Why is 2 double root of the derivative?

A polynomial function P(x) with degree 5 increases in the interval $(-\mathrm{\infty},1)$ and $(3,\mathrm{\infty})$ and decreases in the interval (1,3). Given that ${P}^{\prime}(2)=0$ 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: ${P}^{\prime}(x)=5(x-1)(x-3)(x-2)(x-\alpha )$.

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.

A polynomial function P(x) with degree 5 increases in the interval $(-\mathrm{\infty},1)$ and $(3,\mathrm{\infty})$ and decreases in the interval (1,3). Given that ${P}^{\prime}(2)=0$ 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: ${P}^{\prime}(x)=5(x-1)(x-3)(x-2)(x-\alpha )$.

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.