A fixed point iteration formula for $2x^3-4x^2+x+1=0$ can be derived: $$x_{r+1}=4x_r^2-2x_r^3-1$$
Starting with the initial value of 2:
$$x_0=2$$ $$x_1=-1$$ $$x_2=5$$ $$x_3=-151$$ $$x_4=6977105$$
Instead of converging to the root the sequence diverges.
However, the formula seems to work for 1.5.
Why does the sequence converge for 1.5 but diverge for 2?
On
If $f(x, h(x)) = 0 $ then $x_{r+1} = g(x_r)$ will be the iterative equation iff the $|g'(x)| \in (0, 1) \forall x\in \text{domain}$ as commented by @Peter which in the above function for $2x^3−4x^2+x+1=0$ your iterative equation $g(x) = 4x^2_r−2x^3_r−1$ which is giving you answer for $x = 1.5$ but if I choose $f(x) = 2x^3 - 4x^2 + x + 1 = 0$ so if I add x to both sides and $x = 2x^3 - 4x^2 + 2x + 1 = g(x)$ this also diverges for $x = 1.5 $
so basically it's the choice of your function that decides the fate of $x$ whether it'll be destined to the root or will be thrown away from the roots.
Observe from graph that if you will start from $x_0=2$, $4x^2-2x^3-1|_{x=2}<0$ and for all other iteration you can observe that you will go down and down on the graph of $4x^2-2x^3-1$. Hence iteration scheme will diverge.