## Precalculus (6th Edition) Blitzer

By performing long division, we get: x+1\overset{{{x}^{4}}-{{x}^{3}}+{{x}^{2}}-x+1}{\overline{\left){\begin{align} & \,\,\,\,\,\,\,\,\,\,{{x}^{5}}+1 \\ & \frac{\begin{align} & +{{x}^{5}}+{{x}^{4}} \\ & -\,\,\,\,\,\,- \\ \end{align}}{\,\,\,\,\,\,\,\,\,\,\,\,0-{{x}^{4}}+1} \\ & \frac{\begin{align} & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-{{x}^{4}}-{{x}^{3}} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\,\,\,\,\,\,\,+ \\ \end{align}}{\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{x}^{3}}+1} \\ & \frac{\begin{align} & \,\,\,\,\,\,\,\,\,+{{x}^{3}}+{{x}^{2}} \\ & \,\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,- \\ \end{align}}{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0-{{x}^{2}}+1} \\ & \frac{\begin{align} & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-{{x}^{2}}-x \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\,\,\,\,\,\,\,\,+ \\ \end{align}}{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0+x+1} \\ & \frac{\begin{align} & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+x+1 \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-\,\,\,\,- \\ \end{align}}{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0} \\ \end{align}}\right.}} Thus, it is observed that the quotient is ${{x}^{4}}-{{x}^{3}}+{{x}^{2}}-x+1$ but not ${{x}^{4}}+1$.