<s>[INST] Quadratic substitution question: applying substitution $p=x+\frac1x$ to $2x^4+x^3-6x^2+x+2=0$ By using the substitution $p=x+\frac{1}{x}$, show that the equation $$2x^4+x^3-6x^2+x+2=0$$
reduces to $2p^2+p-10=0$.
I can't think of anything that produces a useful result, I tried writing p as $p=\frac{x^2+1}{x}$ and finding areas to substitute but have come with no progress. Could someone offer a slight hint on how to proceed?
Thanks
 [/INST] $$
\begin{align}
2x^4+x^3-6x^2+x+2 & = x^2\Big(2x^2 + x - 6 + \frac1x + \frac{2}{x^2}\Big) \\[12pt]
& = x^2 \Big(2\left(x+\frac1x\right)^2 + \left(x+\frac1x\right) - 10 \Big) \\[12pt]
& = x^2(2p^2 +p-10).
\end{align}
$$
This equals $0$ only if $x=0$ or the second factor equals $0$.  But clearly $x=0$ is not one of the solutions.
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