Solving a polynomial with modular arithmetic.

Question

$x^5+ax^4+bx^3+cx^2+dx+e=0$

Assume that $a,b,c,d,e$ are not arbitrary and that they are known.

I was wondering if it were possible to reduce or 'simplify' this using some modular arithmetic.

It would become something along the lines of $$(1+zm)x^5+(a+zn)x^4+(b+zo)x^3+(c+zp)x^2+(d+zq)x+e+zr\equiv0\pmod{z}$$

And I could choose $m,n,o,p,q,r,z$ to transform the given polynomial into another polynomial that I had already solved.

The solution to the above could then be used to find the exact solutions through trial and error, however, the exact solutions could first be approximated, with the exact solutions found afterwards as a subset of the solutions produced by the above polynomial?

Or perhaps we could attempt to solve $f(x)=0$ with $f(x)\equiv g(x)\pmod{(x-r)}$ and attempt to have $g(x)=0$. (I imagine this wouldn't work, but who knows?)

Is it possible to do? Can it make a seemingly unfactorable polynomial factorable?

@BenS. Um, I don't think I quite know? I can say I'll accept solutions $x\in\mathbb{C}$, but I haven't learned much field theory and group theory or related subjects. — Simply Beautiful Art, Mar 15 '16 at 00:54
For example, the integers modulo $p$ form a field. Sometimes polynomials can be factored over some of these fields but not over $\mathbb{R}$ or $\mathbb{Q}$, but every polynomial can be factored completely over $\mathbb{C}$. For example, $x^2+1$ can be factored as $(x+1)^2$ over $\mathbb{Z}/2\mathbb{Z}$, but not factored in $\mathbb{R}$. — Ben Sheller, Mar 15 '16 at 01:01
@BenS. That's not true. Not all polynomials can be factored completely over $\mathbb{C}$, that's why we have Galois Theory. If I am correct. — Simply Beautiful Art, Mar 15 '16 at 01:05
See the second sentence here: https://en.wikipedia.org/wiki/Fundamental_theorem_of_algebra — Ben Sheller, Mar 15 '16 at 01:07
Galois theory has more to do with the symmetries of the roots. — Ben Sheller, Mar 15 '16 at 01:08
Ahhh...yes, that is true. Not all have a closed form involving just addition, subtraction, multiplication, and $n$-th roots. But over $\mathbb{C}$ every polynomial can nonetheless be factored completely, though figuring out exactly what the factors are is the difficult part. — Ben Sheller, Mar 15 '16 at 01:09
@BenS. Well, my question is whether or not using modular arithmetic can help. — Simply Beautiful Art, Mar 15 '16 at 01:09
Yes, we can factor polynomials over $,\Bbb Z,$ by using modular factorizations, Hensel and CRT lifting. This is a standard technique in computer algebra, e.g. see the citations here. — Bill Dubuque, Aug 28 '17 at 15:30

Solving a polynomial with modular arithmetic.

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