How can I prove that $f$ doesn't have all real roots $\forall a\in\mathbb{C}$ We have $f=x^4+ax^3+4x^2+1\in\mathbb{C}[x]$ with $x_1,x_2,x_3,x_4\in\mathbb{C}$.
We need to prove that $\color\red{\forall a\in\mathbb{C}},f$ doesn't have all real roots. How can I begin to solve this exercise.

Here is what I've tried:
$$\sum_{k=1}^4 x_k^2< 0\Rightarrow\:f\:doesn't\:have\:all\:real\:roots$$ 


*

*Therefore $$\left(\sum_{k=1}^4 x_k\right)^2-2\left(\sum_{1\leq k<i\leq 4}x_k x_i\right)=a^2-8$$


$\Rightarrow a^2-8<0\Rightarrow a\in(-\sqrt{8},\sqrt{8})$

But what I proved was that $f$ doesn't have all real roots for $a\in(-\sqrt{8},\sqrt{8})$. I don't have ideea how can I prove that  $f$ doesn't have all real roots $\color\red{\forall a\in\mathbb{C}}$. 

 A: Since $0$ is not a root, we can as well consider the “reverse” polynomial
$$
g(x)=x^4+4x^2+ax+1
$$
whose roots are the reciprocals of the roots of $f$.
If $g$ has four distinct real roots, its derivative must vanish in three points; now
$$
g'(x)=4x^3+8x+a
$$
and the second derivative must vanish in two distinct points; but
$$
g''(x)=12x^2+8
$$
has no real root.
This also settles the case of multiple roots, of course (with just a bit of work).
A: Consider the case $a>0$. Then $f$ is increasing on $[0,\infty)$, so there are no positive zeros. Thus, if $f$ had four real zeros, they would have to be negative, and then by Rolle's theorem, $f'$ would have to have three negative zeros. But $f'(0)=0$, so...
Can you take it from there?
A: This is just a start.
If $a$ is complex, there are no real zeros. So we'll assume $a$ real.
Given any polynomial with $n$ real zeros, $f'(x)$ has at least $n-1$ real zeros, $f''(x)$ has $n-2$ real zeros, etc.
In this case, the second derivative is:
$$12x^2+6ax+8$$
which, by dividing by by $3$, has the same number of roots as:
$$4x^2+2ax+\frac{8}{3}= \left(2x+\frac{a}{2}\right)^2 + \frac{8}{3}-\frac{a^2}{4}$$
So you need $\frac{a^2}{4}>\frac{8}{3}$ or $|a|>\sqrt{\frac{32}{3}}$ to have two real roots for the second derivative.
