Can any polynomials in the rational field be decomposed like this

Question

I' ve learned that the following examples can be used to decompose a factor in this way:

x^5 - 5 x + 12 = (x - a) (a^4 - (5 a^3)/8 + (7 a^2)/8 + 1/8 (5 a^2 - 12 a) + 
   1/8 (5 a^3 - 12 a^2) + ((5 a^4)/16 - a^3/8 + (7 a^2)/16 - 
      3/16 (5 a^2 - 12 a) + 1/16 (12 a^3 - 5 a^4) + 
      1/8 (12 a^2 - 5 a^3) - (9 a)/4) x^2 + ((5 a^4)/16 + a^3/4 + (
      5 a^2)/16 + 1/16 (12 a - 5 a^2) + 1/16 (12 a^3 - 5 a^4) + a/2 + 
      1/4 (12 - 5 a) - 3) x + a x^3 + a/4 + 1/4 (5 a - 12) + x^4 - 2)

Can any polynomials f(x) in the field of rational numbers be factorized into the form of (x - a) g (x, a)? Besides a, other coefficients of g (x, a) should also be in the rational number field.

In the case of $x^5-5 x+12$, we can know the algebraic relations of his five roots (The letter a is a root of equation $x^5−5x+12$):

$x^5-5 x+12=(x-a) \left(x-\frac{1}{8} \left(-a^4-a^3-a^2-2 \sqrt{2} \sqrt{3 a^3-2 a^2+a+4}-a+4\right)\right) \left(x-\frac{1}{8} \left(-a^4-a^3-a^2+2 \sqrt{2} \sqrt{3 a^3-2 a^2+a+4}-a+4\right)\right) \left(x-\frac{1}{8} \left(a^4+a^3+a^2-2 \sqrt{2} \sqrt{a^4+a^2+6 a-8}-3 a-4\right)\right) \left(x-\frac{1}{8} \left(a^4+a^3+a^2+2 \sqrt{2} \sqrt{a^4+a^2+6 a-8}-3 a-4\right)\right)$

I used the function in this link to find the relationship between a polynomial Galois group and a root set.

Answer 1

It seems to me that you are asking two separate questions. One is whether every $f$ can be factored as $(x-\alpha)g(x)$ where the coefficients of $g$ are simple expressions in (rationals and) $\alpha$. The other is whether every $f$ can be factored as $(x-\alpha)(x-p_1(\alpha))\cdots(x-p_{n-1}(\alpha))$ where each $p_i$ is a radical expression in $\alpha$.

The answer to the first question is Yes. Let $K={\bf Q}(\alpha)$. Then $f$ has a zero in $K$, so it factors over $K$, and one factor is $x-\alpha$, and the other is a polynomial $g(x)$ with coefficients in $K$, so the coefficients of $g$ are polynomials in $\alpha$ (of degree less than the degree of $\alpha$).

The answer to the second question is Yes if the degree of $f$ is five (or less), but in general it's No if the degree of $f$ is larger. If the degree of $f$ is five, then $f(x)=(x-\alpha)g(x)$, where $g$ is a polynomial of degree four, whose coefficients are polynomials of degree at most four in $\alpha$. Since $g$ has degree four, its roots can be expressed in radicals in its coefficients, which is to say, radicals of polynomials in $\alpha$. That's where stuff like $\sqrt{\alpha^4+\alpha^2+6\alpha-8}$ comes from in your example. But if the degree of $f$ exceeds five, then the degree of $g$ is at least five, and (in general) there will be no radical expression for the roots of $g$ in terms of radicals of polynomials in $\alpha$.

Answer 2

Here:

     In[5]:= qpoly = x^5 - 5 x + 12;
                rt = First[x /. Solve[qpoly == 0, x]]
               fax = Factor[qpoly, Extension -> rt] /. rt -> a
               (% /. a -> Root[#1^5 - 5 #1 + 12 &, 1]) // Simplify
     Out[5]=(1/16)*(x - a)*(-a^4 - a^3 - a^2 + (-a^4 - a^3 - a^2 + 3*a + 4)*x - 5*a + 4*x^2 + 8)*(-(2*a^3) + (a^4 + a^3 + a^2 + a - 4)*x - 2*a + 4*x^2 - 4)
            x^5-5 x+12

Answer 3

Every (monic) polynomial $f(x)\in \Bbb{Q}[x]$ factorizes into $$f(x) = \prod_{j=1}^{\deg(f)} (x-\beta_j)=\prod_{j=1}^{\deg(f)} (x-g_j(\alpha)), \qquad g_j\in \Bbb{Q}[x]$$ $\beta_j$ are the roots, they are algebraic numbers, they generate $f$'s splitting field $$K=\Bbb{Q}(\beta_1,\ldots,\beta_{\deg(f)})$$

The primitive element theorem gives $$K=\Bbb{Q}(\alpha)=\Bbb{Q}[\alpha]$$ Thus each $\beta_j$ is a polynomial in $\alpha$.

The main problem is that in general $\alpha$ is not defined by some $n$-th roots, when it is we say that $f$ is radical. Every polynomial of degree $\le 4$ is radical. The unsovable quintic theorem shows that most polynomials of degree $\ge 5$ are not radical.

For non-radical polynomials, $\alpha$ (as well as the roots of $f$) is only defined by its minimal polynomial, we don't have any simpler expression for it.

For your quintic polynomial, you have exploited the fact that $[\Bbb{Q}(\beta_1,\beta_j):\Bbb{Q}(\beta_1)]\le 4$ to obtain that $\beta_j$ is radical in $\beta_1$, obtaining $$f(x) = \prod_{j=1}^{\deg(f)} (x-f_j(\beta_1))$$ where $f_j$ is an expression with a variable $t$, some rational numbers, additions, multiplications and $n$-th roots.