So I'm going over my practice midterms (which all seem to have solutions like this one),
Can anyone help clarify this for me? I understand that you multiply by the reciprocal to get to line two. But after that I'm completely lost, I don't understand how:
$$x^{2} + 1 - [(x + h)^{2} + 1]$$
can become:
$$(x-(x+h))(x+x+h)$$
and so forth, I'm sorry if this is a stupid question the solution doesn't seem to explain it very well.
This is one of those "verification left to the reader" moments.
If it helps, we can use the intermediate step that $$x^2-1-[(x+h)^2+1]=x^2-(x+h)^2,$$ so the conclusion follows from the fact that $a^2-b^2=(a-b)(a+b).$
Too, they didn't explain how they got from the third line to the fourth, but since $$(x-(x+h))(x+(x+h))=-h(2x+h),$$ that should be fairly straightforward. You'll run into this kind of thing a lot. If you're ever uncertain how they got there, just see if you can get there through some intermediate steps.
Well, $x^2+1-[(x+h)^2+1]=x^2-(x+h)^2=(x-(x+h))(x+(x+h))$. The last equality comes from the difference of two squares.
This is easier to see if we work backward:
$$\begin{align*}(x-(x+h))(x+x+h) & = (x-(x+h))(x+(x+h)) \\ & = x^2 -(x+h)^2 \\ & = x^2 + 1 -(x+h)^2 - 1\\ & = (x^2 + 1) -[(x+h)^2 + 1] \end{align*} $$
And, as others have pointed out, the trick on the first line here is to see a difference of two squares. This is fairly simple working backward like this, but you might want to manually expand all of the terms yourself, and then simplify and factor again.
