Answer:
Explanation:
First one is simple enough. completing the square is simply taking the b term, or the term with just x as a variable, cutting it in half and squaring it then taking this new term and adding it to both sides of the equation, or adding it and subtracting it on one side so it zeroes out, you then take the x^2 and x term and the new ADDED term, not the subtracted one, and it works out to be (x+b/2)^2 + c - (b/2)^2.
That may have been confusing to follow so I will work it out. You do want to make sure you are starting with just x^2 as the first term. if it is not x^2 and there is a number in front of it you will have to factor it out.
So, we start with the first one
-x^2-8x-31
First, take not that we don't just have x^2 for the first term, we have -x^2, so we need to factor out a -1 from everything.
-(x^2+8x+31)
Secondwe take the coefficient with just x on it and add and subtract it to this one part of this equation, so that winds up being +(8/2)^2 - (8/2)^2 or + 16 - 16
-(x^2+8x+16-16+31)
Now, just looking at x^2+8x+16 we can factor it into (x+(8/2)^2. If you need help with that part let me know and I can give an explanation. But when you complete the square you will always get something in the form (x+b/2)^2 where b is that coefficient with just an x by it from when we started.
-((x+4)^2+15)
Now at the end we redistribute whatever we factored out, if we did. In this case we factored out -1 so we redistribute it.
-(x+4)^2-15
There you go, it is in the form that was requested, called vertex form. Not sure what to do with that second one though.
Also I just wanted to repeat, make sure you only have x^2 as your first term. If instead for this one you had, since this got me a lot when I was first learning it. If the original problem was -5 x^22 - 8x - 31 for instance we'd have to get rid of that -5.
-5x^2-8x-31
-5(x^2 + (8/5)x + 31/5)
That is what you would have gotten instead. It doesn't look pretty but you're going to want to keep your eye out and make sure you have just x^2 at the front.