Factoring a degree 4 polynomial without power of 2 term For my hobby, I'm trying to solve $x$ for $ax^4 + bx^3 + dx + e = 0$. (note there's no $x^2$) I hope there is a simple solution.
I'm trying to write it as $(fx + g)(hx^3+i) = 0$
It follows that
$fh=a; gh=b; if=d; gi=e$
At first sight it looks promising with 4 equations and 4 unknowns ($f,g,h,i$). Unfortunately when substituting them you'll find a dependency so that this only works when $db=ae$. Is there an easy solution for the more general case?

I suppose that you are searching a decomposition of the given polynomial in factors with real coefficients.
First note that your condition $db=ae$ means 
$$
\frac{a}{d}=\frac{b}{e}=k
$$
so the polynomial is obviously decomposable as:
$$
kdx^4+kex^3+dx+e=kx^3(dx+e)+dx+e=(dx+e)(kx^3+1)
$$
Also note that a degree $4$ polynmial without the $x^2$ term can be decomposable in other forms, as:
$$
3x^4+5x^3+5x+3=(x^2+1)(3x^2+5x-3)
$$
So, in general, also for a quartic equation of the given form, the solutions can be found only using the (not simple) general  methods.
