Factorise polynomial with real and complex roots How would you go about finding the roots of the polynomial: 
$$x^4 +5x^3+4x^2+6x-4=0.$$
I attempt to form two quadratics e.g.
$$(x^2+ax+b)(x^2+cx+d)$$
and then tried to expand, collect like terms, and solve for the coefficients simultaneously, but this results in a hard system of equations. 
I was wondering if there was a better approach to this question. 
Any help would be greatly appreciated. 

I like the following way.
For all real value of $k$ we have:
$$x^4+5x^3+4x^2+6x-4=\left(x^2+\frac{5}{2}x+k\right)^2-\left(\left(2k+\frac{9}{4}\right)x^2+(10k-6)x+k^2+4\right).$$
Now, we'll choose a value of $k$, for which $2k+\frac{9}{4}>0$ and $$(5k-3)^2-\left(2k+\frac{9}{4}\right)(k^2+4)=0.$$
Easy to see that $k=0$ is valid.
Thus,$$x^4+5x^3+4x^2+6x-4=\left(x^2+\frac{5}{2}x\right)^2-\left(\frac{9}{4}x^2-6x+4\right)=$$
$$=\left(x^2+\frac{5}{2}x\right)^2-\left(\frac{3}{2}x-2\right)^2=(x^2+x+2)(x^2+4x-2).$$
Can you end it now?
