Question of solving the Equation of the Fifth Degree. 539 
in which the roots are supposed to be all unequal, and the 
coefficients D and E to be, both of them, different from 0, 
and, 2nd, an equation of the form 
ep = Oa + F(a), \sennsvsieneceansiann (2.) 
in which f (x) denotes any rational function of x, whether in- 
tegral or fractional, 
M! 2! + M"2*" + &e. 
J (2) = , xz x! 3 eeeecsece (3.) 
K’2* + K"2* + &e. 
and if, in the result of this elimination, which will always be 
an equation of the fifth degree in y, of the form 
0= PLA Y4BD P4+CY+4D' y+E, «.. (4.) 
we suppose that the coefficients are such as to satisfy, znde- 
pendently of Q, the second as well as the first of the two con- 
ditions 
Be! Ose WHO Silasctgapnciss Sasa » (5.) 
in virtue of the values of the constants 
MM’, MY, on tly polly woe K/, KY, wee 2t'y 20!', wee vee  (6.) 
in the rational function f(z); Isay that then those constants 
(6.) must be such as to admit of our reducing that rational 
function to the form 
SF (2) = Ge+(2?+Dx+E). 9 (@),  occoee (7.) 
g being some new constant, and ¢ (x) being some new rational 
function of x, which does not contain the polynome 2°+ Dx 
+E as a divisor. 
Demonstration.—Let 2x, %22%3%4%; denote the five roots of 
the equation (1.), which are supposed to be all unequal among 
themselves, and different from 0; and let us put for abridge- 
ment 
fle) — fe) =h | 
F (#2) — 2S (2s) = as | 
Fes) — F(x) = Ie» F seseonneetennan (8.) 
F(t) =F (ts) = has | 
LE) = 4 Q+9=Q. | 
