12x^5+6x^3+8x^2 = 2x^2(6x^3+3x+4)
Let f(x) = 6x^3+3x+4.
This is way too messy to solve, but for the record...
Use Cardano's method to solve f(x) = 0.
Let x = u + v
f(x) = 6(u+v)^3 + 3(u+v) + 4
=6u^3+6v^3+(18uv+3)(u+v) + 4
Let v = -1/6u.
Then 18uv+3 = 0 and
f(x) = 6u^3-1/(36u^3)+4
If f(x) = 0 then:
6u^3-1/(36u^3)+4 = 0
Multiply through by 36u^3 to get:
216(u^3)^2+144(u^3)-1 = 0
From the quadratic formula:
u^3 = (-144 +-sqrt(144^2+4*216))/(2*216)
=(-2^4 3^2+-sqrt(2^8 3^4 + 2^5 3^3))/(2^4 3^3)
=-1/3 +-(2^2 3 sqrt(2^4 3^2 + 2*3))/(2^4 3^3)
=-1/3 +-sqrt(150)/36
=-1/3 +-(5sqrt(6))/36
Since this derivation has been symmetric in u and v:
Let u = root(3)(-1/3 +(5sqrt(6))/36)
and v = root(3)(-1/3 -(5sqrt(6))/36)
The roots of f(x) = 0 are:
x_1 = u + v
x_2 = omega u + omega^2 v
x_3 = omega^2 u + omega v
where omega = -1/2 + i sqrt(3)/2
x_1 is the real root of f(x) = 0
x_2 and x_3 are a pair of complex conjugate roots.
Hence f(x) = 6(x - x_1)(x - x_2)(x - x_3)
and 12x^5+6x^3+8x^2 = 12x^2(x - x_1)(x - x_2)(x - x_3)
