Question: [![enter image description here](https://i.stack.imgur.com/XWIA0.jpg)](https://i.stack.imgur.com/XWIA0.jpg)

[![enter image description here](https://i.stack.imgur.com/KmZLY.jpg)](https://i.stack.imgur.com/KmZLY.jpg)

How is $p\ '(x)$ written in that solution? I understand that $p\ '(x)$ will be some $4$ degree polynomial but how did we find out that $2$ and $-2$ will be roots of $p\ '(x)$?

Answer: 
By the Division Algorithm, we know that there are some degree $2$ polynomials $Q\_1, Q\_2$ such that:
$$
P(x) = Q\_1(x) \cdot (x - 2)^3 - 2 = Q\_2(x) \cdot (x + 2)^3 + 2
$$

Differentiating each equation with product rule, we get:
\begin{align\*}
P'(x) &= Q\_1'(x) \cdot (x - 2)^3 + Q\_1(x) \cdot 3(x - 2)^2
= (x - 2)^2 \cdot (\underbrace{(x - 2) Q\_1'(x) + 3Q\_1(x)}\_{\textsf{some polynomial of degree $2$}}) \\ \\
P'(x) &= Q\_2'(x) \cdot (x + 2)^3 + Q\_2(x) \cdot 3(x + 2)^2
= (x + 2)^2 \cdot (\underbrace{(x + 2) Q\_2'(x) + 3Q\_2(x)}\_{\textsf{some polynomial of degree $2$}})
\end{align\*}

But then since $(x - 2)^2$ and $(x + 2)^2$ are coprime factors of $P'(x)$ and since $\deg P' = 4$, it follows that:
$$
P'(x) = K(x - 2)^2(x + 2)^2
$$
for some constant $K$, as desired.
Hint: Take derivative of that polynomial to get $16 - 8 x^2 + x^4=(x^2-4)^2$.