Esercizio
$\frac{2x^4-8x^3+4x^2-11x-1}{x^2-1}$
Soluzione passo-passo
1
Dividere $2x^4-8x^3+4x^2-11x-1$ per $x^2-1$
$\begin{array}{l}\phantom{\phantom{;}x^{2}-1;}{\phantom{;}2x^{2}-8x\phantom{;}+6\phantom{;}\phantom{;}}\\\phantom{;}x^{2}-1\overline{\smash{)}\phantom{;}2x^{4}-8x^{3}+4x^{2}-11x\phantom{;}-1\phantom{;}\phantom{;}}\\\phantom{\phantom{;}x^{2}-1;}\underline{-2x^{4}\phantom{-;x^n}+2x^{2}\phantom{-;x^n}\phantom{-;x^n}}\\\phantom{-2x^{4}+2x^{2};}-8x^{3}+6x^{2}-11x\phantom{;}-1\phantom{;}\phantom{;}\\\phantom{\phantom{;}x^{2}-1-;x^n;}\underline{\phantom{;}8x^{3}\phantom{-;x^n}-8x\phantom{;}\phantom{-;x^n}}\\\phantom{;\phantom{;}8x^{3}-8x\phantom{;}-;x^n;}\phantom{;}6x^{2}-19x\phantom{;}-1\phantom{;}\phantom{;}\\\phantom{\phantom{;}x^{2}-1-;x^n-;x^n;}\underline{-6x^{2}\phantom{-;x^n}+6\phantom{;}\phantom{;}}\\\phantom{;;-6x^{2}+6\phantom{;}\phantom{;}-;x^n-;x^n;}-19x\phantom{;}+5\phantom{;}\phantom{;}\\\end{array}$
$2x^{2}-8x+6+\frac{-19x+5}{x^2-1}$
Risposta finale al problema
$2x^{2}-8x+6+\frac{-19x+5}{x^2-1}$