16x^3-44x^2-42x=2x(8x^2-22x-21)=2x(4x+3)(2x-7). I worked out the factors by trying factors of 8 against factors of 21. The factors of 8 are (8,1) and (2,4); the factors of 21 are (21,1) and (3,7). The next step was to use both pairs of factors for each number in combinations: {(8,1) (21,1)} {(8,1) (3,7)} {(2,4) (21,1)} {(2,4) (3,7)}. If we represent the pairs that we're looking for by {(a,8/a) (b,21/b)}, then, ignoring whether the result is positive or negative, either ab-8/a*21/b=22 or a*21/b-b*8/a=22. By trying out each combination I discovered that a=4 and b=7 because 4*7-2*3=28-6=22. The final step is to consider the minus sign. Because we have -22, the expression requires 7 to be negative and therefore 3 to be positive so that we get -28+6=-22. Hence we arrive at the above factorisation. Another way to factorise is to use the quadratic formula on 8x^2-22x-21. x=(22+sqrt(484+672))/16=(22+sqrt(1156))/16=(22+34)/16=56/16=7/2 and -12/16=-3/4. These correspond to the factors 2x-7 and 4x+3.