prove that a polynomial is lower bounded I need help with this question from Data-Structure course.
I need to prove that the following polynomial is lower bounded by  $n^k $, meaning I need to show that: 
$$ p(n) = b_kn^k - b_{k-1}n^{k-1}-b_0n = \Omega(n^k) $$
So I have switched the Omega to inequlity and now I need to show that there exist constant $c$ and $n_0$ that solve the following inequality:
$$ b_kn^k - b_{k-1}n^{k-1}-b_0 n \ge c \cdot n^k $$ 
holds for $n \ge n_0$. It is also given that $$ b_k , b_{k-1} , b_0 > 0 \,.$$
 A: For any polynomial $$P(x)=\sum_{k=0}^d p_kx^k$$
and $x>0$, we have
$$\frac{P(x)}{x^d}=\sum_{k=0}^d\frac{p_k}{x^{d-k}}=p_d+\frac{p_{d-1}}x+\frac{p_{d-2}}{x^2}+\cdots\frac{p_0}{x^d}.$$
Then if $p_d>0$, we can find an $x$ large enough such that 
$$-\frac{p_d}d<\frac{p_{d-1}}x,\frac{p_{d-2}}{x^2},\cdots\frac{p_0}{x^d}<\frac{p_d}d,$$
for instance $$x>\dfrac{p_d}d\max(|p_{d-1}|,|p_{d-2}|,\cdots|p_0|).$$
Then, with this value of $x$ and all larger,
$$(1-\frac{d-1}d)p_d<\frac{P(x)}{x^d}<(1+\frac{d-1}d)p_d$$
which reads
$$cx^d<P(x)<c'x^d,$$
expressing $$P(x)=\Theta(x^d).$$
A: Let us assume you are interested in positive $n$. Since there are explicit negative signs, I also assume the $b_k \ge 0$. Now set:
$$ n_0 = \max \left(\frac{3b_{k-1}}{b_k}, \sqrt[k-1]{3\frac{b_{0}}{b_k}},1\right)\,.$$
They relate to the second monomial, the last one, and the $1$ avoids divisions by zero. Basically, the $3$ factors are meant to end up with something like:
$$ p(n) \ge b_k n^k  -\frac{b_k}{3} n^k -  \frac{b_k}{3} n^k\,,$$
to get a constructive bound. Other choices are possible.
So if $n \ge n_0$ , then 
$$b_{k-1} n^{k-1} = b_{k-1} \frac{n^{k}}{n}\le b_{k-1} \frac{n^{k}}{n_0} \le \frac{b_k}{3} n^k\,, $$ 
and similarly 
$$b_{0} n = b_0 \frac{n^{k}}{n^{k-1}}\le b_0 \frac{n^{k}}{n_0^{k-1}} \le \frac{b_k}{3} n^k \,.$$ 
Thus $p(n) \ge \frac{b_k}{3}n^k$.
Finally, for $n_0$ as above, you can choose $$c = \frac{b_k}{3}\,.$$
On the notation side, it  could be interesting to suppose $k \ge 2$, and a $b_1 n$ would be more consistent with your notations.
