The problem given is as follows.
Problem: For each of the following polynomial functions $f$, find an integer $n$ such that $f(x) = 0$ for some $x$ between $n$ and $n+1$.
Let's consider only one given polynomial because I am interested in the idea of the solution and not the exact solution itself.
$f(x) = x^5 - 5x^4 + 2x + 1$.
Question: Is there any clever argument that helps me to find the number $n$ efficiently? In the answer book there are only numerical answers.
The way I tried to solve the problem: The idea is that if I know that for some $n$ it is true that $f(n) < 0$ and that $f(n+1) > 0$ then that must be the answer to the problem because by the Intermediate value theorem there is an $x$ between $n$ and $n+1$ such that $f(x) = 0$. Similarly, if $f(n+1) < 0$ and $f(n) > 0$ then that $n$ is the answer to the problem.
Number of integers is infinite (unfortunately) and therefore I tried to restrict the values by the method that is used in the book to prove some theorems.
If $x \neq 0$ then $f(x) = x^5\left(1-\frac{5}{x} + \frac{2}{x^4} + \frac{1}{x^5}\right) = x^5 g(x)$. Using the properties (probably axioms) of the real numbers I can show that if $x>0$ then $x^5>0$ and similarly if $x<0$ then $x^5<0$. Now I want to consider $|x|$ large enough so that for those $x$ it is true that $g(x) > 0$, because then if I take $x<0$ then $f(x) < 0$ and if I take $x>0$ then $f(x) > 0 $.
To make $g(x) > 0$ I want that the part that depends on $x$ is larger than $-1$ or equivalently that it's absolute value is smaller than $1$. Now, I can find that if I take $|x| > \textrm{max}(1, 8) = 8 $ then
$\left| \frac{-5}{x} + \frac{2}{x^4} + \frac{1}{x^5} \right| \leq \frac{5}{|x|} + \frac{2}{|x^4|} + \frac{1}{|x^5|} < \frac{5}{|x|} + \frac{2}{|x|} + \frac{1}{|x|} = \frac{8}{|x|} < 1$.
Now I know that if $|x| > 8$ then $f(x) < 0$ for negative $x$ and $f(x) > 0$ for positive $x$.
That means, I don't have to consider values of $n$ smaller than $-9$ because I know that for such $n$ it is true that $f(n) < 0$ and $n+1$ is smaller than $-8$ and therefore it is also true that $f(n-1) < 0$. Similarly, I understand that I do not have to consider $n$ that are larger than $8$.
That still leaves me with values of $n \in \mathbb{Z} \cap [-9, 8]$.
But now, I don't even know what $(-9)^5$ is... Any help is appreciated! Thanks!
As I wrote, since $f$ has positive coefficients, $f$ has no positive roots. Moreover, $$f(x) = x^4(x+5) + (2x+1)$$ so the left term and the right term should counter-balance each other, and the left term is negative over $x < -5$, so that won't work either. You are now looking over $[-5,-1]$, a much more manageable task.
If you think about it, even at $x=4$, the first term is overwhelmingly positive, so you need $f(-5)$ to make it $0$ and turn the entire $f(-5)<0$...