# Relation between polynomial division and derivative

algebra-precalculusderivativespolynomialsreal-analysis

$$P(x)$$ 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?

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. $$a=-3$$, $$b=6$$, $$c=-2$$.