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?

$\, 0=(4\alpha^2\!-\!1)^2\!-\!(2\alpha)^2\! =\, \color{#0a0}4\alpha(4\alpha^3\!-3\alpha)\!+\!\color{#c00}1\,\Rightarrow\, \alpha(4\alpha^3\!-\!3\alpha) \,=\, {-}\dfrac{\color{#c00}1}{\color{#0a0}4}\,=\, $ product of roots
