Condition that all 3 roots of $az^3+bz^2+cz+d=0$ have negative real part The problem is - 'find the condition that all 3 roots of $f(z)=az^3+bz^2+cz+d=0$ have negative real part, where $z$ is a complex number'. 
The answer - '$a,b$, and $d$ have the same sign.'
Honestly, I have no clue about how to proceed. Here is what I tried- $ f'(z)=3az^2+2bz+c$, which at extrema gives the roots as $z=\frac{-2b+/-\sqrt{(4b^2-12ac)}}{6a}$. If the real part is negative, then $\frac{-2b}{6a}<0$, which implies that $a,b$ have the same sign. I am not sure if what I have done is right, and have no idea about proving the rest of it. Please help.
Thanks in advance!!

Ok, I think I have it.
Since $z$ is a complex number, $f(z)$ must have a complex root. As they occur in conjugate pairs, 2 out of 3 roots of f(z) must be imaginary. The remaining one shall be purely real (and obviously rational). Now, all three have negative real parts. So, lets name the roots-$\alpha=-x+iy, \beta=-x-iy, \gamma=-k$, where $x,y,k>0$. Now,since $\alpha+\beta+\gamma=-b/a$; we get (after substituting and a trivial simplification), 
$-(2x+k)=-b/a$. As both x and k are positive, we get $b/a>0$ i.e. $b$ and $a$ are of same sign.(Which can also be arrived at with differentiation, as in my question)
Now, $\alpha\beta\gamma=-d/a$, which after substitution and simplification gives-
$(-k)(x^2+y^2)=-d/a$.
As $x^2+y^2$ is always positive, multiplying with negative number $(-k)$ gives the LHS as negative. Thus, $-d/a<0$, implying $d/a>0$, and hence, $a,d$ have same sign.
Combining above two results, $a,b,d$ have the same sign.
