Question 35930
Wow!!  That's a tough one to do by my usual trial and error method.  So, let's try setting it equal to zero, and solve by the quadratic formula.  The numbers will be HUGE so I'll try to use the Quadratic Equation solver:

{{{135x2 - 222x + 91= 0}}}

*[invoke solve_quadratic_equation 135, -222, 91]


The solutions are x=0.866666 . . .  and x= 0.7777777 . . .

With a calculator you can easily convert these repeating decimals to fractions, and you might be able to get the second one in your head.  I get x=13/15, and the second one is x = 7/9.


Now what this means is 
x=13/15
15x=13
(15x-13)=0


Also, 
x=7/9
9x=7
(9x-7)=0


Therefore the factored form for the trinomial is as follows:

{{{135x2 - 222x + 91}}}
{{{(15x-13)(9x-7) }}}


My other idea would be to graph this on a graphing calculator that can find the zeros (or roots) of a graph.  Determine the decimal values of these roots and convert them to fractions, proceeding in the same way, but without solving by the quadratic formula.


What a GREAT problem!  If you want to re-post this question, and see if someone else has any better ideas on it, I will refrain from answering it a second time in order to allow a second opinion.  (And I will not have my feelings hurt when I see it posted again!!)


R^2 at SCC