Factor a polynomial over the reals

Question

I can't seem to find a function in Mathematica that factors a polynomial over the reals. Obviously, Factor doesn't work since it works over the integers, over $\mathbb{Z}_p$, or some algebraic fields extensions of the type $\mathbb{Q}(\alpha_1, \ldots, \alpha_n)$ where $\alpha_1, \ldots, \alpha_n$ are algebraic numbers. My question is general, although it arouse from my need to factor over $\mathbb{R}$ the polynomial $$ f(\lambda) = \lambda^4 + 17\lambda^2+12. $$ Of course, in this particular case I can find the roots and factor it myself.

Answer 1

One way is to find the roots, separate into real and complex, further separate the complex ones into conjugate pairs, then reform as a factorization (taking into account the leading coefficient). I'll illustrate with the given example.

poly = x^4 + 17 x^2 + 12;
roots = x /. NSolve[poly];
rroots = Select[rts, FreeQ[#, Complex] &]
croots = Complement[roots, rroots];
topquad = Select[croots, Im[#] > 0 &]
mult = Coefficient[poly, x, Exponent[poly, x]]
rfax = x - rroots
cfax = x^2 - 2*x*Re[topquad] + Re[topquad]^2 + Im[topquad]^2

(* Out[54]= {}

Out[56]= {0. + 0.859018 I, 0. + 4.03263 I}

Out[57]= 1

Out[58]= {}

Out[59]= {0.737913 + x^2, 16.2621 + x^2} *)

Putting together the result:

Apply[Times, Join[{mult}, rfax, cfax]]

(* Out[61]= (0.737913 + x^2) (16.2621 + x^2) *)

Answer 2

My knowledge in this area is quite lacking, so there are probably better ways. The following function will force-factor anything, and will return Root objects if necessary.

factor[poly_, x_] :=
 Module[{n, nreal},
  n = Exponent[poly, x];
  nreal = CountRoots[poly, x];
  Times @@ Join[
    Table[(x - ToRadicals@Root[poly & /. x -> #, i]), {i, nreal}],
    Table[
     With[{r1 = ToRadicals@Root[poly & /. x -> #, i], 
       r2 = ToRadicals@Root[poly & /. x -> #, i + 1]},
      (x^2 - Expand[(r1 + r2)] x + Expand[r1 r2])
      ],
     {i, nreal + 1, n, 2}
     ]
    ]
  ]

This is relying on the ordering used by Root, described under "Details". Real roots come first, then complex conjugate pairs.

Example:

factor[x^4 + 17 x^2 + 12, x]
(* (17/2 - Sqrt[241]/2 + x^2) (17/2 + Sqrt[241]/2 + x^2) *)

Any coefficient in the result will be real, regardless of whether it includes terms that look imaginary in its expression. Example:

factor[x^4 + 1, x]
(* (1 - ((-1)^(1/4) - (-1)^(3/4)) x + x^2) (1 - (-(-1)^(1/4) + (-1)^(3/4)) x + x^2) *)

(-1)^(1/4) has an imaginary part, but (-1)^(1/4) + (-1)^(3/4), which is the actual coefficient of x, doesn't.

Expect not very useful results for cases like this:

factor[x^5 + 2 x^4 + x^3 + x^2 + x + 1, x]
(* (x - Root[1 + #1 + #1^2 + #1^3 + 2 #1^4 + #1^5 &, 1]) (x^2 +
    Root[1 + #1 + #1^2 + #1^3 + 2 #1^4 + #1^5 &, 2] Root[
     1 + #1 + #1^2 + #1^3 + 2 #1^4 + #1^5 &, 3] - 
   x (Root[1 + #1 + #1^2 + #1^3 + 2 #1^4 + #1^5 &, 2] + 
      Root[1 + #1 + #1^2 + #1^3 + 2 #1^4 + #1^5 &, 3])) (x^2 + 
   Root[1 + #1 + #1^2 + #1^3 + 2 #1^4 + #1^5 &, 4] Root[
     1 + #1 + #1^2 + #1^3 + 2 #1^4 + #1^5 &, 5] - 
   x (Root[1 + #1 + #1^2 + #1^3 + 2 #1^4 + #1^5 &, 4] + 
      Root[1 + #1 + #1^2 + #1^3 + 2 #1^4 + #1^5 &, 5])) *)

This is why yarchik's solution is better.

Answer 3

Factor[x^4 + 17 x^2 + 12, Extension -> Sqrt[241]]
(*-(1/4) (-17 + Sqrt[241] - 2 x^2) (17 + Sqrt[241] + 2 x^2)*)

The answer to another example can be obtained as follows:

f = 180 + 1027 x^2 + 666 x^4 + 27 x^6 
Factor[f, Extension -> { (-293657 + 20 I Sqrt[4568243])^(1/3)}] /. (-293657 + 20 I Sqrt[4568243]) -> a // Simplify

Finally a small remark on mathematical side. Sometimes algebraic equations of higher orders (>4) can be solved in radicals. Otherwise solutions for $n=5,6,7$ are known in terms of more general functions, see for instance the wiki article on the septic equation. Very simple solutions are known for trinomial equations $x^n+x-q=0$ in terms of hypergeometric functions ($n\in\mathbb{N}$). As far as I know, these solutions and more general solutions are not directly implemented in MA. Your two examples are of course solvable because of the substitution $\lambda^2=x$.