Polynomial Equations: Roots, Theorems & Methods

Definition and Degree

A linear equation has one root and a quadratic up to two, each found by a formula.

2x - 6 = 0

gives

x = 3

;

x^2 - 4 = 0

gives

x = \pm 2

. The pattern suggests that the higher the power of

x

, the more roots become possible.

Raising the highest power of x raises the ceiling on the number of roots — though an equation need not reach it.

equation	highest power of x	real roots	how many
2x − 6 = 0	1	3	1
x² − 4 = 0	2	−2, 2	2
x³ − 3x = 0	3	−√3, 0, √3	3
x³ + x = 0	3	0	1 — the other two are complex
x⁴ − 5x² + 4 = 0	4	−2, −1, 1, 2	4

x^3 - 3x = 0

reaches three real roots, while

x^3 + x = x(x^2 + 1) = 0

has only one: the factor

x^2 + 1

never vanishes for real

x

.

The definition

A polynomial equation in one variable is an equation

a_n x^n + \cdots + a_1 x + a_0 = 0

with

a_n \neq 0

; the highest power

n

is its degree.

A polynomial equation in one variable has the general form

Definition and Degree

a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0 = 0

where

a_n \neq 0

and

n

is a positive integer. The integer

n

is the degree of the equation, and

a_n

is the leading coefficient. The coefficients

a_0, a_1, \dots, a_n

are typically real numbers, though the theory extends naturally to complex coefficients.

So

x^3 - 3x = 0

is a polynomial equation of degree

3

, and

x^4 - 5x^2 + 4 = 0

one of degree

4

.

The degree caps the number of roots: at most n real roots, and exactly n when complex roots are counted with multiplicity.

The degree governs the maximum number of roots.

THEOREM

A polynomial of degree

n

has at most

n

real roots, and when counted with multiplicity over the complex numbers, it has exactly

n

roots.

Counted with multiplicity, a cubic may have one or three real roots, a quartic may have zero, two, or four, and the possible counts depend on the specific coefficients involved.

x³ − 3x = 0: degree 3, three real roots

The degree of the equation is the highest power present, and it caps the number of crossings with the level line: this cubic reaches its cap of three. A quadratic could show at most two, a linear equation exactly one. Raise the degree and count how many crossings become possible on the equation visual explorer.

The theorems below say how many roots exist over the complex numbers; the picture shows the real ones.

Nothing is lost: linear and quadratic equations are the first two members of the family.

Linear equations correspond to

n = 1

and quadratic equations to

n = 2

. Both have complete, closed-form solutions and are treated on their own pages. This page addresses degree three and above, where new phenomena emerge: more roots, more intricate factorizations, and eventually the impossibility of a universal solving formula.

What is lost: from degree five on, no general formula exists.

Cubic and quartic equations still have formulas, but they are long and rarely used by hand. For degree five and higher no formula in radicals can exist at all, as degree five and beyond explains.

The rest of the page covers the fundamental theorem of algebra, the factor and remainder theorems, the rational root theorem, solving cubics and quartics, degree five and beyond, root multiplicity, and the relationships between roots and coefficients.

The Fundamental Theorem of Algebra

The Fundamental Theorem of Algebra states that every polynomial of degree

n \geq 1

with complex coefficients has at least one root in

\mathbb{C}

. Repeated application of this fact yields a stronger conclusion: every such polynomial has exactly

n

roots in

\mathbb{C}

when counted with multiplicity.

The consequence for factoring is immediate. A degree-

n

polynomial

P(x)

with leading coefficient

a_n

decomposes completely as

The Fundamental Theorem of Algebra

P(x) = a_n(x - r_1)(x - r_2) \cdots (x - r_n)

where

r_1, r_2, \dots, r_n

are the

n

roots (not necessarily distinct). Over the complex numbers, every polynomial splits into linear factors. No irreducible quadratic or higher-degree factors remain.

Over the real numbers, the picture is slightly different.

THEOREM

Complex roots of a polynomial with real coefficients always appear in conjugate pairs: if

r = a + bi

is a root, then

\overline{r} = a - bi

is also a root.

Each conjugate pair combines into a quadratic factor

x^2 - 2ax + (a^2 + b^2)

with real coefficients and negative discriminant. A real polynomial therefore factors into a product of linear factors (from real roots) and irreducible quadratic factors (from conjugate pairs).

Degree 4, four roots, two conjugate pairs

z⁴ + 1 = 0 has real coefficients but no real root, since z⁴ is never negative for real z. In ℂ it has exactly four roots, one per degree, as the theorem promises. They sit in mirror pairs across the real axis, (1 + i)/√2 with (1 − i)/√2 and (−1 + i)/√2 with (−1 − i)/√2: each pair multiplies out to a real quadratic factor, here x² − √2x + 1 and x² + √2x + 1.

Real roots appear on the real axis; every non-real root brings its conjugate with it.

The Factor Theorem and Remainder Theorem

THEOREMRemainder Theorem

The Remainder Theorem connects evaluation and division: when a polynomial

P(x)

is divided by the binomial

(x - r)

, the remainder equals

P(r)

. In symbols,

P(x) = (x - r)Q(x) + P(r)

, where

Q(x)

is the quotient polynomial of degree one less than

P(x)

.

THEOREMFactor Theorem

The Factor Theorem is the special case where

P(r) = 0

. If substituting

r

into

P(x)

yields zero, then the remainder is zero, which means

(x - r)

divides

P(x)

evenly. Conversely, if

(x - r)

is a factor, then

P(r) = 0

.

Finding roots and finding factors are two expressions of the same problem.

This connection creates a degree-reduction strategy. Once a root

r

is identified — by inspection, by the rational root theorem, or by other means — dividing

P(x)

by

(x - r)

produces a quotient

Q(x)

of degree

n - 1

. The roots of

Q(x)

are the remaining roots of

P(x)

. Repeating the process reduces the degree one step at a time until the quotient is quadratic, at which point the quadratic formula finishes the job.

Pitfall: Using the wrong sign of r

Dividing

x^3 + 2x^2 - 5x - 6

by

(x + 1)

with

r = 1

gives the remainder

P(1) = -8

, so the factor seems to fail. Since

x + 1 = x - (-1)

,

r = -1

: the remainder is

P(-1) = 0

and the quotient is

x^2 + x - 6

.

Polynomial division can be carried out using long division or synthetic division. The polynomial calculator does both and shows every step. Synthetic division is faster but limited to divisors of the form

(x - r)

, which is exactly the form needed when factoring out a known root.

In synthetic division, the coefficients of

P(x)

are written in a row and the number

r

from the divisor

(x - r)

is placed beside them. Each step multiplies by

r

and adds the next coefficient. The last number produced is the remainder

P(r)

, so the same short table both tests a candidate and, when the remainder is zero, delivers the quotient.

Each root found peels off one factor, leaving a smaller polynomial that holds all the remaining roots.

The Rational Root Theorem

The Rational Root Theorem narrows the search for rational roots of a polynomial with integer coefficients. It states: if

\frac{p}{q}

is a rational root in lowest terms, then

p

divides the constant term

a_0

and

q

divides the leading coefficient

a_n

.

This produces a finite, testable list of candidates. For the polynomial

2x^3 - 3x^2 - 8x + 12 = 0

, the constant term is

12

and the leading coefficient is

2

. The divisors of

12

are

\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12

, and the divisors of

2

are

\pm 1, \pm 2

. The candidate rational roots are all fractions

\frac{p}{q}

formed from these:

The Rational Root Theorem

\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12, \pm \frac{1}{2}, \pm \frac{3}{2}

Pitfall: Swapping the roles of p and q

Building the list as divisors of

2

over divisors of

12

gives

\pm 1, \pm 2, \pm \frac{1}{2}, \pm \frac{1}{3}, \dots, \pm \frac{1}{12}

, and the root

\frac{3}{2}

is missing. The numerator divides the constant term:

3

divides

12

and

2

divides

2

, so

\frac{3}{2}

is a candidate.

Some of these fractions repeat values already on the list, such as

\frac{2}{2} = 1

and

\frac{6}{2} = 3

, so each value is written only once, and sixteen candidates remain. Each one is tested by substitution, and the search can stop at the first candidate that gives zero, since one root is enough to begin reducing the degree.

Testing

x = 2

:

P(2) = 2(8) - 3(4) - 8(2) + 12 = 16 - 12 - 16 + 12 = 0

. So

x = 2

is a root,

(x - 2)

is a factor, and dividing reduces the cubic to a quadratic

2x^2 + x - 6 = 0

, which the quadratic formula or further factoring can handle.

Pitfall: Dropping the root that was divided out

Solving only the quotient

2x^2 + x - 6 = 0

gives

x = \frac{3}{2}

or

x = -2

, two roots for a cubic. The root found by testing stays in the answer:

x = 2

,

x = \frac{3}{2}

and

x = -2

.

Each division lowers the degree by one, so the work on this cubic ends with a quadratic. That quadratic is a new, smaller equation: its roots are the roots of the cubic that remain once the factor

(x - 2)

has been taken out. Multiplying back,

(x - 2)(2x^2 + x - 6)

returns the original polynomial.

The theorem guarantees nothing about irrational or complex roots. A polynomial like

x^3 - 2 = 0

has no rational roots at all — its only real root is

\sqrt[3]{2}

. The rational root theorem is a sieve, not a guarantee, and when every candidate fails, other methods are needed.

Candidates versus roots

The section's cubic 2x³ − 3x² − 8x + 12. The theorem lists sixteen possible rational roots; the graph crosses the axis at only three of them, −2, 3/2 and 2. Testing the list by substitution finds x = 2 first; dividing out (x − 2) leaves 2x² + x − 6 = (2x − 3)(x + 2), which gives the other two.

The theorem shortens the search but never guarantees that any candidate is a root.

Solving Cubic Equations

A cubic equation

ax^3 + bx^2 + cx + d = 0

can be transformed into a reduced (depressed) cubic by eliminating the

x^2

term. The substitution

x = t - \frac{b}{3a}

produces an equation of the form

Solving Cubic Equations

t^3 + pt + q = 0

where

p

and

q

depend on the original coefficients. This depressed form is the starting point for Cardano's formula, which expresses the roots through cube roots and square roots of expressions involving

p

and

q

.

Fact: Real roots through imaginary numbers

Cardano's formula writes a root of

x^3 - 15x - 4 = 0

as

\sqrt[3]{2 + 11i} + \sqrt[3]{2 - 11i}

, which is just

4

, since

(2 \pm i)^3 = 2 \pm 11i

. Bombelli worked this out in 1572: a quadratic needing

\sqrt{-1}

could be dismissed as unsolvable, but this cubic plainly has the root

4

.

Cardano's formula yields all three roots, but its practical use is limited. The expressions under the radicals often involve nested roots that are difficult to simplify. Worse, when all three roots are real — the so-called casus irreducibilis — the formula necessarily passes through complex intermediate values to reach real answers. The formula is correct but computationally awkward.

The practical approach to most cubics encountered in coursework begins with the rational root theorem. If a rational root exists, polynomial division reduces the cubic to a quadratic, and the remaining roots follow from the quadratic formula. When no rational root exists, numerical methods such as Newton's method or graphical estimation provide the real roots to any required precision.

COROLLARY

Every cubic with real coefficients has at least one real root.

This is a consequence of the intermediate value theorem: a real cubic changes sign as

x

ranges from

-\infty

to

+\infty

, so it must cross zero somewhere.

To solve a specific polynomial equation with each step shown, use the polynomial equation solver.

Solving Quartic Equations

A quartic equation

ax^4 + bx^3 + cx^2 + dx + e = 0

is the highest degree for which a general closed-form solution in radicals exists. Ferrari's method, developed in the sixteenth century, reduces the quartic to a resolvent cubic. Solving the cubic provides a key quantity that splits the quartic into two quadratics, each solvable by the quadratic formula.

The procedure works in full generality, but the algebra is heavy. Even for modest numerical examples, the resolvent cubic and the subsequent quadratic splitting produce expressions that are lengthy to simplify by hand. For this reason, Ferrari's method is more important as a theoretical milestone than as a practical tool.

In practice, quartic equations that appear in coursework are usually structured to admit simpler approaches. Bi-quadratic equations

ax^4 + bx^2 + c = 0

reduce directly to quadratics via the substitution

u = x^2

. Quartics with a rational root are handled by the rational root theorem and degree reduction. Quartics that factor into two quadratics — recognizable by grouping or by inspection — split without invoking the full Ferrari machinery.

Pitfall: Taking only the positive square root

Back-substituting

u = 1

and

u = 4

in

x^4 - 5x^2 + 4 = 0

as

x = 1

and

x = 2

gives only two roots. Each positive

u

gives two values,

x = \pm\sqrt{u}

, so the roots are

x = \pm 1

and

x = \pm 2

.

With

u = x^2

, the quadratic in

u

is solved first, and its solutions are values of

x^2

, not of

x

. Each one is then turned back into values of

x

by solving

x^2 = u

. The sign of

u

decides what that step yields: a negative

u

yields no real

x

at all, since no real square is negative.

Counted with multiplicity, a quartic with real coefficients can have zero, two, or four real roots. Unlike cubics, a quartic is not guaranteed to have any real roots at all.

Four, two or no real roots

Three bi-quadratics, each solved by u = x². x⁴ − 5x² + 4 gives u = 1 or 4, so x = ±1, ±2: four real roots. x⁴ − 2x² − 3 gives u = 3 or −1, and only u = 3 yields real x = ±√3: two real roots. x⁴ + 1 never reaches 0: no real roots at all, unlike a cubic, which always crosses the axis at least once.

Real roots of a quartic come in even numbers when counted with multiplicity: four, two or none.

Degree Five and Beyond

The Abel–Ruffini theorem, proved in the early nineteenth century, establishes that no formula composed of arithmetic operations and radicals can solve every polynomial equation of degree five or higher. This does not mean that specific quintic equations are unsolvable — many factor, many have rational roots, and many yield to substitution. It means that no single formula analogous to the quadratic formula exists that accepts five arbitrary coefficients and returns the roots.

Fact: A single quintic beyond radicals

Some single equations are out of reach too: no root of

x^5 - x - 1 = 0

can be written in radicals, since its Galois group is

S_5

. Its one real root, about

1.1673

, is still computable to any precision.

The impossibility is not a matter of insufficient ingenuity. Galois theory provides the precise criterion: a polynomial equation is solvable by radicals if and only if its Galois group is a solvable group. The symmetric group

S_n

is solvable for

n \leq 4

and not solvable for

n \geq 5

, which is why degree four is the last degree admitting a universal radical formula.

For equations of any degree, the practical toolkit consists of the rational root theorem (to find and factor out rational roots), synthetic division (to reduce the degree step by step), and the quadratic formula (to handle the final quadratic quotient). When these algebraic tools run out — when no rational roots exist and no convenient factorization appears — numerical methods step in. Newton's method, the bisection method, and related algorithms approximate real roots to arbitrary precision, and these are the techniques actually used in computation for large-degree polynomials.

Tool	Purpose	When applicable	Limit
Rational Root Theorem	enumerate candidate rational roots	integer (or rational) coefficients	finds only rational roots
Synthetic / polynomial division	factor out (x − r) and reduce degree by 1	a confirmed root r is known	requires a root in advance
Quadratic formula	solve the quadratic quotient exactly	degree has been reduced to 2	degree-2 only
Numerical methods (Newton, bisection)	approximate real roots to any precision	algebraic tools have run out	approximate, not exact

From degree five on, no formula in radicals covers every equation, yet every root exists and can be computed to any precision.

Root Multiplicity

A root

r

of the polynomial

P(x)

has multiplicity

k

if

(x - r)^k

divides

P(x)

but

(x - r)^{k+1}

does not. A root of multiplicity

1

is called a simple root, multiplicity

2

a double root, multiplicity

3

a triple root, and so on.

RULE

The sum of all multiplicities across all roots equals the degree of the polynomial.

The equation

x^3 - 3x^2 + 3x - 1 = 0

factors as

(x - 1)^3 = 0

, so

x = 1

is a triple root and the three roots (counted with multiplicity) are all equal to

1

.

Multiplicity has both algebraic and geometric significance. Algebraically, it affects how the root behaves under perturbation: a simple root of

P(x) = 0

moves smoothly when the coefficients change slightly, while a multiple root tends to split into several nearby roots. Geometrically, multiplicity determines how the graph of

y = P(x)

interacts with the

x

-axis at the root. At a simple root, the graph crosses the axis cleanly. At a double root, the graph touches the axis and turns back without crossing — the root is a local extremum. At a triple root, the graph crosses the axis but flattens as it does so, producing an inflection point on the axis.

The general pattern is: odd multiplicity means the graph crosses the axis, even multiplicity means it touches and reverses direction. Higher multiplicity in either case means greater flattening at the crossing or touching point.

Multiplicity	Name	Graph behavior at the root
1	simple root	crosses the x-axis cleanly
2	double root	touches the axis and turns back (local extremum)
3	triple root	crosses with an inflection on the axis
even k ≥ 2	(general)	touches and turns back; more flattening as k grows
odd k ≥ 3	(general)	crosses with flattening; more flattening as k grows

x³ = 0: one root of multiplicity three

The curve flattens against the axis at the origin and passes through without the usual crossing angle: the root x = 0 appears three times in the factorisation x · x · x, and the flattening is what a triple root looks like. A double root would touch and turn back instead. Compare the touch of x² = 0 with this crossing on the equation visual explorer.

Counted with multiplicity, the roots always add up to the degree.

Relationships Between Roots and Coefficients

Vieta's formulas, introduced for the quadratic case on the quadratic equations page, extend to polynomials of any degree.

Vieta's Formulas (General)

r_1 + r_2 + \cdots + r_n = -\frac{a_{n-1}}{a_n} \qquad r_1 r_2 \cdots r_n = (-1)^n \frac{a_0}{a_n}

Learn more about this formula: Vieta's Formulas (General) →

Browse all algebra formulas →

For the monic polynomial

x^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0 = 0

with roots

r_1, r_2, \dots, r_n

, the coefficients encode symmetric combinations of the roots:

Relationships Between Roots and Coefficients

r_1 + r_2 + \cdots + r_n = -a_{n-1}

Relationships Between Roots and Coefficients

r_1 r_2 + r_1 r_3 + \cdots + r_{n-1}r_n = a_{n-2}

Relationships Between Roots and Coefficients

r_1 r_2 r_3 + \cdots = -a_{n-3}

continuing with alternating signs down to

Relationships Between Roots and Coefficients

r_1 \cdot r_2 \cdots r_n = (-1)^n a_0

Pitfall: Dropping the sign in the product of roots

Reading the product of the roots of

2x^3 - 3x^2 - 8x + 12 = 0

as

\frac{12}{2} = 6

drops the alternating sign. For degree

3

the product is

(-1)^3 \cdot \frac{12}{2} = -6

, and indeed

(-2) \cdot \frac{3}{2} \cdot 2 = -6

.

Each formula equates a coefficient to an elementary symmetric polynomial in the roots. The first gives the sum of all roots, the second the sum of all pairwise products, the third the sum of all triple products, and the last the product of all roots.

These relationships hold whether the roots are real or complex, rational or irrational. They allow certain questions about the roots to be answered without solving: the sum and product of the roots are readable directly from the coefficients. For a non-monic polynomial

a_n x^n + \cdots + a_0 = 0

, dividing through by

a_n

first converts it to monic form, or equivalently, each formula acquires

a_n

in its denominator.

Symmetric quantity	In terms of roots	In terms of coefficients
Sum of roots	r₁ + r₂ + ⋯ + rₙ	−aₙ₋₁ ⁄ aₙ
Sum of pairwise products	Σᵢ<ⱼ rᵢ rⱼ	aₙ₋₂ ⁄ aₙ
Sum of triple products	Σᵢ<ⱼ<ₖ rᵢ rⱼ rₖ	−aₙ₋₃ ⁄ aₙ
⋮ (alternating signs)	⋮	⋮
Product of all roots	r₁ · r₂ ⋯ rₙ	(−1)ⁿ a₀ ⁄ aₙ

A polynomial's coefficients already hold the sum and the product of its roots, so both are known before any root is found.

Summary by Degree

The table below collects the picture by degree: what the polynomial is called, whether a closed-form formula in radicals exists, the maximum number of real roots, and the technique most commonly applied in practice. The Fundamental Theorem of Algebra guarantees that the roots exist over

\mathbb{C}

for every row; what changes across degrees is whether they can be reached by a finite formula in radicals.

Degree	Name	Closed-form formula?	Max real roots	Primary technique
1	Linear	✓	1	x = −b ⁄ a
2	Quadratic	✓	2	quadratic formula / discriminant
3	Cubic	✓ (Cardano)	3	RRT + division → quadratic in practice
4	Quartic	✓ (Ferrari)	4	bi-quadratic substitution, RRT, or factoring into two quadratics
≥ 5	Quintic and higher	✗ (Abel–Ruffini)	n	RRT + repeated division; numerical methods when no rational roots

Formula Summary

This summary collects the formulas of the page in one table. Each row gives the formula's name with a line on where it comes from and a link to its section, the statement itself, and what it says in words.

The first rows describe the equation itself: its general form, and the complete factorization that the fundamental theorem of algebra guarantees over the complex numbers. The depressed cubic follows, the reduced form that Cardano's method starts from. The remaining rows are Vieta's formulas, which tie the roots to the coefficients, first for any leading coefficient and then for a monic polynomial.

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Formula	Statement	What it says
General formEvery polynomial equation can be written this way, as definition and degree shows.	$a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0 = 0$	the degree is the highest power with a nonzero coefficient
Complete factorizationThe consequence of the fundamental theorem of algebra.	$P(x) = a_n(x - r_1)(x - r_2) \cdots (x - r_n)$	a degree- $n$ polynomial has exactly $n$ complex roots, counted with multiplicity
Depressed cubicThe substitution $x = t - \frac{b}{3a}$ in solving cubic equations leads here, the starting point for Cardano.	$t^3 + pt + q = 0$	the cubic with its squared term removed
Sum of the rootsVieta's formulas for any degree, from roots and coefficients.	$\displaystyle r_1 + r_2 + \cdots + r_n = -\frac{a_{n-1}}{a_n}$	read off the two leading coefficients
Product of the rootsThe other end of the same list of formulas in roots and coefficients.	$\displaystyle r_1 r_2 \cdots r_n = (-1)^n \frac{a_0}{a_n}$	read off the constant and leading coefficients
Monic: sum of rootsWith $a_n = 1$ , each coefficient is a symmetric combination of the roots (roots and coefficients).	$r_1 + r_2 + \cdots + r_n = -a_{n-1}$	for leading coefficient $1$
Monic: pairs and triplesThe pattern continues down the coefficients.	$r_1 r_2 + r_1 r_3 + \cdots + r_{n-1}r_n = a_{n-2}$ , $r_1 r_2 r_3 + \cdots = -a_{n-3}$	products of two roots, then of three, with alternating signs
Monic: product of rootsThe last formula of the list.	$r_1 \cdot r_2 \cdots r_n = (-1)^n a_0$	the constant term, with sign $(-1)^n$

Polynomial equation formulas·All algebra formulasLearn Math Class

Together the rows let a polynomial be read in both directions. The coefficients give the sum and the product of the roots without solving anything, and known roots rebuild the polynomial through its factorization.

Common Pitfalls

Five mistakes come up again and again when polynomial equations 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.

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1Using the wrong sign of r

The divisor is written as

(x + a)

, and the number

a

is used as

r

in synthetic division or in the Remainder Theorem, although both are stated for a divisor of the form

(x - r)

.

Wrong turn

Error here

To divide

P(x) = x^3 + 2x^2 - 5x - 6

by

(x + 1)

, synthetic division is run with

r = 1

. The coefficients

1, 2, -5, -6

become

1, 3, -2, -8

, so the remainder is

-8

and

(x + 1)

is rejected as a factor.

Right path

x + 1 = x - (-1)

, so

r = -1

. The coefficients become

1, 1, -6, 0

: the remainder is

0

and the quotient is

x^2 + x - 6 = (x + 3)(x - 2)

.

x^3 + 2x^2 - 5x - 6 = (x + 1)(x + 3)(x - 2)

The roots are

x = -1

,

x = -3

and

x = 2

.

Why it happensThe plus sign in the divisor is copied as it is written, but the theorems subtract

r

, and

x + 1

is

x

minus the number

-1

.

How to checkSet the divisor equal to zero.

x + 1 = 0

gives

x = -1

, the number that belongs in the division. The wrong remainder is a clue too:

-8

is

P(1) = 1 + 2 - 5 - 6

, which tests the factor

(x - 1)

, not

(x + 1)

.

2Swapping the roles of p and q

The Rational Root Theorem pairs two lists of divisors, one from the constant term and one from the leading coefficient. The lists are put the wrong way round, so the candidates are built as divisors of

a_n

over divisors of

a_0

.

Wrong turn

Error here

For

2x^3 - 3x^2 - 8x + 12 = 0

, divisors of

2

over divisors of

12

give

\pm 1, \pm 2, \pm \frac{1}{2}, \pm \frac{1}{3},

\pm \frac{2}{3}, \pm \frac{1}{4}, \pm \frac{1}{6}, \pm \frac{1}{12}

Testing this list finds

x = 2

and

x = -2

, but the third root is not on it.

Right path

The numerator

p

divides the constant term

12

, and the denominator

q

divides the leading coefficient

2

.

\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12, \pm \frac{1}{2}, \pm \frac{3}{2}

This list contains all three roots:

x = -2

,

x = \frac{3}{2}

and

x = 2

.

Why it happensBoth coefficients are split into divisors in exactly the same way, and nothing in that arithmetic shows which list belongs on top.

How to checkTry the recipe on a polynomial whose roots are known.

(x - 2)(x - 3) = x^2 - 5x + 6

has the roots

2

and

3

, which divide the constant term

6

. The swapped recipe would allow only

\pm 1, \pm \frac{1}{2}, \pm \frac{1}{3}, \pm \frac{1}{6}

and miss both.

3Dropping the root that was divided out

Once a root has been found and divided out, the work moves on to the quotient. The roots of the quotient are given as the answer, and the root that started the reduction is left behind.

Start

2x^3 - 3x^2 - 8x + 12 = (x - 2)(2x^2 + x - 6)

2x^2 + x - 6 = (2x - 3)(x + 2) = 0

\implies x = \frac{3}{2} \text{ or } x = -2

Wrong turn

Error here

The answer is given as

x = \frac{3}{2}

or

x = -2

.

Right path

The cubic is the product of all three factors.

(x - 2)(2x - 3)(x + 2) = 0

\implies x = 2, \; x = \frac{3}{2} \text{ or } x = -2

Why it happensThe quadratic quotient looks like a complete new problem, and solving it feels like the last step. The first root was found by testing rather than by solving, so it never appears in the final line of working.

How to checkCount the roots against the degree. Each factor

(x - r)

removed by division accounts for one root, and the quotient holds the rest; here one division and two roots of the quotient make the three that a cubic has. Substituting confirms it:

P(2) = 16 - 12 - 16 + 12 = 0

.

4Taking only the positive square root

In a bi-quadratic equation, the substitution

u = x^2

gives a quadratic in

u

. Its solutions are turned back into values of

x

with the positive square root only, and half of the roots are lost.

Start

x^4 - 5x^2 + 4 = 0

\implies u^2 - 5u + 4 = (u - 1)(u - 4) = 0

\implies u = 1 \text{ or } u = 4

Wrong turn

Error here

x = \sqrt{1} = 1 \text{ or } x = \sqrt{4} = 2

Right path

Each value of

u

is a value of

x^2

, and

x^2 = u

has two solutions when

u > 0

.

x^2 = 1 \implies x = \pm 1

x^2 = 4 \implies x = \pm 2

The quartic has four real roots:

-2

,

-1

,

1

and

2

.

Why it happens

\sqrt{4} = 2

is correct as the value of the square root, and the step from

x^2 = 4

to

x = \sqrt{4}

looks like the same operation. But solving

x^2 = 4

asks for every number whose square is

4

, and there are two.

How to checkA bi-quadratic has only even powers of

x

, so

x

and

-x

always give the same value; the opposite of every root is a root too. Substituting

x = -1

gives

1 - 5 + 4 = 0

.

5Dropping the sign in the product of roots

Vieta's formulas carry alternating signs: the product of the roots is

(-1)^n \frac{a_0}{a_n}

and the sum is

-\frac{a_{n-1}}{a_n}

. The sign factor is dropped and the product is read straight from the coefficients as

\frac{a_0}{a_n}

.

Wrong turn

Error here

For

2x^3 - 3x^2 - 8x + 12 = 0

, the product of the roots is taken as

\frac{12}{2} = 6

.

Right path

The degree is

n = 3

, so

r_1 r_2 r_3 = (-1)^3 \cdot \frac{12}{2} = -6

The roots

-2

,

\frac{3}{2}

and

2

confirm it:

(-2) \cdot \frac{3}{2} \cdot 2 = -6

. Their sum is

-\frac{-3}{2} = \frac{3}{2}

, and indeed

-2 + \frac{3}{2} + 2 = \frac{3}{2}

.

Why it happensFor a quadratic, the product of the roots is

\frac{c}{a}

with no visible sign, because

(-1)^2 = 1

. The habit carries over to odd degrees, where the sign factor is

-1

.

How to checkBuild a cubic from simple roots and compare.

(x - 1)(x - 2)(x - 3) = x^3 - 6x^2 + 11x - 6

has roots with product

6

, while its constant term is

-6

.

Common pitfalls in polynomial equations·/algebra/equations/polynomialLearn Math Class

Four of the five slips end with fewer roots than the equation has: a factor is rejected because of the wrong sign of

r

, a list of candidates misses a root, a root that was divided out is forgotten, or the negative square roots are dropped. A count catches them: an equation of degree

n

has at most

n

roots, and when the working finds fewer, it is worth asking where the others went. The fifth slip, the sign in the product of roots, is caught by building a small polynomial from known roots and comparing.

Polynomial Equations FAQ

Does Abel–Ruffini mean fifth-degree equations cannot be solved?

+

No, and this is the usual misreading. Plenty of quintics are solvable: many factor, many have rational roots, many yield to substitution. What the theorem rules out is a single formula in arithmetic and radicals that takes five arbitrary coefficients and returns the roots. Many individual equations, such as

x^5 - 32 = 0

, are solvable in radicals; but no universal formula exists, and some specific equations, such as

x^5 - x - 1 = 0

, have no solution in radicals at all.Read more →

How does root multiplicity show up on the graph?

+

Parity decides. At odd multiplicity the graph crosses the axis; at even multiplicity it touches and turns back without crossing, making the root a local extremum. Higher multiplicity flattens the curve near the root either way, so a triple root crosses but flattens into an inflection point sitting on the axis.Read more →

If there is no formula for high degrees, how are these equations actually solved?

+

By reduction, then approximation. The rational root theorem finds any rational roots, synthetic division factors them out and drops the degree, and the quadratic formula finishes whatever quadratic remains. When no rational root exists and nothing factors, numerical methods take over: Newton's method and bisection approximate the real roots to any precision required.Read more →

Can a quartic equation have no real roots?

+

Yes.

x^4 + 1 = 0

has none, since

x^4

is never negative for real

x

; its four roots are two complex conjugate pairs. Counted with multiplicity, a real quartic has zero, two or four real roots: with

u = x^2

,

x^4 - 5x^2 + 4 = 0

gives four,

\pm 1

and

\pm 2

, and

x^4 - 2x^2 - 3 = 0

gives two,

\pm\sqrt{3}

. A cubic is different: it always crosses the axis at least once.Read more →

How can Cardano's formula reach a real root through imaginary numbers?

+

Because the imaginary parts cancel. For

x^3 - 15x - 4 = 0

the formula gives

\sqrt[3]{2 + 11i} + \sqrt[3]{2 - 11i}

, and since

(2 \pm i)^3 = 2 \pm 11i

, the two cube roots are

2 + i

and

2 - i

, whose sum is the real root

4

. Bombelli worked this out in 1572. When all three roots of a cubic are real, the casus irreducibilis, this detour through complex values cannot be avoided.Read more →

What happens to a double root when the coefficients change slightly?

+

It splits into two nearby roots, which may be real or complex, whereas a simple root only shifts a little. The equation

(x - 1)^2 = 0

has the double root

1

; changing the right side to

10^{-6}

gives the two roots

0.999

and

1.001

, and changing it to

-10^{-6}

gives

1 \pm 0.001i

. A change of one millionth in a coefficient moves the roots by a thousandth.Read more →

Polynomial Equations

Higher-Degree Equations and the Limits of Exact Solutions

Key Terms

Equation Structure

Solutions

Formulas Used on This Page

Definition and Degree

The Fundamental Theorem of Algebra

The Factor Theorem and Remainder Theorem

The Rational Root Theorem

Solving Cubic Equations

Solving Quartic Equations

Degree Five and Beyond

Root Multiplicity

Relationships Between Roots and Coefficients

Summary by Degree

Formula Summary

Common Pitfalls

Polynomial Equations FAQ

Does Abel–Ruffini mean fifth-degree equations cannot be solved?

How does root multiplicity show up on the graph?

If there is no formula for high degrees, how are these equations actually solved?

Can a quartic equation have no real roots?

How can Cardano's formula reach a real root through imaginary numbers?

What happens to a double root when the coefficients change slightly?