Working through the problem of solving the cubic, the question arises, suppose I remove the offset? That is, I have
which is the same same as
...shifted to the right or left by some unknown number of units.
Writing , the first root is 0, and the last two roots are given by the quadratic equation.
If we write g(x) = f(x-t), is it possible to solve for the mystery offset t and the new mystery coefficients?

(Alas, the answer is no.)
I have not offered a proof, but I am satisfied that solving for t or the new coefficients are simply restatements of the problem: find the zeros of a cubic polynomial. One good way to approach proofs is to hunt for an absolute assertion ... which can then be falsified by a single counterexample. Here, I might ask, What form must f(x) have if g(x) = f(x-t) simplifies to solving a quadratic? (More: https://www.dpmms.cam.ac.uk/~wtg10/cubic.html)
But what is this business with Newton? By differentiating f(x), we can quickly obtain a description of the curve: its local minima/maxima and inflection point. Using this information, we can set up a simple root-finding system for order 3 polynomials...
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Roots of the Cubic Equation
Solution 1 (Iterative): http://tube.geogebra.org/material/show/id/143036
Solution 2 (Algebraic): http://tube.geogebra.org/material/show/id/143424