$P(x)$ 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?

## Best Answer

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 itâ€™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. $a=-3$, $b=6$, $c=-2$.