Final answer to the problem
$x^{4}+x^{3}+x^{2}+x+1+\frac{1}{x-1}$
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Step-by-step Solution
How should I solve this problem?
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- Write in simplest form
- Solve by quadratic formula (general formula)
- Find the derivative using the definition
- Simplify
- Find the integral
- Find the derivative
- Factor
- Factor by completing the square
- Find the roots
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1
Divide $x^5$ by $x-1$
$\begin{array}{l}\phantom{\phantom{;}x\phantom{;}-1;}{\phantom{;}x^{4}+x^{3}+x^{2}+x\phantom{;}+1\phantom{;}\phantom{;}}\\\phantom{;}x\phantom{;}-1\overline{\smash{)}\phantom{;}x^{5}\phantom{-;x^n}\phantom{-;x^n}\phantom{-;x^n}\phantom{-;x^n}\phantom{-;x^n}}\\\phantom{\phantom{;}x\phantom{;}-1;}\underline{-x^{5}+x^{4}\phantom{-;x^n}\phantom{-;x^n}\phantom{-;x^n}\phantom{-;x^n}}\\\phantom{-x^{5}+x^{4};}\phantom{;}x^{4}\phantom{-;x^n}\phantom{-;x^n}\phantom{-;x^n}\phantom{-;x^n}\\\phantom{\phantom{;}x\phantom{;}-1-;x^n;}\underline{-x^{4}+x^{3}\phantom{-;x^n}\phantom{-;x^n}\phantom{-;x^n}}\\\phantom{;-x^{4}+x^{3}-;x^n;}\phantom{;}x^{3}\phantom{-;x^n}\phantom{-;x^n}\phantom{-;x^n}\\\phantom{\phantom{;}x\phantom{;}-1-;x^n-;x^n;}\underline{-x^{3}+x^{2}\phantom{-;x^n}\phantom{-;x^n}}\\\phantom{;;-x^{3}+x^{2}-;x^n-;x^n;}\phantom{;}x^{2}\phantom{-;x^n}\phantom{-;x^n}\\\phantom{\phantom{;}x\phantom{;}-1-;x^n-;x^n-;x^n;}\underline{-x^{2}+x\phantom{;}\phantom{-;x^n}}\\\phantom{;;;-x^{2}+x\phantom{;}-;x^n-;x^n-;x^n;}\phantom{;}x\phantom{;}\phantom{-;x^n}\\\phantom{\phantom{;}x\phantom{;}-1-;x^n-;x^n-;x^n-;x^n;}\underline{-x\phantom{;}+1\phantom{;}\phantom{;}}\\\phantom{;;;;-x\phantom{;}+1\phantom{;}\phantom{;}-;x^n-;x^n-;x^n-;x^n;}\phantom{;}1\phantom{;}\phantom{;}\\\end{array}$
2
Resulting polynomial
$x^{4}+x^{3}+x^{2}+x+1+\frac{1}{x-1}$
Final answer to the problem
$x^{4}+x^{3}+x^{2}+x+1+\frac{1}{x-1}$
