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?

If $\alpha$ is a root then
$$
4 \alpha^2 + 2 \alpha - 1 = 0.
$$
Therefore
$$
\begin{eqnarray}
4 \Big( -\tfrac{1}{2} - \alpha \Big)^2 + 2 \Big( -\tfrac{1}{2} - \alpha \Big) - 1 &=&
4 \Big( \tfrac{1}{4} + \alpha + \alpha^2 \Big)
 - 2 \Big( \tfrac{1}{2} + \alpha \Big) - 1\\
&=& 4 \alpha^2 + 2 \alpha - 1\\
&=& 0.
\end{eqnarray}
$$
So if $\alpha$ is a root, then $- \tfrac{1}{2} - \alpha$ is also a root, but
$$
\begin{eqnarray}
4 \alpha^3 - 3 \alpha &=& \alpha \Big( 4\alpha^2 - 3 \Big)\\
&=& \alpha \Big( 1 - 2 \alpha - 3 \Big)\\
&=& - \Big( 2 \alpha^2 + 2 \alpha \Big)\\
&=& - \Big( \frac{1}{2} - \alpha + 2 \alpha \Big)\\
&=& - \frac{1}{2} - \alpha,
\end{eqnarray}
$$
So $ 4 \alpha^3 - 3\alpha$ is also a root.
