Exercice
$\left(x^{3}-9x^{2}+21x+4\right):\left(x-5\right)$
Solution étape par étape
1
Diviser $x^3-9x^2+21x+4$ par $x-5$
$\begin{array}{l}\phantom{\phantom{;}x\phantom{;}-5;}{\phantom{;}x^{2}-4x\phantom{;}+1\phantom{;}\phantom{;}}\\\phantom{;}x\phantom{;}-5\overline{\smash{)}\phantom{;}x^{3}-9x^{2}+21x\phantom{;}+4\phantom{;}\phantom{;}}\\\phantom{\phantom{;}x\phantom{;}-5;}\underline{-x^{3}+5x^{2}\phantom{-;x^n}\phantom{-;x^n}}\\\phantom{-x^{3}+5x^{2};}-4x^{2}+21x\phantom{;}+4\phantom{;}\phantom{;}\\\phantom{\phantom{;}x\phantom{;}-5-;x^n;}\underline{\phantom{;}4x^{2}-20x\phantom{;}\phantom{-;x^n}}\\\phantom{;\phantom{;}4x^{2}-20x\phantom{;}-;x^n;}\phantom{;}x\phantom{;}+4\phantom{;}\phantom{;}\\\phantom{\phantom{;}x\phantom{;}-5-;x^n-;x^n;}\underline{-x\phantom{;}+5\phantom{;}\phantom{;}}\\\phantom{;;-x\phantom{;}+5\phantom{;}\phantom{;}-;x^n-;x^n;}\phantom{;}9\phantom{;}\phantom{;}\\\end{array}$
$x^{2}-4x+1+\frac{9}{x-5}$
Réponse finale au problème
$x^{2}-4x+1+\frac{9}{x-5}$