You first reserve hours for lots of fun being frustrated at computationally intensive trial and error. Seriously, it's hard to factor something like that, especially with that leading 18 coefficient. There are various tricks (there's even a formula, like the quadratic formula, for cubics (Cardano's Formula) - but it's rather messy), but none will make you happy unless you get lucky and one of them works prefectly. In general, it sucks.

If it's going to factor over the integers (meaning it has a rational solution) then it will have a root which is a ratio of a divisor of 100 divided by a divisor of 18.

Divisors of 100: {1, 2, 4, 5, 10, 20, 25, 50, 100}

Divisors of 18: {1, 2, 3, 6, 9, 18}

Thus plug each of these:

{1, 2, 4, 5, 10, 20, 25, 50, 100}

and {1/2, 5/2, 25/2}

and {1/3, 2/3, 4/3, 5/3, 10/3, 20/3, 25/3, 50/3, 100/3}

and {1/6, 5/6, 25/6} and {1/9, 2/9, 4/9, 5/9, 10/9, 20/9, 25/9, 50/9, 100/9}

and {1/18, 5/18, 25/18}

into this and check if any of them gives you 0.

Then you're HALF done, because you have to check all the same numbers, but with a minus sign in front.

And when you're done plugging those 72 different fractions into that cubic, if none of them give you 0, then all you can say is that "either I made a computational mistake evaluating at least one of them - which is a virtual certainty - or there are no rational roots to that cubic, and so it can't factor nicely over the integers".

Have fun.

Such approaches can work, but virtually never do. Unless you're fantastic at factoring, I'd suggest not wasting your time even trying to get it to work like this.

I'll give you an example where it works, but again, in my experience this rarely seems to be a beneficial thing to spend too much time looking for (unless you're some kind of polynomial multiplying savant):

.