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. 
 A: Note that $x^4-4=(x^2+2)(x^2-2)$. This suggests that you might try to express your polynomial as$$(x^2+ax+2)(x^2+bx-2)=x^4+(a+b)x^3+abx^2+2(-a+b)x-4.$$It is now easy to see that all it takes is to choose $a=1$ and $b=4$.
A: Since there are no rational roots (a rational root should be a divisor of $4$), your attempt of factorization as
$$x^4 +5x^3+4x^2+6x-4=(x^2+ax+b)(x^2+cx+d)$$
is a good idea.
Hint. By expanding the right-hand side and by comparing it with the the left-hand side, we have that $bd=-4$. Assuming that $b$ and $d$ are integers, by symmetry, you can try the following couples:
$$(b,d)\in\{(4,-1), (2,-2), (-4,1)\}$$
and then solve the corresponding systems with respect to the remaining coefficients $c$ and $d$. 
P.S. If you are lucky you will start with the couple $(2,-2)$. Be careful though, there are polynomials, which are "similar" to the given one, such that the other couples work:
i) $(4,-1)$ for $x^4+2x^2-5x-4=(x^2+x+4)(x^2-x-1)$.
ii) $(-4,1)$ for $x^4-4x^2+5x-4=(x^2+x-4)(x^2-x+1)$.
A: 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?
