Answer
True.
Work Step by Step
If c is a zero, then $(x-c)$ is a factor of P (Factor theorem).
Then, $\displaystyle \frac{P(x)}{(x-c)}$ reduces to a polynomial $P_{1}(x)$, not defined for c.
Any zero (not c!) of $P_{1}$ is a zero of $P. $
The statement is true, all the OTHER zeros of P are zeros of $\displaystyle \frac{P(x)}{(x-c)}=P_{1}(x).$
