Part of the proof of the fundamental theorem of algebra Consider the polynomial function: $f(x)=c_0+c_1x+c_2 x^2+...+c_{n-1}x^{n-1}+x^n$, where $n\geq 1$, $c_0\neq0$, $x$ and $c_0,c_1,c_2,...,c_{n-1}$ are complex numbers. Let $C=max(|c_0|,|c_1|,...,|c_{n-1}|)>0$. Given that $|f(x)|\ge |x|^n-nC|x|^{n-1}>0 $  if $|x|\ge nC+1 := P_0$ .  $|f(x)|$ is continuous and there exists closed and compact disc Q:={$ x: |x|\le P_0$}. Set thus $\inf_{x\in Q} |f(x)|$ exists at some $x_0 \in Q$. Say that $x_0$ may be on the boundary of Q. Prove that you can find $P_1 > P_0$ so that $|f(x)|\ge 1+|f(x_0)|$  for $|x|\ge P_1$ and conclude that $x_0$ is in the interior of Q.
Tried to prove by contradiction and assume that there's no such $P_1 > P_0$ such that $|f(x)|\ge 1+|f(x_0)|$  , then for all $P_1 > P_0$, $|f(x)|< 1+|f(x_0)|$ , but didn't know how to move on or the idea might be wrong. 
 A: $f$ is a polynomial. For large enough $|x|$, $f(0) \approx x^n$, which goes to $\infty$ as $|x| \to \infty$. For any value $M > 0$ (such as $1 + |f(x_0)|$), you can certainly find a $P$ such that if $|x| > P$, then $|f(x)| > M$. 
I don't think you are interpreting this right. The fact that $|f(x)| \ge 1 + |f(x_0)|$ for $|x| \ge P_1$ does not imply that $x_0$ is in the interior of this particular set $Q$. What it does imply is that you can enlarge $Q$ to the place (radius $P_1$) where its infimum must lie within its interior.
A: (1).Actually $|f(x)|>0$ for $|x|\geq 1+C$ which is sharper than $|x|\geq 1+ n C$ for $n>1$.............. (2). You have correctly established that $f(z)$ attains a minimum value, that is,  $\exists z_0 \forall z (|f(z_0)|\leq |f(z)|) . $ The next step is to show by contradiction that $f(z_0)=0 .$ One way is to take the least $j>0$ such that $\frac {d^j f}{dx^j}(z_0)\ne 0$. Then there exists  a constant $A\ne 0$ and a polynomial $g$ such that, for all $x$ we have $f(x)=f(z_0)+(x-z_0)^j(A+(x-z_0)g(x))$. Suppose $f(z_0)\ne 0$. If $0<|x-z_0|<r$  when $r$ is arbitrarily small, and if $A(x-z_0)^j/f(z_0)$ is a negative real number, what is $|f(x)|$ compared to $|f(z_0)|$ ?
