[Math] Find Big-O Estimate of $f(n) = n^3 + 4n^2 \log(n) + n + 1$

Question

So to show the big-O complexity I'm trying to find a C and a K such that $f(n) \le C g(n)$ for all $n > k$

To start solving $f(n) = n^3 + 4n^2 \log(n) + n + 1$

Fix $k = 1$, so $n > 1$

$f(n) = n^3 + 4n^2 \log(n) + n + 1 \le n^3 + 4n^3 \log(n) + n^3 + n^3$

$f(n) = n^3 + 4n^2 \log(n) + n + 1 \le 3n^3 + 4n^3 \log(n)$

But I'm not sure what to do next or if I even did that last step right. Any help is appreciated, thanks in advance.

Answer 1

Hint

$$f(n) = n^3 + 4n^2 \log(n) + n + 1 \leq n^3 +4n^3+n^3+n^3$$