For question 17, a polynomial
![](https://latex.codecogs.com/png.latex?\bg_white P(x) = 0)
can inly have a multiple root if there are any
![](https://latex.codecogs.com/png.latex?\bg_white x)
-values for which
![](https://latex.codecogs.com/png.latex?\bg_white P(x) = P'(x) = 0)
, using the multiple root theorem. If you attempt to solve the equation
![](https://latex.codecogs.com/png.latex?\bg_white P(x) = P'(x))
, you should find it has only one solution,
![](https://latex.codecogs.com/png.latex?\bg_white x = 0)
. However,
![](https://latex.codecogs.com/png.latex?\bg_white P(0) = 1 \neq 0)
and so the only potential multiple root isn't a root at all, and so there are no multiple roots.
In fact, if
![](https://latex.codecogs.com/png.latex?\bg_white P(x))
is extended to an infinite polynomial, it is the polynomial version of
![](https://latex.codecogs.com/png.latex?\bg_white e^x)
, as indicated by the fact that you then have
![](https://latex.codecogs.com/png.latex?\bg_white P'(x) = P'(x))
and thus there are no zeroes of
![](https://latex.codecogs.com/png.latex?\bg_white P(x) = 0)
at all.