For what value of $(a+b)$ will all roots of $f(x)=x^4-8x^3+ax^2+bx+16$ be positive? I was thinking of using Descartes' rule of signs, from which I find there are at most 2 positive roots and 2 negative roots of the given equation.
Also, $f(\infty)>0$ and $f(0)>0$ imply that either there are no real roots or 2 real roots in $(0,\infty)$. Similar is the case for $(-\infty,0)$. 
If I could factor $f'(x)=4x^3-24x^2+2ax+b$ and find its roots, then the real and distinct roots of $f(x)$ will be separated by those of $f'(x)$. How do I solve the problem?

Let us consider first the case when $b=-32$.
$$f_1(x)=x^4-8x^3+ax^2-32x+16=0\iff a=\frac{-x^4+8x^3+32x-16}{x^2}$$
Now, let
$$g(x):=\frac{-x^4+8x^3+32x-16}{x^2}$$
Then,
$$g'(x)=\frac{-2 (x-2)^3 (x+2)}{x^3}$$
So, $g(x)$ is increasing for $x\lt -2$ or $0\lt x\lt 2$, and is decreasing for $-2\lt x\lt 0$ or $x\gt 2$ with $\lim_{x\to\pm\infty}g(x)=-\infty,\lim_{x\to 0^{\pm}}g(x)=-\infty$.
Now considering the graph of $y=g(x)$ gives that
$$\begin{align}&\text{all roots of $f_1(x)$ are positive}\\&\implies \text{all $x$ coordinates of the intersection points of $y=a$ with $y=g(x)$ are positive}\\&\implies a=24\ (=g(2))\\&\implies a+b=-8\end{align}$$
By the way, for $a=28, b=-36$ where $a+b=-8$, 
$$f_2(x)=x^4-8x^3+28x^2-36x+16$$
$$f'_2(x)=4x^3-24x^2+56x-36=4(x-1)\left(\left(x-\frac 52\right)^2+\frac{11}{4}\right)$$
so $f_2(x)$ is decreasing for $x\lt 1$ and is increasing for $x\gt 1$ with $f_2(1)=1\gt 0$ from which $f_2(x)\gt 0$ follows. Hence, $f_2(x)$ has no real roots.
Therefore, there are no $a+b$ such that all roots of $f(x)$ are positive.
