Consider:
$$\frac{1}{2m}(\hat p (\hat p \hat x-\hat x \hat p) + (\hat p \hat x-\hat x \hat p) \hat p)f,$$ where $\hat p=$ momentum operator and $\hat x=$ position operator.
Can I do the following?
$$\frac{1}{2m}(\hat p (\hat p \hat x(f)-\hat x \hat p(f)) + (\hat p \hat x(f)-\hat x \hat p(f)) \hat p)$$
In case you're interested, I'm trying to figure out what $\frac{1}{2m}[\hat p^2,\hat x]$ is.
Using the identity \begin{align} [\hat x, \hat p] = i\hbar\delta \end{align} we see that \begin{align} [\hat p^2, \hat x] =&\ \hat p^2\hat x-\hat x \hat p^2 =\hat p(\hat x\hat p-i\hbar \delta) - (\hat p \hat x+i\hbar\delta)\hat p\\ =&\ \hat p \hat x\hat p-i\hbar\hat p -\hat p \hat x \hat p -i\hbar\hat p = -2i\hbar p. \end{align}
Moreover, if you want to use test functions, then it will go as follows \begin{align} [p^2, x] \varphi =&\ p[p(x(\varphi))]-x[p(p(\varphi))] \\ =&\ p[x(p(\varphi))-i\hbar\varphi]-p[x(p(\varphi))]-i\hbar p[\varphi]\\ =&\ -2i\hbar p[\varphi]. \end{align}
Additional: To help you understand the process more, let us write it more explicitly as follows \begin{align} \left[\frac{d^2}{dx^2}, x\right]f(x)=& \left(\frac{d^2}{dx^2}x- x\frac{d^2}{dx^2}\right)f(x)\\ =&\ \frac{d^2}{dx^2}[xf(x)]- x\frac{d^2}{dx^2}f(x). \end{align}