Qiiestion of solving the Equation of the Fifth Degree. 31 



and the product of the two other conditions (67.) and (66.) 

 will become 



f2 + hf + i = 0, (74.) 



so that the product of all the three conditions becomes 



f3+ (g + h)f2 + (gh + i)f+gi = 0; (75.) 

 and the symmetric functions f, g + h, gh + i, gi, may be 

 expressed as follows, 



f = — a34-3a6-3c, (76.) 



g + b=c6— 3c, : (77.) 



gh + i = a^c— 4.0^6?— 2aZ>c + 3c% (78.) 



g\ = a^d—^a^hd + '^a'^cd-\-abc"—c^. (79.) 

 Again, the proposed equation of the fifth degree, 



^5 + B^-^ + D.r + E = 0, (58.) 



must be exactly divisible by the biquadratic equation (66.), 

 because all the roots of the latter are also roots of the former; 

 and therefore we must have 



B = h—a-, D = d-a-h, E = —ad, (80.) 



and c = ah (81.) 



This relation c = ab reduces the expressions (76.)«..(79.) 

 to the following, 



f = -o% -| 



g + h=-2a6, I , 



gh+i = a*Z> + «"6- — ia" rf, { 



gi = a^d; J 



and thereby reduces the condition (75.), that is, the product 

 of the three conditions (66.) (67.) (68.), to the form 



—a' — 3a''b-a^b~ + 5a^d = 0, (83.) 



which gives either 



a = 0, (84.) 



or else 



a'* + 3a-6+i'' = 5rf, (85.) 



and therefore, by (80.), either 



E = 0, (60.) 



or else 



5D = B2 (61.) 



Thus, when we set aside these two particular cases, we see 

 by (45.), that under the circumstances supposed in the enun- 

 ciation of the theorem, the i\mc\\o\\ f{x) — q x vanishes, for 

 every value of x which makes the polynome.*'^ + B.«"' + Djr + E 

 vanish ; and that therefore if we set aside the third and only 

 remaining case of exception, namely, the case in which the 



