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i'm trying to see if a quartic equation I obtained can be factored into simpler forms, such as the product of two quadratics. The problem is that their coefficients are some complex expressions in terms of several parameters. I wonder how to do this factorization. Or is there any better way to see how the solutions look at? Maybe looking at the formulas for the closed-form solutions of a quadratic equation?

Here is a simplified version of the equation:

a1*t^4 + a2*t^3 + a3*t^2 + a4*t + a5

One of the coefficients is shown below to demonstrate their complexity: enter image description here

The expression is attached below:

1/4 (n . Subscript[f, 1])^2 - 
 3/4 (Subscript[v, 3] . Subscript[f, 1])^2 + 
 t (n . Subscript[f, 1] Subscript[v, 2] . Subscript[f, 1] + 
    n . Subscript[f, 1] Subscript[v, 2] . Subscript[v, 3] Subscript[v,
       3] . Subscript[f, 1]) + 
 t^3 (-n . Subscript[f, 1] Subscript[v, 2] . Subscript[f, 1] + 
    2 n . Subscript[f, 1] Subscript[v, 2] . Subscript[f, 
      1] (Subscript[v, 2] . Subscript[v, 3])^2 + 
    n . Subscript[f, 1] Subscript[v, 2] . Subscript[v, 3] Subscript[v,
       3] . Subscript[f, 1]) + 
 t^2 (-(1/2) (n . Subscript[f, 1])^2 + (Subscript[v, 2] . Subscript[f,
       1])^2 - 
    Subscript[v, 2] . Subscript[f, 1] Subscript[v, 2] . Subscript[v, 
      3] Subscript[v, 3] . Subscript[f, 1] - 
    1/2 (Subscript[v, 3] . Subscript[f, 1])^2) + 
 t^4 (1/4 (n . Subscript[f, 1])^2 - (n . Subscript[f, 
       1])^2 (Subscript[v, 2] . Subscript[v, 3])^2 + (Subscript[v, 
       2] . Subscript[f, 1])^2 (Subscript[v, 2] . Subscript[v, 3])^2 -
     Subscript[v, 2] . Subscript[f, 1] Subscript[v, 2] . Subscript[v, 
      3] Subscript[v, 3] . Subscript[f, 1] + 
    1/4 (Subscript[v, 3] . Subscript[f, 1])^2)

where the dot represents inner product of two three-dimensional vectors, and $n$ is normal to $v_2$ and $v_3$; moreover, $f_1$ is a vector that is not normal to $n$.

Notice that all the vectors involved are unit vectors. An numerical example can be produced using the following code:

nNum = Normalize[RandomReal[{-1, 1}, 3]]
v2Num = Normalize[Cross[RandomReal[{-1, 1}, 3], nNum]]
v3Num = Normalize[Cross[RandomReal[{-1, 1}, 3], nNum]]
f1Num = RandomReal[{-1, 1}, 3]
{n, Subscript[v, 2], Subscript[v, 3], Subscript[f, 1]} = {nNum, v2Num,
   v3Num, f1Num}

Here is an example of the vectors::

{n, Subscript[v, 2], Subscript[v, 3], Subscript[f, 1]}={{-0.159132, -0.385389, -0.908929}, {-0.808778, 
  0.578873, -0.103846}, {-0.87546, 0.480643, -0.0505222}, {-0.471107, 
  0.714036, 0.725794}}

However, I'm not trying to see whether it can be factored for a given set of these values; instead, I want to see if the factorization is possible regardless of their specific values.

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  • $\begingroup$ For general coefficients, there is the general solution only, with highest coefficient normalized to 1 try $$Solve[Subscript[a, #] & /@ Range[0, 4] . t^Range[0, 4] == 0 /. {Subscript[a, 4] -> 1}, t]$$ $\endgroup$
    – Roland F
    Aug 23, 2023 at 9:31
  • $\begingroup$ How can we know if it can be factored when you did not provide all coefficients? (in Mathematica code not image) $\endgroup$ Aug 23, 2023 at 10:23
  • $\begingroup$ @azerbajdzan, please see the expression in the updated post; I'm trying to figure out if there is any way to simplify the process of solving this equation. Currently, I'm guessing that it might be factored as two quadratic polynomials, or it can be rewritten as a quadratic equation in terms of another quadratic polynomial, but don't have any clue yet. $\endgroup$
    – larry
    Aug 23, 2023 at 11:05
  • 1
    $\begingroup$ Why not just use Factor? $\endgroup$ Aug 23, 2023 at 16:04
  • 1
    $\begingroup$ Factor handles multivariate polynomials just fine. $\endgroup$ Aug 23, 2023 at 19:24

1 Answer 1

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tu is the polynomial in the question and if you substitute the following you see that the polynomial cannot be factored into nothing simpler than roots in the form of genearal quartic formula which means that the original polynomial (without specific values) cannot be factored too.

tu /. {n -> {2, 0, 0}, Subscript[v, 2] -> {0, 1, 0}, 
  Subscript[v, 3] -> {0, 0, 1}, Subscript[f, 1] -> {1, 1, 2}}
% == 0 // Solve // ToRadicals

(*-2 + 2 t - 3 t^2 - 2 t^3 + 2 t^4*)

Edit:

Usually by saying a polynomial can be factored is meant this:

1 + 2 x - 3 x^2 - 2 x^3 + 2 x^4 == (-1 + x) (1 + x) (-1 - 2 x + 2 x^2) // FullSimplify

(* True *)

And usually by saying a polynomial cannot be factored is meant that it cannot be expressed as product of expressions with integers coefficients.

So even though the following polynomial cannot be factored over integers or rationals or any simple extensions it still can be factored, for example, as follows:

-2+2 x-3 x^2-2 x^3+2 x^4==1/8 (4 x^2+x (-2+2 Sqrt[5+2 (-19+2 Sqrt[97])^(1/3)-2 (19+2 Sqrt[97])^(1/3)])-Sqrt[(11+(667-24 Sqrt[97])^(1/3)+(667+24 Sqrt[97])^(1/3))]+(-19+2 Sqrt[97])^(1/3)-(19+2 Sqrt[97])^(1/3)-1) (4 x^2+x (-2-2 Sqrt[5+2 (-19+2 Sqrt[97])^(1/3)-2 (19+2 Sqrt[97])^(1/3)])+Sqrt[(11+(667-24 Sqrt[97])^(1/3)+(667+24 Sqrt[97])^(1/3))]+(-19+2 Sqrt[97])^(1/3)-(19+2 Sqrt[97])^(1/3)-1)//FullSimplify
(* True *)

enter image description here

If you do not care about complicated nested roots then all quartics can be factored, but I think it is not what you mean by factoring your polynomial. I guess you want some nice coefficients without roots, but it is impossible for your polynomial - just as it is impossible for my polynomial in the example.

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