Wilder's Algebraic Solution. ger a! 
Proposition fourth. Let o(xypq ete.) be a factor of 
9’ (xypq ete.) +0" (xypq etc.), independently of x, y, p, ete.;_ 
then I say that if ¢” (xypq ete.) written for y, eee 
viene etc.) =0, it, will also make 9 (xypq ete.)=0, and 
sieves etc.) = bt) = writing y'=9'" (apyete. ‘% we shall have 
ny'pq ere.) + 9"(xy'pgq ete.) _ 
WW, Pq etc.) 
ete 
or better ox Hele fi ie.) 0 ; since therefore 9(xy'pq ete.) di- 
vides the whole vey Pq ete.)-+e" (xy'pq ete.), and the part 
e(xy'pq etc.), it must also divide the remainder 9’ («y’pq etc.) ; 
ge (ay'pq ete.) 
o(xy'pq ete. ) ory Pd te.) © 
o(xy'pg ete.)=0; but o(ry’pgq ete. ak a fates of 9!(xy ‘pq etc.), 
therefore, 9'(«y'pq etc.) =0. 
Further, it may not be amiss to observe, that a function is 
equal to the continued product of all its factors ; and if a is 
a root of (z" +1)=0, different from unity, then the roots of 
this equation are, 
1 
consequently; + 
Let us now return to the second example, ol continue to 
denote by x3, ¢(bc), we have then, 
sotto) ? 
“2=ay 
TP a es sapocotiad 
i 
_ (aa =p) 
_(ate 242) 
; thus (c) admits of but. 
three simple ice. 
The continued product (y= 2? ) otf yy 
ar? + 
ate) ought to reproduce. ©: 3 and accordingly we have 
is 3 Sa a = 
2 
on 
hat | ay | r&-Lps 
ye ata +p ta") + yee 
= = (by having 
[ee serie) | 
atx? 
