Understanding why the polynomial $p(n) = \sum_{i=0}^{k} a_in^i$ is in $\Theta(n^k)$
The starting point is the inequality $p(n) \geq a_kn^k - An^{k-1}$. We want to deduce from this inequality that for large enough $n$, $p(n) \geq bn^k$ for some $b &gt; 0$. So take any $b &gt; 0$ satisfying $b &lt; a_k$. Since $$p(n) \geq a_k n^k - An^{k-1} = \left(a_k - \frac{A}{n}\right) n^k,$$ we have $p(n) \geq bn^k$ as long as $a_k - A/n \geq b$. As $n \to \infty$, $A/n \to 0$, and so it is clear that the inequality $a_k - A/n \geq b$ holds for large enough $n$. If we want an explicit lower bound on $n$, we can do some algebra to obtain the equivalent inequality $n \geq A/(a_k - b)$.