132 METHODS OF ABBREVIATION. [CHAP. IX. 



to the final product. Thus, the resulting factors being y + z 

 and , if to their product yt + zt we add the terms x and yw, 



we have 



x + yw + yt + zt, 



as an expression equivalent to the product of the given factors 

 x + y + z and x + yw + t ; equivalent namely in the process of 

 elimination. 



Let us consider, first, the case in which the two factors have 

 a common term a?, and let us represent the factors by the expres- 

 sions x + P, x + Q, supposing P in the one case and Q in the 

 other to be the sum of the positive terms additional to x. 



Now, 



(x + P) (x + Q) = x + xP + xQ + PQ. (1) 



But the process of elimination consists in multiplying certain 

 factors together, and equating the result to 0. Either then the 

 second member of the above equation is to be equated to 0, or it 

 is a factor of some expression which is to be equated to 0. 



If the former alternative be taken, then, by the last Propo- 

 sition, we are permitted to reject the terms xP and#Q, inasmuch 

 as they are positive terms having another term # as a factor. 

 The resulting expression is 



which is what we should obtain by rejecting x from both factors, 

 and adding it to the product of the factors which remain. 



Taking the second alternative, the only mode in which the 

 second member of (1) can affect the final result of elimination 

 must depend upon the number and nature of its constituents, 

 both which elements are unaffected by the rejection of the terms 

 xP and xQ. For that development of x includes all possible con- 

 stituents of which # is a factor. 



Consider finally the case in which one of the factors contains 

 a term, as xy, divisible by a term, x 9 in the other factor. 



Let x + P and xy + Q be the factors. Now 



(x + P) (xy + Q) = xy+ xQ + xyP + PQ. 



But by the reasoning of the last Proposition, the term xyP may be 

 rejected as containing another positive term xy as a factor, whence 

 we have 



