aarjaneiro/planetmath huggingface dataset:
irreducibility of binomials with unity coefficients

irreducibility of binomials with unity coefficients

Let

n

be a

positive integer

.  We consider the possible

factorization

of the

binomial

x

n

+

1

.

•

If

n

has no

odd prime

factors, then the binomial

x

n

+

1

is

irreducible

(

http://planetmath.org/

Irreducible Polynomial

).  Thus,

x

+

1

,

x

2

+

1

,

x

4

+

1

,

x

8

+

1

and so on are irreducible polynomials (i.e. in the field

ℚ

of their

coefficients

).  N.B., only

x

+

1

and

x

2

+

1

are in the field

ℝ

; e.g. one has

x

4

+

1

=

(

x

2

-

x

⁢

2

+

1

)

⁢

(

x

2

+

x

⁢

2

+

1

)

.

•

If

n

is an

odd number

, then

x

n

+

1

is always

divisible

by

x

+

1

:

x

n

+

1

=

(

x

+

1

)

(

x

n

-

1

-

x

n

-

2

+

x

n

-

3

-

+

⋯

-

x

+

1

)

(1)

This is usable when

n

is an odd prime number, e.g.

x

5

+

1

=

(

x

+

1

)

⁢

(

x

4

-

x

3

+

x

2

-

x

+

1

)

.

•

When

n

is not a

prime number

but has an odd prime factor

p

, say

n

=

m

⁢

p

,  then we write

x

n

+

1

=

(

x

m

)

p

+

1

and apply the idea of (1); for example:

x

12

+

1

=

(

x

4

)

3

+

1

=

(

x

4

+

1

)

⁢

[

(

x

4

)

2

-

x

4

+

1

]

=

(

x

4

+

1

)

⁢

(

x

8

-

x

4

+

1

)

There are similar results for the binomial

x

n

+

y

n

, and the corresponding to (1) is

x

n

+

y

n

=

(

x

+

y

)

(

x

n

-

1

-

x

n

-

2

y

+

x

n

-

3

y

2

-

+

⋯

-

x

y

n

-

2

+

y

n

)

,

(2)

which may be verified by performing the

multiplication

on the right hand .

Title

irreducibility of binomials with unity coefficients

Canonical

name

IrreducibilityOfBinomialsWithUnityCoefficients

Date of creation

2013-03-22 15:13:08

Last modified on

2013-03-22 15:13:08

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

14

Author

pahio (2872)

Entry type

Result

Classification

msc 12D05

Classification

msc 13F15

Related topic

FactoringASumOrDifferenceOfTwoCubes

Related topic

PrimeFaxtorsOfXn1

Related topic

PrimeFactorsOfXn1

Related topic

ExpressibleInClosedForm

Generated on Fri Feb 9 16:36:59 2018 by

LaTeXML
polynomial long division

polynomial long division

Given two

polynomials

a

⁢

(

x

)

and

b

⁢

(

x

)

polynomial (long)

division

is a method for calculating

a

⁢

(

x

)

/

b

⁢

(

x

)

that is, finding the polynomials

q

⁢

(

x

)

and

r

⁢

(

x

)

such that

a

⁢

(

x

)

=

b

⁢

(

x

)

⁢

q

⁢

(

x

)

+

r

⁢

(

x

)

.

Here is an example to show the method.Let

a

⁢

(

x

)

=

x

4

-

2

⁢

x

3

+

5

and

b

⁢

(

x

)

=

x

2

+

3

⁢

x

-

2

.
The method looks very

similar

to integer division since a polynomial

∑

i

=

0

n

c

i

⁢

x

i

is somewhat similar to an integer

∑

i

=

0

n

c

i

⁢

10

i

In the initial setting we only write the coefficients, notice that

a

⁢

(

x

)

=

x

4

-

2

⁢

x

3

+

0

⁢

x

2

+

0

⁢

x

+

5

. It will then be

In the next step we se that

1

/

1

=

1

and we multiply 1 3 -2 with 1 and then subtract the result.

Then we move down the next number, in this case a zero, and

-

5

/

1

=

-

5

so we get -5, and multiply by -5 and subtract

as a final result we get

The result is

q

⁢

(

x

)

=

1

-

5 17

, which

translates

to

q

⁢

(

x

)

=

x

2

-

5

⁢

x

+

17

and

r

⁢

(

x

)

=

-

61

⁢

x

+

39

.

It is also possible to write the entire polynomial, that is, writing all the

x

i

’s. Like this

Title

polynomial long division

Canonical name

PolynomialLongDivision

Date of creation

2013-03-22 14:19:59

Last modified on

2013-03-22 14:19:59

Owner

rm50 (10146)

Last modified by

rm50 (10146)

Numerical id

7

Author

rm50 (10146)

Entry type

Definition

Classification

msc 12D05

Related topic

LongDivision

Generated on Fri Feb 9 16:37:01 2018 by

LaTeXML
proof of factor theorem

proof of factor theorem

Suppose that

f

⁢

(

x

)

is a

polynomial

with real or

complex

coefficients of degree

n

-

1

. Since

f

is a polynomial, it is infinitely

differentiable

. Therefore,

f

has a

Taylor expansion

about

a

. Since

f

(

n

)

⁢

(

x

)

=

0

, the terminates after the

n

-

1

th

term. Also, the

n

th

remainder

of the

Taylor series

vanishes

; i.e. (

http://planetmath.org/Ie

),

R

n

⁢

(

x

)

=

f

(

n

)

⁢

(

y

)

n

!

⁢

x

n

=

0

. Thus, the

function

is equal to its Taylor series. Hence,

f

⁢

(

x

)

=

∑

k

=

0

n

-

1

f

(

k

)

⁢

(

a

)

k

!

⁢

(

x

-

a

)

k

=

f

⁢

(

a

)

+

∑

k

=

1

n

-

1

f

(

k

)

⁢

(

a

)

k

!

⁢

(

x

-

a

)

k

=

f

⁢

(

a

)

+

(

x

-

a

)

⁢

∑

k

=

1

n

-

1

f

(

k

)

⁢

(

a

)

k

!

⁢

(

x

-

a

)

k

-

1

=

f

⁢

(

a

)

+

(

x

-

a

)

⁢

∑

k

=

0

n

-

2

f

(

k

+

1

)

⁢

(

a

)

(

k

+

1

)

!

⁢

(

x

-

a

)

k

.

If

f

⁢

(

a

)

=

0

, then

f

⁢

(

x

)

=

(

x

-

a

)

⁢

∑

k

=

0

n

-

2

f

(

k

+

1

)

⁢

(

a

)

(

k

+

1

)

!

⁢

(

x

-

a

)

k

. Thus,

f

⁢

(

x

)

=

(

x

-

a

)

⁢

g

⁢

(

x

)

, where

g

⁢

(

x

)

is the polynomial

∑

k

=

0

n

-

2

f

(

k

+

1

)

⁢

(

a

)

(

k

+

1

)

!

⁢

(

x

-

a

)

k

. Hence,

x

-

a

is a factor of

f

⁢

(

x

)

.

Conversely, if

x

-

a

is a factor of

f

⁢

(

x

)

, then

f

⁢

(

x

)

=

(

x

-

a

)

⁢

g

⁢

(

x

)

for some polynomial

g

⁢

(

x

)

. Hence,

f

⁢

(

a

)

=

(

a

-

a

)

⁢

g

⁢

(

a

)

=

0

.

It follows that

x

-

a

is a factor of

f

⁢

(

x

)

if and only if

f

⁢

(

a

)

=

0

.

Title

proof of factor theorem

Canonical name

ProofOfFactorTheorem

Date of creation

2013-03-22 12:39:54

Last modified on

2013-03-22 12:39:54

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

8

Author

Wkbj79 (1863)

Entry type

Proof

Classification

msc 12D05

Classification

msc 12D10

Generated on Fri Feb 9 16:37:03 2018 by

LaTeXML
proof of rational root theorem

proof of rational root theorem

Let

p

⁢

(

x

)

∈

ℤ

⁢

[

x

]

. Let

n

be a

positive integer

with

deg

⁡

p

⁢

(

x

)

=

n

. Let

c

0

,

…

,

c

n

∈

ℤ

such that

p

⁢

(

x

)

=

c

n

⁢

x

n

+

c

n

-

1

⁢

x

n

-

1

+

⋯

+

c

1

⁢

x

+

c

0

.

Let

a

,

b

∈

ℤ

with

gcd

⁡

(

a

,

b

)

=

1

and

b

>

0

such that

a

b

is a root of

p

⁢

(

x

)

. Then

0

=

p

⁢

(

a

b

)

=

c

n

⁢

(

a

b

)

n

+

c

n

-

1

⁢

(

a

b

)

n

-

1

+

⋯

+

c

1

⋅

a

b

+

c

0

=

c

n

⋅

a

n

b

n

+

c

n

-

1

⋅

a

n

-

1

b

n

-

1

+

⋯

+

c

1

⋅

a

b

+

c

0

.

Multiplying through by

b

n

and rearranging yields:

c

n

⁢

a

n

+

c

n

-

1

⁢

a

n

-

1

⁢

b

+

⋯

+

c

1

⁢

a

⁢

b

n

-

1

+

c

0

⁢

b

n

=

0

c

0

⁢

b

n

=

-

c

n

⁢

a

n

-

c

n

-

1

⁢

a

n

-

1

⁢

b

-

⋯

-

c

1

⁢

a

⁢

b

n

-

1

c

0

⁢

b

n

=

a

⁢

(

-

c

n

⁢

a

n

-

1

-

c

n

-

1

⁢

a

n

-

2

⁢

b

-

⋯

-

c

1

⁢

b

n

-

1

)

Thus,

a

|

c

0

b

n

and, by hypothesis,

gcd

⁡

(

a

,

b

)

=

1

. This implies that

a

|

c

0

.

Similarly:

c

n

⁢

a

n

+

c

n

-

1

⁢

a

n

-

1

⁢

b

+

⋯

+

c

1

⁢

a

⁢

b

n

-

1

+

c

0

⁢

b

n

=

0

c

n

⁢

a

n

=

-

c

n

-

1

⁢

a

n

-

1

⁢

b

-

⋯

-

c

1

⁢

a

⁢

b

n

-

1

-

c

0

⁢

b

n

c

n

⁢

a

n

=

b

⁢

(

-

c

n

-

1

⁢

a

n

-

1

-

⋯

-

c

1

⁢

a

⁢

b

n

-

1

-

c

0

⁢

b

n

-

1

)

Therefore,

b

|

c

n

a

n

and

b

|

c

n

.

Title

proof of rational root theorem

Canonical name

ProofOfRationalRootTheorem

Date of creation

2013-03-22 13:03:53

Last modified on

2013-03-22 13:03:53

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

11

Author

Wkbj79 (1863)

Entry type

Proof

Classification

msc 12D05

Classification

msc 12D10

Generated on Fri Feb 9 16:37:06 2018 by

LaTeXML
sextic equation

sextic equation

The

sextic Equation

is the

univariate polynomial

of the sixth degree:

x

6

+

a

⁢

x

5

+

b

⁢

x

4

+

c

⁢

x

3

+

d

⁢

x

2

+

e

⁢

x

+

f

=

0

.

Joubert showed in 1861 that this

polynomial

can be reduced without any form of accessory irrationalities to the 3-parameterized resolvent:

x

6

+

a

⁢

x

4

+

b

⁢

x

2

+

c

⁢

x

+

c

=

0

.

This polynomial was studied in great detail by Felix Klein and Robert Fricke in the 19th century and it is diectly related to the

algebraic

aspect of Hilbert’s 13th Problem. Its

solution

has been reduced (by Klein) to the solution of the so-called Valentiner Form problem, a ternary form problem which seeks the ratios of the

variables

involved in

the invariant system of the Valentiner Group of order 360. It can also be solved with a class of generalized hypergeometric series, by Birkeland’s approach to

algebraic equations

. Scott Crass has given an explicit solution to the Valentiner problem by purely iterational methods, see

Title

sextic equation

Canonical

name

SexticEquation

Date of creation

2013-03-22 13:55:29

Last modified on

2013-03-22 13:55:29

Owner

mathcam (2727)

Last modified by

mathcam (2727)

Numerical id

12

Author

mathcam (2727)

Entry type

Definition

Classification

msc 12D05

Synonym

sextic polynomial

Generated on Fri Feb 9 16:37:08 2018 by

LaTeXML
synthetic division

synthetic division

Synthetic division

is simply a shorthand for

long division

,
when one of the factors is linear.
So when an equation such as

y

=

2

⁢

x

2

+

10

⁢

x

+

12

is given and a factor of

x

+

3

is known, we can do the following:

2

10

12

First, we can use

-

3

since

-

3

is a solution if

x

+

3

=

0

.

-

3

2

10

12

Carry down the first

coefficient

:

-

3

2

10

12

2

Multiply

2

by

-

3

:

-

3

2

10

12

-

6

2

Add up

10

and

-

6

:

-

3

2

10

12

-

6

2

4

Multiply

4

by

-

3

:

-

3

2

10

12

-

6

-

12

2

4

0

This means that, by

factoring

the equation by

x

+

3

, we will get another factor as

2

⁢

x

+

4

.

Synthetic is simply an

variation

of the long division — a version that’s manipulated and simplified to shorten the calculations.

Title

synthetic division

Canonical name

SyntheticDivision

Date of creation

2013-03-22 14:07:27

Last modified on

2013-03-22 14:07:27

Owner

mathwizard (128)

Last modified by

mathwizard (128)

Numerical id

7

Author

mathwizard (128)

Entry type

Definition

Classification

msc 12D05

Generated on Fri Feb 9 16:37:09 2018 by

LaTeXML
y^2=x^3-2

y

2

=

x

3

-

2

We want to solve the equation

y

2

=

x

3

-

2

over the

integers

.

By writing

y

2

+

2

=

x

3

we can factor on

ℤ

⁢

[

-

2

]

as

(

y

-

i

⁢

2

)

⁢

(

y

+

i

⁢

2

)

=

x

3

.

Using

congruences

modulo

8

, one can show that both

x

,

y

must be odd, and it can also be shown that

(

y

-

i

⁢

2

)

and

(

y

+

i

⁢

2

)

are relatively prime (if it were not the case, any

divisor

would have even norm, which is not possible).

Therefore, by

unique factorization

, and using that the only units (

http://planetmath.org/UnitsOfQuadraticFields

) on

ℤ

⁢

[

-

2

]

are

1

,

-

1

, we have that each factor must be a cube.

So let us write

(

y

+

i

⁢

2

)

=

(

a

+

b

⁢

i

⁢

2

)

3

=

(

a

3

-

6

⁢

a

⁢

b

2

)

+

i

⁢

(

3

⁢

a

2

⁢

b

-

2

⁢

b

3

)

⁢

2

Then

y

=

a

3

-

6

⁢

a

⁢

b

2

and

1

=

3

⁢

a

2

⁢

b

-

2

⁢

b

3

=

b

⁢

(

3

⁢

a

2

-

2

⁢

b

2

)

. These two equations imply

b

=

±

1

and thus

a

=

±

1

, from where the only possible solutions are

x

=

3

,

y

=

±

5

.

References

1

Esmonde, Ram Murty;

Problems in

Algebraic Number Theory

. Springer.

Title

y

2

=

x

3

-

2

Canonical name

Y2X32

Date of creation

2013-03-22 14:52:05

Last modified on

2013-03-22 14:52:05

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

10

Author

CWoo (3771)

Entry type

Application

Classification

msc 12D05

Classification

msc 11R04

Synonym

y

2

+

2

=

x

3

Synonym

finding integer solutions to

y

2

+

2

=

x

3

Related topic

UFD

Generated on Fri Feb 9 16:37:11 2018 by

LaTeXML
a

a

asw

Title

a

Canonical name

A

Date of creation

2013-03-22 19:00:57

Last modified on

2013-03-22 19:00:57

Owner

Altair (23757)

Last modified by

Altair (23757)

Numerical id

6

Author

Altair (23757)

Entry type

Topic

Classification

msc 12D10

Generated on Fri Feb 9 16:37:13 2018 by

LaTeXML
Cardano’s derivation of the cubic formula

Cardano’s derivation of the cubic formula

To solve the

cubic polynomial

equation

x

3

+

a

⁢

x

2

+

b

⁢

x

+

c

=

0

for

x

, the first step is to apply the Tchirnhaus

transformation

x

=

y

-

a

3

. This reduces the equation to

y

3

+

p

⁢

y

+

q

=

0

, where

p

=

b

-

a

2

3

q

=

c

-

a

⁢

b

3

+

2

⁢

a

3

27

The next step is to substitute

y

=

u

-

v

, to obtain

(

u

-

v

)

3

+

p

⁢

(

u

-

v

)

+

q

=

0

(1)

or, with the terms collected,

(

q

-

(

v

3

-

u

3

)

)

+

(

u

-

v

)

⁢

(

p

-

3

⁢

u

⁢

v

)

=

0

(2)

From equation (

2

), we see that if

u

and

v

are chosen so that

q

=

v

3

-

u

3

and

p

=

3

⁢

u

⁢

v

, then

y

=

u

-

v

will

satisfy

equation (

1

), and the

cubic equation

will be solved!

There remains the matter of solving

q

=

v

3

-

u

3

and

p

=

3

⁢

u

⁢

v

for

u

and

v

. From the second equation, we get

v

=

p

/

(

3

⁢

u

)

, and substituting this

v

into the first equation yields

q

=

p

3

(

3

⁢

u

)

3

-

u

3

which is a

quadratic equation

in

u

3

. Solving for

u

3

using the

quadratic formula

, we get

u

3

=

-

27

⁢

q

+

108

⁢

p

3

+

729

⁢

q

2

54

=

-

9

⁢

q

+

12

⁢

p

3

+

81

⁢

q

2

18

v

3

=

27

⁢

q

+

108

⁢

p

3

+

729

⁢

q

2

54

=

9

⁢

q

+

12

⁢

p

3

+

81

⁢

q

2

18

Using these values for

u

and

v

, you can back–substitute

y

=

u

-

v

,

p

=

b

-

a

2

/

3

,

q

=

c

-

a

⁢

b

/

3

+

2

⁢

a

3

/

27

, and

x

=

y

-

a

/

3

to get the expression for the first root

r

1

in the

cubic formula

. The second and third roots

r

2

and

r

3

are obtained by performing

synthetic division

using

r

1

, and using the quadratic formula on the remaining quadratic factor.

Title

Cardano’s

derivation

of the cubic formula

Canonical name

CardanosDerivationOfTheCubicFormula

Date of creation

2013-03-22 12:10:28

Last modified on

2013-03-22 12:10:28

Owner

djao (24)

Last modified by

djao (24)

Numerical id

12

Author

djao (24)

Entry type

Proof

Classification

msc 12D10

Related topic

FerrariCardanoDerivationOfTheQuarticFormula

Generated on Fri Feb 9 16:37:15 2018 by

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conjugated roots of equation

conjugated roots of equation

The rules

w

1

+

w

2

¯

=

w

1

¯

+

w

2

¯

and

w

1

⁢

w

2

¯

=

w

1

¯

⁢

w

2

¯

,

concerning the

complex conjugates

of the sum and product of two

complex numbers

, may be by induction generalised for arbitrary number of complex numbers

w

k

. Since the complex conjugate of a

real number

is the same real number, we may write

a

k

⁢

z

k

¯

=

a

k

⁢

z

¯

k

for real numbers

a

k

(

k

=

0

,

1

,

2

,

…

)

. Thus, for a

polynomial

P

⁢

(

x

)

:=

a

0

⁢

x

n

+

a

1

⁢

x

n

-

1

+

…

+

a

n

we obtain

P

⁢

(

z

)

¯

=

a

0

⁢

z

n

+

a

1

⁢

z

n

-

1

+

…

+

a

n

¯

=

a

0

⁢

z

¯

n

+

a

1

⁢

z

¯

n

-

1

+

…

+

a

n

=

P

⁢

(

z

¯

)

.

I.e., the values of a polynomial with real coefficients computed at a complex number and its complex conjugate are
complex conjugates of each other.

If especially the value of a polynomial with real coefficients vanishes at some complex number

z

, it vanishes also at

z

¯

.  So the roots of an

algebraic equation

P

⁢

(

x

)

=

0

with real coefficients are pairwise complex conjugate numbers.

Example.

The roots of the

binomial equation

x

3

-

1

=

0

are

x

=

1

,

x

=

-

1

±

i

⁢

3

2

,  the third

roots of unity

.

Title

conjugated roots of equation

Canonical name

ConjugatedRootsOfEquation

Date of creation

2013-03-22 17:36:51

Last modified on

2013-03-22 17:36:51

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

7

Author

pahio (2872)

Entry type

Topic

Classification

msc 12D10

Classification

msc 30-00

Classification

msc 12D99

Synonym

roots of algebraic equation with real coefficients

Related topic

PartialFractionsOfExpressions

Related topic

QuadraticFormula

Related topic

ExampleOfSolvingACubicEquation

Generated on Fri Feb 9 16:37:20 2018 by

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cubic formula

cubic formula

The three roots

r

1

,

r

2

,

r

3

of a

cubic polynomial

equation

x

3

+

a

⁢

x

2

+

b

⁢

x

+

c

=

0

are given by

r

1

=

-

a

3

+

(

-

2

⁢

a

3

+

9

⁢

a

⁢

b

-

27

⁢

c

+

(

2

⁢

a

3

-

9

⁢

a

⁢

b

+

27

⁢

c

)

2

+

4

⁢

(

-

a

2

+

3

⁢

b

)

3

54

)

1

/

3

+

(

-

2

⁢

a

3

+

9

⁢

a

⁢

b

-

27

⁢

c

-

(

2

⁢

a

3

-

9

⁢

a

⁢

b

+

27

⁢

c

)

2

+

4

⁢

(

-

a

2

+

3

⁢

b

)

3

54

)

1

/

3

r

2

=

-

a

3

-

1

+

i

⁢

3

2

⁢

(

-

2

⁢

a

3

+

9

⁢

a

⁢

b

-

27

⁢

c

+

(

2

⁢

a

3

-

9

⁢

a

⁢

b

+

27

⁢

c

)

2

+

4

⁢

(

-

a

2

+

3

⁢

b

)

3

54

)

1

/

3

+

-

1

+

i

⁢

3

2

⁢

(

-

2

⁢

a

3

+

9

⁢

a

⁢

b

-

27

⁢

c

-

(

2

⁢

a

3

-

9

⁢

a

⁢

b

+

27

⁢

c

)

2

+

4

⁢

(

-

a

2

+

3

⁢

b

)

3

54

)

1

/

3

r

3

=

-

a

3

+

-

1

+

i

⁢

3

2

⁢

(

-

2

⁢

a

3

+

9

⁢

a

⁢

b

-

27

⁢

c

+

(

2

⁢

a

3

-

9

⁢

a

⁢

b

+

27

⁢

c

)

2

+

4

⁢

(

-

a

2

+

3

⁢

b

)

3

54

)

1

/

3

-

1

+

i

⁢

3

2

⁢

(

-

2

⁢

a

3

+

9

⁢

a

⁢

b

-

27

⁢

c

-

(

2

⁢

a

3

-

9

⁢

a

⁢

b

+

27

⁢

c

)

2

+

4

⁢

(

-

a

2

+

3

⁢

b

)

3

54

)

1

/

3

Title

cubic formula

Canonical name

CubicFormula

Date of creation

2013-03-22 12:10:25

Last modified on

2013-03-22 12:10:25

Owner

djao (24)

Last modified by

djao (24)

Numerical id

10

Author

djao (24)

Entry type

Theorem

Classification

msc 12D10

Synonym

cubic equation

Related topic

QuarticFormula

Related topic

GaloisTheoreticDerivationOfTheQuarticFormula

Related topic

FerrariCardanoDerivationOfTheQuarticFormula

Related topic

FundamentalTheoremOfGaloisTheory

Generated on Fri Feb 9 16:37:24 2018 by

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derivation of quadratic formula

derivation of quadratic formula

Suppose

A

,

B

,

C

are

real numbers

, with

A

≠

0

, and suppose

A

⁢

x

2

+

B

⁢

x

+

C

=

0

.

Since

A

is nonzero, we can divide by

A

and obtain the equation

x

2

+

b

⁢

x

+

c

=

0

,

where

b

=

B

A

and

c

=

C

A

.
This equation can be written as

x

2

+

b

⁢

x

+

b

2

4

-

b

2

4

+

c

=

0

,

so completing the square, i.e., applying the

identity

(

p

+

q

)

2

=

p

2

+

2

⁢

p

⁢

q

+

q

2

, yields

(

x

+

b

2

)

2

=

b

2

4

-

c

.

Then, taking the square root of both sides, and solving for

x

, we obtain
the solution

formula

x

=

-

b

2

±

b

2

4

-

c

=

B

2

⁢

A

±

B

2

4

⁢

A

2

-

C

A

=

-

B

±

B

2

-

4

⁢

A

⁢

C

2

⁢

A

,

and the

derivation

is completed.

A slightly less intuitive but more aesthetically pleasing approach to this derivation can be achieved by multiplying both sides of the equation

a

⁢

x

2

+

b

⁢

x

+

c

=

0

by

4

⁢

a

, resulting in the equation

4

⁢

a

2

⁢

x

2

+

4

⁢

a

⁢

b

⁢

x

+

b

2

=

b

2

-

4

⁢

a

⁢

c

,

in which the left-hand side can be expressed as

(

2

⁢

a

⁢

x

+

b

)

2

. From here, the proof is identical.

Title

derivation of quadratic formula

Canonical name

DerivationOfQuadraticFormula

Date of creation

2013-03-22 11:56:44

Last modified on

2013-03-22 11:56:44

Owner

mathcam (2727)

Last modified by

mathcam (2727)

Numerical id

12

Author

mathcam (2727)

Entry type

Proof

Classification

msc 12D10

Related topic

QuadraticFormula

Related topic

QuadraticEquationInMathbbC

Generated on Fri Feb 9 16:37:26 2018 by

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example of solving a cubic equation

example of solving a cubic equation

Let us use Cardano’s formulae for solving algebraically the

cubic equation

x

3

+

3

⁢

x

2

-

1

=

0

.

(1)

First apply the

Tschirnhaus transformation

(

http://planetmath.org/CardanosDerivationOfTheCubicFormula

)

x

:=

y

-

1

for removing the quadratic term; from

(

y

-

1

)

3

+

3

⁢

(

y

-

1

)

2

-

1

=

0

we get the simplified equation

y

3

+

3

⁢

y

-

2

=

0

.

(2)

We now suppose that

y

:=

u

+

v

.  Substituting this into (2) and rewriting the equation in the form

(

u

3

+

v

3

-

2

)

+

3

⁢

(

u

⁢

v

+

1

)

⁢

(

u

+

v

)

=

0

,

one can determine

u

and

v

such that

u

3

+

v

3

-

2

=

0

and

u

⁢

v

+

1

=

0

,  i.e.

{

u

3

+

v

3

=

2

,

u

3

⁢

v

3

=

-

1

.

Using the

properties of quadratic equation

, we infer that

u

3

and

v

3

are the roots of the

resolvent equation

z

2

-

2

⁢

z

-

1

=

0

.

Therefore,

u

and

v

satisfy the

binomial equations

u

3

=

1

+

2

,

v

3

=

1

-

2

,

(3)

respectively.  If we choose the real

radicals

u

=

u

0

=

1

+

2

3

and

v

=

v

0

=

1

-

2

3

,  the other solutions

u

,

v

of (3) are

ζ

⁢

u

0

,

ζ

2

⁢

v

0

;

ζ

2

⁢

u

0

,

ζ

⁢

v

0

,

(4)

where

ζ

=

-

1

+

i

⁢

3

2

,

ζ

2

=

-

1

-

i

⁢

3

2

are the

primitive

third

roots of unity

.  One must combine the pairs

(

u

,

v

)

of (4) so that

u

⁢

v

=

u

3

⁢

v

3

3

=

-

1

.

Accordingly, all three roots of the cubic equation (2) are

{

y

1

=

u

0

+

v

0

=

1

+

2

3

+

1

-

2

3

,

y

2

=

ζ

⁢

u

0

+

ζ

2

⁢

v

0

=

-

1

+

i

⁢

3

2

⁢

1

+

2

3

+

-

1

-

i

⁢

3

2

⁢

1

-

2

3

,

y

3

=

ζ

2

⁢

u

0

+

ζ

⁢

v

0

=

-

1

-

i

⁢

3

2

⁢

1

+

2

3

+

-

1

+

i

⁢

3

2

⁢

1

-

2

3

.

The roots of the original equation (1) are gotten via the used substitution equation

x

:=

y

-

1

, i.e. adding

-

1

to the values of

y

.  If we also separate the real (

http://planetmath.org/RealPart

) and

imaginary parts

, we have the following solution of (1):

{

x

1

=

-

1

+

1

+

2

3

+

1

-

2

3

,

x

2

=

-

1

-

1

2

⁢

(

1

+

2

3

+

1

-

2

3

)

+

i

⁢

3

2

⁢

(

1

+

2

3

-

1

-

2

3

)

,

x

3

=

-

1

-

1

2

⁢

(

1

+

2

3

+

1

-

2

3

)

-

i

⁢

3

2

⁢

(

1

+

2

3

-

1

-

2

3

)

.

One of the roots is a

real number

, but the other two are (i.e. non-real)

complex conjugates

of each other.

Title

example of solving a cubic equation

Canonical name

ExampleOfSolvingACubicEquation

Date of creation

2015-11-18 14:15:18

Last modified on

2015-11-18 14:15:18

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

14

Author

pahio (2872)

Entry type

Example

Classification

msc 12D10

Synonym

example of using Cardano’s formulas

Related topic

PolynomialEquationOfOddDegree

Related topic

ConjugatedRootsOfEquation2

Generated on Fri Feb 9 16:37:33 2018 by

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