We begin with #5+34x-7x^2#. I am not used to seeing a problem written this way, so I am going to rewrite it into standard form (#ax^2+bx+c#).
If we do that, we end with #-7x^2+34x+5#. From here, I'm going to do one last thing: I'm going to factor out that negative in front of the #7x^2#. I'm doing that because I don't want to have to deal with it, and because I am not actually changing the expression, just rewriting it. NOW we have something I'm ready to factor: #-(7x^2-34x-5)#.
Let's get started with factoring. Now, if #-(7x^2-34x-5)# is in standard form, then #7# would be #a#, #-34# would be #b#, and #-5# would be #c#. Keep that in mind when i discuss factoring. Anyways, the first thing I always do is find two numbers that can be multiplied to equal the #a*b# and can be added to equal #c#.
So, what does #a*b# equal? For us, that's #7*-5#, which is #-35#. And in our case, #c# is #-34#.
Okay, we know what we're looking for: two numbers that add to #-34# and multiply to #-35#, and I think #1*-35# fits those constraints!
From here, we just write the first and last values with space left for the numbers we just found, like this: #(7x^2+ ____x)+(____x+-5)#. We can insert the #-35# and the #1# wherever we want. I'm going to put the #1# with the #7x^2# and the #-35# with the #-5#, but you can do it any way you want.
We now have #(7x^2+ 1x)+(-35x+-5)#. I see that we can factor out an #x# from the first parentheses and a #-5# from the second. that gives us #x(7x+1)+ -5(7x+1)#.
We now take the factored values (the #x# and the #-5#) and write them alongside the other group (#7x+1#) to give us #(x-5)(7x+1)#.
We are almost done, save for the fact that the whole problem looks like this #-((x+1)(7x+1))#. We multiply in the negative for one of the groups (I picked #7x+1#). That leaves us with #(-7x-1)(x+1)#. Now we are done.
