Question 1014361
<pre>
Since the leading coefficient 3
can only be factored only one way,
I think the trial-and-error factoring 
method is better in this case:

<b>
3x²-31x-60=0
</b>
Since the last term -60 is negative,
we know that the absolute value 
of the middle coefficient, -31, which
is 31, must be the result when the
absolute values of the product of the
outer and inner terms are subtracted. 
<font size=1>[If the last term were +, we would 
know that they would be added].</font>

The factorizations are either of
these forms:

(x A)(3x B) 
(x B)(3x A)

where A and B are as in this chart:

&#8741; A, B &#8741; |B-3A|,|A-3B|
&#8741; 1,60 &#8741;   57    179
&#8741; 2,30 &#8741;   24     88
&#8741; 3,20 &#8741;   11     57
&#8741; 4,15 &#8741;    3     41
&#8741; 5,12 &#8741;    3     <font color="red"><b>31</b></font>
&#8741; 6,10 &#8741;    8     24

How to make that chart:

In the first column we list all the
factor pairs of 60. Then we form
the differences mentally, until we
come to the absolute value of the
coefficient of the middle term, 31. 
In this case it was the red one.  
Often when using this method, we will 
reach it much sooner, and stop 
whenever we reach it.

Now since the 31 came from the second
column we know that the factorization 
is of the form:

(x B)(3x A) 
 
so the factorization is 

(x 12)(3x 5)

Now the product of the outer terms
is 5x and the product of the inner 
terms is 36x, so to get the middle 
term -31x, the 36x must be made
negative and the 5x must be made 
positive.  So the factorization is

(x-12)(3x+5).

It takes some practice to use this
method, but it'll save time when 
either the first or last term can 
be factored only one way.

Edwin</pre>