IV. ( a b) X (e d) = ac be ad-f bd. 



Sit enim c d = 5, erit (a b) X (c d) = (a b) X * = &a *b 

 (29,11.) = (c d) X a (c d) X b} atqui(c d) X a = ac ad, et 

 (c d) X b = bc bd;ergo (a b) X (c d) = (ac ad; (be 

 bd) = ac ad be + bd ( 16 , 2'.). 



His praemissis , sit nunc propositum sequens 



PROBLEMA I. 



Polynomium quodcunque per monomium multiplicare , 

 v. g. a b -f- c d + f" per k. 



RESOLUTIO. 



30. Si fiat a -J- c + f = p et b + d = n , polynomium a b + c d 

 -f f ad formam p n reducitur : igitur (a b + c d + f ) X k = . . . 

 (p _ n) X k = p k n k : sed 



pk = (a + c + f) X k = ak + ck + fk 

 nk = (b + d) X k = bk + dk, (29, I.) : 

 Ergo pk nk = ak + ck + fk bk dk , sive 

 (a _ b 4* c - d + f ) X k = ak - bk + ck - dk + fk (E) 



PROBLEMA II. 



Polynomium quodcumque per aliud polynomium multiplicare , 

 v. g. f g + b. k per l + m n r + s - 



RESOLXJTIO. 



31. Fiatf + h = a,g + k = b,l + m + s = cetn-f-r = d, eritf g 

 + h k = a-~betl+m n r + s = c d, et(f g + h k) 

 x (1 + m_n r + s) = (a b) X (c d) = ac be ad + bd 

 (29, IV) : jam veto 



