Roots of $z^4 + 2z^3 + 20z + 12 = 0$ 
Find the roots of $$z^4 + 2z^3 + 20z + 12 = 0.$$

I still haven't been able to solve it after hours of trying. There are no rational roots and factoring seems impossible. Who of you, mathematical geniuses, can help me? :-)  
 A: There is a more general approach to factoring quartics, which is actually powerful eniugh to identify some factors with irrational coefficients using only rational root testing.  It's basically the standard general method with a resolvent cubic, except this version is implemented without the often combersome process of "reducing" the quartic.
Assume the quartic is rendered as
$x^4+bx^3+cx^2+dx+e=0$
with all coefficients rational.  We also assume here no rational roots.  This is because the method intrinsically does not work on equation with such a root; rather factoring out the rational root and dealing with the remaining cubic would then be more economical.  Here we will put in the values $b=2,c=0,d=20,e=12$ as an example.
Once we find no rational roots of the original quartic, we assume a quadratic factorization of the form
$x^4+bx^3+cx^2+dx+e=(x^2+(r+\sqrt{s})x+t_1)(x^2+(r-\sqrt{s})x+t_2)$
$x^4+2x^3+20x+12=(x^2+(1+\sqrt{s})x+t_1)(x^2+(1-\sqrt{s})x+t_2)$
where $r=b/2=1$ assuring the quartic and cubic tetms are matched.
Next match the quadratic and cubic terms in the product with your quartic, getting a pair of linear equations for $t_1$ and $t_2$ in terms of $s$:
$t_1+t_2+(r^2-s)=c=0$
$t_1+t_2=s-1$  Eq. 1
$r(t_1+t_2)-\sqrt{s}(t_1-t_2)=r(c+s-r^2)-\sqrt{s}(t_1-t_2)=d=20$
$t_1-t_2=(-s-21)/\sqrt{s}$  Eq. 2
Equations 1 abd 2 may be solved for the $t$ variables.  The general formula is getting unwieldy so I just report the result specific to the equation at hand:
$t_1=\dfrac{s-1-(s+21)/\sqrt{s}}{2}$  Eq. 3
$t_2=\dfrac{s-1+(s+21)/\sqrt{s}}{2}$  Eq. 4
Now match the constant terms:
$t_1t_2=e=12$
Upon substituting Eqs. 3 and 4 for our specific example, we find that the square roots cancel out (always) and clearing fractions leads to our resolvent cubic by this method:
$s^3-3s^2-5s-441=0$  Eq. 5
Now we seek rational roots, which in this case are divisors of $441=21^2$.  The search is cut down by noting that for $|s|\in\{1,3\}$ the constant term dominates, too large to fit in with the other terms, while $|s|\ge 21$ the cubic term similarly is overpowering.  We are forced to accept $s\in\{\pm7, \pm9\}$.
There are then four cases to consider:

*

*If a squared rational root applies, in this case it must be $s=+9$, we get a factorization with rational numbers only, giving roots that are quadratic surds (possibly complex).


*If a different positive root works, in this case it would be $s=+7$, the quadratic factors will have real square roots.  Roots of the quartic will appear with nested square roots.


*If we need a negative rational root ($-7$ or $-9$), we will get complex numbers in the quadratic factorization and we will later need to extract the square roots of these complex numbers.  This can be done algebraically, but the results are rather complicated.


*If none of the rational root candidates apply we are forced to use cubic roots that contain cube root radicals like the usual general solution method, too unwieldy for most realistic applications.
With this hierarchy we try $s=+9$ first for our example; succeeding (thus Case 1), we get the $t$ parameters from Eqs. 3 and 4, and the resulting factorization given elsewhere
$x^4+2x^3+20x+12=(x^2+4x+2)(x^2-2x+6)$
and continue from there by the usual methods for quadratic equations.
An attractive feature of this method is it has a hidden geometric significance.  If any of Cases 1, 2, 3 above applies -- if the resolvent cubic has a rational root -- then the quartic roots are constructible by unmarked straight edge and compasses, and the factorization gives the blueprint for such a construction.  If the resolvent cubic fails to have a rational root (Case 4), then the irrational roots of both the cubic and the quartic equations will contain cube roots and we have to use a marked ruler.  Only a single rational root of the quartic, if available, can be rendered without the marks.
A: The rational roots test fail so there is no linear real factor but there may be some $(z^2 + az + b)(z^2 + cz + d)$ where
$a+c = 2$ and $b+ ac + d =0$ and $ad+bc = 20$ and $bd=12$
which .... isn't obvious but.... oh heck.  If we assume $\{b,d\} = \{\pm 1,12\}, \{\pm 2,6\}, \{\pm 3,4\}$ then $ac =-b-d= \mp 13, \mp 8, \mp 7$.  But  $a+c = 2$ so (assuming integers) $a$ and $c$ are opposite signs.  $13$ and $7$ are primes larger than $3$ so a difference of $2$ between their factors is not possible.  SO the nicest is $\{b,d\} = \{2,6\}$ and $ac = -8$.  And as $a+c = 2$ so $a=4; c=-2$.  (That's the nicest; it's not a guarentee.)
So $ad+bc= 4*(2,6)-2*(6,2) = 20$ is solved with $d=6; b= 2$.
Hurray!  $(z^2 +4z +2)(z^2 -2z + 6)=z^4 + 2z^3 + 20z +12$
And we can use quadratic formula to solve:
$z = -2 \pm \sqrt 2$ and $z= 1 \pm \sqrt{5}i$.
