Why is $2$ double root of the derivative?

Question

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.

Answer 1

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 $P'(x)$.

Answer 2

Hint:

$2$ is an inflexion point so necessarily $P''(2) = 0$.

Now $P'(2) = P''(2) = 0$ so $2$ is a double root of $P'$.

Answer 3

Since $2$ is an inflection point, we know that $P'(2)=0$ and $P''(2)=0$. (Convince yourself of this. What happens if $P''(2) \neq 0$?)

$P''(x)$ is then $-15ax^2 + 60ax - 55a + 20x^3 - 90x^2 + 110x - 30$. Plug in $2$ to get

$$P''(2) = 5 a - 10$$

And so $P''(2)$ is $0$ when $a=2$.

EDIT: mechanodroid gives us a really clever shortcut:

If $P′$ has a double root at $2$ then $P′(x)=(x−2)^2 Q(x)$. This immediately gives $P′(2)=0$. Differentiating $P′$ gives: $P″(x)=2(x−2)Q(x)+(x−2)^2 Q′(x)$ so $P″(2)=0$.

This means that knowing that $P'(2) = P''(2) = 0$ is sufficient to conclude that $(x-2)^2|P'(x)$, hence $\alpha=2$.

Answer 4

The fact that $x=2$ is an inflection point implies that the second derivative $P''(2)=0$ as well. That means $2$ should not just be a root of $P'$, but should still be a root if you take the derivative again. That is the case if it is a double root.