#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)#
