This question has been solved. However, I notice that both the solutions (including the comment) are verifying the fact that $4\alpha^3-3\alpha$ is the other root if $\alpha$ is one.
What if I rephrase the question as:-
If $\alpha$ is a root of the equation$4x^2 + 2x – 1 = 0$, express the other root in the form of 4h + 3k where h should be in terms of $\alpha^3$ and k should be in terms of $\alpha$.
Note that if $\alpha$ is a root of $ax^2+bx+c$, then
$$\alpha^2+\tfrac ba \alpha+\tfrac ca=0$$
Thus,
\begin{align} \alpha^3&=\alpha\cdot\alpha^2\\ &=\alpha\cdot(-\tfrac ba \alpha - \tfrac ca)\\ &=-\tfrac ba \alpha^2 - \tfrac ca\alpha\\ &=-\tfrac ba\cdot(-\tfrac ba \alpha - \tfrac ca) - \tfrac ca\alpha\\ &=\tfrac {b^2}{a^2}\alpha + \tfrac {bc}{a^2} - \tfrac ca\alpha\\ &=\tfrac {b^2-ac}{a^2}\alpha + \tfrac {bc}{a^2} \end{align}
So we can find the other root $\beta=-\frac ba-\alpha$ by seeing $$\frac ba=\frac{a}{c}\left(\alpha^3-\tfrac {b^2-ac}{a^2}\alpha\right)$$
and so $$\beta=-\frac{a}{c}\left(\alpha^3-\tfrac {b^2-ac}{a^2}\alpha\right)-\alpha$$ which in the requested form is $$\beta=\tfrac {ab^2-2a^2c}{a^2c}\alpha-\tfrac ac\alpha^3$$
Let $\beta$ the other root. By Vieta's relations, $\alpha+\beta=-\dfrac12$. Now rewrite the equation satisfied by $\alpha$ as $$-\frac12=-2\alpha^2-\alpha,\enspace\text{so that}\quad\beta=-2\alpha^2-2\alpha.$$ Also , multiplying the equation by $\alpha$, you obtain $\;4\alpha^3+2\alpha^2-\alpha=0$, i.e. $\;-2\alpha^2=4\alpha^3-\alpha$. Replace $-2\alpha^2$ in the expression of $\beta$: $$\beta=-2\alpha^2-2\alpha=4\alpha^3-3\alpha.$$
