The question is:
Show that $s(s^2 + 5s + 4) + k$ has only zeros with negative real part, provided that, $k < 20 $.
I know that both roots of $s^2 + as + b$ have negative real part if, and only if, a and b are both positive. How would I apply it to a 3rd degree polynomial?
I am fairly certain that you are mean to assume k is positive and that you need to use the idea of closed-loop tranfer functions.
Well, we are looking at:
$$\text{s}\cdot\left(\text{s}^2+5\cdot\text{s}+4\right)+\text{k}=0\tag1$$
Now, eliminate the quadratic term by substituting $x:=\text{s}+\frac{5}{\text{s}}$, then we end up with:
$$\frac{70}{27}+\text{k}-\frac{13}{3}\cdot x+x^3=0\tag2$$
Change coordinates by substituting $x=\text{y}+\frac{\lambda}{\text{y}}$, where $\lambda$ is a constant value that will be determined later and after that let $\lambda=\frac{13}{9}$ and let $\text{z}:=\text{y}^3$, then we end up with:
$$\frac{2197}{729}+\frac{\text{z}\cdot\left(70+27\cdot\text{k}\right)}{27}+\text{z}^2=0\tag3$$
Which is a quadratic equation in $\text{z}$, so I think you can solve it from here.