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**johnsomeone** $\displaystyle f(x) = \frac{(x-3)(x-2)(x+1)}{(x+1)(x-3)}, x \notin \{-1, 3 \}$ already is peicewise continuous. I assume the problem is actually asking how to make it continuous.

Here are some questions for you. I'm going to define a function on the same domain. Let $\displaystyle g(x) = x-2, x \notin \{-1, 3 \}$.

First Question: What's the relationship, if any, between $\displaystyle f$ and $\displaystyle g$?

Second Question: Can you think of a continuous function defined on all of $\displaystyle \mathbb{R}$ that equals $\displaystyle g$ on $\displaystyle g$'s domain of $\displaystyle x \notin \{-1, 3 \})$?

In other words, find a function $\displaystyle \tilde{g}: \mathbb{R} \rightarrow \mathbb{R}$ such that $\displaystyle \tilde{g}$ is continuous everywhere, and $\displaystyle \tilde{g}(x) = g(x)$ for all $\displaystyle x \notin \{-1, 3 \}$.

What you're dealing with is something called a "removeable discontinuity", for reasons you'll hopefully soon understand. For now, classify these problems under "So Simple They're Confusing".