We know that the zeroes, if they exist, must be factors of 6, which are 1, 2 and 3, so the zeroes may be ±1, ±2, or ±3, and we know the factors of the highest power coefficient 2 are 1 and 2. So we could have ±1/2 or ±3/2 as zeroes. We can try these possibilities fairly quickly, and we need to test the positive and negative versions. If we can reduce the cubic to a quadratic, we know how to find the zeroes of a quadratic, even if we have to use the formula.
We can look for zeroes where x is a small number like 1. Put x=1, and we get 2, because 2-5-1+6=2. Now try x=-1: -2-5+1+6=0, (note that it’s the same set of numbers but the signs on the odd powers of x change), so we quickly discovered that x=-1 is a zero. We can divide by factor x+1 or use synthetic division with the root -1 to give us the quadratic: 2x²-7x+6, which factorises simply: (x-2)(2x-3). So the other zeroes are x=2 and x=3/2. The complete factorisation is (x+1)(x-2)(2x-3). We could have continued to try the other possible zeroes and found them all without bothering with the quadratic, but it would have taken longer.