od Daniel » Četvrtak, 09. Maj 2013, 11:26
[inlmath]P\left(x\right)=Q_1\left(x\right)\left(x-1\right)+1\\
P\left(x\right)=Q_2\left(x\right)\left(x^2+1\right)+x+2\\
P\left(x\right)=Q_3\left(x\right)\left(x-1\right)\left(x^2+1\right)+R\left(x\right)[/inlmath]
Ostatak [inlmath]R\left(x\right)[/inlmath] je neki polinom koji je manjeg stepena od polinoma kojim se deli, u ovom slučaju [inlmath]\left(x-1\right)\left(x^2+1\right)[/inlmath]. Pošto je [inlmath]\left(x-1\right)\left(x^2+1\right)[/inlmath] polinom [inlmath]3.[/inlmath] stepena, [inlmath]R\left(x\right)[/inlmath] je u opštem slučaju polinom [inlmath]2.[/inlmath] stepena:
[inlmath]R\left(x\right)=ax^2+bx+c[/inlmath]
Koeficijente [inlmath]a[/inlmath], [inlmath]b[/inlmath] i [inlmath]c[/inlmath] određujemo koristeći Bezuovu teoremu:
[inlmath]P\left(x\right)=Q_1\left(x\right)\left(x-1\right)+R\left(1\right)\\
\Rightarrow\quad R\left(1\right)=a+b+c=1\qquad\left(1\right)[/inlmath]
[inlmath]P\left(x\right)=Q_2\left(x\right)\left(x^2+1\right)+x+2\quad\Rightarrow\quad P\left(x\right)=Q_2\left(x\right)\left(x+i\right)\left(x-i\right)+x+2\\
\qquad P\left(x\right)=Q_2'\left(x\right)\left(x+i\right)+R\left(-i\right)\quad\Rightarrow\quad R\left(-i\right)=-a-ib+c=-i+2\qquad\left(2\right)\\
\qquad P\left(x\right)=Q_2''\left(x\right)\left(x-i\right)+R\left(i\right)\quad\Rightarrow\quad R\left(i\right)=-a+ib+c=i+2\qquad\left(3\right)[/inlmath]
[inlmath]\left(1\right)[/inlmath], [inlmath]\left(2\right)[/inlmath] i [inlmath]\left(3\right)[/inlmath] čine sistem od [inlmath]3[/inlmath] jednačine s [inlmath]3[/inlmath] nepoznate, čija su rešenja
[inlmath]a=-1\\
b=1\\
c=1[/inlmath]
pa je ostatak
[inlmath]R\left(x\right)=-x^2+x+1[/inlmath]
I do not fear death. I had been dead for billions and billions of years before I was born, and had not suffered the slightest inconvenience from it. – Mark Twain