Step-by-step explanation:

Not in time for the exam, but anyhoo...

If you can't find common factors to extract from an equation like this, the NEXT step (at least when it's neat enough to have integer - or at least rational - COEFFICIENTS) is to attempt to guess rational solutions. Given the condition in the parenthesis above, that is actually fairly easy, as there exists a condition on what rational solutions are possible:

Assume, for the generality of it, that we have the equation

anxn+…+a1x+a0=0  where all coefficients are integers.

Construct the two SETS  A0  consisting of all divisors of  a0  (in this case  A0={1,2,3,6} ), and  A1  consisting of all divisors of  an  (in this case  A1={1} .

If the equation has a rational SOLUTION  ±pq , then  p∈A0,q∈A1 .

Thus, in this case, the only possible rational solutions are 1/1, 2/1, 3/1, 6/1, -1/1, -2/1, -3/1, -6/1. Additionally, we can see that no solution may be negative (that would have given a sum of four negative terms, adding to zero), so we have only four numbers to try: 1,2,3, and 6.

Of course, besides trial and error, we have no way to guarantee that this will provide us with a solution, but we can then say that any solution - beyond what this trial run gave - will be irrational, and may require the general formula for the solution of a cubic equation (in this case).