Solve $x^4+3x^3+6x+4=0$... easier way? So I was playing around with solving polynomials last night and realized that I had no idea how to solve a polynomial with no rational roots, such as $$x^4+3x^3+6x+4=0$$
Using the rational roots test, the possible roots are $\pm1, \pm2, \pm4$, but none of these work.
Because there were no rational linear factors, I had to assume that  the quartic separated into two quadratic equations yielding either imaginary or irrational "pairs" of roots.  My initial attempt was to "solve for the coefficients of these factors".
I assumed that $x^4+3x^3+6x+4=0$ factored into something that looked like this
$$\left(x^2+ax+b\right)\left(x^2+cx+d\right)=0$$
because the coefficient of the first term is one.  Expanding this out I got
$$x^4+ax^3+cx^3+bx^2+acx^2+dx^2+adx+bcx+bd=0$$
$$x^4+\left(a+c\right)x^3+\left(b+ac+d\right)x^2+\left(ad+bc\right)x+bd=0$$
Equating the coefficients of both equations
$$a+c = 3$$
$$b+ac+d = 0$$
$$ad+bc = 6$$
$$bd = 4$$
I found these relationships between the various coefficients.   Solving this system using the two middle equations:
$$\begin{cases} b+a\left(3-a\right)+\frac4b=0 \\ a\frac4b+b\left(3-a\right)=6 \end{cases}$$
From the first equation: $$a = \frac{3\pm\sqrt{9+4b+\frac{16}{b}}}{2}$$
Substituting this into the second equation:
$$\frac{3\pm\sqrt{9+4b+\frac{16}{b}}}{2}\cdot\frac4b+b\cdot\left(3-\frac{3\pm\sqrt{9+4b+\frac{16}{b}}}{2}\right)=6$$
$$3\left(b-2\right)^2 = \left(b^2-4\right)\cdot\pm\sqrt{9+4b+\frac{16}b}$$
$$0 = \left(b-2\right)^2\cdot\left(\left(b+2\right)^2\left(9+4b+\frac{16}b\right)-9\left(b-2\right)^2\right)$$
So $b = 2$ because everything after $\left(b-2\right)^2$ did not really matter in this case.  From there it was easy to get that $d = 2$, $a = -1$ and $c = 4$.  This meant that
$$x^4+3x^3+6x+4=0 \to \left(x^2-x+2\right)\left(x^2+4x+2\right)=0$$
$$x = \frac12\pm\frac{\sqrt7}{2}i,\space x = -2\pm\sqrt2$$
These answers worked!  I was pretty happy at the end that I had solved the equation which had taken a lot of work, but my question was if there was a better way to solve this?

More generally:
Given a quartic equation with the form
$x^4+bx^3+cx^2+(br)x+r^2=0$
you guarantee factors with the form
$(x^2+px+r)(x^2+qx+r)=0$
For then the cubic and linear terms are both matched by rendering
$p+q=b$
and we have only to match the quadratic terms which oeadsto
$pq=c-2r^2$
From these equations $p$ and $q$ are roots of the quadratic equation
$u^2-bu+(c-2r^2)=0.$
Here $b=3,c=0,r=2$, from which
$u^2-3u-4=0.$
Thus $\{p,q\}=\{-1,4\}$ from which we obtain the factorization
$x^4+3x^3+6x+4=(x^2-x+2)(x^2+4x+2).$
The quartic equation is then solved by the roots of these quadratic factors.
