Question 1026580
.
Factor by grouping (sometimes called the ac-method). 
5x^2-x-6
~~~~~~~~~~~~~~~~~~~~~~~~~~~


<pre>
Simple trick: I will present 6 = 5 + 1, and then . . . 

{{{5x^2 - x - 6}}} = {{{5x^2 - x - 5 - 1}}} = {{{(5x^2-5) - (x+1)}}} = {{{5*(x^2-1) - (x+1)}}} = 5*(x-1)*(x+1) - (x+1) = (x+1)*(5x-5-1) = (x+1)*(5x-6). 

Done.
</pre>