How can this cubic inequality be solved? (2015 A-level paper) I'm trying to complete the following question (from the AQA June 2015 FP1 A-level paper):

By first finding a suitable cubic inequality for $k$, find the greatest value of $k$ for which
  $$\sum_{r=k+1}^{60} (3r+2) \log_8 4^r$$
  is greater than 106060.

The "suitable cubic inequality" referred to is this:
$$2k^3 + 5k^2 + 3k - 132000 \lt 0$$
and the final value of $k$ is 39 (since $k$ must be an integer to be used in the summation), but I can't see how to get from the inequality to the solution.
My typical strategy for this type of question is to form an equation equal to 0 in order to find the roots:
$$2k^3 + 5k^2 + 3k - 132000 = 0$$
but in this case that doesn't seem to help, since it's a cubic, not a quadratic. If it was a quadratic, it would be simple to find the roots via the quadratic formula, then form an inequality or two with $k$, and determine the largest possible value of $k$ from there, but I have no idea how to find the roots of a cubic. The constant term also prevents dividing by $k$ to form a quadratic.
It feels like I'm missing something really obvious, apologies if that's the case.
 A: Hint:  it will take some guesswork, since the cubic has no rational roots and the solution must be an integer. Given $2k^3 \lt 2 k^3 + 5 k^2 + 3k \lt 132000$ it follows that $k \lt \sqrt[3]{132000/2}\,$ $= 10 \sqrt[3]{66}\,$ $\simeq 40.4\,$. First try $k=40$ does not satisfy the inequality, but then next lower value $k=39$ works.
A: Note that all the coefficients except the constant are positive, so if you want the largest positive real root of $f(k) = 2k^3 + 5k^2 + 3k - 132000$, it suffices to use rough bounds; e.g., $$2k^3 + 5k^2 + 3k \le 2(k^3 + 3k^2 + 3k + 1) = 2(k+1)^3,$$ hence $k \le (66000)^{1/3} - 1 = 10 (66)^{1/3} - 1$.  A calculator gives this as $k \le 39$, after which point we can test $k = 40$ reasonably easily and determine $f(40) > 0$.  Calculating $f(39)$ seems harder, but it's not too bad, since we can write $$f(k) = 2(k+1)^3 - (k+1)(k+2) - 132000 = k(k+1)(2k+3) - 132000,$$ hence $$f(39) = 39(40)(83)-132000 = -2520.$$
If you're not happy with the idea of using a calculator to compute $66^{1/3}$, we can also do a rough calculation:  $$4^3 = 64 < 66 < 68.921 = (4.1)^3,$$ hence $4 < 66^{1/3} < 4.1$.
