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!!

let us consider three roots out of which two roots are conjugate of each other
z1=a1+ib1 ; z2=a2+ib2 ; z3=a3+ib3
now z1+z2+z3=-(b/a)
since a and b are real b1+b2+b3=0;
a1+a2+a3 is negative (a1,a2,a3 all are negative as mentioned in the question)
so a and b must have same sign
now consider z1z2+z2z3+z1z3=(c/a)
a1a2-b1b2+a2a3-b2b3+a1a3-b1b3+i(a1b2+b1a2+a2b3+b2a3+a1b3+a3b1)=c/a-------(1)
since a and c are real we can ignore about the imaginary part
(b1+b2+b3)^2=b1^2+b2^2+b3^2+2(b1b2+b2b3+b1b3)
therefore (b1^2+b2^2+b3^2)/2=-(b1b2+b2b3+b1b3)-------(2)
hence the l.h.s of the equation (1) is positive hence c and a have same sign
consider z1z2z3=-(d/a)
since we know that two of the roots are conjugates of each other let us assume z1 and z2 are conjugates of each other so z1z2=|z1|^2 and b1=-b2
since b1+b2+b3=0 ===>b3=0
so z1z2z3=-(d/a)
==>|z1|^2z3=-(d/a)
since the real part of z3 is -ve the overall l.h.s is negative hence a and d have same sign
finally to conclude a,b,c and d have same sign
hope this answer is satisfactory if there are any objection feel free to ask and please let me know if you feel this is correct
