Answer
No, the given statement does not make sense.
Work Step by Step
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 ${{x}^{4}}+1$.
