I want to show that there exists a $-K$ such that $g(-K)<0$ where $K\in N$.
My attempt: $g(-n)= (-n)^d+b_{d-1}(-n)^{d-1}+\cdots+b_0$. Show that $g(-n)$ is negative for some large $n$.
\begin{align} & \left| \frac {b_{d-1}(-n)^{d-1}+\cdots+b_0}{(-n)^d} \right| = \frac {|b_{d-1}(-n)^{d-1} + \cdots +b_0|}{n^d} \\ = {} & \frac {|b_{d-1}n^{d-1}-b_{d-2}n^{d-2}+\cdots-b_1 n+b_0|}{n^d}\le \frac {|b_{d-1}n^{d-1}| + |b_{d-2}n^{d-2}| + \cdots+|b_1n|+|b_0|}{n^d} \\ \le {} & \frac {n^{d-1}(|b_{d-1}+|b_{d-2}| + \cdots+|b_1|+|b_0|)}{n^d} = \frac {(|b_{d-1} + |b_{d-2}|+\cdots+|b_1|+|b_0|)}{n}. \end{align}
Therefore, there exists a $K$ such that
$$\left|\frac {b_{d-1}(-K)^{d-1}+\cdots+b_0}{(-K)^d}\right| < 1 \rightarrow -1 < \frac {b_{d-1}(-K)^{d-1}+\cdots+b_0}{(-K)^d} <1 \rightarrow \frac {b_{d-1}(-K)^{d-1} + \cdots+b_0}{-K^d}<1 \rightarrow b_{d-1}(-K)^{d-1}+\cdots+b_0>-K^d,$$
which is not what we want to show. On the other hand,
$$ -1 < \frac {b_{d-1}(-K)^{d-1} + \cdots+b_0}{-K^d} \rightarrow K^d>b_{d-1}(-K)^{d-1} + \cdots+b_0 $$
as required.
Why do these two have different outcome?
Thank you in advance.
Actually, If $a+ib$ is a root of any polynomial with real co-efficient with $b\neq0$ then $a-ib$ is also be a root of the polynomial. As because Let $P(x)=a_nx^n+....+a_1x+a_0$ and if u assume $P(a+ib)=M+iN$ for real $M$ & $N$ so $P(a-ib)=M-iN$ because, $(a+ib)^n=m+in$ for real m & n then $(a-ib)^n=m-in$ for all positive integer & hence if $M+iN=0$ implies $M=0$ & $N=0$ so $a-ib$ is a root of $P(x)$. Now for odd degree polynomial their are odd numbers of root so a root must be real.