\section*{Problem 5 - 041115}

Calculate all common solutions of the two equations

\[
\begin{array}{r}
3 x^{4}+13 x^{3}+20 x^{2}+17 x+7=0 \\
3 x^{4}+x^{3}-8 x^{2}+11 x-7=0
\end{array}
\]

(No approximation methods should be used.)
}

We define polynomials

\[
\begin{aligned}
& p_{1}(x):=3 x^{4}+13 x^{3}+20 x^{2}+17^{x}+7 \\
& p_{2}(x):=3 x^{4}+x^{3}-8 x^{2}+11 x-7
\end{aligned}
\]

Let \(x\) be a solution to the given equations, \(p_{1}(x)=p_{2}(x)=0\). Then we have

\[
\begin{aligned}
0=p_{1}(x)-p_{2}(x)=12 x^{3}+28 x^{2}+6 x+14 & =: p_{3}(x) \\
0=4 p_{2}(x)-(x-2) p_{3}(x)=18 x^{2}+42 x & =: p_{4}(x) \\
0=3 p_{3}(x)-2 x p_{4}(x)=18 x+42 & =: p_{5}(x)
\end{aligned}
\]

Conversely, since \(p_{4}(x)=x p_{5}(x)\), it follows from \(p_{5}(x)=0\) that \(p_{3}(x)=0\) as well (since \(p_{3}(x)=\frac{1}{3}\left(p_{5}(x)+2 x p_{4}(x)\right)\)), and similarly, that \(p_{1}(x)\) and \(p_{2}(x)\) are also zero.

The sought common solutions to the equations are thus precisely the roots of \(p_{5}\), i.e., \(\left\{-\frac{7}{3}\right\}\).