Übung
$\frac{15x^3-31x^2+4}{x-1}$
Schritt-für-Schritt-Lösung
1
Teilen Sie $15x^3-31x^2+4$ durch $x-1$
$\begin{array}{l}\phantom{\phantom{;}x\phantom{;}-1;}{\phantom{;}15x^{2}-16x\phantom{;}-16\phantom{;}\phantom{;}}\\\phantom{;}x\phantom{;}-1\overline{\smash{)}\phantom{;}15x^{3}-31x^{2}\phantom{-;x^n}+4\phantom{;}\phantom{;}}\\\phantom{\phantom{;}x\phantom{;}-1;}\underline{-15x^{3}+15x^{2}\phantom{-;x^n}\phantom{-;x^n}}\\\phantom{-15x^{3}+15x^{2};}-16x^{2}\phantom{-;x^n}+4\phantom{;}\phantom{;}\\\phantom{\phantom{;}x\phantom{;}-1-;x^n;}\underline{\phantom{;}16x^{2}-16x\phantom{;}\phantom{-;x^n}}\\\phantom{;\phantom{;}16x^{2}-16x\phantom{;}-;x^n;}-16x\phantom{;}+4\phantom{;}\phantom{;}\\\phantom{\phantom{;}x\phantom{;}-1-;x^n-;x^n;}\underline{\phantom{;}16x\phantom{;}-16\phantom{;}\phantom{;}}\\\phantom{;;\phantom{;}16x\phantom{;}-16\phantom{;}\phantom{;}-;x^n-;x^n;}-12\phantom{;}\phantom{;}\\\end{array}$
$15x^{2}-16x-16+\frac{-12}{x-1}$
Endgültige Antwort auf das Problem
$15x^{2}-16x-16+\frac{-12}{x-1}$