Finding $c$ and $n_0$ for a big-O bound
$c=5$ is clearly the wrong answer, since for any $n$,
$$cn^3 &lt; 5n^3 + \text{(positive terms)}$$

However, if we take $c&gt;5$ then there will exists some $N_0$ such that for $n&gt;N_0$
$$cn^3 \ge 5n^2+2n^2+22n+6$$

If we let  $g(n)=(c-5)n^3-2n^2-22n-6$, then Since $\lim_{n\to\infty} g(n)=\infty$,  by the limit definition there always exists $N_0$ such that for any $n&gt;N_0$, $g(n) &gt;0$. This implies the above claim. 

To find that $N_0(c)$ you can try to find the roots of $g(n)$. $N_0$ will be the most positive root.