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.
