Existence of polynomials $g$, $h$, with non-negative coefficients, such that $f(x)= \frac{g(x)}{h(x)}$ Suppose $a$ and $b$ are real numbers such that the quadratic polynomial $f(x) = x^2 + ax + b$has no non-negative real roots. Prove that ther exist two polynomials g,h, whose coefficients are non-negative real numbers, such that
$$f(x)= \frac{g(x)}{h(x)}$$
for all real numbers $x$.
Irish MO 2007

If both roots of $f(x)$  are real then take $g(x)=f(x)$ and $h(x)=1$.
If both are complex with non-negative real part, then $f(x)=(x-c+di)(x-c-di)=x^2-2cx+(c^2+d^2)$. Therefore we can take $h(x)=x^2+2cx+(c^2+d^2)$ and $g(x)=f(x)h(x)=\left[x^2+(c^2+d^2)\right]^2+4c^2x^2$.
If both roots have negative real parts then take again $g(x)=f(x)$ and $h(x)=1$.
