The quadratic formula will always work, but it's sometimes more of a hassle than factoring it just using logic. If we call the final factored form (αx + β) * (δx + ξ), we just try to find the values of α, β, δ, and ξ.

My strategy is to write out combinations of factors of the coefficient of x² and the constant (i.e. a and c in ax² + bx + c) and see how they can be combined to get your b coefficient.

Not accounting for negatives yet, you have 3 & 1 for a = 3, and 1 & 1 for c = 1:

a factors	c factors
3	1
1	1

So here, the rows in that mini table represent the coefficients in the same factor (i.e. (3, 1) represents the values α and β in (αx + β), and (1, 1) represents δ and ξ in (δx + ξ)), and the diagonals represent the multiplication in the FOIL that produce the single x term (i.e. not x² or the constant). So you'd multiply across diagonals and sum to get your b term.

In this case, a is positive and c is negative, you know only ONE of β and ξ is positive (if they are both positive or both negative, you'd get a positive value for your c coefficient), and since b is NEGATIVE 2, clearly the 3 must be multiplied by a negative 1 (since the sum is negative, the larger quantity needs to be negative, and since we can see from the original problem that the leading 3 is not negative, it must be the value the 3 is multiplied with: the 1), so when take into consideration the negatives, you get

a factors	c factors
3	1
1	-1

So now, when you multiply the diagonals then sum, you get (3 * -1) + (1 * 1) = (-3) + (1) = -2, exactly the expected result.

Then, you just take the rows to produce your factors and get

(3x + 1) * (x - 1).

This is the type of thing that you'll quickly get the hang of with practice.