136 
the fifth, we can reduce U to the form U 1 , we have (px 0 by putting 
for H 4 H 2 H 2 H 4 the given algebraic expressions 
i i i i 
Y 6 V 6 V 6 V 6 • 
1 > 1 2 > J 3 5 X 4 } 
Y x , &c., being the four roots of U 1 =0. The roots of F=:0 would 
be then given by five biquadratics, hk l m n being now determined 
functions of the coefficients of F=0 through those of F^O, 
(j)X o=C 0 , C 1? <px 2 =C 2 , (px s —C 3 , 0a? 4 =C 4 ; 
and Dr. Hargreave’s device would bear the coveted fruit. 
Without expressing any opinion about the possibility of this 
transformation, I can shew that the equation U=0 is farther con- 
ditioned beyond what has been seen concerning the arrangement 
of its roots in sets of four. We formed at first H 4 H 2 H 3 H 4 on the 
first of the following groups : — 
1234 
1234 
1234 
1234 
2341 
3142 
4312 
3412 
3412 
4321 
2143 
4321 
4123 
2413 
3421 
2143 
If we had used the second instead of the first group, we should 
have formed H 4 H 2 H 3 H 4 , of which H 4 only is in the former set 
of four : if we had used the third, we should have formed 
H 4 H 11 Hg 1 HJ 1 , all fresh values, except H 4 : if we had employed 
the fourth, we should have formed H 4 H 3 Hi Hi 1 , all before written. 
Thus we see that 11=0 is something very different from the general 
equation of its degree. It may possibly break up into quadratics, 
by introducing surds into the coefficients. 
Whichever group we employ, its five derivates give us five more 
expressions of x Q ] that is, there are 24 ways of expressing x 0 , 
exhibiting 24 different quadruplets from the 24 values of H 4 . 
I fear much that all this is rather a proof of the absurdity of 
attempting, than a contribution towards effecting, the algebraic 
resolution of the quintic. But it may be worth the while of some 
one, endued with more faith, to form and to study this U=0 of 
the 24th degree. He would deserve well of science who could 
shew us how to form it by a compendious method. 
When there are only four roots, x 0 x\ x 3 , we obtain, if we 
