In this problem, we will prove that a polynomial of degree $n$ has at most $n$ zeros (Theorem 6 ). Give a reason for each step. Let $P(x)$ be a polynomial of degree $n,$ and suppose that $P$ has $n$ distinct zeros $r_{1}, r_{2}, \ldots, r_{n}$. We will show that it is impossible for $P$ to have any other zeros.
Step
1: We can write $P(x)$ in the form $P(x)=\left(x-r_{1}\right) Q_{1}(x),$ where the degree of $Q_{1}(x)$ is $n-1 .$
Step
3: We can write $Q_{1}(x)$ in the form $Q_{1}(x)=\left(x-r_{2}\right) Q_{2}(x)$, where the degree of $Q_{2}(x)$ is $n-2$.
Step 4: $P(x)=\left(x-r_{1}\right)\left(x-r_{2}\right) Q_{2}(x)$
Step
5: $P(x)=\left(x-r_{1}\right)\left(x-r_{2}\right) \cdots\left(x-r_{n}\right) Q_{n}(x)$, where the
degree of $Q_{n}(x)$ is $0 .$
Step
6: The only zeros of $P$ are $r_{1}, r_{2}, \ldots, r_{n}$