Why is $2$ double root of the derivative? 
A polynomial function $P(x)$ with degree $5$ increases in the interval
  $(-\infty, 1)$ and $(3, \infty)$ and decreases in the interval
  $(1,3)$. Given that $P'(2)= 0$ and $P(0) = 4$, find $P'(6)$.

In this problem, I have recognised that $2$ is an inflection point and the derivative will be of the form: $P'(x)= 5(x-1)(x-3)(x-2)(x-\alpha)$ .
But I am unable to understand why $x=2$ is double root of the derivative (i.e. why is $\alpha =2$?). It's not making sense to me. I need help with that part. 

HINT: 
If $2$ is an inflexion point then it is necessary that $P''(2)=0$, which means that you have a factor of $(x-2)^2$ in $P'(x)$.
