Let $f: R \rightarrow R$ be a function defined by $f(x) = \begin{cases} x^2 - 4x + 3, & \text{if } x < 2 \\ x - 3, & \text{if } x \geq 2 \end{cases}$. Then the number of real numbers $x$ for which $f(x) = 8$ is:
- A$1$
- B$2$
- C$3$
- D$4$
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