192 REV. R. HARLEY ON SYMMETRIC PRODUCTS, AND 
From the first equation we have §, or ®,=0, and the 
second and third give 
; eh Oe 2 R 
5 or (3: =—, and §: or B=—, 
3 or £3 R and #3 or 8 Q 
which satisfy the fourth. For 
Q? R?2 
Tasks alg 
or QRE+ Q*+ R'=0. 
Consequently, when this relation holds, the roots of the 
quintic 
a — 5Qa*-5Ra+EH=0, 
are all included in the formula 
$5, @ pnts, BR. 
Yt ea hal 
25. Again: — When P, Q and R vanish, (11) is satis- 
fied, and ee te 
x=(1)°V -E, 
the solution of the ordinary binomial form. 
26. To return to the resolvent product. Denote 7,(z) 
by @, and put 
DF (+ 2,24) =U ; 
then (Art. 20) 
TT = LX, Ly XX, + VX? 2X, x,—- U=ac—d-U, 
and 64+5U =a' -—5a@b+5ac+50?- 15d. 
Hence, for the equation 
a —5M2t* — 5P2* —- 5Q2*— 5Rr+E=0, 
we have 
77 =5°MQ+38-5R-U, and 
64+5U=5'M'+5*M*P+5°MQ+5°P?+3-5°R. 
Now it is known that the general equation of the fifth 
degree may be put under any one of the six following 
forms, viz., 
x — 5Pa -—5Q2*°-5Re7+E=0 .. . (a) 
2-5Qe?-5R2a+E=0 ... (0) 
