Let $f(x)=a_0^2x^n+a_1x^{n-1}+a_2x^{n-2}+...........+a_n,\,$ where $a_0,a_1,....,a_n \in \Bbb R$ I was stuck on the following problem which I came across during my study of theory of equations:  

Let $f(x)=a_0^2x^n+a_1x^{n-1}+a_2x^{n-2}+...........+a_n,\,$ where the coefficients $a_0,a_1,....,a_n$ are real. If $\,\xi\,$ be greater than any of the real roots of the equation $f(x)=0,\,$ then I have to show that $\,f(\xi)\,$ is positive. 

In the question it was given that $f(x)=a_0^2x^n+a_1x^{n-1}+.....+a_n$, I am not sure whether it would be $f(x)=a_0x^n+a_1x^{n-1}+.....+a_n$ . 
Can someone give explanation about how to tackle it? Thanks and regards to all.
 A: $f(x)=a_0^2x^n+a_1x^{n-1}+.....+a_n$
What do you know about $\lim\limits_{x\to+\infty}f(x)$?

 You know that $\lim\limits_{x\to+\infty}f(x)=+\infty$. Now try using the definition of the limit to show you have a number so that at all the numbers greater than that number, the function is positive.

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 If you use the definition of the limit, you get $\forall A \in \Bbb R, \exists x_0 \in \Bbb R,\forall x \ge x_0, f(x)\ge A $. Now take $A=1$. You have $x_0\in \Bbb R$ so that $\forall x \ge x_0, f(x) \ge 1>0$. What do you know about $f(\xi)$ and what are the two possible cases?

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 Since we know $f(\xi)\not= 0$, either $f(\xi)>0$ or $f(\xi)<0$. Is one of those impossible? If yes, you could try to prove it can't be by supposing it is true a finding a contradiction.

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 Suppose we had $f(\xi)<0$. If $\xi \ge x_0$, $f(\xi)>0$ which is absurd so $\xi < x_0$. Since $f$ is continuous, $\xi<x_0$ and $f(\xi)<0<f(x_0)$, by the intermediate value theorem, we have $x_1 \in (\xi,x_0)$ so that $f(x_1)=0$ so we have a root of $f$ bigger than $\xi$ which is absurd by the definition of $\xi$.

