1
$\begingroup$

This question already has an answer here:

Prove that the following polynomial $$p(x) = \sum_{k=0}^n x^k$$ has no multiple roots

Hint: consider the polynomial $(x-1)\times \sum_{k=0}^n x^k$

I have tried finding the derivative of the polynomial and trying to reach a contradiction by supposing there is a root such that p(x) = p'(x) = 0 but I can't reach any conclusions. The hint leads me nowhere, I don't know how to use that information

Thanks in advance

$\endgroup$

marked as duplicate by user296602, José Carlos Santos, JMP, user99914, jvdhooft Oct 18 '17 at 6:31

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Really, the hint leads you nowhere? Did you try simplifying $(x - 1) p(x)$ at all? Try a couple of particular cases, like $n = 2$ or $3$.... $\endgroup$ – user296602 Oct 17 '17 at 20:27
1
$\begingroup$

It is easy to show that

$$(x-1)p(x)=x^{n+1}-1.$$

The derivative of this product is $$(n+1)x^n$$ which only has roots at $0$, hence no common root with $p$.

$\endgroup$
0
$\begingroup$

Note that

$(x - 1)\sum_0^nx^k = x^{n + 1}- 1, \tag 1$

and that the roots of $x^{n + 1} - 1$ are

$\exp(\dfrac{2k \pi i}{n + 1}), \; 0 \le k \le n, \tag 2$

all distinct. It follows from (1) that every zero of $x^{n + 1} - 1$ except $1$ (corresponding to $k = 0$) is a root of

$\sum_0^n x^k, \tag 3$

and since there are precisely $n$ of the (2) with $k \ne 0$, all possible zeroes of (3) are presented in (2). Thus all roots of (1) are distinct.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.