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