Polynomial Inequalities: Sign Charts & Multiplicity

Definition

Four inequalities look alike, and only one of them is a polynomial inequality.

Each of these asks where an expression is below zero: A,

(x + 1)x(x - 1) < 0

; B,

\frac{x - 1}{x + 2} < 0

; C,

\sqrt{x} - 2 < 0

; D,

|x| - 3 < 0

. Written out, they differ only in a symbol or two. Drawn, they do not.

A · (x + 1)x(x − 1) < 0

B · (x − 1)/(x + 2) < 0

C · √x − 2 < 0

D · |x| − 3 < 0

Each graph is the left side; the blue bar on the axis is where it is below zero. These are four frozen states of the inequality visual explorer.

Sorted, A is the polynomial inequality, and B, C and D are not.

What decides it is what happens to

x

inside each expression, and it shows directly in the graphs.

candidate	what x is doing	what the graph does
A	whole powers only: x³ − x	unbroken, defined everywhere
B	x in a denominator	breaks at the wall x = −2
C	x under a root	starts at 0, nothing to the left
D	x inside bars	has a corner at 0

Each rejected candidate fails at one exact place, and the near-misses do not fail at all.

B fails at

x = -2

, where its denominator is zero: the expression does not exist there, and the graph breaks. C fails left of

0

, where a square root has no real value. D fails at

0

in a quieter way: the bars make it two different formulas glued at a corner. The near-misses pass.

(x + 1)x(x - 1)

is in factored form, but multiplied out it is

x^3 - x

, an ordinary polynomial.

\frac{x^2 + 1}{5} > x

has a fraction, but its denominator is a constant, so it is just a polynomial times

\frac{1}{5}

.

The rule: only whole-number powers of x, multiplied by constants and added.

Such an expression is defined for every real

x

and its graph never breaks, so it can change sign only by passing through zero — only at a root. That is what makes A easy: its roots

-1

,

0

and

1

cut the line into four pieces, and on each piece the sign is fixed.

piece of the line	x + 1	x	x − 1	product
left of −1	−	−	−	− : in
between −1 and 0	+	−	−	+
between 0 and 1	+	+	−	− : in
right of 1	+	+	+	+

The product is negative on two pieces, so A's answer is

x < -1

or

0 < x < 1

— the two blue bars in its graph.

The definition

A polynomial inequality has the form below — or with

<

,

\leq

or

\geq

— where

P(x)

is a polynomial of degree

n \geq 3

.

Definition

P(x) > 0

Its equation counterpart: polynomial equations.

A polynomial inequality is a statement about the sign of a polynomial, and its answer is a set.

It is not a number and not a function. It asks where a polynomial is above or below the axis, and its answer collects those stretches of the line. Linear and quadratic inequalities are the special cases

n = 1

and

n = 2

; from degree three onward the stretches multiply.

THEOREM

As with polynomial equations, the degree

n

sets an upper bound: a polynomial of degree

n

has at most

n

real roots, and these roots create at most

n + 1

intervals on the number line. Within each interval, the polynomial maintains a constant sign.

A's answer: two of its four pieces. The open dots are the roots

-1

,

0

and

1

, where the product is

0

and so not below it.

The rejected candidates are not lost; each has its own way back.

Excluding B, C and D only means the plain polynomial method does not apply to them as they stand.

rejected candidate	how it is still solved
B · rational	same sign chart, with the denominator zeros added as walls
C · radical	restrict to where the root exists first, then compare
D · absolute value	convert to a compound inequality without bars

Each has its own page: rational inequalities, absolute value inequalities. The polynomial case is the one the others are brought back to.

The rest of the page is the method for the polynomial case: the sign chart, reading signs from factored form, how root multiplicity decides whether the sign changes, end behavior, polynomials that do not factor nicely, worked examples, and filling the chart without test points.

The Sign Chart Method

The sign chart is a systematic procedure for solving any polynomial inequality once the real roots are known.

Step 1: Solve

P(x) = 0

to find all real roots. Use factoring, the rational root theorem, or the quadratic formula after degree reduction — whatever combination of techniques produces the roots.

Pitfall: Dividing both sides by x

Dividing

x^3 > 4x

by

x

gives

x^2 > 4

, so

x < -2

or

x > 2

, but

x = -3

fails:

-27 < -12

. Kept as

x(x + 2)(x - 2) > 0

, the root

0

stays on the chart and the answer is

(-2, 0) \cup (2, \infty)

.

Step 2: Order the roots on the number line from smallest to largest. They divide the line into intervals: one to the left of the smallest root, one between each consecutive pair, and one to the right of the largest.

Step 3: In each interval, determine the sign of

P(x)

. Choose any convenient test point within the interval and evaluate the sign — either by substituting into

P(x)

directly or, if

P(x)

is in factored form, by tracking the sign of each factor separately and multiplying.

Step 4: Select the intervals where the sign matches the inequality. For

P(x) > 0

, take the positive intervals. For

P(x) < 0

, take the negative ones.

Step 5: Decide on endpoint inclusion. For non-strict inequalities (

\leq

or

\geq

), include the roots themselves because

P(r) = 0

satisfies

\leq 0

and

\geq 0

. For strict inequalities (

<

or

>

), exclude them.

Step	Action	Tool / detail
1	solve P(x) = 0 to find all real roots	factoring; rational root theorem; quadratic formula after degree reduction; numerical methods if needed
2	order the roots on the number line	smallest to largest; intervals form between consecutive roots
3	determine the sign of P(x) in each interval	test points, factor signs from factored form, or end behavior + multiplicity
4	select intervals matching the inequality	positive intervals for P(x) > 0; negative intervals for P(x) < 0
5	apply endpoint rules	include roots for ≤ ⁄ ≥; exclude for < ⁄ >

Three simple roots, four intervals, alternating signs

With roots at −2, 1 and 5 the axis splits into four intervals, and because each root is simple the sign flips at every one of them: the chart reads −, +, −, + and the strict inequality selects the two shaded intervals. No test points are needed once the roots and their multiplicities are known. Fill the chart for any factored polynomial on the inequality visual explorer.

The method is the same for every degree; only the number of intervals grows.

Factored Form and Sign Analysis

When the polynomial is expressed as a product of linear factors, the sign in each interval can be determined without evaluating

P(x)

numerically. Instead, the sign of each individual factor is tracked, and the signs are multiplied together.

Consider

(x + 2)(x - 1)(x - 5) < 0

. The roots are

-2

,

1

, and

5

, creating four intervals. In each interval, every factor is either positive or negative depending on whether the test point lies to the left or right of that factor's root.

In

(-\infty, -2)

: all three factors are negative. Three negatives multiply to a negative. The product is negative.

In

(-2, 1)

: the factor

(x + 2)

is positive, while

(x - 1)

and

(x - 5)

are negative. One positive and two negatives: the product is positive.

In

(1, 5)

: the factors

(x + 2)

and

(x - 1)

are positive,

(x - 5)

is negative. Two positives and one negative: the product is negative.

In

(5, \infty)

: all three factors are positive. The product is positive.

Pitfall: Misreading a reversed factor

Reading

(5 - x)

in

(x + 2)(x - 1)(5 - x) < 0

like

(x - 5)

gives

(-\infty, -2) \cup (1, 5)

, but

x = 3

gives

5 \cdot 2 \cdot 2 = 20 > 0

. The factor

5 - x

is positive left of

5

, so the answer is

(-2, 1) \cup (5, \infty)

.

The inequality

< 0

selects the negative intervals:

(-\infty, -2) \cup (1, 5)

. No numerical evaluation of

P(x)

was needed — only sign tracking.

This approach is faster than substituting test points into the expanded polynomial, especially for higher-degree expressions with many terms. It requires the polynomial to be in factored form, which is why factoring is typically the first step.

On any interval between roots, a product is negative exactly when an odd number of its factors are negative.

The same sign tracking, drawn: one strip for each factor of

(x + 2)(x - 1)(x - 5)

, each switching from

-

to

+

at its own root, and a last strip for the product, under the graph of the cubic itself.

DownloadPNG2×PDF

sign analysis by factors

Profile of a factored cubic

P(x) = (x + 2)(x - 1)(x - 5)

3roots4intervals

Key points

1Roots

x = −2, 1, 5

Global

Negative on

(-\infty, -2) \cup (1, 5)

an odd number of negative factors

Positive on

(-2, 1) \cup (5, \infty)

an even number of negative factors

Solves

(x + 2)(x - 1)(x - 5) < 0

the negative intervals

Profile of (x + 2)(x − 1)(x − 5)·/algebra/inequalities/polynomialLearn Math Class

Read any column of strips from top to bottom: count the minus signs in the three factor rows and the product row follows, negative for an odd count and positive for an even one. The solution of

< 0

is the two columns where the product is negative,

(-\infty, -2) \cup (1, 5)

, and the graph dips below the axis on exactly those intervals.

Root Multiplicity and Sign Changes

Not every root causes the sign of the polynomial to change. Whether the sign flips at a root depends on the root's multiplicity.

RULE

At a root of odd multiplicity (simple root, triple root, etc.), the polynomial changes sign. The graph crosses the

x

-axis, and the intervals on either side have opposite signs.

A factor

(x - r)^1

is negative to the left of

r

and positive to the right, so crossing it flips the sign of the product.

RULE

At a root of even multiplicity (double root, quadruple root, etc.), the polynomial keeps its sign. The graph touches the

x

-axis and turns back without crossing.

A factor

(x - r)^2

is positive on both sides of

r

(and zero at

r

itself), so crossing it does not flip the sign of the product.

Example

Consider

(x + 1)(x - 3)^2 > 0

. The roots are

x = -1

(multiplicity

1

) and

x = 3

(multiplicity

2

). In

(-\infty, -1)

:

(x + 1)

is negative and

(x - 3)^2

is positive — product is negative. In

(-1, 3)

:

(x + 1)

is positive and

(x - 3)^2

is positive — product is positive. In

(3, \infty)

: both are positive — product is positive. The sign changes at

x = -1

(odd multiplicity) but not at

x = 3

(even multiplicity). The solution is

(-1, 3) \cup (3, \infty)

, which simplifies to

(-1, \infty) \setminus \{3\}

— all

x > -1

except

x = 3

, where the expression equals zero.

Pitfall: Flipping the sign at a double root

Alternating the signs at every root of

(x + 1)(x - 3)^2 > 0

makes

(-1, 3)

negative and gives

(-\infty, -1) \cup (3, \infty)

. The double root

3

keeps the sign, so the answer is

(-1, 3) \cup (3, \infty)

.

This multiplicity rule eliminates the need to test points in every interval. Once the sign is known in any one interval, the sign in each adjacent interval is determined by whether the separating root has odd or even multiplicity.

In practice the chart is filled by walking along the number line one root at a time and carrying the sign across each root in turn. The exponent of each factor is read straight from the factored form, which is why repeated factors are collected into a single power, such as

(x - 3)^2

, before the chart is drawn.

Multiplicity at root r	Example factor	Sign behavior crossing r	Graph behavior at r
Odd (1, 3, 5, …)	(x − r), (x − r)³	flips — adjacent intervals have opposite signs	graph crosses the x-axis
Even (2, 4, 6, …)	(x − r)², (x − r)⁴	preserved — adjacent intervals have the same sign	graph touches the x-axis but does not cross (tangent)

The double root at x = 2 keeps the sign

The factor (x − 2)² is never negative, so the curve touches the axis at 2 and bounces back: the sign chart shows the same sign on both sides of that root, and only the simple root at −3 produces a flip. Even multiplicity touches, odd multiplicity crosses. Compare a simple and a double root side by side on the inequality visual explorer.

End behaviour, treated next, fixes the sign of the first interval; the rest follow by flipping at odd-multiplicity roots and keeping the sign at even ones.

End Behavior

RULE

The sign of a polynomial for very large values of

|x|

is controlled entirely by the leading term

a_nx^n

.

This provides a guaranteed starting point for the sign chart: the sign in the outermost intervals is known before any test points are evaluated.

For even degree with

a_n > 0

: the polynomial is positive for both very large positive and very large negative

x

. Both outer intervals are positive.

For even degree with

a_n < 0

: the polynomial is negative on both outer intervals.

For odd degree with

a_n > 0

: the polynomial is negative far to the left and positive far to the right. The leftmost interval is negative and the rightmost is positive.

For odd degree with

a_n < 0

: the polynomial is positive far to the left and negative far to the right. The signs are reversed from the previous case.

Fact: Odd degree always has a root

Every polynomial of odd degree has at least one real root, because a graph with no breaks cannot pass from one side of the axis to the other without meeting it. So an odd-degree

P(x) > 0

never holds for every

x

, as

x^2 + 1 > 0

does.

In both odd-degree cases the two outer intervals carry opposite signs: far to the left the polynomial lies on one side of the axis, and far to the right on the other. In both even-degree cases the two ends lie on the same side. Between the two ends, the graph of a polynomial runs along the whole number line with no breaks or jumps.

Knowing the outermost signs and applying the multiplicity rule inward determines the sign in every interval without any test-point computation. Start from the rightmost interval (whose sign is the sign of

a_n

for odd degree, or positive for even degree with

a_n > 0

), then flip or preserve the sign at each root moving leftward according to its multiplicity. This technique is faster and less error-prone than substituting numerical test points.

Case	Leftmost interval (x → −∞)	Rightmost interval (x → +∞)
Even degree, aₙ > 0	+	+
Even degree, aₙ < 0	−	−
Odd degree, aₙ > 0	−	+
Odd degree, aₙ < 0	+	−

Far from the origin a polynomial follows its leading term, so its signs at both ends are known before any root is found.

Polynomials That Do Not Factor Over the Rationals

The sign chart method requires knowing the real roots, but it does not require those roots to be rational. When the polynomial has no rational roots — when the rational root theorem produces no valid candidates — irrational or approximate roots still serve as valid boundary points.

Example

The polynomial

x^3 - 2 = 0

has the single real root

x = \sqrt[3]{2} \approx 1.26

. For the inequality

x^3 - 2 > 0

, this root divides the number line into two intervals. Since the leading coefficient is positive and the degree is odd, the polynomial is negative for

x < \sqrt[3]{2}

and positive for

x > \sqrt[3]{2}

. The solution is

(\sqrt[3]{2}, \infty)

.

When multiple irrational roots exist and exact values are unavailable, numerical approximations from graphing or iterative methods (bisection, Newton's method) identify the boundary points to whatever precision is needed. The sign chart is then constructed using these approximate values. The logical structure — roots create intervals, signs are constant within intervals — holds regardless of whether the roots are rational, irrational, or only known approximately.

For polynomials that factor partially — say, one rational root found and a remaining irreducible quadratic with

\Delta < 0

— the irreducible factor contributes no real roots and maintains constant sign. It affects the overall sign in every interval uniformly but creates no new boundary points.

A boundary that is not a rational number

The section's x³ − 2 > 0. Its only real root is ∛2 ≈ 1.26, which the rational root theorem could never find, yet it splits the line exactly like any other root: the curve is below the axis to its left and above it to its right. The solution is (∛2, ∞), with the boundary excluded because the inequality is strict.

Approximate roots work the same way; only the precision of the boundary changes.

Worked Examples

A cubic with three distinct real roots: solve

x^3 - 6x^2 + 11x - 6 \leq 0

. The polynomial factors as

(x - 1)(x - 2)(x - 3)

. The roots

1

,

2

,

3

create four intervals. All three roots have odd multiplicity, so the sign alternates. The leading coefficient is positive and the degree is odd, so the rightmost interval

(3, \infty)

is positive. Moving left:

(2, 3)

is negative,

(1, 2)

is positive,

(-\infty, 1)

is negative. The inequality

\leq 0

selects the negative intervals and includes the roots:

(-\infty, 1] \cup [2, 3]

.

A quartic with a repeated root: solve

(x + 2)^2(x - 1)(x - 4) > 0

. The roots are

-2

(multiplicity

2

),

1

(multiplicity

1

), and

4

(multiplicity

1

). The leading term is

x^4

(positive), so both outermost intervals are positive. Moving inward from the right:

(4, \infty)

is positive. At

x = 4

(odd), the sign flips:

(1, 4)

is negative. At

x = 1

(odd), the sign flips:

(-2, 1)

is positive. At

x = -2

(even), the sign stays:

(-\infty, -2)

is positive. The solution to

> 0

is

(-\infty, -2) \cup (-2, 1) \cup (4, \infty)

— the even-multiplicity root

-2

splits the left interval without changing the sign, and since

P(-2) = 0

, the strict inequality excludes

-2

automatically.

Pitfall: Losing an isolated root

Solving

(x + 2)^2(x - 1)(x - 4) \leq 0

from the negative interval alone gives

[1, 4]

and loses

x = -2

, where the product is

0

. The solution is

\{-2\} \cup [1, 4]

.

The double root

-2

is the one place in this example where the endpoint rule and the multiplicity rule meet. The sign does not change there: the polynomial is positive on both sides of

-2

and touches zero at that single point. Whether a point like this belongs to the answer is decided by the endpoint rule of Step 5, not by the intervals around it.

A polynomial with an irrational root: solve

x^3 - 3x + 1 > 0

. The rational root theorem yields no rational roots. Numerical methods show three real roots near

x \approx -1.88

,

x \approx 0.35

, and

x \approx 1.53

. All have odd multiplicity, so the sign alternates. The leading coefficient is positive, so the pattern from right to left is

+, -, +, -

. The solution is approximately

(-1.88, 0.35) \cup (1.53, \infty)

.

Reading the signs from the ends inward

The section's (x + 2)²(x − 1)(x − 4) > 0. Its leading term x⁴ makes both ends positive. Moving inward from the right, the simple root 4 flips the sign to negative and the simple root 1 flips it back to positive; the double root −2 only touches the axis, so the sign stays positive on both sides. Excluding the roots themselves, the solution is x < −2, −2 < x < 1, or x > 4.

End behavior plus multiplicity fills the whole sign chart without a single test point.

Summary: Filling the Sign Chart Without Test Points

The sign chart for a polynomial inequality of degree three or higher can be filled in entirely without test points by combining the root multiplicity rule (Section 4) with the end behavior of the polynomial (Section 5). The table below collects the resulting workflow: each step lists what is determined at that stage and the tool that supplies it, building the sign chart from the rightmost interval inward.

Step	What you determine	How
1	all real roots, ordered on the number line	factor; rational root theorem; quadratic formula; numerical methods if needed
2	multiplicity at each root	read directly from the factored form (exponent of each factor)
3	sign in the rightmost interval	end behavior — sign of aₙ for odd degree; sign of aₙ for even degree (both ends share it)
4	sign in every other interval, working leftward	flip the sign at each root of odd multiplicity; keep it the same at each root of even multiplicity
5	solution set	select intervals matching the inequality; include roots for ≤ ⁄ ≥, exclude for < ⁄ >

Common Pitfalls

Four mistakes come up again and again when polynomial inequalities are solved by hand. Each one is marked where it happens in the sections above; here each is worked through in full, with the wrong and the right working side by side.

DownloadPNG2×PDF

1Dividing both sides by x

An inequality with

x

in every term, such as

x^3 > 4x

, is simplified by dividing both sides by

x

, as one might in an equation. The sign of

x

is unknown, so it is unclear whether the inequality should flip, and the root

x = 0

disappears from the chart.

Start

x^3 > 4x

Wrong turn

Error here

\implies x^2 > 4

\implies x < -2 \text{ or } x > 2

Right path

Collect everything on one side and factor.

x^3 - 4x > 0

\implies x(x + 2)(x - 2) > 0

The roots

-2

,

0

and

2

are all simple. The degree is odd and the leading coefficient is positive, so from the right the signs are

+

on

(2, \infty)

,

-

on

(0, 2)

,

+

on

(-2, 0)

and

-

on

(-\infty, -2)

. The solution is

(-2, 0) \cup (2, \infty)

.

Why it happensCancelling a common factor is routine in equations, and dividing an inequality by a positive number is safe, so a variable factor gets treated like any other number.

How to checkTest a value from the wrong answer.

x = -3

gives

(-3)^3 = -27

, which is not greater than

4(-3) = -12

. A value the wrong answer left out,

x = -1

, gives

-1 > -4

, which is true.

2Misreading a reversed factor

A factor written as

(5 - x)

instead of

(x - 5)

is given the usual pattern in the sign chart: negative to the left of its root and positive to the right. For

(5 - x)

it is the other way round.

Wrong turn

Error here

(x + 2)(x - 1)(5 - x) < 0

is charted with the signs of

(x + 2)(x - 1)(x - 5)

, which is negative on

(-\infty, -2)

and

(1, 5)

. The answer is given as

(-\infty, -2) \cup (1, 5)

.

Right path

5 - x

is positive for

x < 5

and negative for

x > 5

.

In

(-\infty, -2)

:

(-)(-)(+)

, positive.

In

(-2, 1)

:

(+)(-)(+)

, negative.

In

(1, 5)

:

(+)(+)(+)

, positive.

In

(5, \infty)

:

(+)(+)(-)

, negative.

The solution is

(-2, 1) \cup (5, \infty)

. Equivalently,

5 - x = -(x - 5)

, so the inequality is

(x + 2)(x - 1)(x - 5) > 0

.

Why it happensThe rule "negative left of the root, positive right of it" is learned on factors of the form

(x - r)

and then applied to every factor by habit. In

(5 - x)

the coefficient of

x

is negative, which reverses the pattern.

How to checkSubstitute a value from the answer.

x = 3

gives

(5)(2)(2) = 20

, which is not below zero. End behavior catches it too: the leading term is

-x^3

, so the rightmost interval must be negative.

3Flipping the sign at a double root

The signs are alternated at every root, as in a chart where all roots are simple, without reading the multiplicity of each root.

Wrong turn

Error here

For

(x + 1)(x - 3)^2 > 0

, the rightmost interval

(3, \infty)

is positive. Flipping at

3

makes

(-1, 3)

negative, and flipping at

-1

makes

(-\infty, -1)

positive. The answer is given as

(-\infty, -1) \cup (3, \infty)

.

Right path

3

has multiplicity

2

, so the sign stays positive on

(-1, 3)

. The simple root

-1

flips it, so

(-\infty, -1)

is negative. The solution is

(-1, 3) \cup (3, \infty)

.

Why it happensThe first examples of sign charts usually have only simple roots, where alternating is right every time, so alternation becomes the rule instead of multiplicity.

How to checkTest one point next to the repeated root.

x = 0

gives

(1)(9) = 9 > 0

, so

(-1, 3)

is positive. A value from the wrong answer,

x = -2

, gives

(-1)(25) = -25

.

4Losing an isolated root

For a non-strict inequality, the answer is read only from the chosen intervals and their endpoints. A root of even multiplicity that sits between two intervals of the wrong sign satisfies

P(r) = 0

but is the endpoint of no chosen interval, so it is lost.

Wrong turn

Error here

For

(x + 2)^2(x - 1)(x - 4) \leq 0

, the signs are

+

on

(-\infty, -2)

,

+

on

(-2, 1)

,

-

on

(1, 4)

and

+

on

(4, \infty)

. The only negative interval is

(1, 4)

; with its endpoints, the answer is given as

[1, 4]

.

Right path

P(-2) = 0

, and

0 \leq 0

, so

x = -2

satisfies the inequality. The solution is

\{-2\} \cup [1, 4]

.

Why it happensStep 5 is read as "include the endpoints of the chosen intervals", and

-2

is not an endpoint of any chosen interval. It is a single point where the graph touches the axis.

How to checkFor

\leq

or

\geq

, list every root at the end. Each root makes

P(x) = 0

, so each one satisfies the inequality and must appear in the answer.

Common pitfalls with polynomial inequalities·/algebra/inequalities/polynomialLearn Math Class

All four slips break the sign chart. Dividing by

x

ignores the sign of

x

and loses the root

0

, a reversed factor such as

5 - x

has the opposite sign of

x - 5

, a root of even multiplicity does not change the sign, and with

\leq

every root belongs to the answer, even one where the sign does not change. Testing one value in each interval, and listing the roots at the end, catches each slip.

Polynomial Inequalities FAQ

Why is $x = -2$ part of the solution of $(x + 2)^2(x - 1)(x - 4) \leq 0$ ?

+

Because the product is

0

there, and

0 \leq 0

is true. The double root

-2

only touches the axis: the polynomial is positive on both sides of it, so it is the endpoint of no negative interval, and reading the answer from the chosen intervals gives just

[1, 4]

. The full solution is

\{-2\} \cup [1, 4]

. For

\leq

or

\geq

, list every root at the end, since each one makes

P(x) = 0

.Read more →

Can an odd-degree polynomial inequality be true for every $x$ ?

+

No. The two ends of an odd-degree polynomial lie on opposite sides of the axis, and its graph has no breaks, so it must cross zero at least once: every odd-degree polynomial has a real root. So

P(x) > 0

fails somewhere, unlike

x^2 + 1 > 0

, which holds everywhere. For the same reason the solution is never empty either:

x^3 - 2 > 0

holds on

(\sqrt[3]{2}, \infty)

and fails to the left of

\sqrt[3]{2}

.Read more →

Do you need exact roots to solve a polynomial inequality?

+

No. The sign chart only needs the roots in the right order, and irrational or approximate roots cut the line just as well.

x^3 - 2 > 0

has the single real root

\sqrt[3]{2} \approx 1.26

, which the rational root theorem cannot find, and the solution is

(\sqrt[3]{2}, \infty)

. For

x^3 - 3x + 1 > 0

, numerical methods give roots near

-1.88

,

0.35

and

1.53

, and the solution is approximately

(-1.88, 0.35) \cup (1.53, \infty)

. Only the precision of the boundaries changes.Read more →

What does a factor like $x^2 + 1$ do to a sign chart?

+

It adds no boundaries. A quadratic factor with a negative discriminant has no real roots, so it creates no critical points and keeps one sign on the whole line. If that sign is positive, as for

x^2 + 1

, the factor can be ignored:

(x - 1)(x^2 + 1) > 0

has the same solution as

x - 1 > 0

, namely

x > 1

. If the factor is negative everywhere, such as

-x^2 - 1

, it flips the sign of every interval at once.Read more →

How do you handle a factor like $(5 - x)$ in a sign chart?

+

Reverse the usual pattern:

5 - x

is positive left of

5

and negative right of it, the opposite of

x - 5

. Charting it like

x - 5

turns

(x + 2)(x - 1)(5 - x) < 0

into

(-\infty, -2) \cup (1, 5)

, but

x = 3

gives

5 \cdot 2 \cdot 2 = 20 > 0

. Writing

5 - x = -(x - 5)

and multiplying by

-1

gives

(x + 2)(x - 1)(x - 5) > 0

, with solution

(-2, 1) \cup (5, \infty)

.Read more →

Polynomial Inequalities

Sign Charts, Root Multiplicity, and the Shape of the Solution

Key Terms

Core Concepts

From Polynomials Category

Definition

The Sign Chart Method

Factored Form and Sign Analysis

Profile of a factored cubic

Root Multiplicity and Sign Changes

End Behavior

Polynomials That Do Not Factor Over the Rationals

Worked Examples

Summary: Filling the Sign Chart Without Test Points

Common Pitfalls

Polynomial Inequalities FAQ

Why is $x = -2$ part of the solution of $(x + 2)^2(x - 1)(x - 4) \leq 0$ ?

Can an odd-degree polynomial inequality be true for every $x$ ?

Do you need exact roots to solve a polynomial inequality?

What does a factor like $x^2 + 1$ do to a sign chart?

How do you handle a factor like $(5 - x)$ in a sign chart?

Polynomial Inequalities

Sign Charts, Root Multiplicity, and the Shape of the Solution

Key Terms

Core Concepts

From Polynomials Category

Definition

The Sign Chart Method

Factored Form and Sign Analysis

Profile of a factored cubic

Root Multiplicity and Sign Changes

End Behavior

Polynomials That Do Not Factor Over the Rationals

Worked Examples

Summary: Filling the Sign Chart Without Test Points

Common Pitfalls

Polynomial Inequalities FAQ

Why is x=−2x = -2x=−2 part of the solution of (x+2)2(x−1)(x−4)≤0(x + 2)^2(x - 1)(x - 4) \leq 0(x+2)2(x−1)(x−4)≤0?

Can an odd-degree polynomial inequality be true for every xxx?

Do you need exact roots to solve a polynomial inequality?

What does a factor like x2+1x^2 + 1x2+1 do to a sign chart?

How do you handle a factor like (5−x)(5 - x)(5−x) in a sign chart?

Why is $x = -2$ part of the solution of $(x + 2)^2(x - 1)(x - 4) \leq 0$ ?

Can an odd-degree polynomial inequality be true for every $x$ ?

What does a factor like $x^2 + 1$ do to a sign chart?

How do you handle a factor like $(5 - x)$ in a sign chart?