Esercizio
$\frac{\left(b^3-9b^2+1\right)}{\left(b-4\right)}$
Soluzione passo-passo
1
Dividere $b^3-9b^2+1$ per $b-4$
$\begin{array}{l}\phantom{\phantom{;}b\phantom{;}-4;}{\phantom{;}b^{2}-5b\phantom{;}-20\phantom{;}\phantom{;}}\\\phantom{;}b\phantom{;}-4\overline{\smash{)}\phantom{;}b^{3}-9b^{2}\phantom{-;x^n}+1\phantom{;}\phantom{;}}\\\phantom{\phantom{;}b\phantom{;}-4;}\underline{-b^{3}+4b^{2}\phantom{-;x^n}\phantom{-;x^n}}\\\phantom{-b^{3}+4b^{2};}-5b^{2}\phantom{-;x^n}+1\phantom{;}\phantom{;}\\\phantom{\phantom{;}b\phantom{;}-4-;x^n;}\underline{\phantom{;}5b^{2}-20b\phantom{;}\phantom{-;x^n}}\\\phantom{;\phantom{;}5b^{2}-20b\phantom{;}-;x^n;}-20b\phantom{;}+1\phantom{;}\phantom{;}\\\phantom{\phantom{;}b\phantom{;}-4-;x^n-;x^n;}\underline{\phantom{;}20b\phantom{;}-80\phantom{;}\phantom{;}}\\\phantom{;;\phantom{;}20b\phantom{;}-80\phantom{;}\phantom{;}-;x^n-;x^n;}-79\phantom{;}\phantom{;}\\\end{array}$
$b^{2}-5b-20+\frac{-79}{b-4}$
Risposta finale al problema
$b^{2}-5b-20+\frac{-79}{b-4}$