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
 A: Hint: Try working backwards. Start with:
$$
2\left(x + \dfrac{1}{x}\right)^2 + \left(x + \dfrac{1}{x}\right) - 10 = 0
$$
then try to obtain the original equation by expanding then clearing the denominators.
A: Like this,
as the equation is Reciprocal Equation of the First type with $x\ne0,$
divide either sides by $\displaystyle x^{\frac42}=x^2$ to reduce the degree of the equation by half
$$2x^2+x-6+\frac1x+\frac2{x^2}=0$$
$$\implies 2\left(x^2+\frac1{x^2}\right)+\left(x+\frac1x\right)-6=0$$
$$\implies 2\left\{\left(x+\frac1x\right)^2-2\right\}+\left(x+\frac1x\right)-6=0$$
Reference : Reciprocal Equation is explained here:
Chapter XI of Higher Algebra,Barnard & Child and
Article $568−570$ of Higher algebra, Hall & Knight
A: $$p=x+x^{-1}$$
$$p^2=x^2+2+x^{-2}$$
$$2x^4+x^3-6x^2+x+2=0$$
$$2x^2+x-6+x^{-1}+2x^{-2}=0$$
$$2(x^2+2+x^{-2})+(x+x^{-1})-10=0$$
$$2p^2+p-10=0$$
A: $$
\begin{align}
2x^4+x^3-6x^2+x+2 & = x^2\Big(2x^2 + x - 6 + \frac1x + \frac{2}{x^2}\Big) \\[12pt]
& = x^2 \Big(2\left(x+\frac1x\right)^2 + \left(x+\frac1x\right) - 10 \Big) \\[12pt]
& = x^2(2p^2 +p-10).
\end{align}
$$
This equals $0$ only if $x=0$ or the second factor equals $0$.  But clearly $x=0$ is not one of the solutions.
