Chapter 6 - Factoring, Solving Equations, and Problem Solving - 6.4 - Factoring Trinomials of the Form ax^2+bx+c - Problem Set 6.4 - Page 261: 7

$(4x-5)(3x-4)$

In accordance with the rules of factoring trinomials, we need to find two integers whose product is $240$ and whose sum is $-31$. A little searching determines that these numbers are $-15$ and $-16$. Therefore, we can express the middle term $-31x$ as $(-15x-16x)$ and proceed to factoring by grouping: $12x^{2}-31x+20$ =$12x^{2}-15x-16x+20$ =$3x(4x-5)-4(4x-5)$ =$(4x-5)(3x-4)$