aarjaneiro/planetmath huggingface dataset:
exclusion of integer root

exclusion of integer root

Theorem

.

The equation

p

⁢

(

x

)

:=

a

n

⁢

x

n

+

a

n

-

1

⁢

x

n

-

1

+

…

+

a

0

=

0

with

integer

coefficients

a

i

has no integer roots (

http://planetmath.org/Equation

), if

p

⁢

(

0

)

and

p

⁢

(

1

)

are odd.

Proof.

Make the antithesis, that there is an integer

x

0

such that

p

⁢

(

x

0

)

=

0

.  This

x

0

cannot be even, because else all terms of

p

⁢

(

x

0

)

except

a

0

were even and thus the whole sum could not have the even value 0.  Consequently,

x

0

and also its powers (

http://planetmath.org/GeneralAssociativity

) have to be odd.  Since

2

∣

0

=

p

(

x

0

)

and

2

∤

p

(

0

)

=

a

0

,

there must be among the coefficients

a

n

,

a

n

-

1

,

…

,

a

1

an odd amount of

odd numbers

.  This means that

2

∣

a

n

+

a

n

-

1

+

…

+

a

1

+

a

0

=

p

(

1

)

.

This however contradicts the

assumption

on the parity of

p

⁢

(

1

)

, whence the antithesis is wrong and the theorem .

Title

exclusion of integer root

Canonical name

ExclusionOfIntegerRoot

Date of creation

2013-03-22 19:08:21

Last modified on

2013-03-22 19:08:21

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

5

Author

pahio (2872)

Entry type

Theorem

Classification

msc 12D10

Classification

msc 12D05

Related topic

DivisibilityInRings

Generated on Fri Feb 9 16:37:35 2018 by

LaTeXML
factor theorem

factor theorem

If

f

⁢

(

x

)

is a

polynomial

over a

ring with identity

, then

x

-

a

is a factor if and only if

a

is a root (that is,

f

⁢

(

a

)

=

0

).

This theorem is of great help for finding

factorizations

of higher degree polynomials. As example, let us think that we need to factor the polynomial

p

⁢

(

x

)

=

x

3

+

3

⁢

x

2

-

33

⁢

x

-

35

. With some help of the

rational root theorem

we can find that

x

=

-

1

is a root (that is,

p

⁢

(

-

1

)

=

0

), so we know

(

x

+

1

)

must be a factor of the polynomial. We can write then

p

⁢

(

x

)

=

(

x

+

1

)

⁢

q

⁢

(

x

)

where the polynomial

q

⁢

(

x

)

can be found using long or

synthetic division

of

p

⁢

(

x

)

between

x

-

1

. In our case

q

⁢

(

x

)

=

x

2

+

2

⁢

x

-

35

which can be easily factored as

(

x

-

5

)

⁢

(

x

+

7

)

. We conclude that

p

⁢

(

x

)

=

(

x

+

1

)

⁢

(

x

-

5

)

⁢

(

x

+

7

)

.

Title

factor theorem

Canonical name

FactorTheorem

Date of creation

2013-03-22 12:17:24

Last modified on

2013-03-22 12:17:24

Owner

drini (3)

Last modified by

drini (3)

Numerical id

10

Author

drini (3)

Entry type

Theorem

Classification

msc 12D10

Classification

msc 12D05

Synonym

root theorem

Related topic

Polynomial

Related topic

RationalRootTheorem

Related topic

Root

Related topic

APolynomialOfDegreeNOverAFieldHasAtMostNRoots

Generated on Fri Feb 9 16:37:37 2018 by

LaTeXML
Ferrari-Cardano derivation of the quartic formula

Ferrari-Cardano derivation of the quartic formula

Given a

quartic equation

x

4

+

a

⁢

x

3

+

b

⁢

x

2

+

c

⁢

x

+

d

=

0

, apply the Tchirnhaus

transformation

x

↦

y

-

a

4

to obtain

y

4

+

p

⁢

y

2

+

q

⁢

y

+

r

=

0

(1)

where

p

=

b

-

3

⁢

a

2

8

q

=

c

-

a

⁢

b

2

+

a

3

8

r

=

d

-

a

⁢

c

4

+

a

2

⁢

b

16

-

3

⁢

a

4

256

Clearly a solution to Equation (

1

) solves the original, so we replace the original equation with Equation (

1

). Move

q

⁢

y

+

r

to the other side and

complete

the square on the left to get:

(

y

2

+

p

)

2

=

p

⁢

y

2

-

q

⁢

y

+

(

p

2

-

r

)

.

We now wish to add the quantity

(

y

2

+

p

+

z

)

2

-

(

y

2

+

p

)

2

to both sides, for some unspecified value of

z

whose purpose will be made clear in what follows. Note that

(

y

2

+

p

+

z

)

2

-

(

y

2

+

p

)

2

is a quadratic in

y

. Carrying out this

addition

, we get

(

y

2

+

p

+

z

)

2

=

(

p

+

2

⁢

z

)

⁢

y

2

-

q

⁢

y

+

(

z

2

+

2

⁢

p

⁢

z

+

p

2

-

r

)

(2)

The goal is now to choose a value for

z

which makes the right hand side of Equation (

2

) a

perfect square

. The right hand side is a

quadratic polynomial

in

y

whose

discriminant

is

-

8

⁢

z

3

-

20

⁢

p

⁢

z

2

+

(

8

⁢

r

-

16

⁢

p

2

)

⁢

z

+

q

2

+

4

⁢

p

⁢

r

-

4

⁢

p

3

.

Our goal will be achieved if we can find a value for

z

which makes this discriminant zero. But the above

polynomial

is a

cubic polynomial

in

z

, so its roots can be found using the

cubic formula

. Choosing then such a value for

z

, we may rewrite Equation (

2

) as

(

y

2

+

p

+

z

)

2

=

(

s

⁢

y

+

t

)

2

for some (complicated!) values

s

and

t

, and then taking the

square root

of both sides and solving the resulting

quadratic equation

in

y

provides a root of Equation (

1

).

Title

Ferrari-Cardano derivation of the quartic formula

Canonical name

FerrariCardanoDerivationOfTheQuarticFormula

Date of creation

2013-03-22 12:37:21

Last modified on

2013-03-22 12:37:21

Owner

djao (24)

Last modified by

djao (24)

Numerical id

8

Author

djao (24)

Entry type

Proof

Classification

msc 12D10

Related topic

CardanosDerivationOfTheCubicFormula

Generated on Fri Feb 9 16:37:39 2018 by

LaTeXML
Galois-theoretic derivation of the cubic formula

Galois-theoretic derivation of the cubic formula

We are trying to find the roots

r

1

,

r

2

,

r

3

of the

polynomial

x

3

+

a

⁢

x

2

+

b

⁢

x

+

c

=

0

. From the equation

(

x

-

r

1

)

⁢

(

x

-

r

2

)

⁢

(

x

-

r

3

)

=

x

3

+

a

⁢

x

2

+

b

⁢

x

+

c

we see that

a

=

-

(

r

1

+

r

2

+

r

3

)

b

=

r

1

⁢

r

2

+

r

1

⁢

r

3

+

r

2

⁢

r

3

c

=

-

r

1

⁢

r

2

⁢

r

3

The goal is to explicitly construct a

radical tower

over the field

k

=

ℂ

⁢

(

a

,

b

,

c

)

that contains the three roots

r

1

,

r

2

,

r

3

.

Let

L

=

ℂ

⁢

(

r

1

,

r

2

,

r

3

)

. By

Galois theory

we know that

Gal

⁡

(

L

/

ℂ

⁢

(

a

,

b

,

c

)

)

=

S

3

. Let

K

⊂

L

be the

fixed field

of

A

3

⊂

S

3

. We have a tower of field

extensions

Generated on Fri Feb 9 16:37:46 2018 by

LaTeXML
Galois-theoretic derivation of the quartic formula

Galois-theoretic derivation of the quartic formula

Let

x

4

+

a

⁢

x

3

+

b

⁢

x

2

+

c

⁢

x

+

d

be a general

polynomial

with four roots

r

1

,

r

2

,

r

3

,

r

4

, so

(

x

-

r

1

)

⁢

(

x

-

r

2

)

⁢

(

x

-

r

3

)

⁢

(

x

-

r

4

)

=

x

4

+

a

⁢

x

3

+

b

⁢

x

2

+

c

⁢

x

+

d

. The goal is to exhibit the

field extension

ℂ

⁢

(

r

1

,

r

2

,

r

3

,

r

4

)

/

ℂ

⁢

(

a

,

b

,

c

,

d

)

as a

radical extension

, thereby
expressing

r

1

,

r

2

,

r

3

,

r

4

in terms of

a

,

b

,

c

,

d

by

radicals

.

Write

N

for

ℂ

⁢

(

r

1

,

r

2

,

r

3

,

r

4

)

and

F

for

ℂ

⁢

(

a

,

b

,

c

,

d

)

. The

Galois group

Gal

⁡

(

N

/

F

)

is the

symmetric group

S

4

, the

permutation group

on the four elements

{

r

1

,

r

2

,

r

3

,

r

4

}

, which has a

composition series

1

⊲

ℤ

/

2

⊲

V

4

⊲

A

4

⊲

S

4

,

where:

•

A

4

is the

alternating group

in

S

4

, consisting of the

even
permutations

.

•

V

4

=

{

1

,

(

12

)

⁢

(

34

)

,

(

13

)

⁢

(

24

)

,

(

14

)

⁢

(

23

)

}

is the

Klein four-group

.

•

ℤ

/

2

is the two–element

subgroup

{

1

,

(

12

)

⁢

(

34

)

}

of

V

4

.

Under the

Galois correspondence

, each of these subgroups corresponds
to an intermediate field of the

extension

N

/

F

. We denote these

fixed
fields

by (in increasing order)

K

,

L

, and

M

.

We thus have a tower of field extensions, and corresponding

automorphism groups

:

Generated on Fri Feb 9 16:37:51 2018 by

LaTeXML
irreducible polynomial

irreducible polynomial

Let

f

⁢

(

x

)

=

a

0

+

a

1

⁢

x

+

⋯

+

a

n

⁢

x

n

be a

polynomial

with

complex

coefficients

a

ν

and with the degree (

http://planetmath.org/Polynomial

)

n

>

0

.  If

f

⁢

(

x

)

can not be written as

product

of two polynomials with

positive

degrees and with coefficients in the field

ℚ

⁢

(

a

0

,

a

1

,

…

,

a

n

)

,  then the polynomial

f

⁢

(

x

)

is said to be .  Otherwise,

f

⁢

(

x

)

is

reducible

.

Examples.

All linear polynomials are .  The polynomials

x

2

-

3

,

x

2

+

1

and

x

2

-

i

are (although they split in linear factors in the fields

ℚ

⁢

(

3

)

,

ℚ

⁢

(

i

)

and

ℚ

⁢

(

1

+

i

2

)

, respectively).  The polynomials

x

4

+

4

and

x

6

+

1

are not .

The above definition of polynomial is special case of the more general setting where

f

⁢

(

x

)

is a non-constant polynomial in the

polynomial ring

K

⁢

[

x

]

of a field

K

; if

f

⁢

(

x

)

is not

expressible

as product of two polynomials with positive degrees in the ring

K

⁢

[

x

]

, then

f

⁢

(

x

)

is (in

K

⁢

[

x

]

).

Example.

If

K

is the

Galois field

with two elements (0 and 1), then the

trinomial

x

2

+

x

+

1

of

K

⁢

[

x

]

is (because an equation

x

2

+

x

+

1

=

(

x

+

a

)

⁢

(

x

+

b

)

would imply the two conflicting conditions

a

+

b

=

1

and

a

⁢

b

=

1

).

Title

irreducible polynomial

Canonical name

IrreduciblePolynomial

Date of creation

2013-03-22 14:24:22

Last modified on

2013-03-22 14:24:22

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

18

Author

pahio (2872)

Entry type

Definition

Classification

msc 12D10

Synonym

prime polynomial

Synonym

indivisible polynomial

Related topic

EisensteinCriterion

Related topic

Irreducible

Related topic

Monic2

Defines

irreducible polynomial

Defines

reducible

Generated on Fri Feb 9 16:37:52 2018 by

LaTeXML
multiplicity

multiplicity

If a

polynomial

f

⁢

(

x

)

in

ℂ

⁢

[

x

]

is divisible by

(

x

-

a

)

m

but not by

(

x

-

a

)

m

+

1

(

a

is some

complex number

,

m

∈

ℤ

+

), we say that

x

=

a

is a

zero of the polynomial with

multiplicity

m

or alternatively a

zero of order

m

.

Generalization

of the multiplicity to real (

http://planetmath.org/RealFunction

) and complex functions (by rspuzio):  If the

function

f

is

continuous

on some open set

D

and

f

⁢

(

a

)

=

0

for some

a

∈

D

, then the zero of

f

at

a

is said to be of multiplicity

m

if

f

⁢

(

z

)

(

z

-

a

)

m

is continuous in

D

but

f

⁢

(

z

)

(

z

-

a

)

m

+

1

is not.

If

m

≧

2

, we speak of a

multiple zero

; if

m

=

1

, we speak of a

simple zero

.  If

m

=

0

, then actually the number

a

is not a zero of

f

⁢

(

x

)

, i.e.

f

⁢

(

a

)

≠

0

.

Some

properties

(from which 2, 3 and 4 concern only polynomials):

1.

The zero

a

of a polynomial

f

⁢

(

x

)

with multiplicity

m

is a zero of the

f

′

⁢

(

x

)

with multiplicity

m

-

1

.

2.

The zeros of the polynomial

gcd

⁡

(

f

⁢

(

x

)

,

f

′

⁢

(

x

)

)

are same as the multiple zeros of

f

⁢

(

x

)

.

3.

The

quotient

f

⁢

(

x

)

gcd

⁡

(

f

⁢

(

x

)

,

f

′

⁢

(

x

)

)

has the same zeros as

f

⁢

(

x

)

but they all are .

4.

The zeros of any

irreducible polynomial

are .

Title

multiplicity

Canonical

name

Multiplicity

Date of creation

2013-03-22 14:24:18

Last modified on

2013-03-22 14:24:18

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

14

Author

pahio (2872)

Entry type

Definition

Classification

msc 12D10

Synonym

order of the zero

Related topic

OrderOfVanishing

Related topic

DerivativeOfPolynomial

Defines

zero of order

Defines

multiple zero

Defines

simple zero

Defines

simple

Generated on Fri Feb 9 16:37:54 2018 by

LaTeXML
proof of factor theorem due to Fermat

proof of factor theorem due to Fermat

Lemma (cf.

factor theorem

).

If the

polynomial

f

⁢

(

x

)

:=

a

0

⁢

x

n

+

a

1

⁢

x

n

-

1

+

⋯

+

a

n

-

1

⁢

x

+

a

n

vanishes at

x

=

c

,  then it is divisible by the

difference

x

-

c

, i.e. there is valid the

identic equation

f

⁢

(

x

)

≡

(

x

-

c

)

⁢

q

⁢

(

x

)

(1)

where

q

⁢

(

x

)

is a polynomial of degree

n

-

1

, beginning with the

a

0

⁢

x

n

-

1

.

The lemma is here proved by using only the

properties

of the

multiplication

and

addition

, not the

division

.

Proof.

If we denote

x

-

c

=

y

,  we may write the given polynomial in the form

f

⁢

(

x

)

=

a

0

⁢

(

y

+

c

)

n

+

a

1

⁢

(

y

+

c

)

n

-

1

+

⋯

+

a

n

-

1

⁢

(

y

+

c

)

+

a

n

.

It’s clear that every

(

y

+

c

)

k

is a polynomial of degree

k

with respect to

y

, where

y

k

has the

coefficient

1 and the is

c

k

.  This implies that

f

⁢

(

x

)

may be written as a polynomial of degree

n

with respect to

y

, where

y

n

has the coefficient

a

0

and the on

y

is equal to

a

0

⁢

c

n

+

a

1

⁢

c

n

-

1

+

⋯

+

a

n

-

1

⁢

c

+

a

n

, i.e.

f

⁢

(

c

)

.  So we have

f

⁢

(

x

)

=

a

0

⁢

y

n

+

b

1

⁢

y

n

-

1

+

b

2

⁢

y

n

-

2

+

⋯

+

b

n

-

1

⁢

y

+

f

⁢

(

c

)

=

f

⁢

(

c

)

+

y

⋅

(

a

0

⁢

y

n

-

1

+

b

1

⁢

y

n

-

2

+

⋯

+

b

n

-

1

+

a

n

)

,

where

b

1

,

b

2

,

…

,

b

n

-

1

are certain coefficients.  If we return to the

indeterminate

x

by substituting in the last identic equation

x

-

c

for

y

, we get

f

⁢

(

x

)

≡

f

⁢

(

c

)

+

(

x

-

c

)

⁢

[

a

0

⁢

(

x

-

c

)

n

-

1

+

b

1

⁢

(

x

-

c

)

n

-

2

+

⋯

+

b

n

-

1

]

.

When the powers

(

x

-

c

)

k

are expanded to polynomials, we see that the

expression

in the brackets is a polynomial

q

⁢

(

x

)

of degree

n

-

1

with respect to

x

and with the coefficient

a

0

of

x

n

-

1

.  Thus we obtain

f

⁢

(

x

)

≡

f

⁢

(

c

)

+

(

x

-

c

)

⁢

q

⁢

(

x

)

.

(2)

This result is true independently on the value of

c

.  If this value is chosen such that

f

⁢

(

c

)

=

0

,  then (2) reduces to (1), Q. E. D.

References

1

Ernst Lindelöf

:

Johdatus korkeampaan analyysiin

(‘

Introduction

to Higher

Analysis

’).  Fourth edition. WSOY, Helsinki (1956).

Title

proof of factor theorem due to Fermat

Canonical

name

ProofOfFactorTheoremDueToFermat

Date of creation

2013-03-22 15:40:12

Last modified on

2013-03-22 15:40:12

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

10

Author

pahio (2872)

Entry type

Proof

Classification

msc 12D10

Classification

msc 12D05

Synonym

proof of factor theorem without division

Generated on Fri Feb 9 16:37:57 2018 by

LaTeXML
proof of factor theorem using division

proof of factor theorem using division

Lemma

(cf. factor theorem).

Let

R

be a

commutative ring

with

identity

and let

p

⁢

(

x

)

∈

R

⁢

[

x

]

be a

polynomial

with

coefficients

in

R

. The element

a

∈

R

is a root of

p

⁢

(

x

)

if and only if

(

x

-

a

)

divides

p

⁢

(

x

)

.

Proof.

Let

p

⁢

(

x

)

be a polynomial in

R

⁢

[

x

]

and let

a

be an element of

R

.

1.

First we assume that

(

x

-

a

)

divides

p

⁢

(

x

)

. Therefore, there is a polynomial

q

⁢

(

x

)

∈

R

⁢

[

x

]

such that

p

⁢

(

x

)

=

(

x

-

a

)

⋅

q

⁢

(

x

)

. Hence,

p

⁢

(

a

)

=

(

a

-

a

)

⋅

q

⁢

(

a

)

=

0

and

a

is a root of

p

⁢

(

x

)

.

2.

Assume that

a

is a root of

p

⁢

(

x

)

, i.e.

p

⁢

(

a

)

=

0

. Since

x

-

a

is a

monic polynomial

, we can perform the

polynomial long division

(

http://planetmath.org/LongDivision

) of

p

⁢

(

x

)

by

(

x

-

a

)

. Thus, there exist polynomials

q

⁢

(

x

)

and

r

⁢

(

x

)

such that:

p

⁢

(

x

)

=

(

x

-

a

)

⋅

q

⁢

(

x

)

+

r

⁢

(

x

)

and the degree of

r

⁢

(

x

)

is less than the degree of

x

-

a

(so

r

⁢

(

x

)

is just a constant). Moreover,

0

=

p

⁢

(

a

)

=

0

+

r

⁢

(

a

)

=

r

⁢

(

a

)

=

r

⁢

(

x

)

. Therefore

p

⁢

(

x

)

=

(

x

-

a

)

⋅

q

⁢

(

x

)

and

(

x

-

a

)

divides

p

⁢

(

x

)

.

∎

Title

proof of factor theorem using division

Canonical name

ProofOfFactorTheoremUsingDivision

Date of creation

2013-03-22 15:08:58

Last modified on

2013-03-22 15:08:58

Owner

alozano (2414)

Last modified by

alozano (2414)

Numerical id

8

Author

alozano (2414)

Entry type

Proof

Classification

msc 12D10

Classification

msc 12D05

Generated on Fri Feb 9 16:37:59 2018 by

LaTeXML
properties of quadratic equation

properties of quadratic equation

The

quadratic equation

a

⁢

x

2

+

b

⁢

x

+

c

=

0

or

x

2

+

p

⁢

x

+

q

=

0

with

rational

, real,

algebraic

(

http://planetmath.org/AlgebraicNumber

) or

complex

coefficients

(

a

≠

0

) has the following properties:

•

It has in

ℂ

two roots (which may be equal), since the

complex numbers

form an algebraically closed field containing the coefficients.

•

The sum of the roots is equal to

-

b

a

,  i.e.

-

p

.

•

The

product

of the roots is equal to

c

a

,  i.e.

q

.

Corollary.

If the leading coefficient and the constant are equal, then the roots are

inverse numbers

of each other.

Without solving the equation, the value of any

symmetric polynomial

of the roots can be calculated.

Example.

If one has to

x

1

3

+

x

2

3

, when

x

1

and

x

2

are the roots of the
equation

x

2

-

4

⁢

x

+

9

=

0

,  we have

x

1

+

x

2

=

4

and

x

1

⁢

x

2

=

9

.  Because

(

x

1

+

x

2

)

3

=

x

1

3

+

3

⁢

x

1

2

⁢

x

2

+

3

⁢

x

1

⁢

x

2

2

+

x

2

3

=

(

x

1

3

+

x

2

3

)

+

3

⁢

x

1

⁢

x

2

⁢

(

x

1

+

x

2

)

,

we obtain

x

1

3

+

x

2

3

=

(

x

1

+

x

2

)

3

-

3

⁢

x

1

⁢

x

2

⁢

(

x

1

+

x

2

)

=

4

3

-

3

⋅

9

⋅

4

=

-

44

.

Note.

If one wants to write easily a quadratic equation
with

rational roots

, one could take such one that the sum of the
coefficients is zero (then one root is always 1).  For instance,
the roots of the equation

5

⁢

x

2

+

11

⁢

x

-

16

=

0

are 1 and

-

16

5

.

Title

properties of quadratic equation

Canonical name

PropertiesOfQuadraticEquation

Date of creation

2015-02-12 9:55:41

Last modified on

2015-02-12 9:55:41

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

14

Author

pahio (2872)

Entry type

Result

Classification

msc 12D10

Related topic

VietasFormula

Related topic

ValuesOfComplexCosine

Related topic

IntegralBasisOfQuadraticField

Generated on Fri Feb 9 16:38:01 2018 by

LaTeXML
quadratic formula

quadratic formula

The roots of the

quadratic equation

a

⁢

x

2

+

b

⁢

x

+

c

=

0

a

,

b

,

c

∈

ℝ

,

a

≠

0

are given by the formula

x

=

-

b

±

b

2

-

4

⁢

a

⁢

c

2

⁢

a

.

The number

Δ

=

b

2

-

4

⁢

a

⁢

c

is called the

discriminant

of the equation.
If

Δ

>

0

, there are two different

real roots

,
if

Δ

=

0

there is a single real root,
and if

Δ

<

0

there are no real roots (but two different

complex roots

).

Let’s work a few examples.

First, consider

2

⁢

x

2

-

14

⁢

x

+

24

=

0

.
Here

a

=

2

,

b

=

-

14

, and

c

=

24

.
Substituting in the formula gives us

x

=

14

±

(

-

14

)

2

-

4

⋅

2

⋅

24

2

⋅

2

=

14

±

4

4

=

14

±

2

4

=

7

±

1

2

.

So we have two solutions (depending on whether we take the sign

+

or

-

):

x

=

8

2

=

4

and

x

=

6

2

=

3

.

Now we will solve

x

2

-

x

-

1

=

0

.
Here

a

=

1

,

b

=

-

1

, and

c

=

-

1

, so

x

=

1

±

(

-

1

)

2

-

4

⁢

(

1

)

⁢

(

-

1

)

2

=

1

±

5

2

,

and the solutions are

x

=

1

+

5

2

and

x

=

1

-

5

2

.

Title

quadratic formula

Canonical name

QuadraticFormula

Date of creation

2013-03-22 11:46:15

Last modified on

2013-03-22 11:46:15

Owner

yark (2760)

Last modified by

yark (2760)

Numerical id

13

Author

yark (2760)

Entry type

Theorem

Classification

msc 12D10

Classification

msc 26A99

Classification

msc 26A24

Classification

msc 26A09

Classification

msc 26A06

Classification

msc 26-01

Classification

msc 11-00

Related topic

DerivationOfQuadraticFormula

Related topic

QuadraticInequality

Related topic

QuadraticEquationInMathbbC

Related topic

ConjugatedRootsOfEquation2

Related topic

QuadraticCongruence

Generated on Fri Feb 9 16:38:03 2018 by

LaTeXML
quadratic resolvent

quadratic resolvent

The

quadratic resolvent

of the

cubic equation

y

3

+

p

⁢

y

+

q

=

0

,

(1)

where

p

and

q

are known

complex numbers

, is the auxiliary equation

z

2

+

q

⁢

z

-

(

p

3

)

3

=

0

determining as its roots (

http://planetmath.org/Equation

)

z

1

=

u

3

,

z

2

=

v

3

the numbers

u

and

v

whose sum

y

=

u

+

v

satisfies the
equation (1).  See
example of solving a cubic equation (

http://planetmath.org/exampleofsolvingacubicequation

).

Analogically, a

quartic equation

has a

cubic resolvent

(

resolvent cubic

) equation.

Title

quadratic resolvent

Canonical name

QuadraticResolvent

Date of creation

2014-11-27 16:05:30

Last modified on

2014-11-27 16:05:30

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

9

Author

pahio (2872)

Entry type

Definition

Classification

msc 12D10

Synonym

quadratic resolvent equation

Related topic

CardanosFormulae

Related topic

TchirnhausTransformations

Defines

cubic resolvent

Defines

resolvent equation

Generated on Fri Feb 9 16:38:04 2018 by

LaTeXML
rational root theorem

rational root theorem

Consider the

polynomial

p

⁢

(

x

)

=

a

n

⁢

x

n

+

a

n

-

1

⁢

x

n

-

1

+

⋯

+

a

1

⁢

x

+

a

0

where all the coefficients

a

i

are

integers

.

If

p

⁢

(

x

)

has a

rational

zero

u

/

v

where

gcd

⁡

(

u

,

v

)

=

1

,  then

u

∣

a

0

and

v

∣

a

n

.  Thus, for finding all rational zeros of

p

⁢

(

x

)

, it suffices to perform a finite number of tests.

The theorem is related to the result about

monic polynomials

whose coefficients belong to a

unique factorization domain

. Such theorem then states that any root in the

fraction field

is also in the base domain.

Title

rational root theorem

Canonical name

RationalRootTheorem

Date of creation

2013-03-22 11:46:18

Last modified on

2013-03-22 11:46:18

Owner

drini (3)

Last modified by

drini (3)

Numerical id

13

Author

drini (3)

Entry type

Theorem

Classification

msc 12D10

Classification

msc 12D05

Classification

msc 26A99

Classification

msc 26A24

Classification

msc 26A09

Classification

msc 26A06

Classification

msc 26-01

Classification

msc 11-00

Related topic

FactorTheorem

Generated on Fri Feb 9 16:38:21 2018 by

LaTeXML