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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?

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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.

**Corrected version:**

The roots are nonegative only for $a=24$ and $b=-32$, i.e. $a+b=-8$.

Proof.

Let $x_1,x_2,x_3,x_4$ denote the roots. Then by Vieta’s formula

$$

x_1+x_2+x_3+x_4 = 8,\ \text{ and } x_1x_2x_3x_4 = 16.

$$

If the roots are positive, then

$$

8 = x_1+x_2+x_3+x_4 \geq 4(x_1x_2x_3x_4)^{1/4}=8,

$$

implying $x_1=x_2=x_3=x_4=2$. Since $f(x) = (x-2)^4=x^4-8x^3+24x^2-32x+16$, it follows that $a=24$ and $b=-32$, i.e. $a+b=-8$.

**The previous version (contains a typo in the last line):**

It seems that $x_1,x_2,x_3,x_4$ cannot be all nonnegative for any values of $a$ and $b$.

Proof.

Let $x_1,x_2,x_3,x_4$ denote the roots. Then by Vieta’s formula

$$

x_1+x_2+x_3+x_4 = 8,\ \text{ and } x_1x_2x_3x_4 = 16.

$$

If the roots are positive, then

$$

8 = x_1+x_2+x_3+x_4 \geq 4(x_1x_2x_3x_4)^{1/4}=8,

$$

implying $x_1=x_2=x_3=x_4=2$. Thus, $f(x) = (x-2)^4=x^4-4x^3+\dots\ne x^4-8x^3+\dots$, a contradiction.

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