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Let $P(x)$ be a polynomial of degree $n,$ then, remainder of $\left( \frac{P(x)}{x-a}\right)$ is $P(a)$, this is by the remainder theorem. However, what is the remainder of $ \frac{P(x)}{(x-a)^n}$? Are there any theorems for this?

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  • $\begingroup$ Consider $P(x)=x$, $a=1$, the remainder is still $x$. $\endgroup$ Jul 25, 2020 at 21:05
  • $\begingroup$ How does $\frac{P(x)}{(x-a)^2}$ give the same remainder of $P(a)$? We have $P(x) = P(a) + P'(a)(x-a) + R(x)(x-a)^2$, so the remainder is $P(a) + P'(a)(x-a)$. $\endgroup$ Jul 25, 2020 at 21:05

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Let $P$ be a polynomial of degree $m\ge n$. The remainder upon division by $(x-a)^n$ will be a polynomial of degree $n-1$, call it $R$. Then you have say $P(x)=(x-a)^n Q(x)+R(x)$. When $n=1$ $R$ had degree $0$, a constant say $r_0$ then we get $P(x)=(x-a)Q(x)+r_0$. We substitute the value $x=a$ to compute $r_0$, $P(a)=0+r_0$.

In case $n=2$ the remainder will be of degree $1$ i.e a linear function say $R(x)=r_1x+r_0$. Then we get $P(x)=(x-a)^2Q(x)+(r_1x+r_0)$. Now to compute the coefficients first we put $x=a$ to get $P(a)=r_1a+r_0$, this is a linear equation in two variables ($r_0,r_1$). Differentiating we get $P'(x)=2(x-a)Q(x)+(x-a)^2Q'(x)+r_1$ now substitute $x=a$ in this to get $P'(a)=r_1$.

In both cases the remainder was a polynomial. All you do is compute the coefficients using substitution. So the claim that the remainder is always a constant polynomial $P(a)$ is flat out false, as mentioned in the comments.

Try working out for what polynomials will your claim be true.

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  • $\begingroup$ what is degree of your P? $\endgroup$ Jul 25, 2020 at 22:13
  • $\begingroup$ thanks I added it. $deg(P)$ is anything greater than that of $(x-a)^n$. $\endgroup$
    – tassle
    Jul 25, 2020 at 22:16
  • $\begingroup$ I feel like a nice generalization of this could be made by usage of l'hopitals $\endgroup$ Jul 25, 2020 at 22:19
  • $\begingroup$ @DDD4C4U Sure go ahead and share what you are thinking. $\endgroup$
    – tassle
    Jul 25, 2020 at 22:20
  • $\begingroup$ pl check the new answer that I have put $\endgroup$ Jul 26, 2020 at 4:38
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Consider a polynomial function $P(x)$ of degree 'm', I Taylor expand the polynomial to make it of form,

$$ P(x) = P(a) + P'(a) (x-a) + \frac{ P''(a) (x-a)^2}{2}...\frac{P^{m+1}}{(m+1)!} (x-a)^{m+1} $$

Now, suppose for some $ 0<k<m+1$

$$ \frac{P(x)}{ (x-a)^k } = \frac{\underbrace{\sum_{j=0}^{j=k-1} P^{j}(a) (x-a)^j}}{(x-a)^k} + \sum_{j=0}^{k} \frac{ P^{m+1-j} (a) (x-a)^{m+1-j} }{ (m+1-j)!} $$

The underbraced item is the remainder when dividing by a repeated factor of $ (x-a)^k$, example:

The remainder of $\frac{P(x)}{(x-a)}$ is

$$ \sum_{j=0}^{0}\frac{ P^{j} (a)}{(x-a)^1} (x-a)^{j} = P(a)$$

Q.E.D

Intuition: Group the Taylor polynomial into the part divisible the repeated factor on bottom and non divisible ones, from this we can directly write the remainder.

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