Relation between polynomial division and derivative 
$P(x)$ is a polynomial and it is equal to $2x^3 + 2ax^2 +bx +c$ . It is given that $P(x)$ can be divided by $(x-1)^3$ with zero remainder. Then , what is $c$ ?

This is a basic polynomials question, to reach the solution we use derivative for shortcut. For example, we find $P(1) ,P'(1),P''(1)$ respectively. Then answer is $2$..
My question is why we use derivative, I could not conceive the reason behind the usage of derivative. In first thought, I thought that if $(x-1)^3$ divides $P(x)$ , then $(x-1)^2$ and $(x-1)$ divides $P(x)$, as well. However, I could not see any relation with derivative. Can you enlighten me?
 A: That is much easier to solve. As degree of $P$ is $3$ $P$ needs to be a constant multiple of $(x-1)^3$. The lead coefficient gives us that this constant factor is $2$ and thus $c=2(-1)^3 = -2$.
But for your question: You can easily proove that $x_0$ is only a double root of a polynomial if it’s also a root of the derivative, and more general, $x_0$ is a root with muliplicity $n$ if and only if it is a root of the derivative of multiplicity $n-1$.
Thus $x-1$ must also be a root of $P'$ and of $P''$. This means that $c$ has to be chosen in such a way, that $(x-1)$ is a common factor of $P,P',P''$.
If we choose $x=1$ then $(x-1)=0$, so
$$ 0=P(1) = 2+2a+b+c$$
$$ 0=P'(1) = 6+4a+b $$
$$ 0=P''(1) = 12+4a $$
Now you could solve this system, i.e. $a=-3$, $b=6$, $c=-2$.
A: So, for all $x$ we have $$2x^3 + 2ax^2 +bx +c =k(x-1)^3\;\;\;\;\;\;(\spadesuit)$$ for some constatnt $k$. Clearly we see that if we put in it $x=1$ we get $$2+2a+b+c =0$$
Now let's take a look what happend if we take the derivative of $(\spadesuit)$, we get:
$$6x^2+4ax+b = 3k(x-1)^2\;\;\;\;  (\diamondsuit)$$ which is valid also for all $x$, so in particular, for $x=1$ we get:
$$6+4a+b=0$$ and for the last time, if we again take the derivative of $(\diamondsuit)$ we get: $$12x+4a=6k(x-1)$$ which is again valid for all $x$ and thus also for $x=1$...
A: Assume a zero of order $p$ for the function $f(x)$.
This implies that we can write $f(x) = (x-x_0)^pg(x)$
Now, look at the derivative.
$f'(x) = p(x-x_0)^{p-1}g(x) + (x-x_0)^pg'(x)$
For $p=2$ we see that the second order zero will result in $f'(x=x_0)=0$.
You can repeat the same logic for higher order zeros.
