Can any polynomials in the rational field be decomposed like this 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.

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

