Since the leading coefficient 3 can only be factored only one way, I think the trial-and-error factoring method is better in this case: 3x²-31x-60=0 Since the last term -60 is negative, we know that the absolute value of the middle coefficient, -31, which is 31, must be the result when the absolute values of the product of the outer and inner terms are subtracted. [If the last term were +, we would know that they would be added]. The factorizations are either of these forms: (x A)(3x B) (x B)(3x A) where A and B are as in this chart: ∥ A, B ∥ |B-3A|,|A-3B| ∥ 1,60 ∥ 57 179 ∥ 2,30 ∥ 24 88 ∥ 3,20 ∥ 11 57 ∥ 4,15 ∥ 3 41 ∥ 5,12 ∥ 3 31 ∥ 6,10 ∥ 8 24 How to make that chart: In the first column we list all the factor pairs of 60. Then we form the differences mentally, until we come to the absolute value of the coefficient of the middle term, 31. In this case it was the red one. Often when using this method, we will reach it much sooner, and stop whenever we reach it. Now since the 31 came from the second column we know that the factorization is of the form: (x B)(3x A) so the factorization is (x 12)(3x 5) Now the product of the outer terms is 5x and the product of the inner terms is 36x, so to get the middle term -31x, the 36x must be made negative and the 5x must be made positive. So the factorization is (x-12)(3x+5). It takes some practice to use this method, but it'll save time when either the first or last term can be factored only one way. Edwin