Esercizio
$\frac{x^4+2x^3+x^2-1}{x^2+3}$
Soluzione passo-passo
1
Dividere $x^4+2x^3+x^2-1$ per $x^2+3$
$\begin{array}{l}\phantom{\phantom{;}x^{2}+3;}{\phantom{;}x^{2}+2x\phantom{;}-2\phantom{;}\phantom{;}}\\\phantom{;}x^{2}+3\overline{\smash{)}\phantom{;}x^{4}+2x^{3}+x^{2}\phantom{-;x^n}-1\phantom{;}\phantom{;}}\\\phantom{\phantom{;}x^{2}+3;}\underline{-x^{4}\phantom{-;x^n}-3x^{2}\phantom{-;x^n}\phantom{-;x^n}}\\\phantom{-x^{4}-3x^{2};}\phantom{;}2x^{3}-2x^{2}\phantom{-;x^n}-1\phantom{;}\phantom{;}\\\phantom{\phantom{;}x^{2}+3-;x^n;}\underline{-2x^{3}\phantom{-;x^n}-6x\phantom{;}\phantom{-;x^n}}\\\phantom{;-2x^{3}-6x\phantom{;}-;x^n;}-2x^{2}-6x\phantom{;}-1\phantom{;}\phantom{;}\\\phantom{\phantom{;}x^{2}+3-;x^n-;x^n;}\underline{\phantom{;}2x^{2}\phantom{-;x^n}+6\phantom{;}\phantom{;}}\\\phantom{;;\phantom{;}2x^{2}+6\phantom{;}\phantom{;}-;x^n-;x^n;}-6x\phantom{;}+5\phantom{;}\phantom{;}\\\end{array}$
$x^{2}+2x-2+\frac{-6x+5}{x^2+3}$
Risposta finale al problema
$x^{2}+2x-2+\frac{-6x+5}{x^2+3}$