AND IMAGINARY ROOTS OR EQUATIONS. 
591 
when g, )] are both positive, and 
S<-2( i > + l)f( 16bis ), i. e. < r<(p+l) 2 -( i? +l) 3 -2(_p + l)f 
when g, ri are both negative. 
If now we substitute (p -\-l) 2 -\-{p -\-lf — 2(j?+l)* for <r in A, I say that the resulting 
value will be positive whatever positive value be given to (^ + 1); in fact, if we write 
p—v 2 — 1, and make <r=—v 4 -\-2v 5 —v 6 , so that A becomes a function of the twelfth degree 
in v, this function is what the discriminant of the equation in x, y becomes when we 
have s=)]=v; but in the antecedent Lemma it has been shown that this discriminant is 
only negative when the two equal quantities g or q, or, which is the same thing, when v 
lies between 1 and ^ ; hence A is positive when v is negative, and consequently when 
Thus A, a quadratic function in <r, and its discriminant are respectively + and — for 
this value of a, as well as for <r= — oo . Hence no real root of a lies between such value 
of <r and — qq , and consequently A must be always positive when g and n are both negative. 
Hence, if A is negative, we must have l>g?j >ro 5 s>0 ; n> 0. But our criteria give 
g 4 — g^ 2 >0, v\ 4 — ye 2 >0, 
which, when g>0, q>0, imply s 3 >tf, > g 2 , and consequently g??> 1, which is in con- 
tradiction to the inequality 1 > s/j. Hence when these criteria are satisfied the determi- 
nant is necessarily positive , and all the roots are imaginary, which completes the proof 
of Newton’s rule for equations of the fifth degree. 
(10) It follows as a corollary to the Lemma employed in the preceding investigation, 
that if in Awe write <r=— (v 2 — i/ 3 ) 2 andy)=^ 2 — 1, and distinguish this particular value by 
the symbol (A), then (A) ought to break up into the product of odd powers of v— 1, 
of some even power of and of a factor incapable of changing its sign, and remain- 
ing always positive. This may be easily verified ; for dividing (A) by (v — l) 4 , we obtain 
1296»' 8 (648(v+1) 2 -1-24(v 2 -1Xj'+1) 2 )/+96(i/ 2 -1) 2 ( v +1) 4 +176(^ 2 -1X«'+1) 4 +81(v+ 1) 4 ; 
and collecting the terms 1296v 8 — 648^ 6 (>'+1) 2 +81 (j'+ 1) 4 whose sum contains the factor 
( v — 1), we have 
(^)5 = 64:S(v 7 -\-V 6 -\-V 5 -\-V 4 -\-V 3 -\-V 2 + V-\-l) 
— 1296( ^q-^+Z+^+^+vH-l) 
— 648( > 5 +i> 4 +v , +v 2 +j' + 1) 
T- 81( v 3 -|-5v 2 +lL+15) 
- 24(»' 7 + 3 j > 6 + 3* 5 - f v 4 ) 
+ 96(v 7 H- 5v 6 + 9^ 5 -f- 5^ 4 — 5v 3 - 9v 2 -5v— 1) 
+ 176( 10y 3 -t-10i/ 2 +5v+l) 
= 72(V— 24(V i — 328^ 5 + 40v 4 + 65^ + 5^ 2 -5i/- 1. 
Hence 
( A )= ( v ~ l) 5 (2v - 1) 3 { 90r 4 + 105* 3 + 49* 2 + 1L + 1 } 
=(v-l) 5 (2y-l) 3 (3v+l) 2 {10^ 2 +5t'+l}; 
(i6 bis ) jj. ^ 0 £ course understood that (p+1)^ is to be taken positive. 
