9. Compute the sum of all real numbers $x$ such that
$$
2 x^{6}-3 x^{5}+3 x^{4}+x^{3}-3 x^{2}+3 x-1=0
$$
Answer: $\quad-\frac{1}{2}$
Solution: The carefully worded problem statement suggests that repeated roots might be involved (not to be double counted), as well as complex roots (not to be counted). Let $P(x)=2 x^{6}-3 x^{5}+3 x^{4}+x^{3}-3 x^{2}+3 x-1$. Now, $a$ is a double root of the polynomial $P(x)$ if and only if $P(a)=P^{\prime}(a)=0$. Hence, we consider the system
$$
\begin{aligned}
P(a)=2 a^{6}-3 a^{5}+3 a^{3}+a^{3}-3 a^{2}+3 a-1 & =0 \\
P^{\prime}(a)=12 a^{5}-15 a^{4}+12 a^{3}+3 a^{2}-6 a+3 & =0 \\
\Longrightarrow 3 a^{4}+8 a^{3}-15 a^{2}+18 a-7 & =0 \\
37 a^{3}-57 a^{2}+57 a-20 & =0 \\
a^{2}-a+1 & =0
\end{aligned}
$$

We have used polynomial long division to deduce that any double root must be a root of $a^{2}-a+1$ ! With this information, we can see that $P(x)=\left(x^{2}-x+1\right)^{2}\left(2 x^{2}+x-1\right)$. The real roots are easily computed via the quadratic formula, leading to an answer of $-\frac{1}{2}$. In fact the repeated roots were complex.