Question 8427
There a several ways to solve this. The easiest is to use the quadratic equation, however I suspect, given the results I get from the quadratic equation, that you are actually required to factor the problem, so, how do you do this?<br>
First, notice that you result will have to have the form<br>
{{{(ax + b)(cx + d) = acx^2 + (ad + bc)x + bd}}}.<br>
The goal is to find the values for a, b, c and d. To do this notice that according to the equation,:<br>
ac = 6,
ad + bc = 14 and
bd = -12<br>
There are four possible ways to get ac = 6<br>
a&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;c&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;ac
6&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;1&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;6
1&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;6&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;6
3&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;2&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;6
2&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;3&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;6

There are twelve ways to get db = -12

d&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;b&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;db
&nbsp;12&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;-1&nbsp;&nbsp;&nbsp;&nbsp;-12
-12&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;1&nbsp;&nbsp;&nbsp;-12
&nbsp;&nbsp;1&nbsp;&nbsp;&nbsp;-12&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;-12
&nbsp;-1&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;12&nbsp;&nbsp;&nbsp;&nbsp;-12
&nbsp;6&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;-2&nbsp;&nbsp;&nbsp;&nbsp;-12
-6&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;2&nbsp;&nbsp;&nbsp;&nbsp;-12
&nbsp;2&nbsp;&nbsp;&nbsp;&nbsp;-6&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;-12
-2&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;6&nbsp;&nbsp;&nbsp;&nbsp;-12
&nbsp;4&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;-3&nbsp;&nbsp;&nbsp;&nbsp;-12
-4&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;3&nbsp;&nbsp;&nbsp;&nbsp;-12
&nbsp;3&nbsp;&nbsp;&nbsp;&nbsp;-4&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;-12
-3&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;4&nbsp;&nbsp;&nbsp;&nbsp;-12


With these numbers, there are forty-eight combinations we can used to find {{{ad + bc}}}. We could list them, or we try to examine the numbers in these tables. As it happens, 6*3 - 1*4 = 14, which is the result we are looking for. This means, the solution is {{{(6x - 4)(x + 3)}}}