Existence of polynomials $g$, $h$, with non-negative coefficients, such that $f(x)= \frac{g(x)}{h(x)}$ [closed]

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

closed as off-topic by user147263, Daniel W. Farlow, Adam Hughes, Claude Leibovici, Gabriel RomonApr 3 '15 at 16:49

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Community, Daniel W. Farlow, Adam Hughes, Claude Leibovici, Gabriel Romon
If this question can be reworded to fit the rules in the help center, please edit the question.

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 $h(x)=1$.
• This is correct, but you are reusing $a$ and $b$ incorrectly, they are already the coefficients of $f(x)$ - please switch to $c \pm di$ – ivancho Apr 3 '15 at 2:46
Let $$g(x)=f(x)$$ $$h(x)=1$$