Question 703275
Of course, I set the example to zero, found the greatest fact and this is my result: 3(2x^2-3x-20)=0 was the result which I then factored further by grouping and the result was: 3(2x+5)(x-4)=0 then I set each factor to zero and got the correct answer of x=-5/2 and x=4 which the example showed as the possible results, however, it is confusing to me why we disregard the 3 which is another factor. Can you give me some thoughts regarding this. 

Good question!
So if we have two (or more) numbers multiplied together that equal 0, then one or more of those factors must equal zero right?
Well what if we had 3x = 0.  3 time a number equals zero.  Then the only solution is x = 0.  The 3 doesn't give us any solutions because 3 never equals zero.
You factored correctly:  3(2x + 5)(x - 4) = 0
You have three numbers that multiply to give zero, 
Then we set each factor to zero
3 = 0, 2x + 5 = 0, or x - 4 = 0
The first equation doesn't give us any solutions, because it is a contradiction, 3 never equals 0.
The only options are that either the second number (2x + 5) or the third number (x - 4) equals 0 (or both).