There are two methods.  The trial and error method and the "ac"-method.
When a trinomial starts or ends with a prime number or 1, the trial and
error method is easier.  In your problem both 2 and 5 are prime, so
the trial-and-error method is easier.  I'll show it here.  However,
if you want to learn the ac-method, go here:

http://www.algebra.com/my/change_this_name32371.lesson?content_action=show_dev

Factor 2x²+9x-5

It's the reverse of FOIL

The first coefficient is 2.

The only way to factor that first coefficient, 2, is as 2·1

The constant (ignoring signs) is 5.

The only way to factor 5 is  as 5·1

So the incomplete factorization without the signs 
has to be one of these two:

(2x   5)(x   1)  or  (2x   1)(x   5)

-----
For (2x  5)(x   1), if we used FOIL, the OUTERS and INNERS
would be 2x and 5x, but there is NO way to assign signs
to them so that when they are combined they would give 
the middle term +9x.  So that is not the right form.

For (2x  1)(x   5), if we used FOIL the OUTERS and INNERS
would be 10x and 1x, and there IS a way to assign signs
to them so that when they are combined they would give 
the middle term +9x.  So we have to assign the 10x a +
sign and the 1x a - sign.  The 10x comes from the OUTERS
so we put a + before the 5, and we put a - sign before 
the 1:

Answer:  (2x - 1)(x + 5)

Check by multiplying it out by FOIL:

     2x² + 10x - 1x - 5
       2x² + 9x - 5

That's the original trinomial, so we are correct.


Edwin