Problem of quadratic equation If $\alpha$ be a root of $ 4x^2 +2x -1 = 0 $ , prove that the other root is $4\alpha^3  - 3\alpha$ . I have tried to do it but of no success.[$4\alpha^3 -2\alpha$ = $\dfrac {-1}{2}$  and $4\alpha^4 - 3\alpha^2$ = $\dfrac {-1}{4}$ ] .How to prove it?

Someone will probably show you a smart way to do this, but ...I guess you could just use that $\alpha^2 = 1 - 2\alpha$ and then
$$\begin{align}
4(4\alpha^3 - 3\alpha)^2 + &2(4\alpha^3 - 3\alpha) - 1 \\ &= 4(4\alpha^6 + 9\alpha^2 - 24\alpha^4) + 8\alpha^3 - 6\alpha - 1 \\
&= 16\alpha^6 + 36\alpha^2 - 96\alpha^4 + 8\alpha^3 - 6\alpha -1\\
&= 16(1-2\alpha)^3 + 36(1 - 2\alpha) - 96(1-2\alpha)^2 + 8\alpha(1-2\alpha) - 6\alpha -1 \\
&= \dots
\end{align}
$$
