How do you factor #2x^2-5x+3#?

Factoring small polynomials, such as the one in this problem, is actually easier than it seems.

In this equation there are two key parts.
The #2x^2# and the #3#.

When I try to factor a small polynomials I always look at the coefficient of the largest degree term, so the #2# in #2x^2#, and the constant, the #3#.

From these two numbers we can do a number of things.

The #2# in #2x^2# can tell us that the factored form will look something like #(2x+- _ )(x+- _)#.
Since #1 & 2# are the only integers that you can multiply to get #2#.

The constant, #3#, can also tell us something about the factored equation.
Think about what numbers you can multiply to get three, which will be #1&3#.
And because the #3# is actually #-3# the numbers are actually #-1&-3#, #-1&3# or #1&-3#.

Now the rest is just a simple guess and check.
When you get used to factoring small polynomials you can start doing this in your head. But for now you should check every possibility just to be sure.

So the possible factored form of this equation to get the #2x^2# and #3# are such:

#(2x-1)(x-3)#
#(2x+1)(x-3)#
#(2x-1)(x+3)#
#(2x-3)(x-1)#
#(2x+3)(x-1)#
#(2x-3)(x+1)#

And the one that works is #(2x+1)(x-3)#.

Hope this helped, sorry for such a long explanation. I'm still new to Socratic :3