Generalized remainder theorem for powers of linear factors

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?

• Consider $P(x)=x$, $a=1$, the remainder is still $x$. Jul 25, 2020 at 21:05
• 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 $P(a) + P'(a)(x-a)$. Jul 25, 2020 at 21:05

2 Answers

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.

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

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

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

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.