1.Teorema impartirii cu rest
Fiind date doua polinoame oarecare cu coeficienti complecsi f si g cu g<>0, atunci exista doua plinoame cu coeficienti complecsi q si r a .i.
f = gq+r unde grad r < grad g (1)
In plus polinoamele q si r sunt unice satisfacand proprietatea (1)
f = deimpartit
g impartitor
q cat
r = rest
Demonstratie
1.Existenta
f = an Xn + an-1 X n-1 +...+a1 X+a0 C[x]
g= bm Xm +bm-1 X m-1 +...+b1 X +b0 C[x]
grad f = n
grad g = m
1.n < m
q
f=0*g+f
2.n >= m
an / bm
an Xn / bm Xm
q1= (an / bm) * X n-m
f (an / bm) * X n-m ) *g + f1 (1)
grad f1 = n1 <grad f = n
an Xn + an-1 X n-1 +...+a1 X+a0/ : bmXm
f1= an1 Xn1 + an1-1 X n1-1 +...+a11 X+a01
Daca gr. f1 =n1
i) gr f1 < gr g STOP
ii) daca gr f1 >= gr g
f1 (an1 / bm) * X n1-m ) *g + f2 (2)
gr f2=n2 <n1 < n
i) gr n2<m STOP
ii) gr n2>=m
f2 (an2 / bm) * X n2-m ) *g + f3 (3)
p pasi
p
fp (anp / bm) * X np-m ) *g + fp+1 (p+1)
gr f p+1<m STOP
f1 f ( (an / bm) * X n-m ) *g /
f2 f1 (an1 / bm) * X n1-m ) *g /
f3 f2- ( (an2 / bm) * X n2-m ) *g / +
/
f p+1 f p (anp / bm) * X np-m) *g /
f p+1 = f - g ((an / bm) * X n-m + (an1 / bm) * X n1-m +...+
(anp / bm) * X np-m )
f = fp +g ((an / bm) * X n-m + (an1 / bm) * X n1-m +...+
(anp / bm) * X np-m )
q = ((an / bm) * X n-m + (an1 / bm) * X n1-m +...+
(anp / bm) * X np-m )
r = f p+1
Gr f p+1< m
