Question 989174
{{{acx^2+bcx+adx+bd}}}
{{{acx^2+(bc+ad)x+bd}}}


{{{system(ac=6,bc+ad=1,bd=-12)}}}


A better way is to check separate trial combinations based on factorizations for 6 and factorizations for -12.


<pre>
(2x 2)(3x 6)
(2x 6)(3x 2)
(2x 3)(3x 4)
(2x 4)(3x 3)
(2x 1)(3x 12)
(2x 12)(3x 1)
but then there are six other combinations to check also!....
</pre>