Exercise
$\frac{\left(2x^3-8x^2-19x+49\right)}{x-5}$
Step-by-step Solution
1
Divide $2x^3-8x^2-19x+49$ by $x-5$
$\begin{array}{l}\phantom{\phantom{;}x\phantom{;}-5;}{\phantom{;}2x^{2}+2x\phantom{;}-9\phantom{;}\phantom{;}}\\\phantom{;}x\phantom{;}-5\overline{\smash{)}\phantom{;}2x^{3}-8x^{2}-19x\phantom{;}+49\phantom{;}\phantom{;}}\\\phantom{\phantom{;}x\phantom{;}-5;}\underline{-2x^{3}+10x^{2}\phantom{-;x^n}\phantom{-;x^n}}\\\phantom{-2x^{3}+10x^{2};}\phantom{;}2x^{2}-19x\phantom{;}+49\phantom{;}\phantom{;}\\\phantom{\phantom{;}x\phantom{;}-5-;x^n;}\underline{-2x^{2}+10x\phantom{;}\phantom{-;x^n}}\\\phantom{;-2x^{2}+10x\phantom{;}-;x^n;}-9x\phantom{;}+49\phantom{;}\phantom{;}\\\phantom{\phantom{;}x\phantom{;}-5-;x^n-;x^n;}\underline{\phantom{;}9x\phantom{;}-45\phantom{;}\phantom{;}}\\\phantom{;;\phantom{;}9x\phantom{;}-45\phantom{;}\phantom{;}-;x^n-;x^n;}\phantom{;}4\phantom{;}\phantom{;}\\\end{array}$
$2x^{2}+2x-9+\frac{4}{x-5}$
Final answer to the exercise
$2x^{2}+2x-9+\frac{4}{x-5}$