Research Article 2017 iMedPub Journals International Journal of Applied Science - Research and Review www.imedpub.com ISSN 2394-9988 Vol. 4 No. 2:15

DOI: 10.21767/2394-9988.100065

Radical Solution of the General Samuel Bonaya Buya*

Sextic Equation Ngao Girls Secondary School, P.O. Box 12, Garsen - 80201, Kenya

Abstract A factorization method is given for solving the general sextic equation without *Corresponding author: the Tschirnhausen transformation. Three solvable factorized forms consisting of Samuel Bonaya Buya auxiliary quartic and quadratic factors. Two of these forms have one dependent parameter and are therefore not suitable for the purpose of this paper. The third  [email protected] factorized form selected has six independent parameters, which can be correlated to the parameters of the original sextic equation. Mathematics/Physics Teacher, Ngao Girls Secondary School, P.O. Box 12, Garsen - Keywords: Sextic equation; Polynomials; Factorized 80201, Kenya.

Received: October 20, 2017; Accepted: November 21, 2017; Published: December Tel: 0722825673 04, 2017

Citation: Buya SB (2017) Radical Solution of the General Sextic Equation. Appl Sci Res Rev Vol. 4 No.2:15 Introduction factorized the quartic equation to two quadratic factors. He then got a way to correlate the unknown coefficients of the quadratic The work of Abel (1826) and Galois (1832) have shown that factors to the known coefficients of the general quartic equation. general polynomials of degree five and greater cannot be solved Higher degree polynomial should be reducible if a rational or in radicals. Galois gave a more rigorous proof of the impossibility radical formulae can be established connecting the unknown to of solving these higher degree equations using group theory. After known parameters. Galois Theory failed to explore connections the work of these two researchers, there have been attempts to obtain algebraic solution using symbolic coefficients. In modern that may exist between known and unknown parameters in algebra, for example, the general quintic equation has been possible solvable factorized forms that may exist in higher degree solved using the Bring Radicals, while the general sextic equation general polynomial equations. Galois Theory instead looks into has been solved in terms of Kampe de Feriet Functions. solvable connections between the roots and coefficients and attempts to solve the fundamental symmetric equations. The The genesis to the modern approach to solving algebraic equations can be traced back to work of Francois Viète, a French Galois resolvent magnifies the problem of achieving radical mathematician. It was Francois Viète who set into course the solutions general algebraic equations of degree five and above to path of solving higher degree polynomial equations by relating its levels of impossibility. coefficients to the roots via elementary symmetric equations. Louis The problem of obtaining radical solution of higher degree Lagrange introduced Resolvents and they were systematically polynomial equations needs to be to be lessened through used by Galois as a tool for seeking specific solution of higher some intermediate connections that will generate lower degree degree polynomials. The Lagrange and Galois resolvents were auxiliary solvable factors. This calls for some forms of solvable successful in establishing a unified approach to solving quadratic, decomposition of higher degree polynomial equations. cubic, and quartic equations, but failed to achieve formulae for general algebraic equations greater than four. In 2011, Edward Thabo Motlotle achieved a radical solution of the Bring-Jerrard quintic equation with the aid of Newton’s Either the approach employed by Galois and the forerunners sum formula. The problem with the method employed by him of the theory was inappropriate in achieving a general radical is that it is very involving and dampens any hope of achieving solution of higher degree polynomials or such a solution is radical solution of higher degree polynomial equations. Between impossible. Other possible approaches for obtaining radical 2013 and 2017, Buya SB made attempts to contribute further to solution need to be fully explored before coming to conclusions these solutions. In his yet to be published work, Buya SB further of impossibility. Impossibility theorems can be arrived at using showed that radical solution of the general quintic equation is dead end approaches to solutions. possible without a Tschirnhausen. These results contradict Abel’s impossibility theorem and Galois Theory. In an attempt to solve the general quartic equation, Lodvico Ferrari

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In this research a contribution will be made in deriving radical method is a most appropriate method for solving higher degree solution of the general sextic equation. Many have attempted algebraic equations, provided the appropriate factorized form is to obtain algebraic solution of sextic equations. One noteworthy selected. work was done by Raghavendra G. Kulkarni, in which he In this paper, we show a possible factorized form that can be used attempted the solution of the general sextic equation by method to obtain a general radical solution of the general sextic equation. of decomposition. However, he failed to correlate all the unknown The auxiliary factors derived are quadratic and quartic. The coefficients, he introduced to the real coefficients of the general unknown coefficients introduced in these factors are correlated sextic equation. He attributed the inability to correlate the to the original parameters before obtaining the solution of coefficients to Abel’s impossibility theorem. He then introduced auxiliary equations. This means a radical solution of the unknown supplementary coefficients that resulted in some original coefficients has first to be to be sought for in terms offinite coefficients losing their independence. Thus, he failed to achieve addition, subtraction, multiplication, and division of the known a general solution of the general sextic equation. coefficients of the original equation [1-6]. The radical solution of In paper entitled “Trinomial sextic equation, its algebraic solution the general sextic equation then follows. to solvable factorized form,” Buya SB showed that the factorization Methodology Factorized form one The general sextic equation takes the form:

65432 1 x++++++= ax5 ax 4 ax 3 ax 2 ax 10 a 0

The factorization method is a good candidate method for solving the general sextic equation algebraically. The big task is to identify a solvable form of such an equation.

A solvable factorized form of the above equation can be given by:

4cc22 c a0 ((x+ a25 − a ())())()0 a 15 − ac − c x + a15 − ac x + c x ++= ax5 2 aa00 a0 c Expanding equation 2:

65cc 4 cc c 32 x+ ax5 +((()))(( a 25 − a a 15 −− ac c x + a5 a 25 − a ())()) a 15 −−+− ac c a15 ac x + ax 2 ++= ax 1 a 00 3 aa00 aa00 a0 Equating x4 coefficients

c ca(2− a 5 ( a 1 − ac 5 )) −= c aa40 4 a0

cc 5 aaaaacca5( 25− ( 15 − )) −= 3 aa00

For a3 to be independent then:

a22a5 aa a a 2 32−0 + 15 +− 0 0 = 6 c3() 2 c 23 ca3 0 aa50 a 0 a 5 a 5 Obtain the Cardano solution of c.

To obtain the dependent value of a4 then substitute the solution of 6 into 4

Take: a22 a aa1 =−−05 5 7 A 32() aa50 a 0 2 This article is available in: http://www.imedpub.com/applied-science-research-and-review/ 2017 International Journal of Applied Science - Research and Review ISSN 2394-9988 Vol. 4 No. 2:15

a = 0 8 B 2 a5 a2 = − 0 Ca3 3 9 a5

12 AB 12 AB 11 c=3 ()()() AC3 −++ AC3 −+− 2 AB 23 −− 2 27 3 4 27 3 27 3 10 12 AB 12 AB 11 A 3 −()()()AC3 −++ AC3 −+− 2 AB 23 −− 2 27 3 4 27 3 27 3 3

a a22 a aa 0((−− 0 5 15 ) 12 aaaaaaaa2 2 23 2 a 2 −−−0515550003 − ((32 ( ) 3a3 ) 2 27 aa50 a 0 3 a5 c = 3 a a22 a aa 0((−− 0 5 15 ) 12 a222322 a aa a a a a a 11 a22 a aa a + −−−0 5 153 5 5 0 0− 0 2 −−−−0 5 1523 0 (((32 ) 3a3 ) (((322 ) ) 4 27 aa50 a 0 3 a5 27 3 aa50 a 0 a 5 a a22 a aa 0((−− 0 5 15 ) 12 a2 aaaaaaa2 23 2 a 2 11 −−((0 ()51555000−−3 −a ) 2 27 a3 aa233 a3 − 5 00 5 3 a a22 a aa 0((−− 0 5 15 ) 12 a222322 a aa a a a a a 11 a22 a aa a + −−−0 5 153 5 5 0 0− 0 2 −−−−0 5 1523 0 (((32 ) 3a3 ) (((322 ) ) 4 27 aa50 a 0 3 a5 27 3 aa50 a 0 a 5 a22 a aa1 05()− 5 aa32 a + 50 0 3 c ca(25− a ( a 1 −− ac 5 )) c a0 12 a4 = a0

a4 is a dependent parameter The auxiliary quadratic equation of the above sextic equation is given by: a x2 + ax +=0 0 5 c In this case two of the roots of the sextic equation are given by:

2 a0 −±aa55 −4 x = c 1,2 2

In this research, we are interested in obtaining a radical solution of the general sextic equation without a Tschirnhausen transformation.

The a4 coefficient is not independent. Therefore, the above factorization falls short of this requirement. © Under License of Creative Commons Attribution 3.0 License 3 2017 International Journal of Applied Science - Research and Review ISSN 2394-9988 Vol. 4 No. 2:15

Factorized form two The general sextic equation takes the form: 65432 1 x++++++= ax5 ax 4 ax 3 ax 2 ax 10 a 0 The factorization method is a good candidate method for solving the general sextic equation algebraically. The big task is to identify a solvable form of such an equation.

A solvable factorized form of the above equation can be given by: a (x43+ bx ++ dx e)( x 2 + ( a − b ) x +0 ) = 0 2 5 e

a0 a 03 a ba 0 3 ba=3 → = →= e e eb a3 aa ba()−+ b03 = a → ba () −+ b = a →− b32 ab + ab −= a 0 4 5 eb4 5 4 5 43 Obtain the Cardano solution of 4

12 aa 12 aa 11 b=−3 ()()() aa3 +++54 aa3 +++54 2 aa 23 −− 2 2753 3 4 27 53 3 27 3 54 4 12 aa 12 aa 11 a 3 ()()()aa3+++54 aa3 +++54 2 aa 23 −+5 2 2753 3 4 27 53 3 27 3 54 3

ba0 a2 − ae2 − a3 5 da()52−+= b e a →= d →= d ab55−− ab

ba0 a2 − a0 a33 a ba 0 6 d+ eaba()5 −=→= 11 a ×+()ab5 − e ab53− b a

a1 is a dependent parameter. The formulae obtained as a result of the above factorization fall short of obtaining radical solution of the general sextic equation without a Tschirnhausen transformation.

Factorized form 3 The general sextic equation takes the form:

65432 x++++++= ax5 ax 4 ax 3 ax 2 ax 10 a 0 1

The factorization method is a good candidate method for solving the general sextic equation algebraically. The big task is to identify a solvable form of such an equation.

A solvable factorized form of the above equation can be given by: ba()− b a a3 b a 2+ −502 + −3 + ++ 0 −22 + − +0 = 2 ((x a45)()bx dxe 33 ( ) )(x ( a bx ) ) 0 2(a5 − b ) 3 be 27b e 2 e Equating coefficients:

ba()5− b a00 a 2(a4− )() abb 5 −+ − () abda 53 −+= 3 2(ab5 − ) e e

22 ba( − b) aa1  a −5 00 − + 0 −+ −+= 4 2a4  (abdabea55) ( ) 2 29(a5 − b) e e be  4 This article is available in: http://www.imedpub.com/applied-science-research-and-review/ 2017 International Journal of Applied Science - Research and Review ISSN 2394-9988 Vol. 4 No. 2:15

2 3 ba( − b) 1 a − 5 −+0 + −= 5 a4 (ab5)  eaba( 51) 29(a5 − b) be

If in equation 3:

ba()5− b a 00 a221 a 0 da=−−2(4 ) − () + () 6 2(a5 − b ) e e9 be Then

ea= 2 7 2 ba()50− b 1 a 3 a4−()()() ab5 −+ + aaba25 −= 1 2(a52− b ) 9 ba b 1 a 230 8 (a4− )()()() ab 5 −+ + aaba25 −= 1 29ba2

43 a5 2 240a 3 b−4( b a4 + )4( + b aa45 ++ a 2 )4( ba 1 − aa 25 − aa 54)() − 9 59a2

If we let we take:

a Aa=−+()5 10 4 5

B=4( aa45 + a 2 ) 11

2 C=−−4( a1 aa 25 aa 54) 12

4 a0 3 D = − () 13 9 a2 Then, the equation 9 takes the form: b432+ Ab + Bb + Cb += D 0 14 A Take bu= − 15 4 Then:

AAAA ()()()()u−+−+−+−+=432 Au Bu Cu D 0 16 4444 After regrouping the terms, the equation 16 takes the depressed quartic equation form: 42 u+ pu + qu += r 0 17

a5 2 2 32(aa++ a ) 3( a + ) 83BA− 45 2 4 18 p = = 5 88

aa5532 3 −+(a ) + 16( a + ( aa + a ) + 32( a − aa − aa) A−+48 AB C 4 4 45 2 1 25 54 19 q = = 55 88 −+3A42 256 D − 64 CD + 16 A r = = 256

aa5041024 3 1024 23a0 a 52 20 −+3(a4 ) − ( ) + (a1 − aa 25 − aa 54)( ) + 16( a4 + ) 59aa22 9 5 256 Equation 17 can be written in the form 21 below to enable determination of the roots of u.

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r (u22++ eu f )( u −+ eu ) = 0 21 f Where:

r 2 p=−+ ef 22 f er q= − fe 23 f Multiplying 22 by e and rearranging: er pe+= e3 + fe 25 f Adding 25 to 23:

3 2er e+ pe += q 26 f Subtracting 25 with 23: e3 + pe −= q2 fe 27 Multiplying together 27 and 28:

(e33++ pe q )( e +−= pe q ) 4 e 2 r 28

Equation 28 can be simplified to the solvable form:

e6+2 pe 4 + ( p 2 − 4) r e 22 −= q 0 29

Take:

a 32(aa++ a ) 3( a +5 )2 45 2 4 30 Ep=2 = 5 4

a5 2 32(aa45++ a 2 ) 3( a 4 + ) Fp=−=224( r 5 )= 8 31 aa5041024 3 1024 23a0 a 52 −+3(a4 ) − ( ) + (a1 − aa 25 − aa 54)( ) + 16( a4 + ) 59aa22 9 5 64

a5 32 −+(a4 ) − 4 AB + 32( a1 − a 25 a − a 54 a Gq=−=−22()5 32 8 Then, the equation 29 takes the form:

e64+ Ee ++= F G 0 33

Take ev2 = 34 Then the equation 34 takes the form:

v32+ Ev + Fv += G 0 35 Take: E vw= − 36 3 Then the equation 35 takes the form:

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EEE ()()()0w−32 + Ew − + Fw − += G 37 333 On expanding the equation 37: 12FE w3+() F − Ew 23 + E − += G 0 38 3 27 3 One of the roots of equation 36 is given by:

12FE 12FE 1 1 v=−3 ()()() E3 −+E3 − 2 + FE −23 − 1 2 27 3 4 27 3 27 3 39 12FE 12FE 1 1 E 3 ()()()E3−+E3 − 2 + FE −23 − 2 27 3 4 27 3 27 3 3

One of the roots of e in equation 34 is given by:

12FE 12FE 1 1 3 −()()()E3 −+E3 − 2 + FE −23 − 2 27 3 4 27 3 27 3 e1 = 40 12FE 12FE 1 1 E 3 ()()()E3−+E3 − 2 + FE −23 − 2 27 3 4 27 3 27 3 3

From equation 27: e3 +− pe q f = 41 2e Equation 21 can be written in the form 42 below:

3 22e+− pe q r (u++ eu )(u−+ eu 3 ) = 0 42 2e e+− pe q 2e The roots of u are given by:

3 2 e+− pe q −±ee −2 43 u = e 1,2 2

2 r ee±−4 3 e+− pe q 44 u = 2e 3,4 2

The roots of b in equation 9 are given by:

e3 +− pe q −±ee2 −2 AA 45 bu= −= e − 1,2 1,2 424 A rA b= u −=± ee2 −4 − 3,4 3,4 44e3 +− pe q 46 2e The radical solution of the quantity b enables determination of quantity d in equation 6. The auxiliary quadratic equation of the factorized sextic equation is: © Under License of Creative Commons Attribution 3.0 License 7 2017 International Journal of Applied Science - Research and Review ISSN 2394-9988 Vol. 4 No. 2:15

2 a0 x+−() a5 bx += 0 47 a2 Two of the roots of the sextic equation are: a ba−( ba −− )42 0 55a 48 x = 2 1,2 2

The other roots can be found by solving the auxiliary quartic equation. Conclusion We have achieved the algebraic solution of the general sextic equation. Radical solution of the general sextic equation is possible to achieve. This is a case of violation of Abel’s impossibility theorem and Galois Theory. Acknowledgements I would like to express my since gratitude to Almighty God for making this piece of work successful, and my wife, Julie Ahadi, for her encouragement and service.

References 1 Adamchik VS, Jeffrey DJ (2003) Polynomial transformations of Tschirnhaus, Bring and Jerrard. ACM SIGSAM Bulletin 37: 90-94. 2 Buya SB (2017) The trinomial sextic equation, its algebraic solution by conversion to solvable factorized form. Inter J Applied Sci and Innovations 1: 1-19. 3 Buya SB (2017) The general quintic equation, its solution by factorization to cubic and quadratic factors. Inter J Applied Sci and Innovations 1: 29-36. 4 E.T. Motlotle (2011) The Bring-Jerrard quintic equation, its solution and a formula for the universal gravitational constant. University of South Africa. 5 Cajori F (1991) A history of mathematics. 5th edition, Chelsea Publishing Company, New York, USA. 6 Kulkarni RG (2008) Solving sextic equations. Atlantic Electronic J Mathematics 3: 56-60.

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