If 6x^2 + x - 12 = (ax + b)(cx + d), then |a| + |b| + |c| + |d|	Select
A) 10 B) 12 C) 15 D) 18 E) 20
This is the hard one, definitely a 700+ level question. We need numbers a, b, c, and d such that   6x^2 + x - 12 = (ax + b)(cx + d)   This means that ac = 6, bd = –12, and ad + bc = 1. The a &amp; c pair could be (1, 6) or (2, 3), in some order. The absolute values of the b &amp; d pair could be (1, 12) or (2, 6) or (3, 4), and of course, in each case, one of the two would have to be negative. After some trial and error, we find:   6x^2 + x - 12 = (2x + 3)(3x - 4)   Thus, we see:   |a| + |b| + |c| + |d| = 2 + 3 + 3 + 4= 
<gadget id="calculator">2 + 3 + 3 + 4</gadget>
<output>12</output>
12   Answer= 12
<result>B</result>
If 6x^2 + x - 12 = (ax + b)(cx + d), then |a| + |b| + |c| + |d|