252 Prof. Sylvester on a netso and more 



+ {x-e^) X ( g; ^3 ... g„ ). 



or simply = 2 (x — «) . ( gg e^ ... e„ ^ 



Hence by making 



X % ( gg ^3 ... e,^ — 2 {<?! X ( gg g3 ... gn } } = 0, 



we have an equation for finding the equal roots gj g^. 

 Again, it is easily seen upon the same hypothesis, that 



2 {{x-Cc^) (j:— gg) (x—e^) ... {x—e„) x ( e^e^ ... g„ )} 

 = 2(a;-g2) (^— gy) ... (x-g„)x(g2_£3_iii^n)- 



Hence, to form the equation having the same roots as 

 Ix — a) <pXf we have only to make 



j;''-'^ (fT^TT^) -x"-^ 2 {^2 + ^3 + ..rg7x (£i£3jiii£!i)} 

 ± S {g^ga... g„x( g2g3»- g» )] = 0. _ 



Suppose now in general that we have (r) perfect square 

 factors, so that/jr = ^x. [x—a^)- {x — ac^Y ... {x—ar)'^. 



To form the equation C {x—ai)'{x—a^) ... (x— «»•) = 0, 

 we have only to make 



S {x—e^ .x-eo ... ar — g;.x(^ g^+i gr+2 ••• g« )} = 

 And to obtain T><px x {x-a{) {x—Oc,) ... (jr-«,.) = 



we must make 



2 {(.r-gr+i) (^-<^r+2) ... {x-e„)x (~er+ier+2"- en )} = 0. 



The theory of perfect square factors is not yet complete 

 until it has been shown how to obtain constructively ^ x, 

 and as analogy suggests the complementary part B'^x—a^y 

 , (x—aS- ... {x—Or}^ each in its lowest terms. To effect the 

 latter it miwht be said that it is only necessary to take the square 

 of C (x-fl,) (x—a^) ... (x-«,). It is true the polynomial 

 so formed would contain every pair of equal factors, but not 

 in the lowest terms as regards the coefficients (as we shall 

 presently show). 



To solve this last part of the problem, let it be agreed that 

 two rows of letters inclosed in a parenthesis shall indicate the 

 product of the squares of the differences got by subtracting 



each in the row from each in the other, so that ( ^ j 



{h-cf.ib-df. 



