Finding a sufficient statistic for $\beta$ Let ${Y_1,...,Y_n}$ be independent random variables and $Y_i\sim N(\beta x_i, 1)$ where $x_1,...,x_n$ are fixed known constants, and $\beta$ is an unknown parameter.  
I'm trying to find a sufficient statistic for $\beta$
The issue I'm running into  is that after I find the likelihood I'm left with 2 betas in separate terms in the exponent. I'm not sure how to use the factorization criterion when there are different terms containing the parameter.
 A: The factorization criterion says that $T(y_1, \dots, y_n)$ is a sufficient statistic for $\beta$ iff we can find non-negative functions $h$ and $g$ satisfying $f_{\beta}(y_1,\dots, y_n) = h(y_1,\dots, y_n)g(\beta, T(y_1, \dots, y_n))$. So you should just gather all terms containing $\beta$ together and to look by means of which function $T(y_1,\dots, y_n)$ the resulting function $g(\beta, t)$ depends on $y_1, \dots, y_n$.
In your case the likelihood is
$$f_{\beta}(y_1, \dots, y_n) = (2\pi)^{-\frac{n}{2}}\exp \left(-\frac{1}{2}\sum_{i = 1}^n(y_i - \beta x_i)^2\right) = \\
= \left[(2\pi)^{-\frac{n}{2}}\exp \left(-\frac{1}{2}\sum_{i=1}^n y^2_i\right)\right] \exp\left(-\frac{1}{2}\left(\beta^2\sum_{i=1}^n x^2_i - 2\beta\sum_{i=1}^n y_ix_i\right)\right) = \\= h(y_1,\dots,y_n)g(\beta,T(y_1,\dots, y_n)).$$
Here $T(y_1,\dots,y_n) = \sum_{i=1}^n y_ix_i$ and 
$$g(\beta, t) = \exp\left(-\frac{1}{2}\left(\beta^2\sum_{i=1}^n x^2_i - 2\beta t\right)\right).$$
In case $x_1 = \dots = x_n = 1$ we get $T(y_1, \dots, y_n) = \sum_{i=1}^n y_i = n\bar{y}$ that is indeed a sufficient statistic for $\mu$ in $\mathcal{N}(\mu, \sigma^2)$. In your case $T(y_1, \dots, y_n)$ is (propotional to) the weighted mean.
