Harmonic Oscillator (Quantum Mechanics) Griffiths uses an algebraic "brute force" technique to solve the harmonic oscillator. I'm somewhat confused regarding a few parts. 
$$\frac{1}{2m}[p^2 + (m \omega x)^2] \psi = E \psi$$
$H = \frac{1}{2m}[p^2 + (m \omega x)^2]$
We are about to factor $H$, noting that the numerical equavilant is $u^2 + v^2 = (iu+v)(-iu+v)$. We now define $a_{\pm} = \frac{1}{\sqrt{2 \hbar m \omega }}(\pm ip+m \omega x)$
Skipping through the commutator, I don't understand how we can say: $a_+ a_ - = \frac{1}{\hbar \omega}H + \frac{1}{2}$
Looking at the original equation, we factored $[p^2 + (m \omega x)^2]$, so we can replace this with $a_+ a_ - $. Under this, couldn't we just say that $H = \frac{1}{2m} (a_+ a_ -) $
My second question has to do with the statement (this is a direct quote): Now, here comes the crucial step: I claim that  if $\psi$ satisfies the Schrödinger equation with energy $E$, (that is: $H \psi = E \psi$), then $a_+ \psi$ satisfies the Schrödinger equation with energy $(E+ \hbar \omega)$
I don't understand why we're multiplying $\psi$ by a piece of Hamiltonian, and the by the Hamiltonian again. 
 A: I think perhaps what you're missing is in the "skipping through the commutator" part. Do you understand where we get this equation (try computing it yourself, if not):
$$a_{-}a_{+} = \frac{1}{2 \hbar m \omega}[p^{2} + (m\omega x)^{2}] - \frac{i}{2\hbar}[x,p]$$
Now, the canonical commutator, I'm sure you noticed (as it's boxed on the same page in Griffiths) is $[x,p] = ih$. Insert this into the above equation and note that we now have:
$$a_{-}a_{+} = \frac{1}{2 \hbar m \omega}[p^{2} + (m\omega x)^{2}] + \frac{1}{2}$$
All you need to do from there recognize the first term as $\frac{1}{h\omega}H$.

Looking at the original equation, we factored $[p^{2}+(m\omega x)^{2}]$, so we can replace this with a+a−. Under this, couldn't we just say that $H=\frac{1}{2}(a_{+}a_{−})$

Careful here... remeber that $p$ and $x$ in this expression (and in the Hamiltonian generally) are operators, not scalars. This is why our "intuitive guesses" of $a_{\pm}$ are not exact factors of $[p^{2}+(m\omega x)^{2}]$, and why the canonical commutator above is important.
Edit: I just noticed that Griffiths does include this intermediate step in computing $a_{-}a_{+}$:
$$a_{-}a_{+} = \frac{1}{2 \hbar m \omega}[p^{2} + (m\omega x)^{2}-im\omega(xp-px)]$$
Notice that if $x$ and $p$ were scalars, the rightmost term would be 0, and your intuition about $a_{-}$ and $a_{+}$ being "factors" would be correct. Once you realize they are operators, however, it's obvious that we need to substitute $[x,p] = xp-px = ih$.
