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On the Solution to Octic Equations The Mathematics Enthusiast Volume 4 Number 2 Article 6 6-2007 On the Solution to Octic Equations Raghavendra G. Kulkarni Follow this and additional works at: https://scholarworks.umt.edu/tme Part of the Mathematics Commons Let us know how access to this document benefits ou.y Recommended Citation Kulkarni, Raghavendra G. (2007) "On the Solution to Octic Equations," The Mathematics Enthusiast: Vol. 4 : No. 2 , Article 6. Available at: https://scholarworks.umt.edu/tme/vol4/iss2/6 This Article is brought to you for free and open access by ScholarWorks at University of Montana. It has been accepted for inclusion in The Mathematics Enthusiast by an authorized editor of ScholarWorks at University of Montana. For more information, please contact [email protected]. TMME, vol4, no.2, p.193 On the Solution to Octic Equations Raghavendra G. Kulkarni1 Senior Member IEEE, Deputy General Manager, HMC division, Bharat Electronics Abstract We present a novel decomposition method to decompose an eighth-degree polynomial equation, into its two constituent fourth-degree polynomials, as factors, leading to its solution. The salient feature of the octic equation solved here is that, the sum of its four roots being equal to the sum of the remaining four roots. We derive the condition to be satisfied by coefficients so that the given octic is solvable by the proposed method. Key words: Octic equation; polynomials; factors; roots; coefficients; solvable polynomial equation; polynomial decomposition; solvable octic. 2000 Mathematics Subject Classification:12D05: Polynomials (Factorization) 1. Introduction It is well known from the works of Ruffini, Abel and Galois that the general polynomial equations of degree higher than the fourth cannot be solved in radicals [1 – 4]. This does not mean that there is no algebraic solution to these equations. The algebraic solutions to the general quintic have been obtained with symbolic coefficients. Hermite solved the Bring-Jerrard quintic using elliptic functions, and Klein gave a solution to the principal quintic using hypergeometric functions [5, 6]. The general sextic can be solved in terms of Kampe de Feriet functions, and a restricted class of sextics can be solved in terms generalized hypergeometric functions in one variable, using Klein’s approach to solving the quintic equation [7]. There is not much literature on the solution to polynomial equations beyond sixth-degree, except in the form of numerical methods. We hope this paper will fill this gap to some extent. In this paper we present a novel technique of decomposing a solvable octic equation into constituent fourth-degree polynomial factors, thereby facilitating the determination of all of its roots. The salient feature of the octic solved here is that, the sum of its four roots is equal to the sum of its remaining four roots. The condition required to be satisfied by the coefficients, in 1 Dr. Raghavendra G. Kulkarni, Senior Member IEEE, Deputy General Manager, HMC division, Bharat Electronics Jalahalli Post, Bangalore-560013, INDIA. Phone: +91-80-22195270, Fax: +91-80-28382738, Email: [email protected] The Montana Mathematics Enthusiast, ISSN 1551-3440, Vol. 4, no.2, pp. 193-209 2007©The Montana Council of Teachers of Mathematics TMME, vol4, no.2, p.194 order that the given octic is solvable in radicals with the proposed technique, is derived. At the end of the paper we solve some numerical examples illustrating the applicability of this method. 2. Formulation of equations for solving the octic equation: Let the octic equation for which the solution is sought be: 8 7 6 5 4 3 2 x + a7x + a6x + a5x + a4x + a3x + a2x + a1x + a0 = 0 (1) where a0, a1, a2, a3, a4, a5, a6, and a7 are the real coefficients. In the method proposed here, we attempt to represent the octic (1), in the form as shown below. 4 3 2 2 2 4 3 2 2 2 [(x + b3x + b2x + b1x + b0) – p (x + c3x + c2x + c1x + c0) ]/(1 – p ) = 0 (2) where b0, b1, b2, b3, and c0, c1, c2, c3 are the unknown coefficients of the respective fourth-degree polynomials in the above equation. The parameter, p, is also an unknown to be determined. The merit of representing the octic (1) in the form of (2) is obvious: notice that (2) can be easily factorized as: 4 3 2 4 3 2 {[(x + b3x + b2x + b1x + b0) – p(x + c3x + c2x + c1x + c0)]/(1 – p)} 4 3 2 4 3 2 {[(x + b3x + b2x + b1x + b0) + p(x + c3x + c2x + c1x + c0)]/(1 + p)} = 0 (3) When each factorial term is equated to zero, we obtain the following two fourth-degree polynomial equations. 4 3 2 4 3 2 [(x + b3x + b2x + b1x + b0) – p(x + c3x + c2x + c1x + c0)]/(1 – p) = 0 (4) 4 3 2 4 3 2 [(x + b3x + b2x + b1x + b0) + p(x + c3x + c2x + c1x + c0)]/(1 + p) = 0 (5) The solution to the above quartic equations can be obtained easily by the Ferrari’s method. The roots of the quartic equations, (4) and (5), will be the roots of the given octic (1), when it gets represented in the form of octic (2). To achieve this the coefficients of octic (1) should be same as that of octic (2). However notice that the coefficients of octic (2) are not explicitly written. Therefore we expand and rearrange (2) in descending powers of x as shown below in order to expose the coefficients for comparison. TMME, vol4, no.2, p.195 8 2 2 7 2 2 2 2 6 x + [2(b3 – c3p )/(1 – p )]x + {[(b3 + 2b2) – (c3 + 2c2)p ]/(1 – p )}x 2 2 5 + {2[(b1 + b2b3) – (c1 + c2c3)p ]/(1 – p )}x 2 2 2 2 4 + {[(b2 + 2b0 + 2b1b3) – (c2 + 2c0 + 2c1c3)p ]/(1 – p )}x 2 2 3 2 2 2 2 2 + {2[(b0b3 + b1b2) – (c0c3 + c1c2)p ]/(1 – p )}x + {[(b1 + 2b0b2) – (c1 + 2c0c2)p ]/(1 – p )}x 2 2 2 2 2 2 + [2(b0b1 – c0c1p )/(1 – p )]x + [(b0 – c0 p )/(1 – p )] = 0 (6) Equating the coefficients of (1) and (6), we obtain the following eight equations as given below. 2 2 [2(b3 – c3p )/(1 – p )] = a7 (7) 2 2 2 2 {[(b3 + 2b2) – (c3 + 2c2)p ]/(1 – p )} = a6 (8) 2 2 {2[(b1 + b2b3) – (c1 + c2c3)p ]/(1 – p )} = a5 (9) 2 2 2 2 {[(b2 + 2b0 + 2b1b3) – (c2 + 2c0 + 2c1c3)p ]/(1 – p )} = a4 (10) 2 2 {2[(b0b3 + b1b2) – (c0c3 + c1c2)p ]/(1 – p )} = a3 (11) 2 2 2 2 {[(b1 + 2b0b2) – (c1 + 2c0c2)p ]/(1 – p )} = a2 (12) 2 2 [2(b0b1 – c0c1p )/(1 – p )] = a1 (13) 2 2 2 2 [(b0 – c0 p )/(1 – p )] = a0 (14) There are nine unknowns (b0, b1, b2, b3, c0, c1, c2, c3, and p), but only eight equations [(7) to (14)] to solve. Therefore one more equation has to be introduced so that all the nine unknowns can be determined. However introducing an equation (involving the unknowns) imposes certain condition on the roots (and hence the coefficients) of the octic equation (1). The equation chosen will dictate the type of solvable octic. The equation, we are introducing here is as follows: b3 = a7/2 (15) As we notice from our further analysis given in this paper, the above equation leads to a solvable octic in which the sum of its four roots is equal to the sum of its remaining four roots. Observe that the equation as given by (15) is not the only equation (or condition) to be introduced to solve the octic by this method. There are several ways of choosing the equation, like b2 = c2, or b1 = c1, or, b0 = c0, etc. However the equation introduced will decide the type of solvable octic. Substituting the value of b3 [using (15)] in equation (7), c3 is evaluated as: c3 = a7/2 (16) TMME, vol4, no.2, p.196 The values of b3 and c3 are substituted in equations, (8) to (11), resulting in expressions, (8A) to (11A), respectively, as given below. 2 2 b2 = c2p + F2(1 – p ) (8A) where F2 is given by: 2 F2 = (4a6 – a7 )/8 2 2 2 b1 = c1p + (a5/2)(1 – p ) – (a7/2)(b2 – c2p ) (9A) 2 2 2 2 2 2 b0 = c0p + (a4/2)(1 – p ) – [(b2 – c2 p )/2] – (a7/2)(b1 – c1p ) (10A) 2 2 2 b0 = c0p + (a3/a7)(1 – p ) – (2/a7)(b1b2 – c1c2p ) (11A) Now we have seven equations, (8A), (9A), (10A), (11A), (12), (13), and (14), with seven unknowns, b0, b1, b2, c0, c1, c2 and p, to be determined from them. We attempt to determine these unknowns through the process of elimination. Using the expression (8A), we eliminate b2 from the equations, (9A), (10A), (11A), and (12), resulting in the following expressions, (9B), (10B), (11B), and (12A), respectively. 2 2 b1 = c1p + F1(1 – p ) (9B) where F1 is: F1 = (a5 – a7F2)/2 2 2 2 2 2 2 b0 = c0p – (a7/2)(b1 – c1p ) + [a4 – F2 + (c2 – F2) p ][(1 – p )/2] (10B) 2 2 2 2 2 b0 = c0p + (a3/a7)(1 – p ) – (2/a7){[c2p + F2(1 – p )]b1 – c1c2p } (11B) 2 2 2 2 2 2 b1 + 2b0[c2p + F2(1 – p )] – (c1 + 2c0c2)p = a2(1 – p ) (12A) Using (9B) we now eliminate b1 from equations (10B), (11B), (12A), and (13), resulting in expressions (10C), (11C), (12B), and (13A) respectively as shown below. 2 2 2 2 2 b0 = c0p + F0(1 – p ) + [(c2 – F2) (1 – p ) p /2] (10C) where F0 is given by: 2 F0 = (a4 – a7F1 – F2 )/2, and TMME, vol4, no.2, p.197 2 2 2 2 b0 = c0p + [(a3 – 2F1F2)/a7](1 – p ) + (2/a7)(1 – p )(c1 – F1)(c2 – F2)p (11C) 2 2 2 2 2 2 2b0(c2 – F2)p + 2F2b0 – 2c0c2p = [a2 – F1 + (c1 – F1) p ](1 – p ) (12B) 2 2 2 b0(c1 – F1)p + F1b0 – c0c1p = (a1/2)(1 – p ) (13A) We are left with five equations, (10C), (11C), (12B), (13A), and (14), involving five unknowns namely b0, c0, c1, c2, and p.
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