x+1)(x^2-9)}\ge 0.$

Since ${x^2+x+1\gt 0}$ for every ${x,}$ the inequality is equivalent to ${x^2-9\ge 0,}$ that is, ${(x+3)(x-3)}\ge 0.$

Answer: $x\in(-\infty,\,-3]\cup[3,\,\infty).$

Rational inequalities

A rational inequality has the unknown in a denominator. After moving all terms to one side and writing them as a single fraction, it has the form ${\frac{P(x)}{Q(x)}}\gt 0$ (or ${\ge,}$ ${\lt,}$ ${\le}$), where ${P}$ and ${Q}$ are polynomials.

Method for rational inequalities

1. Move all terms to one side and bring them to a common denominator.
2. Factor the numerator and the denominator.
3. Find the critical values: the zeros of the numerator and the zeros of the denominator.
4. Mark them on the number line. The zeros of the denominator are always empty circles. The zeros of the numerator are filled circles for ${\ge}$ and ${\le,}$ and empty circles for ${\gt}$ and ${\lt.}$
5. Find the sign on each interval with test values or with the rule of alternating signs. The brackets of the denominator count just like those of the numerator, since ${\frac PQ}$ and ${P\cdot Q}$ always have the same sign.
6. Write the answer. Never include a zero of the denominator.

Example 6

Solve ${\frac{x-3}{x+4}}\ge{\frac{x+2}{x-5}.}$

Algebraic solution

Subtract ${\frac{x+2}{x-5}}$ and bring the two fractions to the common denominator ${(x+4)(x-5)}$:

${\frac{x-3}{x+4}-\frac{x+2}{x-5}}\ge 0$

${\frac{(x-3)(x-5)-(x+2)(x+4)}{(x+4)(x-5)}}\ge 0$

The numerator is ${(x^2-8x+15)}-{(x^2+6x+8)}={{-14x+7},}$ so

${\frac{-14x+7}{(x+4)(x-5)}}\ge 0$

Divide both sides by ${-7.}$ Dividing by a negative number turns the inequality sign around:

${\frac{2x-1}{(x+4)(x-5)}}\le 0$

The critical values are ${x=\frac12}$ (the numerator is zero; a filled circle, since the inequality is not strict) and ${x=-4,}$ ${x=5}$ (the denominator is zero; empty circles). All brackets have the form ${(x-c)}$ up to a positive factor, so from right to left the signs are ${+,}$ ${-,}$ ${+,}$ ${-.}$

We need the intervals marked “−”, together with ${\frac12}$ but without ${-4}$ and ${5.}$

Answer: $x\in(-\infty,\,-4)\cup\left[\frac12,\,5\right).$

As a check, here are test values of ${f(x)}={\frac{x-3}{x+4}-\frac{x+2}{x-5},}$ the difference of the two sides of the original inequality:

Interval	${x}$	${f(x)}$
${(-\infty,\,-4)}$	${-5}$	${7.7}$
${(-4,\,\frac12)}$	${-2}$	${-2.5}$
${(\frac12,\,5)}$	${3}$	${2.5}$
${(5,\,\infty)}$	${6}$	${-7.7}$

Graphical solution

The graph of $y={\frac{x-3}{x+4}-\frac{x+2}{x-5}}$ has breaks at ${x=-4}$ and ${x=5,}$ where the function is not defined; the dashed lines are vertical asymptotes. The curve crosses the ${x}$-axis only once, at ${x=\frac12.}$

The curve is on or above the ${x}$-axis over ${(-\infty,\,-4)}$ and ${\left[\frac12,\,5\right),}$ the same answer. Notice that the function also changes sign at the breaks without ever crossing the axis there. That is why the zeros of the denominator are critical values too.

Example 7

Solve ${\frac{x^2-x-6}{x^2-1}}\ge 0.$

Factor the numerator and the denominator:

${\frac{(x-3)(x+2)}{(x-1)(x+1)}}\ge 0$

The numerator is zero at ${-2}$ and ${3}$ (filled circles); the denominator is zero at ${-1}$ and ${1}$ (empty circles). All four critical values are different, so the signs from right to left are ${+,}$ ${-,}$ ${+,}$ ${-,}$ ${+.}$

Answer: $x\in(-\infty,\,-2]\cup(-1,\,1)\cup[3,\,\infty).$

Common mistakes

Mistake	Wrong	Right
Multiplying by a denominator that contains ${x}$	${\frac1x\lt 1}\Rightarrow{x\gt 1}$	${\frac{x-1}{x}\gt 0}\Rightarrow{x\in(-\infty,\,0)\cup(1,\,\infty)}$
A zero of the denominator included in the answer	${\frac{x+1}{x-2}\ge 0}\Rightarrow{x\in(-\infty,\,-1]\cup[2,\,\infty)}$	${x\in(-\infty,\,-1]\cup(2,\,\infty)}$
Alternating the signs at a double root	${(x-1)^2(x+2)\gt 0}\Rightarrow{x\in(-\infty,\,-2)\cup(1,\,\infty)}$	${x\in(-2,\,1)\cup(1,\,\infty)}$
Forgetting a single point	${x(x-2)^2\le 0}\Rightarrow{x\in(-\infty,\,0]}$	${x\in(-\infty,\,0]\cup\{2\}}$
Cancelling a common factor and forgetting where it is zero	${\frac{x^2-1}{x-1}\gt 0}\Rightarrow{x+1\gt 0}\Rightarrow{x\gt -1}$	${x\in(-1,\,1)\cup(1,\,\infty)}$
Treating ${(2-x)}$ like ${(x-2)}$	${(2-x)(x+3)\gt 0}\Rightarrow{x\in(-\infty,\,-3)\cup(2,\,\infty)}$	${x\in(-3,\,2)}$

Practice problems

Solve the inequalities. Try first, then open the answer.

1. ${(x+3)(x-1)(x-4)}\lt 0$

Roots ${-3,}$ ${1,}$ ${4;}$ signs from right to left ${+,}$ ${-,}$ ${+,}$ ${-.}$ Answer: $x\in(-\infty,\,-3)\cup(1,\,4)$

2. ${x^3\ge 9x}$

${x^3-9x}={x(x-3)(x+3)}\ge 0.$ Answer: $x\in[-3,\,0]\cup[3,\,\infty)$

3. ${x^4-10x^2+9}\le 0$

${x^4-10x^2+9}={(x^2-1)(x^2-9)}={(x+3)(x+1)(x-1)(x-3)}\le 0.$ Answer: $x\in[-3,\,-1]\cup[1,\,3]$

4. ${(x-2)^2(x+5)}\lt 0$

The double root ${2}$ does not change the sign, and at ${x=2}$ the expression is ${0,}$ which is not negative. Answer: $x\in(-\infty,\,-5)$

5. ${x^3+x^2-4x-4}\gt 0$

Group the terms: ${x^2(x+1)-4(x+1)}={(x+1)(x-2)(x+2)}\gt 0.$ Answer: $x\in(-2,\,-1)\cup(2,\,\infty)$

6. ${\frac{x-1}{x+2}}\le 0$

Critical values ${-2}$ (empty circle) and ${1}$ (filled circle). Answer: ${x\in(-2,\,1]}$

7. ${\frac{x+3}{x-1}}\gt 2$

${\frac{x+3}{x-1}-2}={\frac{5-x}{x-1}}\gt{0,}$ that is, ${\frac{x-5}{x-1}}\lt 0.$ Answer: ${x\in(1,\,5)}$

8. ${\frac{x^2-4}{x^2+x}}\ge 0$

${\frac{(x-2)(x+2)}{x(x+1)}}\ge 0.$ Critical values ${-2,}$ ${-1,}$ ${0,}$ ${2;}$ signs from right to left ${+,}$ ${-,}$ ${+,}$ ${-,}$ ${+.}$ Answer: $x\in(-\infty,\,-2]\cup(-1,\,0)\cup[2,\,\infty)$

9. ${\frac{1}{x-1}}\ge{\frac{1}{x+1}}$

${\frac{1}{x-1}-\frac{1}{x+1}}={\frac{2}{(x-1)(x+1)}}\ge 0.$ The numerator is never zero, so the fraction must be positive. Answer: $x\in(-\infty,\,-1)\cup(1,\,\infty)$

10. ${\frac{(x-3)^2}{x^2-4}}\le 0$

The numerator is never negative and equals ${0}$ only at ${x=3.}$ The fraction is negative where ${x^2-4\lt 0.}$ Answer: $x\in(-2,\,2)\cup\{3\}$

You can check your answers with the equation and inequality solver, which shows the sign chart step by step.